Signals and systems
using (Mathematica^®)

EEL-3135

Chapter 2. Linear Time-Invariant Systems

Version 1.07

This Mathematica® notebook was created by 
© Dr. Yu. Vlasov and Dr. J. Andrian
Department of Electrical and Computer Engineering
Florida International University
Miami, FL
USA

This Mathematica® notebook is based on textbook:
“Engineering signals and systems”
by F.T.Ulaby and A.E.Yagle
© 2013 National Technology and Science Press
ISBN: 978-1-934891-16-2

Date

Print[DateString["Day"]," ",DateString["MonthName"] ," ",DateString["Year"]]

Chapter 2. Linear Time-Invariant (LTI) Systems

Table of Contents

Textbook chapter 2 contents

Overview,	p. 31
2-1 Linear Time-Invariant Systems,	p. 31
2-2 Impulse Response,	p. 35
2-3 Convolution,	p. 40
2-4 Graphical Convolution,	p. 46
2-5 Convolution Properties,	p. 49
2-6 Causality and BIBO Stability,	p. 57
2-7 LTI Sinusoidal Response,	p. 60
2-8 Impulse Response of Second-Order LCCDEs,	p. 64
2-9 Car Suspension System,	p. 71
Chapter 2 Summary,	p. 77
Problems,	p. 78

Mathematica notebook table of contents

Overview

2-1. Linear Time-Invariant Systems

2-1.1. Scaling Property

2-1.2. Additivity Property

2-1.3. Linear Differential Equations

2-1.4. Significance of the Linearity Property

2-1.5. Time-Invariant Systems

2-2. Impulse Response

2-2.1. Static and Dynamic Systems

2-2.2. Computing h(t) and (Subscript[y, step])(t) of RC Circuit

2-2.3. Measuring h(t) and (Subscript[y, step])(t) of RC Circuit

2-2.4. Impulse and Ramp Responses from Step Response

2-3. Convolution

2-3.1. Derivation of Convolution Integral

2-3.2. Causal Signals and Systems

2-3.3. Computing Convolution Integrals

2-3.4. Convolution of Two Rectangular Pulses

2-4. Graphical Convolution

2-5. Convolution Properties

2-5.1. Commutative Property

2-5.2. Associative Property

2-5.3. Distributive Property

2-5.4. Causal ⋆ Causal = Causal

2-5.5. Time-Shift Property

2-5.6. Convolution with an Impulse

2-5.7. Width Property

2-5.8. Area Property

2-5.9. Convolution with a Step Function

2-5.10. Differentiation Property

Table 2-1. Convolution properties

Table 2-2. Commonly encountered convolutions

2-6. Causality and BIBO Stability

2-6.1. Causality

2-6.2. BIBO Stability: Definition

2-6.3. BIBO Stability: Condition and Proof

2-6.4. BIBO Stability of System with Decaying Exponentials

2-7. LTI Sinusoidal Response

2-7.1. LTI System Response to Complex Exponential Signals x(t)=((Ae)^(jωt))

2-7.2. Properties of (OverHat[Η])(ω)

2-7.3. (OverHat[Η])(ω) of LCCDE

2-7.4. LTI Response to Sinusoids

2-8. Impulse Response of Second-Order LCCDEs

2-8.1. LCCDE with No Input Derivatives

2-8.2. LCCDE with Input Derivative

2-8.3. Parameters of Second-Order LCCDE

2-8.4. Overdamped Case (ξ > 1)

2-8.5. Underdamped Case (ξ < 1)

2-8.6. Critically Damped Case (ξ = 1)

2-9. Car Suspension System (project)

Objectives

Learn to:

Describe the properties of LTI systems.

Determine the impulse and step responses of LTI systems.

Perform convolution of two functions.

Determine causality and stability of LTI systems.

Determine the overdamped, underdamped, and critically damped responses of second-order systems.

Determine a car's response to various pavement profiles.

Overview

A system is a device or process that accepts as an input a signal x(t) and produces as an output another signal y(t).

It is customary to use the notation x(t) to designate the input and y(t) to designate the output:

Input x(t) -> System -> Output y(t)

Among the various types of systems encountered in science and engineering, linear time-invariant (LTI) systems stand out as the most prominent.

This assertion is supported by the following observations.

Many physical systems, including circuits and mechanical systems with linear elements, are LTI.

The input-output behavior of an LTI system can be completely characterized by determining or observing its response to an impulsive excitation at its input. 
Once this is known, the system’s response to any other input can be computed using convolution (Section 2-3).

An LTI system can alter the amplitude and phase (but not the frequency) of an input sinusoid or complex exponential signal (Section 2-7). This feature is fundamental to the design of systems to filter noisy signals and images (Chapter 6).

The ease with which an LTI system can be analyzed facilitates the inverse process, namely that of designing the system to perform a desired task.

2-1. Linear Time-Invariant Systems

Definition:

A system is linear if it holds the scaling and additivity properties.

2-1.1. Scaling Property

If the response of a system to input x(t) is output y(t), and if this implies that the response to c·x(t) is c·y(t) for any constant c, the system is scalable.

Given:

x(t) -> System -> y(t)

then the system is scalable (has the scaling property) if

c·x(t) -> System -> c·y(t)

Thus, scaling the input also scales the output.

The scaling property is also known as the homogeneity or scalability property.

Consider a system whose output y(t) is linked to its input x(t) by the differential equation

(((d^2)y)/(d(t^2)))+2((dy)/(dt))+3y=4((dx)/(dt))+5x

Upon replacing x(t) with c·x(t) and y(t) with c·y(t) in all terms, we end up with

((d^2)/(d(t^2)))(cy)+2(d/(dt))(cy)+3(cy)=4(d/(dt))(cx)+5(cx)

Since c is constant, we can rewrite the expression as

c[(((d^2)y)/(d(t^2)))+2((dy)/(dt))+3y]=c[4((dx)/(dt))+5x]

which is identical to the original equation (1), but multiplied by the constant c.

Hence, since the response to c·x(t) is c·y(t), the system is scalable and has the scaling property.

However, the system described by the differential equation

(((d^2)y)/(d(t^2)))+2((dy)/(dt))+3y=4((dx)/(dt))+5x+6

is not scalable, because the last term on the right-hand side is independent of both x(t) and y(t), so we cannot factor out c. Therefore the system is not linear.

The scaling property provides a quick mechanism to test if a system is not linear. A good rule of thumb that works often (but not always) is to test whether the system has the scaling property with c = 2.

If doubling the input does not double the output, the system is not linear. But if doubling the input doubles the output, the system is probably (but not necessarily) linear.

2-1.2. Additivity Property

If the system responses to N inputs (Subscript[x, 1])(t), (Subscript[x, 2])(t), ... , (Subscript[x, N])(t) are respectively (Subscript[y, 1])(t), (Subscript[y, 2])(t), ... , (Subscript[y, N])(t) , then the system is additive if

(∑_(i=1)^N)(Subscript[x, i])(t) -> System -> (∑_(i=1)^N)(Subscript[y, i])(t)

That is, the response to the sum is the sum of the responses.

For an additive system, additivity must hold for both finite and infinite sums, as well as for integrals.

The combination of scalability and additivity is also known as the superposition principle.

If we denote (Subscript[y, 1])(t) as the response to (Subscript[x, 1])(t) in Eq. (1), and similarly (Subscript[y, 2])(t) as the response to (Subscript[x, 2])(t), then

(((d^2)(Subscript[y, 1]))/(d(t^2)))+2((d(Subscript[y, 1]))/(dt))+3(Subscript[y, 1])=4((d(Subscript[x, 1]))/(dt))+5(Subscript[x, 1])
(((d^2)(Subscript[y, 2]))/(d(t^2)))+2((d(Subscript[y, 2]))/(dt))+3(Subscript[y, 2])=4((d(Subscript[x, 2]))/(dt))+5(Subscript[x, 2])
Next,if we add the two equations and denote(Subscript[x, 3]) =(Subscript[x, 1])+(Subscript[x, 2])and(Subscript[y, 3]) =(Subscript[y, 1])+(Subscript[y, 2]) ,we get
(((d^2)(Subscript[y, 3]))/(d(t^2)))+2((d(Subscript[y, 3]))/(dt))+3(Subscript[x, 3])=4((d(Subscript[x, 3]))/(dt))+5(Subscript[x, 3])

Hence, the system characterized by differential equation (1) has the additivity property. 
Since it was shown earlier also to have the scaling property, the system is indeed linear.

2-1.3. Linear Differential Equations

Many physical systems are described by a linear differential equation (LDE) of the form

(∑_(i=0)^n)(Subscript[a, i-1])(((d^i)y)/(d(t^i)))=(∑_(i=0)^m)(Subscript[b, i-1])(((d^i)x)/(d(t^i)))

where coefficients (Subscript[a, 0]) to (Subscript[a, n]) and (Subscript[b, 0]) to (Subscript[b, m]) may or may not be functions of time t.

In either case, Eq. (6) represents a linear system, because it has both the scaling and additivity properties.

If the coefficients are time-invariant (i.e., constants), then the equation is called a linear, constant coefficient, differential equation (LCCDE), and the system it represents is not only linear but time-invariant as well.

Time invariance is discussed in Section 2-1.5.

2-1.4. Significance of the Linearity Property

To appreciate the significance of the linearity property, consider the following scenario.

We are given a linear system characterized by a third-order differential equation.

The system is excited by a complicated input signal x(t).

Our goal is to obtain an analytical solution for the system's output response y(t) .

Fundamentally, we can pursue either of the following two approaches:

Option 1: Direct “brute-force” approach: This involves solving the third-order differential equation with the complicated signal x(t). While feasible, this may be mathematically demanding and cumbersome.

Option 2: Indirect, linear-system approach, comprised of the following steps:

(a) Synthesize x(t) as the linear combination of N relatively simple signals  (Subscript[x, 1])(t), (Subscript[x, 2])(t), ... , (Subscript[x, N])(t) :
x(t) =  (Subscript[x, 1])(t) + (Subscript[x, 2])(t) + ... + (Subscript[x, N])(t)  
with signals  (Subscript[x, 1])(t) to (Subscript[x, N])(t)  chosen so that they yield straightforward solutions of the third-order differential equation.

(b) Solve the differential equation N times, once for each of the N input signals acting alone, yielding outputs  (Subscript[y, 1])(t) through (Subscript[y, N])(t) .

(c) Apply the linearity property to obtain y(t) by summing the outputs  (Subscript[y, 1])(t) through (Subscript[y, N])(t) :
y(t) =  (Subscript[y, 1])(t) + (Subscript[y, 2])(t) + ... + (Subscript[y, N])(t)

Even though the second approach entails solving the differential equation N times, the overall solution may prove significantly more tractable than the solution associated with option 1.

The procedure will be demonstrated using numerous examples scattered throughout the book.

Systems described by LCCDEs encompass all circuits containing ideal resistors, inductors, capacitors, op-amps, and dependent sources, so long as the circuit contains only a single independent current or voltage source acting as the input.

The output is any individual voltage or current in the circuit, since (by definition) a system has only one input and one output.

Limiting the circuit to a single independent source (the input) may seem overly restrictive, as many circuits have more than one independent source.

Fortunately, the limitation is circumvented by applying the superposition principle. To analyze a circuit with N independent sources and zero initial conditions (i.e., all capacitors are uncharged and all inductors have zero currents flowing through them):

(a) We set all but one source to zero and then analyze the (now linear) circuit using that source as the input. A voltage source set to zero becomes a short circuit, and a current source set to zero becomes an open circuit. We designate (Subscript[y, 1])(t) as the circuit response to source (Subscript[x, 1])(t) acting alone.

(b) We then rotate the non-zero choice of source among all of the remaining independent sources, setting all other sources to zero each time. The process generates responses  (Subscript[y, 2])(t) to (Subscript[y, N])(t)  , corresponding to sources  (Subscript[x, 2])(t) to (Subscript[x, N])(t) , respectively.

(c) We add responses (Subscript[y, 1])(t) through (Subscript[y, N])(t) .

We shall see in Section 4-1 that non-zero initial conditions are equivalent to additional independent sources.

Example 2-1. Superposition

Apply the superposition principle to determine the current I through resistor (Subscript[R, 2]) in the circuit of Fig. 2-2(a).

Fig. 2-2(a). Original circuit

(Subscript[I, 0])=6;(Subscript[V, 0])=45;(Subscript[R, 1])=5;(Subscript[R, 2])=10;

Solution:

(a) The circuit contains two sources, (Subscript[I, 0]) and (Subscript[V, 0]). 
We start by transforming the circuit into the sum of two new circuits: one with (Subscript[I, 0]) alone and another with (Subscript[V, 0]) alone, as shown in parts (b) and (c) of Fig. 2-2, respectively.

Fig. 2-2(b). Source (Subscript[I, 0]) alone generates (Subscript[I, 1])

Fig. 2-2(c). Source (Subscript[V, 0]) alone generates (Subscript[I, 2])

The current through (Subscript[R, 2]) due to (Subscript[I, 0]) alone is labeled (Subscript[I, 1]), and that due to (Subscript[V, 0]) alone is labeled (Subscript[I, 2])

Application of current division in the circuit of Fig. 2-2(b) gives

(Subscript[I, 1])=((Subscript[R, 1])/((Subscript[R, 1])+(Subscript[R, 2])))(Subscript[I, 0])

In the circuit of Fig. 2-2(c), Ohm’s law gives

(Subscript[I, 2])=((-(Subscript[V, 0]))/((Subscript[R, 1])+(Subscript[R, 2])))

Hence, by superposition, total current I through resistor (Subscript[R, 2])

(Subscript[I, tot])=(Subscript[I, 1])+(Subscript[I, 2])

2-1.5. Time-Invariant Systems

A system is time-invariant if delaying the input signal x(t) by any constant T generates the same output y(t), but delayed by exactly T.

x(t) -> System -> y(t)

x(t – T) -> System -> y(t – T)

Fig23a=Plot[(1-(E^(-t)))UnitStep[t],{t,-2,10},PlotStyle->{RGBColor[0,0,0.5],Thickness[0.008]},AxesLabel->{t,"Input x(t)"},PlotRange->{-0.2,1.2}];
Fig23c=Plot[(1-(E^(-(t-2))))UnitStep[t-2],{t,-2,10},PlotStyle->{RGBColor[0.5,0,0],Thickness[0.008]},AxesLabel->{t,"Input x(t-T)"},PlotRange->{-0.2,1.2}];
Fig23b=Plot[Sin[8t]*(E^(-t/3))UnitStep[t]UnitStep[5-t],{t,-2,10},Exclusions->None,PlotStyle->{RGBColor[0,0,0.5],Thickness[0.008]},AxesLabel->{t,"Output y(t)"},PlotRange->{-1.2,1.2}];
Fig23d=Plot[Sin[8(t-2)]*(E^(-(t-2)/3))UnitStep[t-2]UnitStep[5-(t-2)],{t,-2,10},Exclusions->None,PlotStyle->{RGBColor[0.5,0,0],Thickness[0.008]},AxesLabel->{t,"Output y(t-T)"},PlotRange->{-1.2,1.2}];
GraphicsGrid[{{Fig23a,Fig23c},{Fig23b,Fig23d}},Frame->All,ImageSize->Medium]

Figure 2-3. Delaying a signal x(t) by T seconds at the input to a time-invariant system results in an equal delay at the output.

Physically, a system is time-invariant if it has no internal clock.

If the input signal is delayed, the system has no way of knowing it, so it accepts the delayed input and delivers a correspondingly delayed output.

How can one tell if a system is time-invariant?

A good rule of thumb that almost always works is:

If t appears explicitly only in the expression for x(t) or in derivatives or limits of integration in the equation describing the system, the system is likely time-invariant.

Examples of time-invariant systems:

(a)(Subscript[y, 1])(t)=3(((d^2)x)/(d(t^2)))

(b)(Subscript[y, 2])(t)=sin[x(t)]

(c)(Subscript[y, 3])(t)=((x(t+2))/(x(t-1))), where x (t+2) is x (t) advanced by 2 seconds and x (t-l) is x(t) delayed by 1 second.

Note that while all three systems are time-invariant, systems (b) and (c) are not linear.

Examples of non-time-invariant systems:

(d)(Subscript[y, 4])(t)=t×x(t)

(e)(Subscript[y, 5])(t)=x((t^2))

(f)(Subscript[y, 6])(t)=x(-t)

Note that systems (d) to (f) all are linear but not time-invariant, and systems (e) and (f) are not time-invariant even though they do pass the rule-of-thumb test stated earlier – namely, that t is part of the expression of x(t) and not in a multiplying function or coefficient. 
The forms of systems (e) and (f) are the two exceptions to the rule of thumb.

For the remainder of the course (textbook), we will deal with LTI systems exclusively.

2-2. Impulse Response

The impulse response h(t) of a system is (logically enough) the response of the system to an impulse δ(t).

Similarly, the step response (Subscript[y, step])(t) is the response of the system to a unit step u(t).

In symbolic form:

δ(t) -> LTI -> h(t) 
and
u(t) -> LTI -> (Subscript[y, step])(t)

The significance of the impulse response is that, if we know h(t) for an LTI system, we can compute the response to any other input x(t) using the convolution integral derived in the next Section 2-3.

The significance of the step response is that, in many physical situations, we are interested in how well a system “tracks” (follows) a step input. We will use the step response extensively in control-related applications in Chapter 4.

2-2.1. Static and Dynamic Systems

A system for which the output y(t) at time t depends only on the input x(t) at time t is called a static or memoryless system .

An example of such a system is

y(t)=((sin[x(t)])/((x^2)(t)+1))

To compute the output of a memoryless system from its input, we just plug into its formula. Parenthetically, we note that this system is not linear.

The only memoryless system that is also LTI is y(t) = a x(t) for any constant a. We remind the reader that the affine system y(t) = a x(t) + b is not a linear system, although a plot of its formula is a straight line. In the physical world, most systems are dynamic in that the output y(t) at time t depends on past (or future) as well as present values of the input x(t). Such systems often involve integrals and derivatives of either the input, the output, or both.

A system may physically exist, and its output may be measured in response to physically realizable inputs, such as a gradual step function (Section 1-4.1) or a sinusoid, or it may be described by a mathematical model (usually in the form of a differential equation constructed from an analysis of the physics, chemistry, or biology governing the behavior of the system). In either case, the system behavior is characterized by its impulse response h(t).

To develop familiarity with what the impulse response means and represents, we will use a simple RC-circuit to derive and demonstrate the physical interpretation of its impulse response h(t).

2-2.2. Computing h(t) and (Subscript[y, step])(t) of RC Circuit

Consider the RC circuit shown in Fig. 2-5(a).

The input is the voltage source x(t), and the output is the voltage y(t) across the capacitor, which is initially uncharged, so y((SuperMinus[0]))=0.

The system in this case is the circuit inside the dashed box.

Figure 2-5(a). RC circuit.

To distinguish between the state of the system (or its output response) before and after the introduction of the input excitation, we denote t=(SuperMinus[0]) and t=0 as the times immediately before and immediately after the introduction of x(t), respectively.

The conditions prevailing at t=(SuperMinus[0]) are called initial conditions.

Accordingly, y((SuperMinus[0])) represents the initial condition of the system’s output.

Application of Kirchhoff’s voltage law around the loop yields

R i(t)+y(t)=x(t)

For the capacitor, the current through it, i(t), is related to the voltage across it, y(t), by

i(t)=C((dy(t))/(dt))

Substituting Eq. (9) in Eq. (8) and then dividing all terms by RC leads to

((dy(t))/(dt))+(1/RC)y(t)=(1/RC)x(t)

Impulse response

To compute the impulse response , we label x(t) = δ(t) and y(t) = h(t) and obtain

((dh(t))/(dt))+(1/RC)h(t)=(1/RC)δ(t)

Next, we introduce the time constant (Subscript[τ, c])=RC and multiply both sides of the differential equation by the integrating factor (e^(t/(Subscript[τ, c]))). The result is

((dh(t))/(dt))(e^(t/(Subscript[τ, c])))+(1/(Subscript[τ, c]))h(t)(e^(t/(Subscript[τ, c])))=(1/(Subscript[τ, c]))δ(t)(e^(t/(Subscript[τ, c])))

The left side of Eq. (12) is recognized as

((dh(t))/(dt))(e^(t/(Subscript[τ, c])))+(1/(Subscript[τ, c]))h(t)(e^(t/(Subscript[τ, c])))=(d/(dt))(h(t)(e^(t/(Subscript[τ, c]))))
and the sampling property of the impulse function given by Eq.(27) reduces the right-hand side of Eq.(12) to
(1/(Subscript[τ, c]))δ(t)(e^(t/(Subscript[τ, c])))=(1/(Subscript[τ, c]))δ(t)

Incorporating these two modifications in Eq. (12) leads to

(d/(dt))(h(t)(e^(t/(Subscript[τ, c]))))=(1/(Subscript[τ, c]))δ(t)

Integrating both sides from (SuperMinus[0]) to t gives

(∫_(SuperMinus[0])^t)(d/(dτ))(h(τ)(e^(τ/(Subscript[τ, c]))))dτ=(1/(Subscript[τ, c]))(∫_(SuperMinus[0])^t)δ(τ)dτ

where τ is a dummy variable of integration and has no connection to the time constant (Subscript[τ, c]). We will use τ as a dummy variable throughout this chapter.

By Eq. (1.24), we recognize the integral in the right-hand side of Eq. (15) as the step function u(t). 
Hence,

(Subscript[(h(τ)(e^(τ/(Subscript[τ, c])))), (SuperMinus[0])]^t)=(1/(Subscript[τ, c]))u(t)

or

h(t)(e^(t/(Subscript[τ, c])))-h((SuperMinus[0]))=(1/(Subscript[τ, c]))u(t)

The function h(t) represents y(t) for the specific case when the input is δ(t).

We are told that the capacitor was initially uncharged, so y((SuperMinus[0])) = 0 for any input.

Hence, h((SuperMinus[0]))=0, and Eq. (16) reduces to

h(t)=(1/(Subscript[τ, c]))(e^((-t)/(Subscript[τ, c])))u(t)
impulse response of the RC circuit

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Figure 2-5(b). Impulse response h(t) of RC circuit.

The plot displayed in Fig. 2-5(b) indicates that the capacitor voltage h(t) jumps instantaneously at t = 0, which contradicts the natural behavior of a capacitor. This contradiction is a consequence of the fact that a pure δ(t) is a physically unrealizable singularity function. However, as discussed in the next subsection, we can approximate an impulse by a finite-duration pulse and then use it to measure the impulse response of the circuit.

Step response

To compute the step response of the circuit, we start with the same governing equation given by Eq. (10), but this time we label x(t) = u(t) and y(t) = (Subscript[y, step])(t).

The procedure leads to

(Subscript[y, step])(t)=(1-(e^((-t)/(Subscript[τ, c]))))u(t)   – step response of the RC circuit

As the initially uncharged capacitor builds up charge, the voltage across it builds up monotonically from zero at t = 0 to 1 V as t -> ∞ [Fig. 2-5(c)].

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Figure 2-5(c). Step response (Subscript[y, step])(t) of RC circuit.

Once we know the step response of an LTI system, we can apply the superposition principle to compute the response of the system to any input that can be expressed as a linear combination of scaled and delayed step functions .

2-2.3. Measuring h(t) and (Subscript[y, step])(t) of RC Circuit

Suppose we have a physical system for which we do not have an LCCDE model. To physically measure its impulse and step responses, we need to be able to apply impulse and step signals at its input terminals.

As ideal impulse and step signals are physically unrealizable, we resort to techniques that offer reasonable approximations.

In general, the approximate techniques used for representing u(t) and δ(t) must be tailormade to suit the system under consideration.

We will illustrate the procedure by measuring the step and impulse responses of an RC circuit [Fig. 2-5(a)] with (Subscript[τ, c])=RC=2 s.

Step response

A unit step function at the input terminals can be represented by a 1-V dc voltage source connected to the RC circuit via a switch that is made to close at t = 0 [Fig. 2-6(a)].

Figure 2-6(a). Simulating u(t) inputs through the use of switches.

In reality, the process of closing the switch has a finite (but very short) switching time (Subscript[τ, s]). That is, it takes the voltage at the input terminals (Subscript[τ, s]) seconds to transition between zero and 1 volt.

For the RC circuit, its time constant (Subscript[τ, c]) is a measure of its reaction time, so if (Subscript[τ, s]) ≪ (Subscript[τ, c]), the circuit will sense the switching action as if it were instantaneous, in which case the measured output voltage (Subscript[y, step])(t) would closely match the computed expression given by Eq. (18),

(Subscript[y, step])(t)=(1-(e^((-t)/2)))u(t)

where we have replaced (Subscript[τ, c]) with 2 s.

Impulse response

A unit impulse δ(t) can be simulated by a rectangle model consisting of two step functions:

x(t)=(1/T)(u(t)-u(t-T)) , when T -> ∞

The rectangular pulse has a duration of T and its amplitude (in volts) is always chosen such that it is numerically equal to 1/T, thereby maintaining the area of the pulse at 1.

Physically, the rectangle excitation is realized by connecting a voltage source of magnitude (1/T) volts to the circuit via a switch that connects the RC circuit to the source at t = 0 and then connects it to a short circuit at t = T [Fig. 2-6(b)].

Figure 2-6(b). Simulating δ(t) input through the use of switches.

Since the RC circuit is LTI, the superposition principle applies. In view of Eq. (19), which is the response to the input excitation u(t) , the response (Subscript[y, pulse])(t) to the pulse excitation modeled by Eq. (20) is

(Subscript[y, pulse])(t)=(1/T)((Subscript[y, step])(t)-(Subscript[y, step])(t-T))=(1/T){(1-(e^(-t/2)))u(t)-(1-(e^(-(t-T)/2)))u(t-T)}

Our goal is to answer the following pair of questions:

Is it possible to measure the impulse response of the RC circuit by exciting it with a rectangular pulse of amplitude (1/T) and duration T?

Are there any constraints on the choice of the value of T?

To explore the answers to these two questions, we display in Fig. 2-7 plots of (Subscript[y, pulse])(t) – all computed in accordance with Eq. (21) – for three different values of T, namely, 5 s, 0.05 s, and 0.01 s.

The first case was chosen to illustrate the response of the circuit to a pulse whose duration T is greater than its own time constant (Subscript[τ, c]) (i.e., T = 5 s and (Subscript[τ, c]) = 2 s).

T = 5 s

Between t = 0 and t = 5 s, the capacitor charges up to a maximum voltage of (1-(e^(-5/2)))/5=0.184 V [Fig. 2-7(a)], almost to the level of the input voltage (0.2 V).

Upon disconnecting the voltage source and replacing it with a short circuit at t = T = 5 s, the capacitor starts to discharge towards zero volts as t -> ∞.

During both the charge-up and discharge periods, the rates of increase and decay are governed by the time constant (Subscript[τ, c]) = 2 s.

(Subscript[y, pulse])[t_,T_]:=(1/T)((1-(E^(-t/2)))*UnitStep[t]-(1-(E^(-(t-T)/2)))*UnitStep[t-T])

Show[
{
Plot[Evaluate[Table[(Subscript[y, pulse])[t,T],{T,{5,0.005,0.001}}]],{t,-0.1,10},Frame->True,Axes->False,Exclusions->None,PlotStyle->{Thick}],
Graphics[{
{Thickness[0.003],RGBColor[1,0,0],Dashing[0.005],
Line[{{5,0},{5,0.184}}]},
{Thickness[0.003],RGBColor[1,0,0],Dashing[0.005],
Line[{{-0.2,0.1835},{5,0.1835}}]},
Text["T=5s; (Subscript[y, max])=0.1835",{5,0.22}],
{Arrowheads[0.025],Arrow[{{1,0.485},{0.01,0.4988}}]},
{Thickness[0.003],RGBColor[1,0,0],Dashing[0.005],
Line[{{-0.2,0.4988},{0,0.4988}}]},
Text["T=0.01s; (Subscript[y, max])=0.4988",{2.8,0.48}],
{Arrowheads[0.025],Arrow[{{1,0.425},{0.05,0.4938}}]},
{Thickness[0.003],RGBColor[1,0,0],Dashing[0.005],
Line[{{-0.2,0.4938},{0,0.4938}}]},
Text["T=0.05s; (Subscript[y, max])=0.4938",{2.8,0.42}]
}
]}]

Zoom - in:

Plot[Evaluate[Table[(Subscript[y, pulse])[t,T],{T,{5,0.005,0.001}}]],{t,-0.01,0.02},Frame->True,Axes->False,Exclusions->None,PlotStyle->{Thickness[0.007]}]

Figure 2-7: Plots of (Subscript[y, pulse])(t), as expressed by Eq. (21), for three values of T: 5 s, 0.05 s, and 0.01 s

T = 0.05 s

Next, we examine the case of T = 0.05 s; the pulse duration now is much shorter than the time constant (T/(Subscript[τ, c])=0.025) .

Also, the pulse amplitude is 1/T = 20 V (compared with only 0.2 V for the previous case).

The response charges up to 20(1-(e^(-0.05/2)))=0.494 V = 0.494 V at t = 0.05 s and then discharges to zero as t -> ∞.

The plot in Fig. 2-7(b) exhibits a close resemblance to the theoretical response shown earlier in Fig. 2-5(b). For the theoretical response, the peak value is 1/(Subscript[τ, c]) = 1/2 = 0.5 V.

T = 0.01 s

As the pulse duration is made shorter still, the measured response approaches both the shape and peak value of the theoretical response.

The plot in Fig. 2-7(c) (corresponding to T = 0.01 s) exhibits a peak value of 100(1-(e^(-0.01/2)))=0.499  V, which is very close to the theoretical peak value of 0.5 V.

Hence, the combined answer to the questions posed earlier is yes, the impulse response of the RC circuit can indeed be determined experimentally, provided the excitation pulse used at the input is of unit area and its duration is much shorter than the circuit’s time constant (Subscript[τ, c]).

This statement can be generalized:

The response of an RC circuit to a sufficiently fast pulse of any shape is 
(A/(Subscript[τ, c]))(e^(-t/(Subscript[τ, c])))u(t)
where A is the area under the pulse. 
“Sufficiently fast” means that the duration of the pulse T ≪ (Subscript[τ, c])

2-2.4. Impulse and Ramp Responses from Step Response

We now extend the procedure we used to determine h(t) of the RC circuit to LTI systems in general:

Physically measure the step response (Subscript[y, step])(t).

Differentiate it to obtain h(t)

h(t)=((d(Subscript[y, step])(t))/(dt))

To demonstrate the validity of Eq. (22), we start with the definition of the step response given by Eq. (7b).

We then use the time-invariance property of LTI systems to delay both the input and output by a constant delay ϵ:

u(t-ϵ) -> LTI -> (Subscript[y, step])(t-ϵ)

Next, using the additivity property of LTI systems, we subtract Eq. (2.23) from Eq. (2.7b), and using the scaling property, we multiply the result by 1/ϵ:

((u(t)-u(t-ϵ) )/ϵ)-> LTI -> (((Subscript[y, step])(t) -(Subscript[y, step])(t-ϵ) )/ϵ)

The two steps were justified by the additivity and scaling properties of LTI systems.

Finally, if we let ϵ -> 0, the lefthand side of Eq. (24) becomes ((du(t))/(dt)), and the right-hand side becomes  ((d(Subscript[y, step]))/(dt)) .

Moreover, from Eq. (1.23), ((du(t))/(dt))=δ(t).

Hence,

((du(t))/(dt))=δ(t)-> LTI -> h(t)=((d(Subscript[y, step])(t))/(dt))

Example 2-2. RC Circuit Impulse Response from Step Response

Find 
(a) the impulse response of the RC circuit in Fig. 2-5(a) by applying Eq. (22) and 
(b) the circuit response to a ramp-function input x(t) = r(t).

Figure 2-5(a). RC circuit.

Solution:

(a) Using the expression for (Subscript[y, step])(t) given by Eq. (18), the impulse response of the RC circuit is

Note: HeavisideTheta[t] is used here instead of UnitStep[t] because of its correct differentiation

τc22=.;ystep22[t_]:=(1-(E^((-t)/(τc22))))HeavisideTheta[t]

ystep22[t]

h[t_]:=ystep22'[t]

h[t]

Let (Subscript[τ, c])=τc22 = 2 s. Hence, the step response is:

τc22=2;Plot[ystep22[t],{t,-2,10},Exclusions->None,PlotStyle->{Thickness[0.007],Blue},AxesLabel->{t,"(Subscript[y, step])(t)"}]

The impulse response h(t) of a system is

Plot[h[t],{t,-2,10},Exclusions->None,PlotStyle->{Thickness[0.007],Blue},AxesLabel->{t,"h(t)"}]

h(t)=(1/(Subscript[τ, c]))(e^(-(t/(Subscript[τ, c]))))u(t)

where we used 
δ(t) = ((du(t))/(dt)) and x(t)·δ(t) = x(0)·δ(t) for any x(t)
Because the step response (Subscript[y, step])(t) was available to us, the process leading to Eq. (26) was quite painless and straightforward (in contrast with the much longer differential equation solution contained in Section 2-2.2).

(b) From Eq. (7b), we have

u(t) -> LTE -> (Subscript[y, step])(t)

The ramp function r(t) is related to the step function u(t) by Eq. (1.18) as

r(t)=(∫_(-∞)^t)u(τ)dτ

Integration is a linear operation, so its application to both sides of Eq. (27) gives

r(t)=(∫_(-∞)^t)u(τ)dτ-> LTE -> (Subscript[y, ramp])(t)

(Subscript[y, ramp])(t)=(t-(1-(e^(-(t/(Subscript[τ, c]))))) (Subscript[τ, c])) u(t)

(Subscript[y, ramp])[t_]:=(∫_(-∞)^t)(1-(E^((-τ)/(τc))))UnitStep[τ]\[DifferentialD]τ

(Subscript[y, ramp])[t]

Exercise 2-4

Determine the impulse response of a system whose step response is

(Subscript[y, step])(t)=({{{, ({{0, t⩽0}, {t, 0⩽t⩽1}, {1, t⩾1}})}})

Solution:

ystep24[t_]:=t*HeavisideTheta[t]-(t-1)*HeavisideTheta[t-1]

Plot[ystep24[t],{t,-0.2,2},PlotStyle->{Thickness[0.007],RGBColor[0,0,0.5]}]

ystep24'[t]

Plot[ystep24'[t],{t,-0.2,2},PlotStyle->{Thickness[0.007],RGBColor[0,0,0.5]},Exclusions->None]

Answer:

h(t) = u(t) - u(t - 1)

Exercise 2-5

The RC circuit of Fig. 2-5(a) is excited by

x(t) = (1 - 1000t) [u(t) - u(t - 0.001)]

Figure 2-5(a). RC circuit.

Compute the capacitor voltage y(t) for t > 0.001 s, given that (Subscript[τ, c])=1 s.

Solution:

Time constant:

τc25=1

Input signal x(t) is a very short wedge-shaped pulse

xinput25[t_]:=(1 - 1000*t)*(HeavisideTheta[t] -HeavisideTheta[t - 0.001])

xinput25[t]

Plot[xinput25[t],{t,-0.001,0.003},Exclusions->None,PlotRange->All,PlotStyle->Thickness[0.008]]

The response of an RC circuit to a sufficiently fast pulse of any shape is 
(A/(Subscript[τ, c]))(e^(-t/(Subscript[τ, c])))u(t)
where A is the area under the pulse. “Sufficiently fast” means that the duration of the pulse T ≪ (Subscript[τ, c])

Pulse area:

A25=(∫_(-∞)^∞)xinput25[t]\[DifferentialD]t

Circuit output signal:

youtput25[t_]:= (A25/(τc25)) (E^((-t)/(τc25))) UnitStep[t]

Plot[youtput25[t],{t,-0.1,3},Exclusions->None,PlotRange->All]

Answer:

y(t)=0.0005(e^(-t)) u(t)

2-3. Convolution

The response y(t) of an LTI system with impulse response h(t) to any input x(t) can be computed explicitly using the convolution integral

y(t)=(∫_(-∞)^∞)x(τ)h(t-τ)dτ=h(t)⋆x(t)

All initial conditions must be zero.

The convolution operation denoted by ⋆ in Eq. (30) is not the same as multiplication.

The term convolution refers to the convoluted nature of the integral.

Note the following:

(a) τ is a dummy variable of integration.

(b) h(t - τ) is obtained from the impulse response h(t) by (1) replacing t with the variable of integration τ , (2) reversing h(τ) along the τ axis to obtain h(-τ), and (3) delaying h(-τ) by time t to obtain h(t - τ) = h(-(τ - t)).

(c) As will be demonstrated later in this section, convolution is commutative; we can interchange x and h in Eq. (30).

According to the convolution property represented by Eq. (30), once h(t) of an LTI system has been determined, the system’s response can be readily evaluated for any specified input excitation x(t) by performing the convolution integration.

The integration may be carried out analytically, graphically, or numerically – all depending on the formats of x(t) and h(t).

Multiple examples are provided in the latter part of this section.

2-3.1. Derivation of Convolution Integral

The derivation of the convolution integral follows directly from the definition of the impulse response and the properties of LTI systems.

We need only the five steps outlined in Fig. 2-8.

Step 1. From the definition of the impulse response given by Eq. (7a)

δ(t) -> LTI -> h(t)

Step 2. According to the time-invariance property of LTI systems, delaying the input δ(t) by a constant τ will delay the output h(t) by the same constant τ

δ(t-τ) -> LTI -> h(t-τ)

Step 3. From the scaling property of LTI systems, if the input has an amplitude x(τ), the output will scale by the same factor

x(τ)δ(t-τ) -> LTI -> x(τ)h(t-τ)

Step 4. According to Eq. (33), an impulsive excitation x((Subscript[τ, 1])) δ(t – (Subscript[τ, 1])) at a specific value of τ (namely, (Subscript[τ, 1])) will generate a corresponding output x((Subscript[τ, 1]))h(t – (Subscript[τ, 1])) .
Similarly, a second impulsive excitation of area x((Subscript[τ, 2])) at τ = (Subscript[τ, 2])

x((Subscript[τ, 1]))δ(t-(Subscript[τ, 1])) -> LTI -> x((Subscript[τ, 1]))h(t-(Subscript[τ, 1])) 
x((Subscript[τ, 2]))δ(t-(Subscript[τ, 2])) -> LTI -> x((Subscript[τ, 2]))h(t-(Subscript[τ, 2]))

If the LTI system is excited by the sum of both inputs, the additivity property assures us that the output will be equal to the sum of the outputs in Eqs. (34).

For any input function x(τ) extending over -∞ < τ < ∞, the input and output of the LTI system become definite integrals

(∫_(-∞)^∞)x(τ)δ(t-τ)dτ -> LTI -> (∫_(-∞)^∞)x(τ)h(t-τ)dτ

Step 5. From the sampling property of impulses

(∫_(-∞)^∞)x(τ)δ(t-τ)dτ=x(t)

Hence, Eq. (35) simplifies to

x(t) -> LTI -> y(t)=(∫_(-∞)^∞)x(τ)h(t-τ)dτ

As a shorthand notation, the convolution integral is represented by an asterisk ⋆, so output y(t) is expressed as

y(t)=x(t)⋆h(t)=(∫_(-∞)^∞)x(τ)h(t-τ)dτ

By changing variables inside the integral of Eq. (38) from τ to (t – τ), the convolution can be expressed by the equivalent form

y(t)=h(t)⋆x(t)=(∫_(-∞)^∞)h(τ)x(t-τ)dτ

Equivalency of the integrals in Eqs.(38) and (39) implies that the convolution operation is commutative:

y(t)=x(t)⋆h(t)=h(t)⋆x(t)

δ(t) -> LTI -> y(t)=h(t) 
⇓
δ(t-τ) -> LTI -> y(t)=h(t-τ) 
⇓
x(τ)δ(t-τ) -> LTI -> y(t)=x(τ)h(t-τ) 
⇓
(∫_(-∞)^∞)x(τ)δ(t-τ)dτ -> LTI ->y(t)=(∫_(-∞)^∞)x(τ)h(t-τ)dτ
⇓
x(t)-> LTI ->y(t)=(∫_(-∞)^∞)x(τ)h(t-τ)dτ

Figure 2-8. Derivation of the convolution integral for a linear time-invariant system.

2-3.2. Causal Signals and Systems

The integral given by Eq. (38) implies that the output y(t) at time t depends on all excitations x(t) occurring at the input, including those that will occur at times later than t.

Since this cannot be true for a physically realizable system, the upper integration limit should be replaced with t instead of ∞.

Also, if we choose our time scale such that no excitation exists before t = 0, we can then replace the lower integration limit with zero.

Hence, for causal signals and systems, Eq. (38) becomes

y(t)=u(t)(∫_0^t)x(τ)h(t-τ)dτ=
=u(t)(∫_0^t)x(t-τ)h(τ)dτ

Multiplication of the integral by unit step u(t) is a reminder that the convolution of two causal functions is itself causal.

Note that the response y(t) provided by the convolution integral assumes that all initial conditions of the system are zero.

If the system has non-zero initial conditions, its total response will be the sum of two responses: one due to the initial conditions and another due to the input excitation x(t).

Through several examples, we will show in Chapter 3 how to account properly for the two contributions of the total response.

2-3.3. Computing Convolution Integrals

Given a system with an impulse response h(t), the output response y(t) can be determined for any specified input excitation x(t) by computing the convolution integral given by Eq. (38) or by Eq. (41) if x(t) is causal.

The process may be carried out using different methods:

(a) analytically, by performing the integration to obtain an expression for y(t), and then evaluating y(t) as a function of time t

(b) graphically, by simulating the integration using plots of the waveforms of x(t) and h(t)

(c) analytically with simplification, by taking advantage of the expedient convolution properties outlined in Section 2-5 to simplify the integration algebra

(d) numerically on a digital computer

As we will see shortly through Examples 2-3 and 2-4, the analytical approach is straightforward, but the algebra often is cumbersome and tedious.

The graphical approach (Section 2-4) allows the user to “visualize” the convolution process, but it is slow and repetitious.

The two approaches offer complementary insight; the choice of one over the other depends on the specifics of x(t) and h(t).

After learning to use approaches (a) and (b), the reader will appreciate the advantages offered by approach (c); invoking the convolution properties simplifies the algebra considerably.

Finally, numerical integration always is a useful default option, particularly if x(t) is a measured signal or difficult to model mathematically.

Example 2-3. RC Circuit Response to Triangular Pulse

An RC circuit with (Subscript[τ, c]) = RC = 1 s is excited by the triangular pulse shown in Fig. 2-9(a). 
The capacitor is initially uncharged.

Determine the output voltage across the capacitor.

Triangle01[t_]:=UnitTriangle[t]

Plot[Triangle01[t-1],{t,-1,3}, PlotStyle->{Thickness[0.007],RGBColor[0,0,0.5]},AxesLabel->{t,"(Subscript[v, in])(t)"},Exclusions->None]

Figure 2-9(a) Triangular pulse – input signal

Solution:

The input signal, measured in volts, is given by

(Subscript[v, in])(t)=({{{, ({{0, t⩽0}, {t, 0⩽t⩽1 s}, {2-t, 1⩽t⩽2 s}, {0, t⩾2 s}})}})

and according to Eq. (17), the impulse response for (Subscript[τ, c])=1 is

h(t)=(1/(Subscript[τ, c]))(e^((-t)/(Subscript[τ, c])))u(t)=(e^(-t))u(t)

According to Eq. (41), the output is given by

(Subscript[v, out])(t)=(Subscript[v, in])(t)⋆h(t)=(∫_0^t)(Subscript[v, in])(τ)h(t-τ)dτ

h(t-τ)=(e^(-(t-τ)))u(t-τ)=({{{, ({{0, t<τ}, {(e^(-(t-τ))), t>τ}})}})

The convolution integral will be evaluated for each of the four time segments associated with (Subscript[v, in])(t) separately.

() t < 0

The lowest value that the integration variable τ can assume is zero. Therefore, when t < 0, t < τ and h(t – τ) = 0.
Consequently,

(Subscript[v, out])(t) = 0  for t < 0

() 0 ⩽ t ⩽ 1 s

h(t-τ)=(e^(-(t-τ))),   (Subscript[v, in])(τ)=τ

(Subscript[v, out])(t)=(∫_0^t)τ (e^(-(t-τ)))dτ=(e^(-t))+t-1,   for 0⩽t ⩽1 s

() 1 s ⩽ t ⩽ 2 s

(Subscript[v, in])(t)=({{{, ({{τ, 0⩽ τ ⩽1 s}, {2-τ, 1 s ⩽τ ⩽2 s}})}})

and

(Subscript[v, out])(t)=(∫_0^1)τ (e^(-(t-τ)))dτ+(∫_1^t)(2-τ) (e^(-(t-τ)))dτ=
=(1-2e)(e^(-t))-t+3,   for 1 s ⩽t ⩽2 s

() t ⩾ 2 s

(Subscript[v, out])(t)=(∫_0^1)τ (e^(-(t-τ)))dτ+(∫_1^2)(2-τ) (e^(-(t-τ)))dτ=
=(1-2e+(e^2))(e^(-t)),   for t ⩾2 s

The cumulative response covering all four time segments is

(Subscript[v, out])(t) = 0,  for t < 0
(Subscript[v, out])(t)=(e^(-t))+t-1,   for 0⩽t ⩽1 s
(Subscript[v, out])(t)=(1-2e)(e^(-t))-t+3,   for 1 s ⩽t ⩽2 s
(Subscript[v, out])(t)=(1-2e+(e^2))(e^(-t)),   for t ⩾2 s

The cumulative response is displayed in Fig. 2-9(b).

vout23[t_]:=((E^(-t))+t-1)*UnitBox[t-0.5]+((1-2*E)*(E^(-t))-t+3)*UnitBox[t-1.5]+((1-2*E+(E^2))*(E^(-t)))*UnitStep[t-2]

vout23Max=MaxValue[vout23[t],t];
tmax23=ArgMax[vout23[t],t];

Show[
{
Plot[vout23[t],{t,-1,7}, PlotStyle->{Thickness[0.007],RGBColor[0,0,0.5]},AxesLabel->{"t (s)","(Subscript[v, out])(t)"},PlotRange->All,Exclusions->None],
Graphics[{
{Thickness[0.003],RGBColor[1,0,0],Dashing[0.005],
Line[{{0,vout23Max},{tmax23,vout23Max}}]},
{Thickness[0.003],RGBColor[1,0,0],Dashing[0.005],
Line[{{tmax23,0},{tmax23,vout23Max}}]},
Text[(* t *)
Style[
Row[{Style["t =",10],ToString[PaddedForm[N@Chop[tmax23],{2,2}]]}],TextAlignment->Left,FontSize->10],{tmax23,-0.04}],
Text[(* t *)
Style[
Row[{Style["(Subscript[v, out Max]) =",10],ToString[PaddedForm[N@Chop[vout23Max],{2,3}]]}],TextAlignment->Left,FontSize->10],{tmax23,vout23Max+0.02}]
}
]
}
]

Figure 2-9(b) Output response to a triangular pulse exciting an RC circuit with (Subscript[τ, c]) = 1 s

Note that the four expressions for (Subscript[v, out])(t) provide identical values at junctions between adjoining time intervals.

The convolution of two signals that contain no impulses must be a continuous signal. 
This is a good check when computing the convolution integral over multiple intervals.

2-3.4. Convolution of Two Rectangular Pulses

Given the rectangular waveforms x(t) and h(t) shown in Fig. 2-10, our goal is to evaluate their convolution.

To that end, we start by expressing x(t) and h(t) in terms of step functions:

x(t)=A(u(t)-u(t-(Subscript[T, 1]))) 
and
h(t)=B(u(t)-u(t-(Subscript[T, 2])))

Example of x(t) and h(t):

x234[t_,A_,T1_]:=A*(HeavisideTheta[t]-HeavisideTheta[t-T1]);
h234[t_,B_,T2_]:=B*(HeavisideTheta[t]-HeavisideTheta[t-T2])

Plot of x(t) and h(t):

A=5;T1=1.5;B=2.5;T2=3;
Show[
{
Plot[{x234[t,A,T1],h234[t,B,T2]},{t,-1,4}, PlotStyle->{{Thickness[0.008],RGBColor[0,0,0.8]},{Thickness[0.008],RGBColor[0.8,0,0]}},AxesLabel->{t,"x(t), h(t)"},Exclusions->None],
Graphics[{
Text[Style["x(t)",10],{T1+0.2,A-0.2}],
Text[Style["A",10],{-0.3,A}],
Text[Style["h(t)",10],{T2+0.2,B-0.2}],
Text[Style["B",10],{-0.3,B}]
}
]
}
]

Figure 2-10(a). x(t) and h(t) waveforms

To evaluate the second-line convolution integral given in Eq. (41), we need expressions for x(t – τ) and h(τ).

y(t)=u(t)(∫_0^t)x(t-τ)h(τ)dτ

Upon replacing the argument (t) in Eq. (43a) with (t – τ) and replacing t in Eq. (43b) with τ, we have

x(t)=A(u(t)-u(t-(Subscript[T, 1]))) 
h(t)=B(u(t)-u(t-(Subscript[T, 2])))

⇓

x(t-τ)=A(u(t-τ)-u(t-(Subscript[T, 1])-τ)) 
h(τ)=B(u(τ)-u(τ-(Subscript[T, 2])))

Inserting these expressions into Eq. (41) leads to

y(t)=u(t)(∫_0^t)x(t-τ)h(τ)dτ=
=A B u(t){(∫_0^t)u(t-τ)u(τ)dτ-(∫_0^t)u(t-τ)u(τ-(Subscript[T, 2]))dτ-(∫_0^t)u(t-(Subscript[T, 1])-τ)u(τ)dτ+(∫_0^t)u(t-(Subscript[T, 1])-τ)u(τ-(Subscript[T, 2]))dτ}

We will now examine each term individually.

Keeping in mind that the integration variable is τ, and because the upper integration limit is t, the difference (t – τ) is never smaller than zero (over the range of integration). Also, (Subscript[T, 1]) and (Subscript[T, 2]) are both non-negative numbers.

Term 1

Since over the range of integration, both u(τ) and u(t – τ) are equal to 1, the integral in the first term simplifies to

u(t)(∫_0^t)u(t-τ)u(τ)dτ= u(t)[(∫_0^t)dτ]=t u(t)

Term 2

The unit step function u(τ – (Subscript[T, 2])) is equal to zero unless τ > (Subscript[T, 2]), requiring that the lower integration limit be replaced with (Subscript[T, 2]) and the outcome be multiplied by u(t – (Subscript[T, 2])):

(∫_0^t)u(t-τ)u(τ-(Subscript[T, 2]))dτ= [(∫_(Subscript[T, 2])^t)dτ]u(t-(Subscript[T, 2]))=(t-(Subscript[T, 2]))u(t-(Subscript[T, 2]))

Term 3

The step function u(t – (Subscript[T, 1]) – τ) is equal to zero, unless τ < t – (Subscript[T, 1]), requiring that the upper limit be replaced with t - (Subscript[T, 1]).

Additionally, to satisfy the inequality, the smallest value that t can assume at the lower limit (τ = 0) is t = (Subscript[T, 1]).

Consequently, the outcome of the integration should be multiplied by u(t – (Subscript[T, 1])):

(∫_0^t)u(t-(Subscript[T, 1])-τ)u(τ)dτ= [(∫_0^(t-(Subscript[T, 1])))dτ]u(t-(Subscript[T, 1]))=(t-(Subscript[T, 1]))u(t-(Subscript[T, 1]))

Term 4

To accommodate the product

u(t-(Subscript[T, 1])-τ)u(τ-(Subscript[T, 2]))

we need to 
(a) change the lower limit to (Subscript[T, 2]), 
(b) change the upper limit to (t – (Subscript[T, 1])), and 
(c) multiply the outcome by u(t – (Subscript[T, 1]) – (Subscript[T, 2])):

(∫_0^t)u(t-(Subscript[T, 1])-τ)u(τ-(Subscript[T, 2]))dτ= 
=[(∫_(Subscript[T, 2])^(t-(Subscript[T, 1])))dτ]u(t-(Subscript[T, 1])-(Subscript[T, 2]))=
=(t-(Subscript[T, 1])-(Subscript[T, 2]))u(t-(Subscript[T, 1])-(Subscript[T, 2]))

Collecting the results given by Eqs. (46) through (49) gives

y(t)=A B [t u(t)-(t-(Subscript[T, 2]))u(t-(Subscript[T, 2]))-(t-(Subscript[T, 1]))u(t-(Subscript[T, 1]))+(t-(Subscript[T, 1])-(Subscript[T, 2]))u(t-(Subscript[T, 1])-(Subscript[T, 2]))]

The waveform of y(t) is displayed in Fig. 2-10(b).

Building on the experience gained from the preceding example, we can generalize the result as follows.

y234[y_]:=Convolve[x234[t,A,T1],h234[t,B,T2],t,y];

Show[
{
Plot[y234[y],{y,-1,5}, PlotStyle->{Thickness[0.008],RGBColor[0.5,0,0]},AxesLabel->{t,"y(t)"},Exclusions->None,PlotRange->All],
Graphics[{
{Thickness[0.003],RGBColor[1,0,0],Dashing[0.005],
Line[{{0,A*B*T1},{T1,A*B*T1}}]},
{Thickness[0.003],RGBColor[1,0,0],Dashing[0.005],
Line[{{T1,0},{T1,A*B*T1}}]},
{Thickness[0.003],RGBColor[1,0,0],Dashing[0.005],
Line[{{T2,0},{T2,A*B*T1}}]},
Text[Style["A·B·(Subscript[T, 1])",10],{-0.5,A*B*T1}],
Text[Style["y(t)=x(t)*h(t)",10],{(T1+T2)/2,A*B*T1*1.05}],
Text[Style["(Subscript[T, 1])",10],{T1,-2}],
Text[Style["(Subscript[T, 2])",10],{T2,-2}],
Text[Style["(Subscript[T, 1])+(Subscript[T, 2])",10],{T1+T2,-2}]
}
]
}
]

Figure 2-10(b) The convolution of the two rectangular waveforms x(t) and h(t) shown in 2-10(a).

Building on the experience gained from the preceding example, we can generalize the result as follows.

Convolution Integral

For functions x(t) and h(t) given by

x(t)=(Subscript[f, 1])(t)u(t-(Subscript[T, 1])) 
and
h(t)=(Subscript[f, 2])(t)u(t-(Subscript[T, 2]))

where (Subscript[f, 1])(t) and (Subscript[f, 2])(t) are any constants or time-dependent signals and (Subscript[T, 1]) and (Subscript[T, 2]) are any non-negative numbers, their convolution is

y(t)=x(t)⋆h(t)=u(t)(∫_0^t)x(t-τ)h(τ)dτ=
=(∫_0^t)(Subscript[f, 1])(t-τ)(Subscript[f, 2])(τ)u(t-(Subscript[T, 1])-τ)u(τ-(Subscript[T, 2])) dτ=
=[(∫_(Subscript[T, 2])^(t-(Subscript[T, 1])))(Subscript[f, 1])(t-τ)(Subscript[f, 2])(τ) dτ]u(t-(Subscript[T, 1])-(Subscript[T, 2]))

The convolution result represented by Eq. (52) will prove useful and efficient when analytically evaluating the convolution integral of signals described in terms of step functions.

Example 2-4. RC Circuit Response to Rectangular Pulse

Given the RC circuit shown in Fig. 2-11 (a), determine the output response to a 1 s long rectangular pulse. The pulse amplitude is 1 V.

Plot[UnitBox[t-0.5],{t,-0.5,1.5}, PlotStyle->{Thickness[0.007],RGBColor[0,0,0.5]},AxesLabel->{"t (s)","(Subscript[v, in])(t) (V)"},Exclusions->None]

Figure 2-11(a). RC lowpass filter and input pulse signal (Subscript[v, in])(t)

Solution:

The time constant of the RC circuit is (Subscript[τ, c]) = RC = (0.5×(10^6))×(10^(-6)) = 0.5 s.

In view of Eq. (17), the impulse response of the circuit is

h(t)=(1/(Subscript[τ, c]))(e^((-t)/(Subscript[τ, c])))u(t)=2(e^(-2t))u(t)

The input voltage is

(Subscript[v, in])(t)=[u(t)-u(t-1)]

Application of Eq. (41) to the expressions for h(t) and (Subscript[v, in])(t), given respectively by Eqs. (53) and (54), gives

(Subscript[v, out])(t)=(Subscript[v, in])(t)⋆h(t)=u(t)(∫_0^t)(Subscript[v, in])(τ)h(t-τ)dτ=
=u(t)(∫_0^t)[u(τ)-u(τ-1)] ×2(e^(-2(t-τ)))u(t-τ)dτ=
=u(t)(∫_0^t)2(e^(-2(t-τ)))u(τ)u(t-τ)dτ-u(t)(∫_0^t)2(e^(-2(t-τ)))u(τ-1)u(t-τ)dτ

Upon application of the recipe described by Eq. (52), (Subscript[v, out])(t) becomes

(Subscript[v, out])(t)=(Subscript[v, in])(t)⋆h(t)=
=[(∫_0^t)2(e^(-2(t-τ)))dτ]u(t)-[(∫_1^t)2(e^(-2(t-τ)))dτ]u(t-1)=
=(Subscript[[(2/2)(e^(-2(t-τ)))], 0]^t)u(t)-(Subscript[[(2/2)(e^(-2(t-τ)))], 1]^t)u(t-1)=
=[1-(e^(-2t))]u(t)-[1-(e^(-2(t-1)))]u(t-1)

where we reintroduced the unit step functions u(t) and u(t – 1) associated with the two integration terms.
Figure 2-11(b) displays the temporal waveform of (Subscript[v, out])(t)

x24[t_]:=UnitBox[t-0.5];
h24[t_]:=2 (E^(-2 t))UnitStep[t];

Show[
{
Plot[{x24[t],h24[t]},{t,-0.5,2.5}, PlotStyle->Thickness[0.007],AxesLabel->{"t (s)","(Subscript[v, in])(t), h(t) (V)"},Exclusions->None],
Graphics[{
Text[Style["(Subscript[v, in])(t)",10],{0.5,1.1}],
Text[Style["h(t)",10],{0.2,1.8}]
}
]
}
]

Input pulse signal (Subscript[v, in])(t) and impulse response h(t)

y24[y_]:=Convolve[x24[t],h24[t],t,y]

Show[
{
Plot[{y24[y],UnitBox[y-0.5]},{y,-0.5,2.5}, PlotStyle->Thickness[0.007],AxesLabel->{"t (s)","(Subscript[v, in])(t), (Subscript[v, out])(t) (V)"},Exclusions->None,PlotRange->All],
Graphics[{
Text[Style["(Subscript[v, in])(t)",10],{0.5,1.05}],
Text[Style["(Subscript[v, out])(t)=(Subscript[v, in])(t)⋆h(t)",10],{1.7,0.6}]
}
]
}
]

Figure 2-11(b). Input pulse signal (Subscript[v, in])(t) and output response (Subscript[v, out])(t)

2-4. Graphical Convolution

The convolution integral given by

y(t)=(∫_0^t)x(τ)h(t-τ)dτ

can be evaluated graphically by computing it at successive values of t.

Graphical Convolution Algorithm
Step 1. On the τ-axis, display x(τ) and h(-τ) with the latter being a folded image of h(τ) about the vertical axis.
Step 2. Shift h(-τ) to the right by a small increment t to obtain h(t – τ) = h(–(τ – t)).
Step 3. Determine the product of x(τ) and h(t – τ) and integrate it over the τ-domain from τ = 0 to τ = t to get y(t). The integration is equal to the area overlapped by the two functions.
Step 4. Repeat steps 2 and 3 for each of many successive values of t to generate the complete response y(t).

By way of illustration, let us consider the RC circuit of Example 2-4.

Plots of the input excitation (Subscript[v, in])(t) and the impulse response h(t) are displayed in Fig. 2-12(a).

To perform the convolution

(Subscript[v, out])(t)=(Subscript[v, in])(t)⋆h(t)=u(t)(∫_0^t)(Subscript[v, in])(τ)h(t-τ)dτ

we need to plot (Subscript[v, in])(τ) and h(t – τ) along the τ axis.

Show[
{
Plot[{x24[t],h24[t]},{t,-0.5,2.5}, PlotStyle->Thickness[0.007],AxesLabel->{"t (s)","(Subscript[v, in])(t), h(t) (V)"},Exclusions->None],
Graphics[{
Text[Style["(Subscript[v, in])(t)",10],{0.5,1.1}],
Text[Style["h(t)",10],{0.2,1.8}]
}
]
}
]

Figure 2-12(a).  Input excitation (Subscript[v, in])(t) and the circuit impulse response h(t).

Figure 2-12(b) through (e) show these plots at progressive values of t, starting at t = 0 and concluding at t = 2 s.

In all cases, (Subscript[v, in])(τ) remains unchanged, but h(t – τ) is obtained by “folding” the original function across the vertical axis to generate h(–τ) and then shifting it to the right along the τ-axis by an amount t.

The output voltage is equal to the integrated product of (Subscript[v, in])(τ) and h(t – τ), which is equal to the shaded overlapping areas in the figures.

At t = 0 (Fig. 2-12(b)), no overlap exists; hence, (Subscript[v, out])(0) = 0.

Sliding h(t – τ) by t = 0.5 s, as shown in Fig. 2-12(c), leads to (Subscript[v, out])(0.5) = 0.63.

Sliding h(t – τ) further to the right leads to a greater overlap, reaching a maximum at t = 1 s (Fig. 2-12(d)).

Beyond t = 1 s, the overlap is smaller in area, as illustrated by part (e) of Fig. 2-12, which corresponds to a shift t = 2 s.

If the values of (Subscript[v, out])(t), as determined through this graphical integration method at successive values of t, are plotted as a function of t, we would get the same circuit response curve shown earlier in Fig. 2-11(b).

Show[
{
Plot[{x24[τ],h24[-τ]},{τ,-1,3}, PlotStyle->Thickness[0.007],AxesLabel->{"τ","(Subscript[v, in])(τ), h(-τ)"},Exclusions->None],
Graphics[{
Text[Style["(Subscript[v, in])(τ)",10],{0.5,1.1}],
Text[Style["h(-τ)",10],{0.2,1.9}]
}
]
}
]

Figure 2-12(b). Convolution  (Subscript[v, in])(τ)*h(t – τ) at t = 0

Show[
{
Plot[{x24[τ],h24[-τ+0.5]},{τ,-1.0,3}, PlotStyle->Thickness[0.007],AxesLabel->{"τ","(Subscript[v, in])(τ), h(t-τ)"},Exclusions->None],
Graphics[{
Text[Style["(Subscript[v, in])(τ)",10],{1,1.1}],
Text[Style["h(0.5-τ)",10],{0.8,1.9}]
}
]
}
]

Figure 2-12(c). Convolution  (Subscript[v, in])(τ)*h(t – τ) at t = 0.5

Circuit response at t = 0.5 equals to overlap area:

(Subscript[v, out])(0.5)=(∫_0^0.5)(Subscript[v, in])(τ) h(0.5-τ)dτ=0.63

(∫_0^0.5)x24[τ] h24[0.5-τ]\[DifferentialD]τ

Show[
{
Plot[{x24[τ],h24[-τ+1]},{τ,-1.0,3}, PlotStyle->Thickness[0.007],AxesLabel->{"τ","(Subscript[v, in])(τ), h(t-τ)"},Exclusions->None],
Graphics[{
Text[Style["(Subscript[v, in])(τ)",10],{0.5,1.1}],
Text[Style["h(1-τ)",10],{0.7,1.9}]
}
]
}
]

Figure 2-12(d). Convolution  (Subscript[v, in])(τ)*h(t – τ) at t = 1

Circuit response at t = 1 equals to overlap area:

(Subscript[v, out])(1)=(∫_0^1)(Subscript[v, in])(τ) h(0.5-τ)dτ=0.86

(∫_0^1)x24[τ] h24[1-τ]\[DifferentialD]τ

Show[
{
Plot[{x24[τ],h24[-τ+2]},{τ,-1.0,3}, PlotStyle->Thickness[0.007],AxesLabel->{"τ","(Subscript[v, in])(τ), h(t-τ)"},Exclusions->None],
Graphics[{
Text[Style["(Subscript[v, in])(τ)",10],{0.5,1.1}],
Text[Style["h(1-τ)",10],{1.7,1.9}]
}
]
}
]

Figure 2-12(e). Convolution  (Subscript[v, in])(τ)*h(t – τ) at t = 2

Circuit response at t = 2 equals to overlap area:

(Subscript[v, out])(2)=(∫_0^2)(Subscript[v, in])(τ) h(2-τ)dτ=0.12

(∫_0^2)x24[τ] h24[2-τ]\[DifferentialD]τ

Demonstration. Graphical convolution solution for RC circuit of Example 2-4 (Section 2-3.4).

x24a[τ_]:=UnitBox[τ-0.5];
h24a[τ_]:=2 (E^(-2 τ))UnitStep[τ];
y24a[t_]=(∫_0^t)x24a[τ] h24a[t-τ]\[DifferentialD]τ;
Manipulate[
GraphicsColumn[{
Show[{
Plot[{x24a[τ]*h24a[-τ+t]},{τ,-2,3}, Filling->Axis,PlotRange->{{-2,3},{-0.1,2.5}},AxesLabel->{"τ","(Subscript[v, in])(τ), h(t-τ)"},ClippingStyle->False,PlotPoints->100],
Plot[{x24a[τ],h24a[-τ+t]},{τ,-2,3}, PlotStyle->Thickness[0.008],Exclusions->None,ClippingStyle->False,PlotPoints->100],
Graphics[{
Text[Style["(Subscript[v, in])(τ)",10],{0.5,1.1}],
Text[Style["h(t-τ)",10],{t,2.1}]
}]}],
Show[{
Plot[y24a[x],{x,-2,t}, PlotStyle->{Thickness[0.007],Blue},Exclusions->None,AxesLabel->{"t","(Subscript[v, out])(t)"},PlotRange->{{-2,3},{-0.1,1}}],
Graphics[{PointSize[Large],Blue, Point[{t, y24a[t]}]}]
}]
}],
{{t,-1.5,"Time t"},-1.5,4},
Paneled->False,
FrameLabel->{None,None,Style["Demonstration: graphical convolution",16,FontFamily->"Times"]},
ControlPlacement->Bottom,
LabelStyle->Directive[RGBColor[0,0,0.5],12,Bold,FontFamily->"Times"],
Frame->True
]

Example 2-5. Graphical Convolution

Given the waveforms shown in Fig. 2-13(a).
Apply the graphical convolution technique to determine the response y(t) = x(t)⋆h(t).

x25[τ_]:=UnitTriangle[τ-1];
h25[τ_]:=UnitBox[τ-0.5]-UnitBox[τ-1.5];
y25[t_]=(∫_0^t)x25[τ] h25[t-τ]\[DifferentialD]τ;
y25b[y_]:=Convolve[x25[x],h25[x],x,y];

GraphicsColumn[
{
Plot[h25[t],{t,-1,3},PlotRange->{{-1,3},{-1.2,1.2}},PlotStyle->{Thickness[0.007],RGBColor[0,0,0.8]},AxesLabel->{"t","h(t)"}],
Plot[x25[t],{t,-1,3},PlotRange->{{-1,3},{-1.2,1.2}},PlotStyle->{Thickness[0.007],RGBColor[0.8,0,0]},AxesLabel->{"t","x(t)"}]
}]

Figure 2-13(a). x(t) and h(t) for Example 2-5.

Figure 2-13(b) shows waveforms x(τ) and h(–τ), plotted along the τ-axis. The waveform h(–τ) is the mirror image of h(τ) with respect to the vertical axis.

Show[
{
Plot[{x25[τ],h25[-τ]},{τ,-2.5,2.5},PlotRange->{{-2.5,2.5},{-1.2,1.2}},PlotStyle->{{Thickness[0.007],RGBColor[0.8,0,0]},{Thickness[0.007],RGBColor[0,0,0.8]}},AxesLabel->{"τ","x(τ), h(t-τ)"},Exclusions->None],
Graphics[{
Text[Style["x(τ)",10],{1,1.1}],
Text[Style["h(-τ)",10],{-0.5,1.1}]
}
]
}
]

Figure 2-13(b). x(τ) and h(t – τ) for t = 0. y(0) = 0. (Example 2-5)

In Fig. 2-13(c) through (e), h(t – τ) is plotted for t = 1 s, 1.5 s, and 2 s, respectively.

Show[
{
Plot[{x25[τ]*h25[-τ+1]},{τ,-2,3}, Filling->Axis,PlotRange->{{-2,4},{-1.5,1.5}},AxesLabel->{"τ","(Subscript[v, in])(τ), h(t-τ)"},PlotStyle->{Thickness[0],RGBColor[0.9,0.9,0.9]}],
Plot[{x25[τ],h25[1-τ]},{τ,-2.5,2.5},PlotRange->{{-2.5,2.5},{-1.2,1.2}},PlotStyle->{{Thickness[0.007],RGBColor[0.8,0,0]},{Thickness[0.007],RGBColor[0,0,0.8]}},AxesLabel->{"τ","x(τ), h(t-τ)"},Exclusions->None],
Graphics[{
Text[Style["x(τ)",10],{1.2,1.1}],
Text[Style["h(-τ)",10],{0.5,1.1}]
}
]
}
]

Figure 2-13(c). x(τ) and h(t – τ) for t = 1 s. y(1) = 0.5. (Example 2-5)

Show[
{
Plot[{x25[τ]*h25[-τ+1.5]},{τ,-2,4}, Filling->Axis,PlotRange->{{-2,4},{-1.5,1.5}},AxesLabel->{"τ","(Subscript[v, in])(τ), h(t-τ)"},PlotStyle->{Thickness[0],RGBColor[0.9,0.9,0.9]}],
Plot[{x25[τ],h25[1.5-τ]},{τ,-2.5,2.5},PlotRange->{{-2.5,2.5},{-1.2,1.2}},PlotStyle->{{Thickness[0.007],RGBColor[0.8,0,0]},{Thickness[0.007],RGBColor[0,0,0.8]}},AxesLabel->{"τ","x(τ), h(t-τ)"},Exclusions->None],
Graphics[{
Text[Style["x(τ)",10],{1.2,1.1}],
Text[Style["h(-τ)",10],{0.5,1.1}]
}
]
}
]

Figure 2-13(d). x(τ) and h(t – τ) for t = 1.5 s. y(1.5) = (6/8)-(1/8)=(5/8). (Example 2-5)

Show[
{
Plot[{x25[τ],h25[2-τ]},{τ,-2.5,2.5},PlotRange->{{-2,4},{-1.2,1.2}},PlotStyle->{{Thickness[0.007],RGBColor[0.8,0,0]},{Thickness[0.007],RGBColor[0,0,0.8]}},AxesLabel->{"τ","x(τ), h(t-τ)"},Exclusions->None],
Graphics[{
Text[Style["x(τ)",10],{0.8,1.1}],
Text[Style["h(-τ)",10],{1.5,1.1}]
}
]
}
]

Figure 2-13(e). x(τ) and h(t – τ) for t = 2 s. y(2) = 0. (Example 2-5)

In each case, the shaded area is equal to y(t). We note that when t > 1 s, one of the shaded areas contributes a positive number to y(t) while the other contributes a negative number.

Plot[{x25[t],h25[-t],y25[t]},{t,-3,6},PlotRange->All,PlotStyle->Thickness[0.007],AxesLabel->{"τ, t","x(τ), h(t-τ), y(t)"}]

Figure. Input x(τ), h(t – τ) and the output response y(t) = x(t)⋆h(t) for Example 2-5.

The overall resultant response y(t) generated by this process of sliding h(t – τ) to the right is displayed in Fig. 2-13(f).

Plot[y25b[t],{t,-1,5},PlotRange->{{-1,5},{-1,1}},PlotStyle->Thickness[0.007],AxesLabel->{"t","y(t)"}]

Figure 2-13(f). Solution y(t) for Example 2-5.

Demonstration. Graphical convolution solution for Example 2-5.

x25a[x_]:=UnitTriangle[x-1];
h25a[x_]:=UnitBox[x-0.5]-UnitBox[x-1.5];
y25a[y_]=(∫_0^y)x25a[τ] h25a[y-τ]\[DifferentialD]τ;
Manipulate[
GraphicsColumn[{
Show[{
Plot[{x25a[τ]*h25a[-τ+t]},{τ,-2,5}, Filling->Axis,PlotRange->{{-2,5},{-1.5,1.5}},AxesLabel->{"τ","(Subscript[v, in])(τ), h(t-τ)"},ClippingStyle->False,PlotPoints->50],
Plot[{x25a[τ],h25a[-τ+t]},{τ,-2,5}, PlotStyle->Thickness[0.008],Exclusions->None,ClippingStyle->False,PlotPoints->50],
Graphics[{
Text[Style["(Subscript[v, in])(τ)",10],{0.5,1.1}],
Text[Style["h(t-τ)",10],{t,2.1}]
}]}],
Show[{
Plot[y25a[x],{x,-2,t}, PlotStyle->{Thickness[0.007],Blue},Exclusions->None,AxesLabel->{"t","(Subscript[v, out])(t)"},PlotRange->{{-2,5},{-1,1}}],
Graphics[{PointSize[Large],Blue, Point[{t, y25a[t]}]}]
}]
}],
{{t,-1.5,"Time t"},-1.5,5},
Paneled->False,
FrameLabel->{None,None,Style["Example 2-5: graphical convolution",16,FontFamily->"Times"]},
ControlPlacement->Bottom,
LabelStyle->Directive[RGBColor[0,0,0.5],12,Bold,FontFamily->"Times"],
Frame->True
]

Exercise 2-6. Graphical Convolution

Apply graphical convolution to the waveforms of x(t) and h(t) shown in Fig. E2-6 to determine y(t) = x(t)⋆h(t).

(* Declaration of original pulses *)h26[τ_]:=τ*(UnitStep[τ]-UnitStep[τ-3]);
x26[τ_]:=4*UnitBox[τ-1.5];
(* Concolution of original pulses *)
y26[t_]=(∫_0^t)x26[τ] h26[t-τ]\[DifferentialD]τ;
y26b[y_]:=Convolve[x26[x],h26[x],x,y];

(* Plots of original pulses *)
GraphicsColumn[
{
Plot[h26[t],{t,-0.2,3.5},PlotRange->{{-0.5,4},{-1.2,4}},Exclusions->None,PlotStyle->{Thickness[0.007],RGBColor[0,0,0.8]},AxesLabel->{"t","h(t)"}],
Plot[x26[t],{t,-0.2,3.5},PlotRange->{{-0.5,4},{-1.2,4.2}},Exclusions->None,PlotStyle->{Thickness[0.007],RGBColor[0.8,0,0]},AxesLabel->{"t","x(t)"}]
}]

Figure E2-6. x(t) and h(t) for Exercise 2-6.

Show[
{
Plot[{x26[τ],h26[-τ]},{τ,-4,5},PlotRange->{{-5,6},{-0.2,5}}, PlotStyle->Thickness[0.007],AxesLabel->{"τ","(Subscript[v, in])(τ), h(t-τ)"},Exclusions->None],
Graphics[{
Text[Style["(Subscript[v, in])(τ)",10],{1.5,4.2}],
Text[Style["h(-τ)",10],{-3,3.2}]
}
]
}
]

Figure E2-6(a). x(τ) and h(t–τ) at t = 0 s, overlap = 0.

Show[
{
Plot[{x26[τ],h26[1-τ]},{τ,-4,5},PlotRange->{{-5,6},{-0.2,5}}, PlotStyle->Thickness[0.007],AxesLabel->{"τ","(Subscript[v, in])(τ), h(t-τ)"},Exclusions->None],
Graphics[{
Text[Style["(Subscript[v, in])(τ)",10],{1.5,4.2}],
Text[Style["h(1-τ)",10],{-2,3.2}]
}
]
}
]

Figure E2-6(b). x(τ) and h(t–τ) at t = 1 s, overlap = 0.

Show[
{
Plot[{x26[τ],h26[2-τ]},{τ,-4,5},PlotRange->{{-5,6},{-0.2,5}}, PlotStyle->Thickness[0.007],AxesLabel->{"τ","(Subscript[v, in])(τ), h(t-τ)"},Exclusions->None],
Graphics[{
Text[Style["(Subscript[v, in])(τ)",10],{1.5,4.2}],
Text[Style["h(2-τ)",10],{-1,3.2}]
}
]
}
]

Figure E2-6(c). x(τ) and h(t–τ) at t = 2 s, overlap = 1/2 × 4 = 2.

Show[
{
Plot[{x26[τ],h26[3-τ]},{τ,-4,5},PlotRange->{{-5,6},{-0.2,5}}, PlotStyle->Thickness[0.007],AxesLabel->{"τ","(Subscript[v, in])(τ), h(t-τ)"},Exclusions->None],
Graphics[{
Text[Style["(Subscript[v, in])(τ)",10],{1.5,4.2}],
Text[Style["h(3-τ)",10],{-0.8,3.2}]
}
]
}
]

Figure E2-6(d). x(τ) and h(t–τ) at t = 3 s, overlap = 1.5 × 4 = 6.

Show[
{
Plot[{x26[τ],h26[4-τ]},{τ,-4,5},PlotRange->{{-5,6},{-0.2,5}}, PlotStyle->Thickness[0.007],AxesLabel->{"τ","(Subscript[v, in])(τ), h(t-τ)"},Exclusions->None],
Graphics[{
Text[Style["(Subscript[v, in])(τ)",10],{1.5,4.2}],
Text[Style["h(4-τ)",10],{0.5,3.2}]
}
]
}
]

Figure E2-6(e). x(τ) and h(t–τ) at t = 4 s, overlap = 2.5 × 4 = 10.

Show[
{
Plot[{x26[τ],h26[5-τ]},{τ,-4,5.5},PlotRange->{{-5,6},{-0.2,5}}, PlotStyle->Thickness[0.007],AxesLabel->{"τ","(Subscript[v, in])(τ), h(t-τ)"},Exclusions->None],
Graphics[{
Text[Style["(Subscript[v, in])(τ)",10],{1.5,4.2}],
Text[Style["h(5-τ)",10],{2.8,3.2}]
}
]
}
]

Figure E2-6(e). x(τ) and h(t–τ) at t = 5 s, overlap = 0.

Plot[y26b[t],{t,-1,6},PlotRange->{{-1,7},{-0.1,11}},PlotStyle->Thickness[0.007],AxesLabel->{"t","y(t)"}]

Figure E2-6(g). Output response y(t).

Demonstration. Graphical convolution solution for Exercise 2-6.

Note: function definition code in Exercise 2-6 has to be executed prior to this demonstration.

h26[τ_]:=τ*(UnitStep[τ]-UnitStep[τ-3]);
x26[τ_]:=4*UnitBox[τ-1.5];
(* Concolution of original pulses *)
y26[t_]=(∫_0^t)x26[τ] h26[t-τ]\[DifferentialD]τ;
y26b[y_]:=Convolve[x26[x],h26[x],x,y];
Manipulate[
GraphicsColumn[{
Show[{
Plot[{x26[τ]*h26[-τ+t]},{τ,-5,6}, Filling->Axis,PlotRange->{{-6,7},{-0.5,12}},AxesLabel->{"τ","(Subscript[v, in])(τ), h(t-τ)"},ClippingStyle->False,PlotPoints->50],
Plot[{x26[τ],h26[-τ+t]},{τ,-5,6}, PlotStyle->Thickness[0.008],Exclusions->None,ClippingStyle->False,PlotPoints->50],
Graphics[{
Text[Style["(Subscript[v, in])(τ)",10],{1.5,4.6}],
Text[Style["h(t-τ)",10],{t-3,3.6}]
}]}],
Show[{
Plot[y26[x],{x,-2,t}, PlotStyle->{Thickness[0.007],Blue},Exclusions->None,AxesLabel->{"t","(Subscript[v, out])(t)"},PlotRange->{{-1,7},{-0.5,12}}],
Graphics[{PointSize[Large],Blue, Point[{t, y26[t]}]}]
}]
}],
{{t,-0.5,"Time t"},-0.5,6},
Paneled->False,
FrameLabel->{None,None,Style["Exercise 2-6: graphical convolution",16,FontFamily->"Times"]},
ControlPlacement->Bottom,
LabelStyle->Directive[RGBColor[0,0,0.5],12,Bold,FontFamily->"Times"],
Frame->True
]

2-5. Convolution Properties

Computation of convolutions can be greatly simplified by using the ten properties outlined in this section.

In fact, in many cases the convolutions can be determined without computing any integrals.

Also, to help the user with both the computation and understanding of the convolution operation, we will attach a physical interpretation to each of the ten properties.

2-5.1. Commutative Property

Commutative property states that

y(t)=h(t)⋆x(t)=x(t)⋆h(t)

Its validity can be demonstrated by changing variables from τ to τ' = t - τ in the convolution integral:

y(t)=h(t)⋆x(t)=(∫_(-∞)^∞)h(τ)x(t-τ)dτ=
=(∫_∞^(-∞))h(t-τ')x(τ')(-dτ')=
=(∫_(-∞)^∞)x(τ')h(t-τ')dτ'=
=x(t)⋆h(t)

Physically, the commutative property means that we can interchange the input and the impulse response of the LTI system without changing its output:

x(t) -> h(t) -> y(t) 
⇑⇓
h(t) -> x(t) -> y(t)

Computationally, we can choose to time-shift and reverse whichever of x(t) and h(t) that makes the computation or graphical calculation of the convolution integral easier.

2-5.2. Associative Property

The combined convolution of three functions is the same, regardless of the order in which the convolution is performed:

y(t)=[g(t)⋆h(t)]⋆x(t)=g(t)⋆[h(t)⋆x(t)]

Proving this property in the time domain requires much algebra, but proving it after introducing the Laplace transform in Chapter 3 is straightforward.

Hence, we will accept it at face value for the time being.

Physically, the associative property implies that the impulse response of two LTI systems connected in series is the convolution of their individual impulse responses

To see why, consider two LTI systems with impulse responses (Subscript[h, 1])(t) and (Subscript[h, 2])(t) connected in series:

x(t) -> (Subscript[h, 1])(t) -> z(t) -> (Subscript[h, 2])(t) -> y(t)

where z(t) is an intermediate signal between the two systems.

Separately, the two systems are characterized by

z(t)=x(t)⋆(Subscript[h, 1])(t)
y(t)=z(t)⋆(Subscript[h, 2])(t)

Combining the two parts of Eq.(61) gives

y(t)=[x(t)⋆(Subscript[h, 1])(t)]⋆(Subscript[h, 2])(t)

The associative property allows us to rewrite Eq.(62) as

y(t)=x(t)⋆[(Subscript[h, 1])(t)⋆(Subscript[h, 2])(t)]

and to represent the process symbolically (Fig. 2-14) as

x(t) -> (Subscript[h, 1])(t)⋆(Subscript[h, 2])(t)  -> y(t)

The joint impulse response of two series-connected LTI systems is equivalent to the convolution of their individual impulse responses carried out in either order. 
A series connection of two systems also is known as a cascade connection.

Systems Connected In-Series
x(t) -> (Subscript[h, 1])(t) ->  (Subscript[h, 2])(t) -> y(t) 
⇓
x(t) -> (Subscript[h, 1])(t)⋆(Subscript[h, 2])(t)  -> y(t)

Figure 2-14. The impulse response of two systems connected in series is equivalent to the convolution of their individual impulse responses.

2-5.3. Distributive Property

The distributive property, which follows directly from the definition of convolution, allows us to perform the convolution operation on the sum of multiple input signals as

h(t)⋆[(Subscript[x, 1])(t)+(Subscript[x, 2])(t)+...+(Subscript[x, N])(t)]=h(t)⋆(Subscript[x, 1])(t)+h(t)⋆(Subscript[x, 2])(t)+...+h(t)⋆(Subscript[x, N])(t)

Conversely, if it is possible to model a complicated impulse response h(t) as the sum of two or more simpler impulse responses,

h(t)=(Subscript[h, 1])(t)+(Subscript[h, 2])(t)+...+(Subscript[h, N])(t)

then application of the distributive property serves to simplify computation of the convolution integral by replacing a computation involving the complicated impulse response with multiple, but simpler, computations involving its constituent components.

That is,

x(t)⋆h(t)=x(t)⋆(Subscript[h, 1])(t)+x(t)⋆(Subscript[h, 2])(t)+...+x(t)⋆(Subscript[h, N])(t)

The impulse response of the parallel connection of two or more LTI systems is the sum of their impulse responses.

Systems Connected In-Parallel

⟶ (Subscript[h, 1])(t) ⟶  
↑                        ↓
x(t)     ⟶  (Subscript[h, 2])(t) ⟶ ⊕  y(t) 
↓         ⋯           ↑
⟶ (Subscript[h, N])(t) ⟶

⇓
x(t) ⟶ (Subscript[h, 1])(t)+(Subscript[h, 2])(t)+⋯+(Subscript[h, N])(t) ⟶ y(t)

Figure 2-15. The impulse response h(t) of multiple LTI systems connected in parallel is equivalent to the sum of their individual impulse responses.

In combination, the associative and distributive properties state that for LTI systems:

(1) Systems in series: Impulse responses convolved.
(2) Systems in parallel: Impulse responses added.

Example 2-6. Four Interconnected Systems

Determine the overall impulse response h(t) of the LTI system depicted in Fig. 2-16.

Figure 2-16. System of Example 2-6.

Solution:

Since (Subscript[h, 1])(t) and (Subscript[h, 2])(t) are in series, their joint impulse response (Subscript[h, 1])(t)⋆(Subscript[h, 2])(t).

Similarly, for the lower branch, the joint response is (Subscript[h, 3])(t)⋆(Subscript[h, 4])(t).

The two branches are in parallel, so their combined impulse response is the sum of the impulse responses of the individual branches:

h(t)=(Subscript[h, 1])(t)⋆(Subscript[h, 2])(t)+(Subscript[h, 3])(t)⋆(Subscript[h, 4])(t)

2-5.4. Causal ⋆ Causal = Causal

A signal x(t) is causal if x(t) = 0 for t < 0.

The output y(t) of a real-world causal system can depend on only present and past values of its input signal x(t).

Hence, if its input is zero for t < 0, its output also must be zero for t < 0.

This argument led us in Section 2-3.2 to change the limits of the convolution integral to [0, t] and to multiply the integral by u(t).

y(t)=u(t)(∫_0^t)h(τ)x(t-τ)dτ=u(t)(∫_0^t)x(τ)h(t-τ)dτ

Convolution of a causal signal with another causal signal or with a causal impulse response of an LTI system generates a causal output. 
Moreover, the impulse response of a causal LTI system is causal.

2-5.5. Time-Shift Property

Given the convolution integral

y(t)=h(t)⋆x(t)=(∫_(-∞)^∞)h(τ)x(t-τ)dτ

the convolution of h(t) delayed by (Subscript[T, 1]) and x(t) delayed by (Subscript[T, 2]) is

h(t-(Subscript[T, 1]))⋆x(t-(Subscript[T, 2]))=(∫_(-∞)^∞)h(τ-(Subscript[T, 1]))x(t-(Subscript[T, 2])-τ)dτ

Introducing the dummy variable τ'=t-(Subscript[T, 1]) everywhere inside the integral leads to

h(t-(Subscript[T, 1]))⋆x(t-(Subscript[T, 2]))=(∫_(-∞)^∞)h(τ')x(t-(Subscript[T, 1])-(Subscript[T, 2])-τ')dτ'=
=(∫_(-∞)^∞)h(τ')x(t'-τ')dτ'

where we introduced the time-shifted variable t'=t-(Subscript[T, 1])-(Subscript[T, 2]).

The form of the integral in Eq. (70) matches the definition of the convolution integral given by Eq. (68), except that t has been replaced with t'.

Hence, the output y(t) becomes

y(t')=y(t-(Subscript[T, 1])-(Subscript[T, 2]))

In conclusion, time-shift property

h(t-(Subscript[T, 1]))⋆x(t-(Subscript[T, 2]))=y(t-(Subscript[T, 1])-(Subscript[T, 2]))

Physically, the time-shift property is just a statement of time-invariance:

delaying the input of a time-invariant system and/or if the system has a built-in delay causes the output to be delayed by the sum of the two delays

Computationally, the shift property allows us to delay or advance signals to take advantage of their symmetry or causality to make the convolution operation simpler.

Example 2-7. Convolution of Two Delayed Rectangular Pulses

Compute the convolution of rectangular pulses (Subscript[x, 1])(t) and (Subscript[x, 2])(t) shown in Fig. 2-17(a).

(* Declaration of original pulses *)
x271[τ_]:=(UnitStep[τ-3]-UnitStep[τ-5]);
x272[τ_]:=(UnitStep[τ-1]-UnitStep[τ-2]);
(* Concolution of original pulses *)
y27[t_]=(∫_0^t)x271[τ] x272[t-τ]\[DifferentialD]τ;
y27b[y_]=Convolve[x271[x],x272[x],x,y];

(* Plots of original pulses *)
GraphicsColumn[
{
Plot[x271[t],{t,-0.2,6},PlotRange->{{-0.5,6},{-0.2,1.2}},Exclusions->None,PlotStyle->{Thickness[0.007],RGBColor[0,0,0.8]},AxesLabel->{"t","(Subscript[x, 1])(t)"}],
Plot[x272[t],{t,-0.2,6},PlotRange->{{-0.5,6},{-0.2,1.2}},Exclusions->None,PlotStyle->{Thickness[0.007],RGBColor[0.8,0,0]},AxesLabel->{"t","(Subscript[x, 2])(t)"}]
}]

Figure 2-17(a). Original pulses  (Example 2-7).

Solution:

Our plan is to take advantage of as many convolution properties as possible. We start by shifting  (Subscript[x, 1])(t) and (Subscript[x, 2])(t)  to the left so that they both start at t = 0 (Fig. 2-17(b)).

The shifted pulses are:

(Subscript[x, 1]^('))(t)=(Subscript[x, 1])(t+3)=u(t)-u(t-2)
(Subscript[x, 2]^('))(t)=(Subscript[x, 2])(t+1)=u(t)-u(t-1)

The causality property can be used to find convolution

u(t)⋆u(t)=u(t)(∫_0^t)1dτ=t u(t)=r(t)

(* Declaration of shifted to start pulses *)
x271prime[τ_]:=(UnitStep[τ]-UnitStep[τ-2]);
x272prime[τ_]:=(UnitStep[τ]-UnitStep[τ-1]);
(* Concolution of shifted to start pulses *)
y27prime[t_]=(∫_0^t)x271prime[τ] x272prime[t-τ]\[DifferentialD]τ;
y27bprime[y_]=Convolve[x271prime[x],x272prime[x],x,y];

(* Plots of shifted to start pulses *)
GraphicsColumn[
{
Plot[x271prime[t],{t,-0.2,6},PlotRange->{{-0.5,6},{-0.2,1.2}},Exclusions->None,PlotStyle->{Thickness[0.007],RGBColor[0,0,0.8]},AxesLabel->{"t","(Subscript[x, 1])'(t)"}],
Plot[x272prime[t],{t,-0.2,6},PlotRange->{{-0.5,6},{-0.2,1.2}},Exclusions->None,PlotStyle->{Thickness[0.007],RGBColor[0.8,0,0]},AxesLabel->{"t","(Subscript[x, 2])'(t)"}]
}]

Figure 2-17(b). Pulses shifted to start at t = 0  (Example 2-7).

Convolution y'(t) of shifted pulses  (Subscript[x, 1]^('))(t) and (Subscript[x, 2]^('))(t) is

y'(t)=(Subscript[x, 1]^('))(t)⋆(Subscript[x, 2]^('))(t)=r(t)-r(t-1)-r(t-2)+r(t-3)

which is plotted in Fig. 2-17(c).

Plot[y27bprime[t],{t,-1,8},PlotRange->All,PlotStyle->Thickness[0.007],AxesLabel->{"t","y(t)"}]

Figure 2-17(c). y'(t), convolution of (Subscript[x, 1]^('))(t) and (Subscript[x, 2]^('))(t)  (Example 2-7).

Finally, we apply the time-shift property again to restore the combined shift of 4 s that was applied earlier to  (Subscript[x, 1])(t) and (Subscript[x, 2])(t) .

y(t)=y'(t+4)

The result is shown in Fig. 2-17(d).

Plot[y27b[t],{t,-1,8},PlotRange->All,PlotStyle->Thickness[0.007],AxesLabel->{"t","y(t)"}]

Figure 2-17(d). y'(t) delayed by 4 s gives y(t) = y'(t + 4) (Example 2-7).

2-5.6. Convolution with an Impulse

Convolution of a signal x(t) with an impulse function δ(t - T) simply delays x(t) by T:

x(t)⋆δ(t-T)=u(t)(∫_(-∞)^∞)x(τ)δ(t-T-τ)dτ=x(t-T)

The result follows from the sampling property of the impulse function that δ(t - T - τ) is zero except at τ = t - T.

Example 2-8. Four Interconnected Systems

Compute the overall impulse response h(t) of the system diagrammed in Fig. 2-18.

Figure 2-18. System of Example 2-8.

Solution:

In the top branch, u(t) and δ(t + 1) are in series, so their joint impulse response is

(Subscript[h, 1])(t)=u(t)⋆δ(t+1)=u(t+1)

In the lower branch, u(t) and δ(t + 1) are in series, so their joint impulse response is

(Subscript[h, 2])(t)=-δ(t)⋆u(t)=-u(t)

For the parallel combination of the two branches

h(t)=(Subscript[h, 1])(t)+(Subscript[h, 2])(t)=u(t+1)-u(t)

The output y(t) is x(t) ⋆ h(t)

y(t)=x(t)⋆h(t)=(∫_(-∞)^∞)x(τ)(u(t+1-τ)-u(t-τ))dτ

Using the recipe given by Eq. (52) or, equivalently, the knowledge that a step function is zero if its argument is negative, leads to

y(t)=(∫_t^(t+1))x(τ)dτ

Since y(t) depends on x(t) beyond time t (upper limit is t + 1), the system is noncausal.

2-5.7. Width Property

When a function of time width (Subscript[T, 1]) is convolved with another of time width (Subscript[T, 2]), the width of the resultant function is ((Subscript[T, 1]) + (Subscript[T, 2])) , regardless of the shapes of the two functions.

Width of y(t) = width of x(t) + width of h(t)

This property was noted earlier in connection with the graphical convolution examples.

If h(t) = 0 when t is outside the interval a ⩽ t ⩽ b and x(t) = 0 when t is outside the interval c ⩽ t ⩽ d, then y(t) = 0 when t is outside the interval (a + c) ⩽ t ⩽ (b + d). 
Hence, we have the width property

[width of y(t)]=(b+d)-( a+c)

In Example 2-7, the widths of (Subscript[x, 1])(t) and (Subscript[x, 2])(t) are 2 s and 1 s, respectively. The width of their convolution is 3 s (Fig. 2-17).

2-5.8. Area Property

The area under the convolution y(t) of two functions x(t) and h(t) is the product of the areas under the individual functions.

The area under the output y(t) is obtained by integrating y(t) over (-∞,∞). 
That is,

[Area of y(t)]=(∫_(-∞)^∞)y(t)dt=(∫_(-∞)^∞)((∫_(-∞)^∞)x(t-τ)h(τ)dτ)dt=
=(∫_(-∞)^∞)(∫_(-∞)^∞)x(t-τ)h(τ)dτdt=
=(∫_(-∞)^∞)h(τ)((∫_(-∞)^∞)x(t-τ)dt)dτ=
=((∫_(-∞)^∞)h(τ)dτ)((∫_(-∞)^∞)x(t-τ)dt)=
=[area of h(t)] × [area of x(t)]

In step 3 of the derivation, we interchanged the order of dt and dτ (because they are independent variables), and in step 4 we used the fact that the areas of x(t) and x(t - τ) are the same when evaluated over (-∞, ∞).

2-5.9. Convolution with a Step Function

From the definition of convolution (Eq. (30)), we have

y(t)=x(t)⋆u(t)=
=(∫_(-∞)^∞)x(τ)u(t-τ)dτ=(∫_(-∞)^t)x(τ)dτ

where we used the fact that u(t - τ) = 0 for t - τ < 0 or, equivalently, when τ > t.

Operationally, the right-hand side of Eq. (77) constitutes an ideal integrator, and for a causal signal, the lower limit on the integral should be zero.

An LTI system whose impulse response is equivalent to a unit step function performs like an ideal integrator.

2-5.10. Differentiation Property

Given the convolution

y(t)=(∫_(-∞)^∞)x(τ)h(t-τ)dτ

taking the derivative with respect to time gives

((dy(t))/(dt))=(d/(dt))(∫_(-∞)^∞)x(τ)h(t-τ)dτ=
=(∫_(-∞)^∞)x(τ)((dh(t-τ))/dt)dτ

Hence

((dy(t))/(dt))=x(t)⋆((dh(t))/(dt))

Similarly, we can show that

((dy(t))/(dt))=((dx(t))/(dt))⋆h(t)

Combining these two relations and generalizing to higher order derivatives, we get

(((d^m)x(t))/(d(t^m)))⋆(((d^n)h(t))/(d(t^n)))=(((d^(m+n))y(t))/(d(t^(m+n))))

In a similar fashion, it is easy to show that

(∫_(-∞)^( t))y(τ)dτ=x(t)⋆((∫_(-∞)^t)h(τ)dτ)=
=((∫_(-∞)^t)x(τ)dτ)⋆h(t)

Table 2-1. Convolution properties

({{Convolution integral, y(t)=x(t)⋆h(t)(∫_(-∞)^∞)x(τ)h(t-τ)dτ}, {Causal Systems, Replace lower limit with 0}})

({{ , Property, Description}, {1, Commutative, x(t)⋆h(t)=h(t)⋆x(t)}, {2, Associative, [g(t)⋆h(t)]⋆x(t)=g(t)⋆[h(t)⋆x(t)]}, {3, Distributive, x(t)⋆[(Subscript[h, 1])(t)+...+(Subscript[h, N])(t)]=
x(t)⋆(Subscript[h, 1])(t)+...+x(t)⋆(Subscript[h, N])(t)}, {4, Causal*Causal=Causal, y(t)=u(t)(∫_0^t)h(τ)x(t-τ)dτ}, {5, Time-shift, h(t-(Subscript[T, 1]))⋆x(t-(Subscript[T, 2]))=y(t-(Subscript[T, 1])-(Subscript[T, 2]))}, {6, Convolution with Impulse, x(t)⋆δ(t-T)=x(t-T)}, {7, Width, [width of y(t)]=[width of x(t)]+[width of h(t)]}, {8, Area, [area of y(t)]=[area of x(t)] × [area of h(t)]}, {9, Convolution with u (t), x(t)⋆u(t)=(∫_(-∞)^t)x(τ)dτ}, {10, Differentiation, (((d^m)x(t))/(d(t^m)))⋆(((d^n)h(t))/(d(t^n)))=(((d^(m+n))y(t))/(d(t^(m+n))))}, {11, Integration, (∫_(-∞)^( t))y(τ)dτ=x(t)⋆((∫_(-∞)^t)h(τ)dτ)=
=((∫_(-∞)^t)x(τ)dτ)⋆h(t)}})

Table 2-2. Commonly encountered convolutions

1.   u(t)⋆u(t)=t · u(t)

2.   (e^at)·u(t)⋆u(t)=(((e^at)-1)/a) · u(t)

3.   (e^at)·u(t)⋆(e^bt)·u(t)=(((e^at)-(e^bt))/(a-b)) · u(t)      (a≠b)

4.   (e^at)·u(t)⋆(e^at)·u(t)=t ·(e^at)· u(t)

5.   t·(e^at)·u(t)⋆(e^bt)·u(t)=(((e^bt)-(e^at)-(a-b)·t·(e^at))/(((a-b))^2))· u(t)      (a≠b)

6.   t·(e^at)·u(t)⋆(e^at)·u(t)=(1/2)·(t^2)·(e^at)·u(t)

7.   δ(t-(Subscript[T, 1]))⋆δ(t-(Subscript[T, 2]))= δ(t-(Subscript[T, 1])-(Subscript[T, 2]))

2-6. Causality and BIBO Stability

Convolution leads to necessary and sufficient conditions for two important LTI system properties:

causality

bounded input / bounded output (BIBO) stability

We derive and illustrate these properties in this section.

2-6.1. Causality

We define a causal system as a system for which the present value of the output y(t) can only depend on present and past values of the input {x(τ), τ ⩽ t}.

For a noncausal system, the present output could depend on future inputs. 
Noncausal systems are also called anticipatory systems, since they anticipate the future.

A physical system must be causal, because a noncausal system must have the ability to see into the future!

For example, 
the noncausal system y(t) = x(t + 2) must know the input two seconds into the future to deliver its output at the present time. This is clearly impossible in the real world.

An LTI system is causal if and only if its impulse response is a causal function:
h(t) = 0 for t < 0.

2-6.2. BIBO Stability: Definition

“BIBO Stable” may sound like a hobbit from The Lord ofthe Rings, but it is an extremely important and desirable property of LTI systems. It can be stated succinctly as: 
“If the input doesn’t blow up, the output won’t either.”

A signal x(t) is bounded if there is a constant C so that

|x(t)|⩽C   for all t

Examples of bounded signals

Bounded signal cos(3t)

Plot[Cos[3t],{t,-1,10}, PlotStyle->{RGBColor[0,0,0.5],Thickness[0.008]},AxesLabel->{t,"cos(3t)"},Exclusions->None,PlotRange->{{-1,11},{-1.2,1.2}}]

Bounded signal  7(e^(-2t))u(t)

Plot[7(E^(-2t))UnitStep[t],{t,-0.5,4}, PlotStyle->{RGBColor[0,0,0.5],Thickness[0.008]},AxesLabel->{t,"7 (E^(-2 t)) u(t)"},Exclusions->None,PlotRange->{{-1,5},{-0.2,8}}]

Bounded signal  (e^(2t))u(1-t)

Plot[(E^(2t))UnitStep[1-t],{t,-3,2}, PlotStyle->{RGBColor[0,0,0.5],Thickness[0.008]},AxesLabel->{t,"(E^(2  t)) u(1 - t)"},Exclusions->None,PlotRange->{{-4,3},{-0.2,8}}]

Examples of unbounded signals

Unbounded signal  (t^2)

Plot[(t^2),{t,-2,2}, PlotStyle->{RGBColor[0,0,0.5],Thickness[0.008]},AxesLabel->{t,"(t^2)"},Exclusions->None,PlotRange->{{-3,3},{-0.2,5}}]

Unbounded signal  (e^(2t))u(t)

Plot[(E^(2t))UnitStep[t],{t,-0.5,1.5}, PlotStyle->{RGBColor[0,0,0.5],Thickness[0.008]},AxesLabel->{t,"(E^(2  t)) u(t)"},Exclusions->None,PlotRange->{{-1,2},{-0.2,25}}]

Unbounded signal  (e^(-t))

Plot[(E^(-t)),{t,-2,4}, PlotStyle->{RGBColor[0,0,0.5],Thickness[0.008]},AxesLabel->{t,"(E^(-t))"},Exclusions->None,PlotRange->{{-2,4},{-0.2,5}}]

Unbounded signal (1/t)

Plot[(1/t),{t,-4,4}, PlotStyle->{RGBColor[0,0,0.5],Thickness[0.008]},AxesLabel->{t,"1/t"},Exclusions->False]

A system is BIBO (bounded input/bounded output) stable if every bounded input x(t) results in a bounded output y(t).

It does not require that an unbounded input result in a bounded output (this would be unreasonable).

Stability is obviously a desirable property for a system.

2-6.3. BIBO Stability: Condition and Proof

How can we ascertain if an LTI system is BIBO stable? 
The answer is provided by the following theorem.

BIBO stability theorem

An LTI system is BIBO stable if and only if its impulse response h(t) is absolutely integrable 
(if  (∫_(-∞)^∞)|h(t)|dt   is finite).

Proof

We will go through the proof of this result in some detail, because it is a useful exercise to learn how to prove mathematical theorems.

A theorem statement that includes the phrase “if and only if” requires a bidirectional proof.

If B represents BIBO stability and A represents absolute integrability of h(t), to prove the bidirectional statement A ⇔ B requires us to prove both A ⇒ B and B ⇒ A.

To prove A ⇒ B, we suppose A is true and use it to prove B is true. 
Alternatively, we can use the contrapositive of A ⇒ B, wherein we suppose B is false ((OverBar[B])) and use it to prove A is false ((OverBar[A])). That is,

A⇒B≡(OverBar[B])⇒(OverBar[A])

“If” Part of Proof: A ⇒ B

Statement A is

A=absolute integrability of h(t)

Statement B is

B=BIBO stability

we suppose A is true and use it to prove A ⇒ B.

Absolute integrability of h(t) implies that there exists a constant L of finite magnitude such that

(∫_(-∞)^∞)|h(t)|dt=L

To prove B, namely, that the system is BIBO stable, we have to suppose that the input x(t) is bounded, which means that there exists a constant M of finite magnitude such that

|x(t)|⩽M

The goal now is to prove that the output y(t) is bounded.

Having properly formulated the problem, it is now clear how to proceed.

Applying the triangle inequality for integrals to the absolute value of the convolution integral given by Eq. (30) leads to

The triangle inequality

If f is an integrable function on a real interval [a,b], then

|(∫_a^b)f(x)dx|⩽(∫_a^b)|f(x)|dx

|y(t)|=|(∫_(-∞)^∞)h(τ)x(t-τ)dτ|⩽(∫_(-∞)^∞)|h(τ)x(t-τ)|dτ⩽(∫_(-∞)^∞)|h(τ)|Mdτ⩽(Underoverscript[M∫, (-∞), ∞])|h(τ)|dτ⩽ML

which states that |y(t)| is bounded by the finite constant LM. Hence, if h(t) is absolutely integrable, a bounded input results in a bounded output, and the system is BIBO stable.

“Only if” Part of Proof: A ⇒ B

We can suppose B (B = BIBO stability) is true and then try to prove A (A = absolute integrability of h(t)) is true, but it is easier to prove the contrapositive.

We suppose h(t) is not absolutely integrable (i.e., A not true), and then identify a bounded-input that results in an unbounded output (B not true). In short, we need a counterexample.

Since we are supposing (∫_(-∞)^∞)|h(t)|dt→∞, we seek a bounded input that would lead to an unbounded output.

A convenient candidate is the noncausal input x(t) given by

x(t)=({{{, ({{((h(-t))/(|h(-t)|))=((|h(-t)|)/(h(-t)))=±1, if h(-t)≠0}, {0, if h(-t)=0}})}})

At t = 0, the corresponding output y(t) is

y(0)=(∫_(-∞)^∞)h(τ)x(0-τ)dτ=(∫_(-∞)^∞)h(τ)((|h(τ)|)/(h(τ)))dτ=(∫_(-∞)^∞)|h(τ)|dτ→∞

This result proves that a system whose impulse response is not absolutely integrable is not BIBO stable ((OverBar[A]) ⇒ (OverBar[B])), which is equivalent to B ⇒ A .

End of proof

Example 2-10. Causality and BIBO Stability

An LTI system is characterized by the impulse response h(t) = (e^(-|t|))

Determine if the system is 
(a) causal
(b) BIBO stable

Plot[(E^(-\[LeftBracketingBar]t\[RightBracketingBar])),{t,-4.5,4.5}, PlotStyle->{RGBColor[0,0,0.5],Thickness[0.008]},AxesLabel->{t,"(E^(-\[LeftBracketingBar]t\[RightBracketingBar]))"},Exclusions->None,PlotRange->{{-5,5},{-0.2,1.2}}]

Figure 2-21. Impulse response of system in Example 2-10.

Solution:

(a) causal

The specified impulse response is a two-sided exponential function (Fig. 2-21). 
Since h(t) ≠ 0 for t < 0, the system is noncausal.

(b) BIBO stable

(∫_(-∞)^∞)|h(t)|dt=(∫_(-∞)^∞)(e^(-|t|))dt=2(∫_0^∞)(e^(-t))dt=2

2(∫_0^∞)(E^(-t))\[DifferentialD]t

Hence, the system is BIBO stable.

2-6.4. BIBO Stability of System with Decaying Exponentials

Consider a causal system with an impulse response

h(t)=((Ce)^(γt))u(t)

where C is a finite constant and γ is, in general, a finite complex coefficient given by

γ∈ℂ ; γ=α+jβ ; α=Re[γ] ; β=Im[γ]

Such a system is BIBO stable if and only if α < 0 (i.e., h(t) is a one-sided exponential with an exponential coefficient whose real part is negative).

To verify the validity of this statement, we test to see if h(t) is absolutely integrable.

Since |(e^(jβt))|=1  and (e^(αt))>0

Abs[(E^I)]

(∫_(-∞)^∞)|h(t)|dt=(∫_0^∞)|C(e^(jβt))(e^(αt))|dt=|C|(∫_0^∞)(e^(αt))dt

α < 0

If α < 0, we can rewrite it as α = -lαl in the exponential

(∫_(-∞)^∞)|h(t)|dt=|C|(∫_0^∞)(e^(-|α|t))dt=((|C|)/(|α|))<∞

Hence, h(t) is absolutely integrable and the system is BIBO stable.

α ⩾ 0

If α ⩾ 0, Eq. (95) becomes

(∫_(-∞)^∞)|h(t)|dt=|C|(∫_0^∞)(e^(-αt))dt→∞

thereby proving that the system is not BIBO stable when α ⩾ 0

By extension, for any positive integer N, an impulse response composed of a linear combination of N exponential signals

h (t)=(∑_(i=1)^N)(Subscript[C, i]) (e^(t (Subscript[γ, i])))u(t)

is absolutely integrable, and its LTI system is BIBO stable, if and only if all of the exponential coefficients (Subscript[γ, i]) have negative real parts (Re[(Subscript[γ, i])] < 0). 
This is a fundamental attribute of LTI system theory.

Some books define an LTI system as marginally stable if one or more of the exponential coefficients (Subscript[γ, i]) have zero real parts and the remaining coefficients have negative real parts.

Accordingly, an LTI system with h(t) = (e^jt)u(t) should be marginally stable, because γ = 0 + j1.

However, a marginally stable system may or may not be BIBO stable.

We can test the validity of the assertion by evaluating the system's output for a causal input signal x(t) = (e^jt)u(t),

y(t)=(∫_(-∞)^∞)x(τ)h(t-τ)dτ=u(t)(∫_0^t)(e^(jτ))(e^(j(t-τ)))dτ=u(t)(e^jt)(∫_0^t)dτ=((te)^jt)u(t)

Since t in the final expression can grow to ∞, the system is not BIBO stable.

Hence, this marginally stable LTI system is not BIBO stable.

2-7. LTI Sinusoidal Response

This section introduces a vitally important property of LTl systems.

The response of an LTI sytem to a complex exponential input signal x(t)=((Ae)^(jωt)) is another complex exponential signal y(t)=(OverHat[Η])(ω)((Ae)^(jωt)), where (OverHat[Η])(ω) is a complex coefficient that depends on ω.

A similar statement applies to sinusoidal inputs and outputs

In Chapter 5, it is shown that a complicated periodic signal x(t) of period (Subscript[T, 0]) and angular frequency (Subscript[ω, 0]) = 2π/(Subscript[T, 0]) can be expressed as a linear combination of an infinite number of complex exponential signals

x (t)=(∑_(n=-∞)^∞)(Subscript[x, n])(e^((Subscript[jnω, 0])t))

with constant coefficients (Subscript[x, n])

A similar process applies to nonperiodic signals, wherein (∑_(n=-∞)^∞) becomes an integral (∫_(-∞)^∞)

According to the sinusoidal-response property, the corresponding output y(t) will assume the form

y (t)=(∑_(n=-∞)^∞)(Subscript[H, n])(Subscript[x, n])(e^((Subscript[jnω, 0])t))

with constant coefficients (Subscript[H, n]).

This sinusoidal-response property is at the heart of how filtering of signals is performed by an LTI system to remove noise, attenuate undesirable frequency components, or accentuate contrast or other features in voice and image signals.

The relevant steps are covered as follows:

In Chapter 2 we will show in later parts of the present section why an LTI system excited by an input signal x(t)=((Ae)^(jωt)) generates an output signal y(t)=(OverHat[Η])(ω)((Ae)^(jωt)), and we will learn how to compute (OverHat[Η])(ω) from h(t).

In Chapter 5 we will learn how to perform the decomposition of periodic signals, as described by Eq. (98). We will also learn how to similarly express nonperiodic signals in terms of frequency integrals.

In Chapter 6 we will design a special LTI system called a Butterworth filter with a special (OverHat[Η])(ω). Additionally, we will use LTI systems to perform multiple filtering examples.

2-7.1. LTI System Response to Complex Exponential Signals x(t)=((Ae)^(jωt))

Let us examine the response of an LTI system to a complex exponential x(t)=((Ae)^(jωt)).

Since the system is LTI, we may (without loss of generality) set A = 1. Hence,

y(t)=(OverHat[Η])(ω)(e^(jωt))

y (t)=h(t)*x(t)=h(t)⋆(e^(jωt))=(∫_(-∞)^∞)h(τ)(e^(jω(t-τ)))dτ=(e^(jωt))(∫_(-∞)^∞)h(τ)(e^(-jωτ))dτ=(OverHat[H])(ω) (e^(jωt))

where the frequency response function (OverHat[Η])(ω) is defined as

(OverHat[H])(ω)=(∫_(-∞)^∞)h(τ)(e^(-jωτ))dτ

The result given by Eq. (100) validates the statement made earlier that the output of an LTI system excited by a complex exponential signal is also a complex exponential signal at the same angular frequency ω.

The frequency response function (OverHat[Η])(ω) bears a one-to-one correspondence to h(t); it is completely specified by the expression for h(t) and the value of frequency ω.

As alluded to earlier in connection with Eqs. (98) and (99), we can express time signals in terms of sums of signals at multiple frequencies. The ability to transform signals between the time domain and the ω-domain will prove very useful in the analysis and design of LTI systems.

It is in the spirit of such transformations that we describe (OverHat[Η])(ω) as the frequency-domain equivalent of h(t).

In symbolic form, Eq. (100) is

(e^(jωt)) ⟶ h(t) ⟶ (OverHat[Η])(ω)(e^(jωt))

In addition to working with (OverHat[Η])(ω) to determine a system’s response to a specified input, we will occasionally find it convenient to also work with its mirror image (OverHat[Η])(-ω). Repetition of the procedure leading to Eq. (101) yields

y(t)=(OverHat[Η])(-ω)(e^(-jωt))

where

(OverHat[H])(-ω)=(∫_(-∞)^∞)h(τ)(e^(jωτ))dτ

In symbolic form

(e^(-jωt)) ⟶ h(t) ⟶ (OverHat[Η])(-ω)(e^(-jωt))

Example 2-11. (OverHat[Η])(ω) of an RC Circuit

Determine the frequency response of an RC circuit characterized by an impulse response h(t)=(e^(-t))u(t). The capacitor contained no charge prior to t = 0 (zero initial condition).

Plot[(E^(-t))*UnitStep[t],{t,-0.5,4.5}, PlotStyle->{RGBColor[0,0,0.5],Thickness[0.008]},AxesLabel->{t,"(E^(-t))u(t)"},Exclusions->None,PlotRange->{{-1,5},{-0.2,1.2}}]

Given impulse response h(t)=(e^(-t))u(t) of an RC circuit in Example 2-11.

Solution:

Application of Eq. (101) leads to

(OverHat[H])(ω)=(∫_(-∞)^∞)h(τ)(e^(-jωτ))dτ=(∫_0^∞)(e^(-τ))(e^(-jωτ))dτ=(1/(1+jω))

H211=(∫_0^∞)(E^(-t))*(E^(-I ω t))\[DifferentialD]t

2-7.2. Properties of (OverHat[Η])(ω)

The frequency response function (OverHat[Η])(ω) is characterized by the following properties:

(OverHat[Η])(ω) is independent of t

(OverHat[Η])(ω) is a function of complex frequency jω, not just ω

(OverHat[Η])(ω) is defined for an input signal (e^(jωt)) that exists for all time, -∞ < t < ∞; i.e., x(t)=(e^(jωt)) , not (e^(jωt))u(t)

If h(t) is real, (OverHat[Η])(ω) is related to (OverHat[Η])(-ω) by

(OverHat[Η])(-ω)=((OverHat[Η])^⋆)(ω)     (for  h(t)∈ℝ)

The condition defined by Eq. (107) is known as conjugate symmetry.

It follows from the definition of (OverHat[Η])(ω) given by Eq. (101), namely,

((OverHat[H])^⋆)(ω)=(∫_(-∞)^∞)(SuperStar[h])(τ)(SuperStar[((e^(-jωτ)))])dτ=(∫_(-∞)^∞)h(τ)(e^(jωτ))dτ=(OverHat[Η])(-ω)

Since h(t) is real, it follows from Eq. (107) that the magnitude |(OverHat[Η])(ω)| is an even function and the phase ∠(OverHat[Η])(ω) is an odd function:

|(OverHat[Η])(ω)|=|(OverHat[Η])(-ω)|
and
∠(OverHat[Η])(ω)=-∠(OverHat[Η])(-ω)

The frequency response function of the RC filter, namely,

(OverHat[Η])(ω)=(1/(1+jω))=((1-jω)/(1+(ω^2)))

satisfies the aforementioned relationships.

2-7.3. (OverHat[Η])(ω) of LCCDE

If an LTI system is described by a linear, constant-coefficient, differential equation (LCCDE), we can determine its frequency response function (OverHat[Η])(ω) without having to know its impulse response h(t).

Consider, for example, the LCCDE

(((d^2)y(t))/(d(t^2)))+2((dy(t))/(dt))+y(t)=4((dx(t))/(dt))+5x(t)

When x(t)=(e^(jωt)), according to Eq. (100), y(t)=(OverHat[Η])(ω)(e^(jωt))

Substituting these expressions for x(t) and y(t) in differential equation Eq. (110) leads to the frequency domain equivalent of Eq. (110):

(((jω))^2)(OverHat[Η])(ω)(e^(jωt))+2(jω)(OverHat[Η])(ω)(e^(jωt))+3(OverHat[Η])(ω)(e^(jωt))=4(jω)(e^(jωt))+5(e^(jωt))

After cancelling (e^(jωt)) (which is never zero) from all terms and then solving for (OverHat[Η])(ω), we get

Simplify[Reduce[(((I ω))^2)Hw (E^(I ω t))+2(I ω)Hw (E^(I ω t))+3Hw (E^(I ω t))==4(I ω)(E^(I ω t))+5(E^(I ω t)),Hw]]

(OverHat[Η])(ω)=((5+4jω)/(-(ω^2)+2jω+3))=((5+4jω)/((3-(ω^2))+2jω))

Note that every (d/(dt)) operator in the time domain translates into a multiplication by jω in the frequency domain

More generally, for any LCCDE of the form

(Subscript[a, n])y+(Subscript[a, n-1])((dy)/(dt))+(Subscript[a, n-2])(((d^2)y)/(d(t^2)))+⋯+(Subscript[a, 0])(((d^n)y)/(d(t^n)))=(Subscript[b, m])x+(Subscript[b, m-1])((dx)/(dt))+(Subscript[b, m-2])(((d^2)x)/(d(t^2)))+⋯+(Subscript[b, 0])(((d^m)x)/(d(t^m)))
or   (∑_(i=0)^n)(Subscript[a, n-i])(((d^i)y)/(d(t^i)))=(∑_(i=0)^m)(Subscript[b, m-i])(((d^i)y)/(d(t^i)))

its (OverHat[Η])(ω) can be determined by substituting x(t)=(e^(jωt)) and y(t)=(OverHat[Η])(ω)(e^(jωt)) in the LCCDE and using ((d^i)/(d(t^i)))(e^(jωt))=(((jω))^i)(e^(jωt)).

The process leads to frequency response function

(OverHat[Η])(ω)=(((∑_(i=0)^m)(Subscript[b, m-i])(((jω))^i))/((∑_(i=0)^n)(Subscript[a, n-i])(((jω))^i)))

In Eqs. (113) and (114), m and n are the orders of the highest derivatives of x(t) and y(t), respectively.

Note that (OverHat[Η])(ω) is a rational function of jω (ratio of two polynomials).

Example 2-12. (OverHat[Η])(ω) of an LCCDE

Compute (OverHat[Η])(1) for the system described by LCCDE

3(((d^3)y(t))/(d(t^3)))+6((dy(t))/(dt))+4y(t)=5((dx(t))/(dt))

Solution:

Apply Eq. (114) with m = 1 and n = 3:

(OverHat[Η])(ω)=(((Subscript[b, 1])(((jω))^0)+(Subscript[b, 0])(((jω))^1))/((Subscript[a, 3])(((jω))^0)+(Subscript[a, 2])(((jω))^1)+(Subscript[a, 1])(((jω))^2)+(Subscript[a, 0])(((jω))^3)))

From the specified differential equation, (Subscript[a, 0]) = 3, (Subscript[a, 1]) = 0, (Subscript[a, 2]) = 6, (Subscript[a, 3]) = 4, (Subscript[b, 0]) = 5, and (Subscript[b, 1]) = 0. Hence,

(OverHat[Η])(ω)=((5(jω))/(4+6(jω)+3(((jω))^3)))

At ω = 1 rad/s,

H212[ω_]:=((5(I ω))/(4+6(I ω)+3(((I ω))^3)))

H212[ω]

H212[1]

Arg[H212[1]]//N

Print["Answer: \[VerticalSeparator](OverHat[Η])(1)\[VerticalSeparator] = ",Abs[H212[1]],"; ∠(OverHat[Η])(1) = ",N[Arg[H212[1]]/Degree,4],"°"]

(OverHat[Η])(ω)=((j5)/(4+j3))=((3+j4)/5)=1(e^(0.93j))=1∠53.13°

2-7.4. LTI Response to Sinusoids

We now examine the response of an LTI system with real-valued h(t) to a pure sinusoid of the form x(t) = cos(ωt).

When analyzing signals and systems, operations involving sinusoids are most easily realized by expressing them in terms of complex exponentials:

x(t)=cos(ωt)=(1/2)((e^(jωt))+(e^(-jωt)))

TrigToExp[Cos[ω*t]]

Using the additivity property of LTI systems, we can determine the system’s response to cos(ωt) by adding the responses expressed by Eqs. (102) and (105):

(e^(jωt)) ⟶ h(t) ⟶ (OverHat[Η])(ω)(e^(jωt))
+			+
(e^(-jωt)) ⟶ h(t) ⟶ (OverHat[Η])(-ω)(e^(-jωt))

Since h(t) is real, Eq. (107) stipulates that

(OverHat[Η])(-ω)=(SuperStar[(OverHat[Η])])(ω)

Complex number in polar coordinate system:

(OverHat[Η])(ω)=|(OverHat[Η])(ω)|(e^(jθ))
(OverHat[Η])(-ω)=|(OverHat[Η])(ω)|(e^(-jθ))

where

θ=∠(OverHat[Η])(ω)

The sum of the two outputs in Eq. (116):

y(t)=  (1/2)((OverHat[Η])(ω)(e^(jωt))+(OverHat[Η])(-ω)(e^(-jωt)))=(1/2)|(OverHat[Η])(ω)|((e^(j(ωt+θ)))+(e^(-j(ωt+θ))))=|(OverHat[Η])(ω)|cos(ωt+θ)

In summary:

When input is x(t)=cos(ωt)
then the output is y(t)=  |(OverHat[Η])(ω)|cos(ωt+θ)

In symbolic form:

cos(ωt) ⟶ h(t) ⟶ |(OverHat[Η])(ω)|cos(ωt+θ)

By applying the scaling and time-invariance properties of LTI systems, this result can be generalized to any cosine function of amplitude A and phase angle ϕ

A cos(ωt+ϕ) ⟶ h(t) ⟶ A |(OverHat[Η])(ω)|cos(ωt+θ+ϕ)

The full implication of this result will become apparent when we express periodic signals as linear combinations of complex exponentials or sinusoids.

Example 2-13. Response of an LCCDE to a Sinusoidal Input

Compute the output y(t) for the system described by symbolic diagram

5 cos(4t+30°) ⟶
((dy(t))/(dt))+3y(t)=3((dx(t))/(dt))+5x(t) ⟶ y(t)

Solution:

Inserting x(t)=(e^(jωt)) and y(t)=(OverHat[Η])(ω)(e^(jωt)) in the specified LCCDE gives

jω(OverHat[Η])(ω)(e^(jωt))+3(OverHat[Η])(ω)(e^(jωt))=3((jωe)^(jωt))+5(e^(jωt))

Solving for (OverHat[Η])(ω) gives

(OverHat[Η])(ω)=((5+3jω)/(3+jω))

The angular frequency ω = 4 rad/s. 
Hence,

H213[ω_]:=((5+3 I ω)/(3+I ω))

H213[ω]

H213[4]

Print["\[VerticalSeparator](OverHat[Η])(4)\[VerticalSeparator] = ",Abs[H213[4]],"; ∠(OverHat[Η])(4) = ",N[Arg[H213[4]]/Degree,4],"°"]

(OverHat[Η])(4)=((5+j12)/(3+j4))=((63+j16)/25)=(13/5)∠14.25°=(13/5)(e^(14.25°))

Application of Eq. (120) with A = 5 and ϕ = 30° gives

y(t)=5×(13/5) ×cos(4t+30°+14.25°) =13cos(4t+44.25°)

Using TransferFunctionModel and OutputResponse commands in Mathematica :

TF213=TransferFunctionModel[((5+3s)/(3+s)),s];
Output213=Simplify[OutputResponse[TF213,5Cos[4 t],t]]

Plot[(12/5) (E^(-3 t))+(63/5) Cos[4 t]-(16/5) Sin[4 t],{t,0,10}]

2-8. Impulse Response of Second-Order LCCDEs

Many physical systems are described by second-order LCCDEs of the form

(((d^2)y(t))/(d(t^2)))+(Subscript[a, 1])((dy(t))/(dt))+(Subscript[a, 2])y(t)=(Subscript[b, 1])((dx(t))/(dt))+(Subscript[b, 2])x(t)

where (Subscript[a, 1]), (Subscript[a, 2]), (Subscript[b, 1]), and (Subscript[b, 2]) are constant coefficients.

In this section, we examine how to determine the impulse response function h(t) for such a differential equation.

2-8.1. LCCDE with No Input Derivatives

For simplicity, we start by considering a version of Eq. (121) without the ((dx(t))/(dt)) term, and then we use the result to treat the more general case in the next subsection.

For (Subscript[b, 1])=0 and (Subscript[b, 2])=1, Eq. (121) becomes

(((d^2)y(t))/(d(t^2)))+(Subscript[a, 1])((dy(t))/(dt))+(Subscript[a, 2])y(t)=x(t)

Step 1. Roots of Characteristic Equation

Assuming y(t) has a general solution of the form y(t) = A (e^st),  substitution in the homogeneous form of Eq. (122) – i.e., with x(t) = 0 – leads to the characteristic equation:

(s^2)+(Subscript[a, 1])s+(Subscript[a, 2])=0

If (Subscript[p, 1]) and (Subscript[p, 2]) are the roots of Eq. (123), then

(s^2)+(Subscript[a, 1])s+(Subscript[a, 2])=(s-(Subscript[p, 1]))(s-(Subscript[p, 2]))

hence,

(Subscript[p, 1])+(Subscript[p, 2])=-(Subscript[a, 1]);   (Subscript[p, 1])·(Subscript[p, 2])=(Subscript[a, 2])

and

(Subscript[p, 1])=-((Subscript[a, 1])/2)+(Sqrt[(((((Subscript[a, 1])/2)))^2)-(Subscript[a, 2])])
(Subscript[p, 2])=-((Subscript[a, 1])/2)-(Sqrt[(((((Subscript[a, 1])/2)))^2)-(Subscript[a, 2])])

Roots[(s^2)+(Subscript[a, 1])s+(Subscript[a, 2])==0,s]

Roots (Subscript[p, 1]) and (Subscript[p, 2]) can be:

(a) real if 	(Subscript[a, 1]^2)>4(Subscript[a, 2])

(b) complex conjugates if 	(Subscript[a, 1]^2)<4(Subscript[a, 2])

(c) identical if 	(Subscript[a, 1]^2)=4(Subscript[a, 2]).

Step 2.  Two Coupled First-Order LCCDEs

The original differential equation given by Eq. (122) now can be  rewritten as

(((d^2)y(t))/(d(t^2)))-((Subscript[p, 1])+(Subscript[p, 2]))((dy(t))/(dt))+(Subscript[p, 1])·(Subscript[p, 2])·y(t)=x(t)
⇓
[(d/(dt))-(Subscript[p, 1])][(d/(dt))-(Subscript[p, 2])]y(t)=x(t)

Furthermore, we can split the second-order differential equation into two coupled first-order equations by introducing an intermediate variable z(t):

((dz(t))/(dt))-(Subscript[p, 1])z(t)=x(t)
((dy(t))/(dt))-(Subscript[p, 2])y(t)=z(t)

These coupled first-order LCCDEs represent a series (or cascade) connection of LTI systems, each described by a first order LCCDE.

In symbolic form, we have

x(t) -> (Subscript[h, 1])(t) -> z(t) 
z(t) -> (Subscript[h, 2])(t) -> y(t)

where (Subscript[h, 1])(t) and (Subscript[h, 2])(t) are the impulse responses corresponding
to Eqs. (128), respectively.

Step 3.   Impulse Response of Cascaded LTI Systems

By comparison with Eq. (10) and its corresponding impulse response, Eq. (17), we conclude that

(Subscript[h, 1])(t)=(e^((Subscript[p, 1])t))u(t)
(Subscript[h, 2])(t)=(e^((Subscript[p, 2])t))u(t)

Using convolution property No.2 in Table 2-1, the impulse response of the series connection of two LTI systems is the convolution of their impulse responses.

Utilizing property No.3 in Table 2-2, the combined impulse response (Subscript[h, c])(t) becomes

(Subscript[h, c])(t)=(Subscript[h, 1])(t)⋆(Subscript[h, 2])(t)=(e^((Subscript[p, 1])t))u(t)⋆(e^((Subscript[p, 2])t))u(t)=[(1/((Subscript[p, 1])-(Subscript[p, 2])))][(e^((Subscript[p, 1])t))-(e^((Subscript[p, 2])t))]u(t)

2-8.2. LCCDE with Input Derivative

We now consider the more general case of a second-order LCCDE that contains a first-order derivative on the input side of the equation

(((d^2)y(t))/(d(t^2)))+(Subscript[a, 1])((dy(t))/(dt))+(Subscript[a, 2])y(t)=(Subscript[b, 1])((dx(t))/(dt))+(Subscript[b, 2])x(t)

By defining the right-hand side of Eq. (132) as an intermediate  variable w(t), the system can be represented as

x(t) -> w(t)=(Subscript[b, 1])((dx(t))/(dt))+(Subscript[b, 2])x(t) -> w(t) 
w(t) -> (Subscript[h, c])(t) -> y(t)

where (Subscript[h, c])(t) is the impulse response given by Eq. (131) for the system with (Subscript[b, 1]) = 0 and (Subscript[b, 2]) = 1.

To determine the impulse response of the overall system, we need to compute the convolution of (Subscript[h, c])(t) with the (yet to be determined) impulse response representing the other box in Eq. (133).

A more expedient route is to use convolution property No.1 in Table 2-1. Since convolution is commutative, we can reverse the order of the two LTI systems in Eq. (133),

x(t) -> (Subscript[h, c])(t) -> υ(t) 
υ(t) -> υ(t)=(Subscript[b, 1])((dυ(t))/(dt))+(Subscript[b, 2])υ(t) -> y(t)

where υ(t) is another intermediate variable created for the sake of convenience.

By definition, when x(t) = δ(t), the output y(t) becomes the impulse response h(t) of the overall system. That is, if we set x(t) = δ(t), which results in υ(t) = (Subscript[h, c])(t) and y(t) = h(t), the system becomes

δ(t) -> (Subscript[h, c])(t) -> (Subscript[h, c])(t) 
(Subscript[h, c])(t) -> h(t)=(Subscript[b, 1])((d(Subscript[h, c])(t))/(dt))+(Subscript[b, 2])(Subscript[h, c])(t) -> h(t)

Finally, the impulse response h(t) of the overall system is

h(t)=(Subscript[b, 1])((d(Subscript[h, c])(t))/(dt))+(Subscript[b, 2])(Subscript[h, c])(t)=[(Subscript[b, 1])(d/(dt))+(Subscript[b, 2])][(1/((Subscript[p, 1])-(Subscript[p, 2])))][(e^((Subscript[p, 1])t))-(e^((Subscript[p, 2])t))]u(t)=(((Subscript[b, 1])(Subscript[p, 1])+(Subscript[b, 2]))/((Subscript[p, 1])-(Subscript[p, 2])))(e^((Subscript[p, 1])t))u(t)-(((Subscript[b, 1])(Subscript[p, 2])+(Subscript[b, 2]))/((Subscript[p, 1])-(Subscript[p, 2])))(e^((Subscript[p, 2])t))u(t)

Having established in the form of Eq. (136) an explicit expression for the impulse response of the general LCCDE given by Eq. (132), we can now determine the response y(t) to any causal input excitation x(t) by evaluating

y(t)=u(t)(∫_0^t)h(τ)x(t-τ)dτ

2-8.3. Parameters of Second-Order LCCDE

Mathematically, our task is now complete. However, we can gain much physical insight into the nature of the system’s response by examining scenarios associated with the three states of roots (Subscript[p, 1]) and (Subscript[p, 2]) [as noted earlier in section 2-8.1 in connection with Eq. (126)]

(a) (Subscript[p, 1]) and (Subscript[p, 2]) are real and distinct

(b) (Subscript[p, 1]) and (Subscript[p, 2]) complex conjugates

(c) (Subscript[p, 1]) and (Subscript[p, 2]) are real and identical

Recall that (Subscript[p, 1]) and (Subscript[p, 2]) are defined in terms of coefficients  (Subscript[a, 1]) and (Subscript[a, 2]) in the LCCDE, so different systems characterized LCCDEs with identical forms but different values of  (Subscript[a, 1]) and (Subscript[a, 2]) may behave quite differently.

Before we examine the three states of (Subscript[p, 1]) and (Subscript[p, 2]) individually it will prove useful to express (Subscript[p, 1]) and (Subscript[p, 2]) in terms of physically meaningful parameters.

To start with, we reintroduce expressions for (Subscript[p, 1]) and (Subscript[p, 2]) given by Eq. (126):

(Subscript[p, 1])=-((Subscript[a, 1])/2)+(Sqrt[(((((Subscript[a, 1])/2)))^2)-(Subscript[a, 2])])
(Subscript[p, 2])=-((Subscript[a, 1])/2)-(Sqrt[(((((Subscript[a, 1])/2)))^2)-(Subscript[a, 2])])

Based on the results of Section 2-6.4, in order for the system described by Eq. (136) to be BIBO stable, it is necessary that the real parts of both (Subscript[p, 1]) and (Subscript[p, 2]) be negative.

We now show that this is true if and only if (Subscript[a, 1])>0 and (Subscript[a, 2])>0. Specifically, we have the following:

(a) If both (Subscript[p, 1]) and (Subscript[p, 2]) are real, distinct and negative Eq. (138) leads to the conclusion that (Subscript[a, 1]^2)>4(Subscript[a, 2]), (Subscript[a, 1])>0 and (Subscript[a, 2])>0.

(b) If (Subscript[p, 1]) and (Subscript[p, 2]) are complex conjugates with negative real parts, it follows that (Subscript[a, 1]^2)<4(Subscript[a, 2]), (Subscript[a, 1])>0 and (Subscript[a, 2])>0.

(c) If (Subscript[p, 1]) and (Subscript[p, 2]) are real, equal and negative, then (Subscript[a, 1]^2)=4(Subscript[a, 2]), (Subscript[a, 1])>0 and (Subscript[a, 2])>0.

The LTI system described by the LCCDE Eq. (132) is BIBO stable if and only if (Subscript[a, 1])>0 and (Subscript[a, 2])>0.

We now introduce three new non-negative, physically meaningful parameters:

α=((Subscript[a, 1])/2)   – attenuation coefficient (Np/s),
(Subscript[ω, 0])=(Sqrt[(Subscript[a, 2])])   – undamped natural frequency (rad/s)
ξ=(α/(Subscript[ω, 0]))=((Subscript[a, 1])/(2(Sqrt[(Subscript[a, 2])])))    – damping coefficient (unitless)

The unit Np is short for nepers, named after the inventor of the
logarithmic scale, John Napier.

The neper (unit symbol Np) is a logarithmic unit for ratios of measurements of physical field and power quantities, such as gain and loss of electronic signals [https://en.wikipedia.org/wiki/Neper]

In view of Eq. (139), (Subscript[p, 1]) and (Subscript[p, 2]) can be written as

(Subscript[p, 1])=-α+(Sqrt[(α^2)-(Subscript[ω, 0]^2)])=(Subscript[ω, 0])[-ξ+(Sqrt[(ξ^2)-1])]
(Subscript[p, 2])=-α-(Sqrt[(α^2)-(Subscript[ω, 0]^2)])=(Subscript[ω, 0])[-ξ-(Sqrt[(ξ^2)-1])]

The damping coefficient ξ plays a critically important role, because its value determines the character of the system’s response to any input x(t).

The system exhibits markedly different responses depending on whether

(a)  ξ>1   ⇒ overdamped response

(b)  ξ=1   ⇒ critically damped response

(c)  ξ<1   ⇒ underdamped response

The three names, overdamped, underdamped, and critically damped, refer to the shape of the system’s response.

Figure 2-22 displays three system step responses, each one of which starts at zero at t = 0 and rises to y = 1 as t → ∞, but the shapes of their waveforms are quite different.

The overdamped response exhibits the slowest path towards y = 1; 
the underdamped response is very fast, but it includes an oscillatory component; 
and the critically damped response represents the fastest path without oscillations.

Clear[ζ,ω,t];SystemModel[ζ_,ω_]:=TransferFunctionModel[ ((ω^2)/((s^2)+2ζ ω s+(ω^2))),s];Show[
{
Plot[{OutputResponse[SystemModel[0.2,6],UnitStep[t],t],OutputResponse[SystemModel[1,6],UnitStep[t],t],OutputResponse[SystemModel[6,6],UnitStep[t],t]}//Evaluate,{t,0,3},PlotRange->All,AxesLabel->{t,"y(t)"},PlotStyle->{Thickness[0.008]}],
Graphics[{
{Thickness[0.003],Dashing[0.005],
Line[{{0,1},{3,1}}]},
Text["underdamped (ξ<1)",{1,1.5}],Text["overdamped (ξ>1)",{2.5,0.6}],
Text["critically damped (ξ=1)",{0.8,0.6}]
}
]}]

Figure 2-22. Comparison of overdamped, underdamped, and critically damped responses. In each case, the response starts at zero at t = 0 and approaches 1 as t → ∞, but the in-between paths are quite different.

2-8.4. Overdamped Case (ξ > 1) (skip)

(skip)

2-8.5. Underdamped Case (ξ < 1) (skip)

(skip)

2-8.6. Critically Damped Case (ξ = 1) (skip)

(skip)

Table 2-3. Impulse and step responses of second-order LCCDE

LCCDE: (((d^2)y(t))/(d(t^2)))+(Subscript[a, 1])((dy(t))/(dt))+(Subscript[a, 2])y(t)=(Subscript[b, 1])((dx(t))/(dt))+(Subscript[b, 2])x(t)

α=((Subscript[a, 1])/2); (Subscript[ω, 0])=(Sqrt[(Subscript[a, 2])]); ξ=(α/(Subscript[ω, 0]))
(Subscript[p, 1])=(Subscript[ω, 0])[-ξ+(Sqrt[(ξ^2)-1])]; (Subscript[p, 2])=(Subscript[ω, 0])[-ξ-(Sqrt[(ξ^2)-1])]

Overdamped Case (ξ > 1)

h(t)=(Subscript[A, 1])(e^((Subscript[p, 1])t))u(t)+(Subscript[A, 2])(e^((Subscript[p, 2])t))u(t)
(Subscript[y, step])(t)=[((Subscript[A, 1])/(Subscript[p, 1]))((e^((Subscript[p, 1])t))-1)+((Subscript[A, 2])/(Subscript[p, 2]))((e^((Subscript[p, 2])t))-1)]u(t)
(Subscript[A, 1])=(((Subscript[b, 1])(Subscript[p, 1])+(Subscript[b, 2]))/((Subscript[p, 1])-(Subscript[p, 2]))); (Subscript[A, 1])=-(((Subscript[b, 1])(Subscript[p, 1])+(Subscript[b, 2]))/((Subscript[p, 1])-(Subscript[p, 2])))

Underdamped Case (ξ < 1)

h(t)=[(Subscript[B, 1])cos ((Subscript[ω, d])t)+(Subscript[B, 2])sin ((Subscript[ω, d])t)](e^(-αt))u(t)
(Subscript[y, step])(t)=(1/((α^2)+(Subscript[ω, d]^2))){[-((Subscript[B, 1])α+(Subscript[B, 2])(Subscript[ω, d]))cos ((Subscript[ω, d])t)+((Subscript[B, 1])(Subscript[ω, d])+(Subscript[B, 2])α)sin ((Subscript[ω, d])t)](e^(-αt))+((Subscript[B, 1])α+(Subscript[B, 2])(Subscript[ω, d]))}u(t)
(Subscript[B, 1])=(Subscript[b, 1]); (Subscript[B, 2])=(((Subscript[b, 2])-(Subscript[b, 1])α)/(Subscript[ω, d])); (Subscript[ω, d])=(Subscript[ω, 0])(Sqrt[1-(ξ^2)])

Critically Damped Case (ξ = 1)

h(t)=((Subscript[C, 1])+(Subscript[C, 2])t)(e^(-αt))u(t)

(Subscript[y, step])(t)=[(((Subscript[C, 1])/α)+((Subscript[C, 2])/(α^2)))(1-(e^(-αt)))-((Subscript[C, 2])/α)t (e^(-αt))]u(t)

(Subscript[C, 1])=(Subscript[b, 1]); (Subscript[C, 2])=(Subscript[b, 2])-(Subscript[b, 1])α

2-9. Car Suspension System (project)

The results of the preceding section will now be used to analyze a car suspension system, which is selected in part because it offers a nice demonstration of how to model both the car suspension system and several examples of input excitations, including driving over a curb, over a pothole, and on a wavy pavement.

x(t) – input, vertical displacement of the pavement, defined relative to a reference ground level.

y(t) – output, vertical displacement of the car chassis from its equilibrium position.

m – one-fourth of the car’s mass, because the car has four wheels

k – spring constant or stiffness of the coil

b – damping coefficient of the shock absorber

Chapter 2 Summary

A system is linear if it has both the scaling and additivity properties.

A system is time-invariant if delaying the input delays the output by the same amount.

A system is LTI if it is both linear and time-invariant.

The impulse response h(t) is the response to an impulse δ(t). The step response (Subscript[y, step])(t) is the response to a step function input u(t).

If a system is LTI and its impulse response is known, its response to any other input can be computed using convolution, if all initial conditions are zero.

If a system is LTI, its response to a complex exponential time signal is another complex exponential time signal at the same frequency.

If a system is LTI, its response to a sinusoid is another sinusoid at the same frequency.

The frequency response function of a system can be computed from the system's LCCDE.

Concept Questions

Concept Question 2-1: 
What three properties must an LTI system have?

Concept Question 2-2: 
Does a system described by a linear differential equation qualify as LTI or just as linear?

Concept Question 2-3: 
How can you measure the impulse response of a real system?

Concept Question 2-4: 
How can you compute the impulse response of a system from its step response?

Concept Question 2-5: 
Describe the process involved in the application of the graphical convolution technique.

Concept Question 2-6: 
What initial conditions does the convolution operation require?

Concept Question 2-7: 
Describe the time-shift property of convolution.

Concept Question 2-8: 
What is the outcome of convolving a signal with a step function? With an impulse function?

Concept Question 2-9: 
What is the area property of convolution?

Concept Question 2-10: 
What constitutes a complete proof of a statement of the form “A is true if and only if B is true”?

Concept Question 2-11: 
Given an expression for the impulse response h(t) of an LTI system, how can you determine if the system is (a) causal and (b) BIBO stable?

Concept Question 2-12: 
If the input to an LTI system is a sinusoid, which of the following attributes can change between the input and the output: amplitude, frequency, or phase?

Concept Question 2-13: 
What kind of function is (OverHat[Η])(ω) for a system described by an LCCDE?

Concept Question 2-14: 
What are the three damping conditions of the impulse response?

Concept Question 2-15: 
How do input derivatives affect impulse responses?

Exercises

Note: answers to exercises are given in the textbook.

Exercise 2-1: 
Does the system  y(t)=(x^2)(t)  have the scaling property?

Exercise 2-2: 
Which of the following systems is linear?
(a) (Subscript[y, 1])(t)=|sin(3t)| x(t)  
(b) (Subscript[y, 2])(t)=a ((dx(t))/(dt))
(c) (Subscript[y, 3])(t)=|x(t)| 
(d) (Subscript[y, 4])(t)=sin[x(t)]

Exercise 2-3: 
Which systems are time-invariant?
(a) y(t)=((dx(t))/(dt))+sin[x(t-1)]   
(b) ((dy(t))/(dt))=2sin[x(t-1)]+3cos[x(t-1)]

Exercise 2-4: 
Determine the impulse response of a system whose step response is
(Subscript[y, step])(t)=({{{, ({{0, t⩽0}, {t, 0⩽t⩽1}, {1, t⩾1}})}})

Exercise 2-5: 
The RC circuit of Fig. 2-5(a) is excited by
x(t)=(1-1000t)[u(t)-u(t-0.001)]  
Compute the capacitor voltage y(t) for t > 0.001 s, given that (Subscript[τ, C])=1 s.

Exercise 2-6: 
Apply graphical convolution to the waveforms of x(t) and h(t) shown in Fig. E2-6 to determine y(t) = x(t)⋆h(t).

(* Declaration of original pulses for Exercise 2-6 *)h26[τ_]:=τ*(UnitStep[τ]-UnitStep[τ-3]);
x26[τ_]:=4*UnitBox[τ-1.5](* Plots of given waveforms x(t) and h(t)*)
GraphicsColumn[
{
Plot[h26[t],{t,-0.2,3.5},PlotRange->{{-0.5,4},{-1.2,4}},Exclusions->None,PlotStyle->{Thickness[0.007],RGBColor[0,0,0.8]},AxesLabel->{Style["t",14,Italic,FontFamily->"Times"],Style["h(t)",14,Italic,FontFamily->"Times"]}],
Plot[x26[t],{t,-0.2,3.5},PlotRange->{{-0.5,4},{-1.2,4.2}},Exclusions->None,PlotStyle->{Thickness[0.007],RGBColor[0.8,0,0]},AxesLabel->{Style["t",14,Italic,FontFamily->"Times"],Style["x(t)",14,Italic,FontFamily->"Times"]}]
}]

Exercise 2-7: 
Evaluate
u(t)⋆δ(t-3)-u(t-4)⋆δ(t+1)

Exercise 2-8: 
Evaluate
(lim_(t→∞))[(e^(-3t))u(t)⋆u(t)]

Exercise 2-9: 
A system’s impulse response is 
h(t)=u(t-1)/(t^2)  
Is the system BIBO stable?

Exercise 2-10: 
A system’s impulse response is 
h(t)=u(t-1)/t  
Is the system BIBO stable?

Exercise 2-11: 
A system’s impulse response is 
h(t)=(3+4j)(e^(-(1-2j)t))u(t)  
Is the system BIBO stable?

Exercise 2-12: 
cos(t) -> h(t)=(e^(-t))u(t) -> ?

Exercise 2-13: 
2 cos(t) -> system -> 2 cos(2t) + 2
Initial conditions are zero. 
Is this system LTI?

Exercise 2-14: 
2 cos(t) -> system -> 0
Can we say that the system is not LTI?

Exercise 2-15: 
Which damping condition is exhibited by h(t) of
(((d^2)y(t))/(d(t^2)))+5((dy(t))/(dt))+4y(t)=2((dx(t))/(dt))

Exercise 2-16: 
For what constant (Subscript[a, 1]) is
(((d^2)y(t))/(d(t^2)))+(Subscript[a, 1])((dy(t))/(dt))+9y(t)=2((dx(t))/(dt))
critically damped?

Temporary items

Demonstration a. Graphical convolution solution for RC circuit of Example 2-4.

x24a[τ_]:=UnitBox[τ-0.5];
h24a[τ_]:=2 (E^(-2 τ))UnitStep[τ];
y24a[t_]=(∫_0^t)x24[τ] h24[t-τ]\[DifferentialD]τ;
Manipulate[
GraphicsColumn[{
Show[{
Plot[{x24a[τ]*h24a[-τ+t]},{τ,-2,3}, Filling->Axis,PlotRange->{{-2,3},{-0.1,2.5}},AxesLabel->{"τ","(Subscript[v, in])(τ), h(t-τ)"},PerformanceGoal->"Quality"],
Plot[{x24a[τ],h24a[-τ+t]},{τ,-2,3}, PlotStyle->Thickness[0.007],Exclusions->None,ClippingStyle->False,PerformanceGoal->"Quality"],
Graphics[{
Text[Style["(Subscript[v, in])(τ)",10],{0.5,1.1}],
Text[Style["h(t-τ)",10],{t,2.1}]
}]}],
Show[{
Plot[y24a[x],{x,-2,t}, PlotStyle->Thickness[0.007],Exclusions->None,ClippingStyle->False,AxesLabel->{"t","(Subscript[v, out])(t)"},PlotRange->{{-2,3},{-0.1,1}},PerformanceGoal->"Quality"],
Graphics[{PointSize[Large],Blue, Point[{t, y24a[t]}]}]
}]
}],
{{t,-1.5,"Time t"},-1.5,4},
Paneled->False,
FrameLabel->{None,None,Style["Graphical convolution",16,FontFamily->"Times"]},
ControlPlacement->Bottom,
LabelStyle->Directive[RGBColor[0,0,0.5],12,Bold,FontFamily->"Times"],
Frame->True
]