[2.4-1 The Convolution Integral](#page-8-0)

← Back to LINEAR SYSTEMS AND SIGNALS Overview

2.4-1 The Convolution Integral

The zero-state response y(t) obtained in Eq. (2.23) is given by an integral that occurs frequently in the physical sciences, engineering, and mathematics. For this reason this integral is given a special name: the convolution integral. The convolution integral of two functions x1(t) and x2(t) is denoted symbolically by x1(t) ∗ x2(t) and is defined as

$$x_1(t) * x_2(t) \equiv \int_{-\infty}^{\infty} x_1(\tau) x_2(t - \tau) d\tau$$

(2.24)

Some important properties of the convolution integral follow.

THE COMMUTATIVE PROPERTY

Convolution operation is commutative; that is, x1(t) ∗ x2(t) = x2(t) ∗ x1(t). This property can be proved by a change of variable. In Eq. (2.24), if we let z = t − τ so that τ = t − z and dτ = −dz, we obtain

$$x_1(t) * x_2(t) = -\int_{-\infty}^{-\infty} x_2(z) x_1(t-z) dz$$

=

$$\int_{-\infty}^{\infty} x_2(z) x_1(t-z) dz$$

=

$$x_2(t) * x_1(t)$$

(2.25)

$$y(t) = \int_{-\infty}^{\infty} x(\tau)h(t, \tau) d\tau$$

where h(t, τ ) is the system response at instant t to a unit impulse input located at τ .

^† In deriving this result we have assumed a time-invariant system. If the system is time-varying, then the system response to the input δ(t−nΔτ ) cannot be expressed as h(t−nΔτ ) but instead has the form h(t,nΔτ ). Use of this form modifies Eq. (2.23) to

THE DISTRIBUTIVE PROPERTY

According to the distributive property,

$$x_1(t) * [x_2(t) + x_3(t)] = x_1(t) * x_2(t) + x_1(t) * x_3(t)$$

\n(2.26)

THE ASSOCIATIVE PROPERTY

According to the associative property,

$$x_1(t) * [x_2(t) * x_3(t)] = [x_1(t) * x_2(t)] * x_3(t)$$

\n(2.27)

The proofs of Eqs. (2.26) and (2.27) follow directly from the definition of the convolution integral. They are left as an exercise for the reader.

THE SHIFT PROPERTY If

$$x_1(t) * x_2(t) = c(t)$$

then

$$x_1(t) * x_2(t - T) = x_1(t - T) * x_2(t) = c(t - T)$$

More generally, we see that

$$x_1(t - T_1) * x_2(t - T_2) = c(t - T_1 - T_2)$$

\n(2.28)

Proof. We are given

$$x_1(t) * x_2(t) = \int_{-\infty}^{\infty} x_1(\tau) x_2(t - \tau) d\tau = c(t)$$

Therefore,

$$x_1(t) * x_2(t-T) = \int_{-\infty}^{\infty} x_1(\tau) x_2(t-T-\tau) d\tau$$ $$= c(t-T)$$

The equally simple proof of Eq. (2.28) follows a similar approach.

CONVOLUTION WITH AN IMPULSE

Convolution of a function x(t) with a unit impulse results in the function x(t) itself. By definition of convolution,

$$x(t) * \delta(t) = \int_{-\infty}^{\infty} x(\tau) \delta(t - \tau) d\tau$$

Because δ(t −τ ) is an impulse located at τ = t, according to the sampling property of the impulse [Eq. (1.11)], the integral here is just the value of x(τ ) at τ = t, that is, x(t). Therefore,

$$x(t) * \delta(t) = x(t)$$

Actually this result was derived earlier [Eq. (2.22)].

172 CHAPTER 2 TIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS

THE WIDTH PROPERTY

If the durations (widths) of x1(t) and x2(t) are finite, given by T¹ and T2, respectively, then the duration (width) of x1(t) ∗ x2(t) is T¹ + T² (Fig. 2.4). The proof of this property follows readily from the graphical considerations discussed later in Sec. 2.4-2.

Figure 2.4 Width property of convolution.

ZERO-STATE RESPONSE AND CAUSALITY

The (zero-state) response y(t) of an LTIC system is

$$y(t) = x(t) * h(t) = \int_{-\infty}^{\infty} x(\tau)h(t-\tau) d\tau$$

\n(2.29)

In deriving Eq. (2.29), we assumed the system to be linear and time-invariant. There were no other restrictions either on the system or on the input signal x(t). Since, in practice, most systems are causal, their response cannot begin before the input. Furthermore, most inputs are also causal, which means they start at t = 0.

Causality restriction on both signals and systems further simplifies the limits of integration in Eq. (2.29). By definition, the response of a causal system cannot begin before its input begins. Consequently, the causal system’s response to a unit impulse δ(t) (which is located at t = 0) cannot begin before t = 0. Therefore, a causal system’s unit impulse response h(t) is a causal signal.

It is important to remember that the integration in Eq. (2.29) is performed with respect to τ (not t). If the input x(t) is causal, x(τ ) = 0 for τ < 0. Therefore, x(τ ) = 0 for τ < 0, as illustrated in Fig. 2.5a. Similarly, if h(t) is causal, h(t − τ ) = 0 for t − τ < 0; that is, for τ > t, as depicted in Fig. 2.5a. Therefore, the product x(τ )h(t − τ ) = 0 everywhere except over the nonshaded interval 0 ≤ τ ≤ t shown in Fig. 2.5a (assuming t ≥ 0). Observe that if t is negative, x(τ )h(t − τ ) = 0 for all τ , as shown in Fig. 2.5b. Therefore, Eq. (2.29) reduces to

$$y(t) = x(t) * h(t) = \begin{cases} \int_0^t x(\tau)h(t-\tau) d\tau & t \ge 0\\ 0 & t < 0 \end{cases}$$

(2.30)

The lower limit of integration in Eq. (2.30) is taken as 0⁻ to avoid the difficulty in integration that can arise if x(t) contains an impulse at the origin. This result shows that if x(t) and h(t) are both causal, the response y(t) is also causal.

Because of the convolution’s commutative property [Eq. (2.25)], we can also express Eq. (2.30) as [assuming causal x(t) and h(t)]

$$y(t) = \begin{cases} \int_0^t h(\tau)x(t-\tau)d\tau & t \ge 0\\ 0 & t < 0 \end{cases}$$

Hereafter, the lower limit of 0⁻ will be implied even when we write it as 0. As in Eq. (2.30), this result assumes that both the input and the system are causal.

EXAMPLE 2.8 Computing the Zero-State Response

For an LTIC system with the unit impulse response h(t) = e−2*^t u*(t), determine the response y(t) for the input

$$x(t) = e^{-t}u(t)$$

Here both x(t) and h(t) are causal (Fig. 2.6). Hence, from Eq. (2.30), we obtain

$$y(t) = \int_0^t x(\tau)h(t-\tau) d\tau \qquad t \ge 0$$

Because x(t) = e−^t u(t) and h(t) = e−2*^t u*(t),

$$x(\tau) = e^{-\tau} u(\tau)$$

and $h(t - \tau) = e^{-2(t - \tau)} u(t - \tau)$

Remember that the integration is performed with respect to τ (not t), and the region of integration is 0 ≤ τ ≤ t. Hence, τ ≥ 0 and t − τ ≥ 0. Therefore, u(τ ) = 1 and u(t − τ ) = 1; consequently,

$$y(t) = \int_0^t e^{-\tau} e^{-2(t-\tau)} d\tau \qquad t \ge 0$$

Because this integration is with respect to τ , we can pull e−2*^t* outside the integral, giving us

$$y(t) = e^{-2t} \int_0^t e^{\tau} d\tau = e^{-2t} (e^t - 1) = e^{-t} - e^{-2t} \qquad t \ge 0$$

Moreover, y(t) = 0 when t < 0 [see Eq. (2.30)]. Therefore,

$$y(t) = (e^{-t} - e^{-2t})u(t)$$

The response is depicted in Fig. 2.6c.

DR ILL 2.5 Computing the Zero-State Response

For an LTIC system with the impulse response h(t) = 6e−^t u(t), determine the system response to the input: (a) 2u(t) and (b) 3e−3*^t u*(t).

ANSWERS

• (a) 12(1−e−^t )u(t)
• (b) 9(e−^t −e−3*^t* )u(t)

DR ILL 2.6 Zero-State Response with Resonance

Repeat Drill 2.5 for the input x(t) = e−^t u(t).

ANSWER

6te−^t u(t)

THE CONVOLUTION TABLE

The task of convolution is considerably simplified by a ready-made convolution table (Table 2.1). This table, which lists several pairs of signals and their convolution, can conveniently determine y(t), a system response to an input x(t), without performing the tedious job of integration. For instance, we could have readily found the convolution in Ex. 2.8 by using pair 4 (with λ¹ = −1 and λ² = −2) to be (e−^t −e−2*^t* )u(t). The following example demonstrates the utility of this table.

EXAMPLE 2.9 Convolution by Tables

Use Table 2.1 to compute the loop current y(t) of the RLC circuit in Ex. 2.4 for the input x(t) = 10e−3*^t u*(t) when all the initial conditions are zero.

The loop equation for this circuit [see Ex. 1.16 or Eq. (1.29)] is

$$(D^2 + 3D + 2)y(t) = Dx(t)$$

The impulse response h(t) for this system, as obtained in Ex. 2.6, is

$$h(t) = (2e^{-2t} - e^{-t})u(t)$$

The input is x(t) = 10e−3*^t u*(t), and the response y(t) is

$$y(t) = x(t) * h(t) = 10e^{-3t}u(t) * [2e^{-2t} - e^{-t}]u(t)$$

Using the distributive property of the convolution [Eq. (2.26)], we obtain

$$y(t) = 10e^{-3t}u(t) * 2e^{-2t}u(t) - 10e^{-3t}u(t) * e^{-t}u(t)$$

= 20[e^-3tu(t) * e^-2tu(t)] - 10[e^-3tu(t) * e^-tu(t)]

Now the use of pair 4 in Table 2.1 yields

$$y(t) = \frac{20}{-3 - (-2)} [e^{-3t} - e^{-2t}] u(t) - \frac{10}{-3 - (-1)} [e^{-3t} - e^{-t}] u(t)$$

= $-20(e^{-3t} - e^{-2t}) u(t) + 5(e^{-3t} - e^{-t}) u(t)$
= $(-5e^{-t} + 20e^{-2t} - 15e^{-3t}) u(t)$

No.	x1(t)	x2(t)	x1(t) ∗ x2(t) = x2(t) ∗ x1(t)
1	x(t)	δ(t −T)	x(t −T)
2	eλt u(t)	u(t)	1−eλt u(t) −λ
3	u(t)	u(t)	tu(t)
4	eλ1t u(t)	eλ2t u(t)	eλ1t −eλ2t u(t) λ1 = λ2 λ1 −λ2
5	eλt u(t)	eλt u(t)	teλt u(t)
6	teλt u(t)	eλt u(t)	1 2eλt t u(t) 2
7	Nu(t) t	eλt u(t)	“N N! eλt N−k N!t λN+1 u(t) − u(t) λk+1(N −k)! k=0
8	Mu(t) t	Nu(t) t	M!N! M+N+1u(t) t (M +N +1)!
9	teλ1t u(t)	eλ2t u(t)	eλ2t −eλ1t +(λ1 −λ2)teλ1t u(t) −λ2)2 (λ1
10	Meλt t u(t)	Neλt t u(t)	M!N! M+N+1eλt t u(t) (N + M +1)!
11	Meλ1t t u(t)	Neλ2t t u(t)	“M (−1)kM!(N M−keλ1t +k)!t −λ2)N+k+1 u(t) k!(M −k)!(λ1 k=0
	λ1 = λ2		“N (−1)kN!(M N−keλ2t +k)!t −λ1)M+k+1 u(t) + k!(N −k)!(λ2 k=0
12	e−αt cos(βt +θ )u(t)	eλt u(t)	cos(θ −φ)eλt −e−αt cos(βt +θ −φ) u(t) (α +λ)2 +β2
			φ = tan−1[−β/(α +λ)]
13	eλ1t u(t)	eλ2t u(−t)	eλ1t u(t)+eλ2t u(−t) Reλ2 > Reλ1 λ2 −λ1
14	eλ1t u(−t)	eλ2t u(−t)	eλ1t −eλ2t u(−t) λ2 −λ1

TABLE 2.1 Select Convolution Integrals

DR ILL 2.7 Convolution by Tables

Use Table 2.1 to show e−2*^t u*(t) ∗ (1−e−^t )u(t) ^{= 1 2 −}e−^{t + 1 2} e−2*^t u*(t).

DR ILL 2.8 Zero-State Response by Convolution Table

Rework Drills 2.5 and 2.6 using Table 2.1.

DR ILL 2.9 Another Zero-State Response by Convolution Table

For an LTIC system with the unit impulse response h(t) = e−2*^t u*(t), determine the zero-state response y(t) if the input x(t) = sin 3t u(t). [Hint: Use pair 12 from Table 2.1.]

ANSWER

1 ¹³ [3e−2*^t* ^{+ √}13 cos(3*^t* ⁻146.32◦)]u(t) or ^{1 13} [3e−2*^t* ^{− √}13 cos(3*^t* ⁺33.68◦)]u(t)

RESPONSE TO COMPLEX INPUTS

The LTIC system response discussed so far applies to general input signals, real or complex. However, if the system is real, that is, if h(t) is real, then we shall show that the real part of the input generates the real part of the output, and a similar conclusion applies to the imaginary part.

If the input is x(t) = xr(t) + jxi(t), where xr(t) and xi(t) are the real and imaginary parts of x(t), then for real h(t)

$$y(t) = h(t) * [x_r(t) + jx_i(t)] = h(t) * x_r(t) + jh(t) * x_i(t) = y_r(t) + jy_i(t)$$

where yr(t) and yi(t) are the real and the imaginary parts of y(t). Using the right-directed-arrow notation to indicate a pair of the input and the corresponding output, the foregoing result can be expressed as follows. If

$$x(t) = x_r(t) + jx_i(t) \implies y(t) = y_r(t) + jy_i(t)$$

then

$$x_r(t) \implies y_r(t) \quad \text{and} \quad x_i(t) \implies y_i(t) \tag{2.31}$$

MULTIPLE INPUTS

Multiple inputs to LTI systems can be treated by applying the superposition principle. Each input is considered separately, with all other inputs assumed to be zero. The sum of all these individual system responses constitutes the total system output when all the inputs are applied simultaneously.

2.4-2 Graphical Understanding of Convolution Operation

The convolution operation can be grasped readily through a graphical interpretation of the convolution integral. Such an understanding is helpful in evaluating the convolution integral of more complex signals. In addition, graphical convolution allows us to grasp visually or mentally the convolution integral’s result, which can be of great help in sampling, filtering, and many other problems. Finally, many signals have no exact mathematical description, so they can be described only graphically. If two such signals are to be convolved, we have no choice but to perform their convolution graphically.

We shall now explain the convolution operation by convolving the signals x(t) and g(t), illustrated in Figs. 2.7a and 2.7b, respectively. If c(t) is the convolution of x(t) with g(t), then

$$c(t) = \int_{-\infty}^{\infty} x(\tau)g(t-\tau) d\tau$$

One of the crucial points to remember here is that this integration is performed with respect to τ so that t is just a parameter (like a constant). This consideration is especially important when we sketch the graphical representations of the functions x(τ ) and g(t−τ ). Both these functions should be sketched as functions of τ , not of t.

The function x(τ ) is identical to x(t), with τ replacing t (Fig. 2.7c). Therefore, x(t) and x(τ ) will have the same graphical representations. Similar remarks apply to g(t) and g(τ ) (Fig. 2.7d).

To appreciate what g(t − τ ) looks like, let us start with the function g(τ ) (Fig. 2.7d). Time reversal of this function (reflection about the vertical axis τ = 0) yields g(−τ ) (Fig. 2.7e). Let us denote this function by φ(τ ):

$$\phi(\tau) = g(-\tau)$$

Now φ(τ ) shifted by t seconds is φ(τ −t), given by

$$\phi(\tau - t) = g[-(\tau - t)] = g(t - \tau)$$

Therefore, we first time-reverse g(τ ) to obtain g(−τ ) and then time-shift g(−τ ) by t to obtain g(t − τ ). For positive t, the shift is to the right (Fig. 2.7f); for negative t, the shift is to the left (Figs. 2.7g, 2.7h).

The preceding discussion gives us a graphical interpretation of the functions x(τ ) and g(t−τ ). The convolution c(t) is the area under the product of these two functions. Thus, to compute c(t) at some positive instant t = t1, we first obtain g(−τ ) by inverting g(τ ) about the vertical axis. Next, we right-shift or delay g(−τ ) by t¹ to obtain g(t¹ − τ ) (Fig. 2.7f), and then we multiply this function by x(τ ), giving us the product x(τ )g(t¹ − τ ) (shaded portion in Fig. 2.7f). The area A¹ under this product is c(t1), the value of c(t) at t = t1. We can therefore plot c(t1) = A¹ on a curve describing c(t), as shown in Fig. 2.7i. The area under the product x(τ )g(−τ ) in Fig. 2.7e is c(0), the value of the convolution for t = 0 (at the origin).

Figure 2.7 Graphical explanation of the convolution operation.

180 CHAPTER 2 TIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS

A similar procedure is followed in computing the value of c(t) at t = t2, where t² is negative (Fig. 2.7g). In this case, the function g(−τ ) is shifted by a negative amount (that is, left-shifted) to obtain g(t2−τ ). Multiplication of this function with x(τ ) yields the product x(τ )g(t2−τ ). The area under this product is c(t2) = A2, giving us another point on the curve c(t) at t = t² (Fig. 2.7i). This procedure can be repeated for all values of t, from −∞ to ∞. The result will be a curve describing c(t) for all time t. Note that when t ≤ −3, x(τ ) and g(t−τ ) do not overlap (see Fig. 2.7h); therefore, c(t) = 0 for t ≤ −3.

SUMMARY OF THE GRAPHICAL PROCEDURE

The procedure for graphical convolution can be summarized as follows:

• 1. Keep the function x(τ ) fixed.
• 2. Visualize the function g(τ ) as a rigid wire frame, and rotate (or invert) this frame about the vertical axis (τ = 0) to obtain g(−τ ).
• 3. Shift the inverted frame along the τ axis by t⁰ seconds. The shifted frame now represents g(t⁰ −τ ).
• 4. The area under the product of x(τ ) and g(t⁰ − τ ) (the shifted frame) is c(t0), the value of the convolution at t = t0.
• 5. Repeat this procedure, shifting the frame by different values (positive and negative) to obtain c(t) for all values of t.

The graphical procedure discussed here appears very complicated and discouraging at first reading. Indeed, some people claim that convolution has driven many electrical engineering undergraduates to contemplate theology either for salvation or as an alternative career (IEEE Spectrum, March 1991, p. 60). Actually, the bark of convolution is worse than its bite. In graphical convolution, we need to determine the area under the product x(τ )g(t − τ ) for all values of t from −∞ to ∞. However, a mathematical description of x(τ )g(t − τ ) is generally valid over a range

Convolution: Its bark is worse than its bite!

of t. Therefore, repeating the procedure for every value of t amounts to repeating it only a few times for different ranges of t.

We can also use the commutative property of convolution to our advantage by computing x(t) ∗ g(t) or g(t) ∗ x(t), whichever is simpler. As a rule of thumb, convolution computations are simplified if we choose to invert (time-reverse) the simpler of the two functions. For example, if the mathematical description of g(t) is simpler than that of x(t), then x(t) ∗ g(t) will be easier to compute than g(t) ∗ x(t). In contrast, if the mathematical description of x(t) is simpler, the reverse will be true.

We shall demonstrate graphical convolution with the following examples. Let us start by using this graphical method to rework Ex. 2.8.

EXAMPLE 2.10 Graphical Convolution of Two Causal Functions

Determine graphically y(t) = x(t) ∗ h(t) for x(t) = e−^t u(t) and h(t) = e−2*^t u*(t).

In Figs. 2.8a and 2.8b we have x(t) and h(t), respectively; and Fig. 2.8c shows x(τ ) and h(−τ ) as functions of τ . The function h(t −τ ) is now obtained by shifting h(−τ ) by t. If t is positive, the shift is to the right (delay); if t is negative, the shift is to the left (advance). Figure 2.8d shows that for negative t, h(t −τ ) [obtained by left-shifting h(−τ )] does not overlap x(τ ), and the product x(τ )h(t −τ ) = 0, so that

$$y(t) = 0 \qquad t < 0$$

Figure 2.8e shows the situation for t ≥ 0. Here x(τ ) and h(t −τ ) do overlap, but the product is nonzero only over the interval 0 ≤ τ ≤ t (shaded interval). Therefore,

$$y(t) = \int_0^t x(\tau)h(t-\tau) d\tau \qquad t \ge 0$$

All we need to do now is substitute correct expressions for x(τ ) and h(t − τ ) in this integral. From Figs. 2.8a and 2.8b, it is clear that the segments of x(t) and g(t) to be used in this convolution (Fig. 2.8e) are described by

$$x(t) = e^{-t} \qquad \text{and} \qquad h(t) = e^{-2t}$$

Therefore,

$$x(\tau) = e^{-\tau}$$

and $h(t - \tau) = e^{-2(t - \tau)}$

Consequently,

$$y(t) = \int_0^t e^{-\tau} e^{-2(t-\tau)} d\tau = e^{-2t} \int_0^t e^{\tau} d\tau = e^{-t} - e^{-2t} \qquad t \ge 0$$

Moreover, y(t) = 0 for t < 0 so that

$$y(t) = (e^{-t} - e^{-2t})u(t)$$

EXAMPLE 2.11 Graphical Convolution: Causal Function and Two-Sided Function

Find c(t) = x(t) ∗ g(t) for the signals depicted in Figs. 2.9a and 2.9b.

Since x(t) is simpler than g(t), it is easier to evaluate g(t) ∗ x(t) than x(t) ∗ g(t). However, we shall intentionally take the more difficult route and evaluate x(t) ∗ g(t).

From x(t) and g(t) (Figs. 2.9a and 2.9b, respectively), observe that g(t) is composed of two segments. As a result, it can be described as

$$g(t) = \begin{cases} 2e^{-t} & \text{segment A} \\ -2e^{2t} & \text{segment B} \end{cases}$$

Therefore,

$$g(t - \tau) = \begin{cases} 2e^{-(t-\tau)} & \text{segment A} \\ -2e^{2(t-\tau)} & \text{segment B} \end{cases}$$

The segment of x(t) that is used in convolution is x(t) = 1 so that x(τ ) = 1. Figure 2.9c shows x(τ ) and g(−τ ).

To compute c(t) for t ≥ 0, we right-shift g(−τ ) to obtain g(t−τ ), as illustrated in Fig. 2.9d. Clearly, g(t − τ ) overlaps with x(τ ) over the shaded interval, that is, over the range τ ≥ 0; segment A overlaps with x(τ ) over the interval (0,t), while segment B overlaps with x(τ ) over (t,∞). Remembering that x(τ ) = 1, we have

$$c(t) = \int_0^\infty x(\tau)g(t-\tau) d\tau$$

=

$$\int_0^t 2e^{-(t-\tau)} d\tau + \int_t^\infty -2e^{2(t-\tau)} d\tau$$

=

$$2(1 - e^{-t}) - 1 = 1 - 2e^{-t} \qquad t \ge 0$$

Figure 2.9e shows the situation for t < 0. Here the overlap is over the shaded interval, that is, over the range τ ≥ 0, where only the segment B of g(t) is involved. Therefore,

$$c(t) = \int_0^\infty x(\tau)g(t-\tau) d\tau = \int_0^\infty -2e^{2(t-\tau)} d\tau = -e^{2t} \qquad t \le 0$$

Therefore,

$$c(t) = \begin{cases} 1 - 2e^{-t} & t \ge 0\\ -e^{2t} & t \le 0 \end{cases}$$

Figure 2.9f shows a plot of c(t).

EXAMPLE 2.12 Graphical Convolution of Two Finite-Duration Functions

Find x(t) ∗ g(t) for the functions x(t) and g(t) shown in Figs. 2.10a and 2.10b.

Here, x(t) has a simpler mathematical description than that of g(t), so it is preferable to time-reverse x(t). Hence, we shall determine g(t) ∗ x(t) rather than x(t) ∗ g(t). Thus,

$$c(t) = g(t) * x(t) = \int_{-\infty}^{\infty} g(\tau) x(t - \tau) d\tau$$

First, we determine the expressions for the segments of x(t) and g(t) used in finding c(t). According to Figs. 2.10a and 2.10b, these segments can be expressed as

$$x(t) = 1$$

and $g(t) = \frac{1}{3}t$

so that

$$x(t - \tau) = 1 \qquad \text{and} \qquad g(\tau) = \frac{1}{3}\tau$$

Figure 2.10c shows g(τ ) and x(−τ ), whereas Fig. 2.10d shows g(τ ) and x(t − τ ), which is x(−τ ) shifted by t. Because the edges of x(−τ ) are at τ = −1 and 1, the edges of x(t − τ ) are at −1 + t and 1 + t. The two functions overlap over the interval (0, 1 + t) (shaded interval) so that

$$c(t) = \int_0^{1+t} g(\tau) x(t-\tau) d\tau = \int_0^{1+t} \frac{1}{3} \tau d\tau = \frac{1}{6} (t+1)^2 \qquad -1 \le t \le 1 \qquad (2.32)$$

This situation, depicted in Fig. 2.10d, is valid only for −1 ≤ t ≤ 1. For t ≥ 1 but ≤ 2, the situation is as illustrated in Fig. 2.10e. The two functions overlap only over the range −1 + t to 1 + t (shaded interval). Note that the expressions for g(τ ) and x(t − τ ) do not change; only the range of integration changes. Therefore,

$$c(t) = \int_{-1+t}^{1+t} \frac{1}{3}\tau \,d\tau = \frac{2}{3}t \qquad \qquad 1 \le t \le 2 \tag{2.33}$$

Also note that the expressions in Eqs. (2.32) and (2.33) both apply at t = 1, the transition point between their respective ranges. We can readily verify that both expressions yield a value of 2/3 at t = 1 so that c(1) = 2/3. The continuity of c(t) at transition points indicates a high probability of a correct answer. Continuity of c(t) at transition points is assured as long as x(t) and g(t) contain no impulse functions.

For t ≥ 2 but ≤ 4, the situation is as shown in Fig. 2.10f. The functions g(τ ) and x(t − τ ) overlap over the interval from −1+t to 3 (shaded interval) so that

$$c(t) = \int_{-1+t}^{3} \frac{1}{3}\tau \,d\tau = -\frac{1}{6}(t^2 - 2t - 8) \qquad 2 \le t \le 4 \qquad (2.34)$$

Both Eqs. (2.33) and (2.34) apply at the transition point t = 2. We can readily verify that c(2) = 4/3 when either of these expressions is used.

For t ≥ 4, x(t − τ ) has been shifted so far to the right that it no longer overlaps with g(τ ) as depicted in Fig. 2.10g. Consequently,

$$c(t) = 0 \qquad \qquad t \ge 4$$

We now turn our attention to negative values of t. We have already determined c(t) up to t = −1. For t < −1, there is no overlap between the two functions, as illustrated in Fig. 2.10h, so that

$$c(t) = 0 \qquad \qquad t \le -1$$

Combining our results, we see that

$$c(t) = \begin{cases} \frac{1}{6}(t+1)^2 & -1 \le t < 1 \\ \frac{2}{3}t & 1 \le t < 2 \\ -\frac{1}{6}(t^2 - 2t - 8) & 2 \le t < 4 \\ 0 & \text{otherwise} \end{cases}$$

Figure 2.10i plots c(t) according to this expression.

THE WIDTH OF CONVOLVED FUNCTIONS

The widths (durations) of x(t), g(t), and c(t) in Ex. 2.12 (Fig. 2.10) are 2, 3, and 5, respectively. Note that the width of c(t) in this case is the sum of the widths of x(t) and g(t). This observation is not a coincidence. Using the concept of graphical convolution, we can readily see that if x(t) and g(t) have the finite widths of T¹ and T² respectively, then the width of c(t) is equal to T¹ +T2. The reason is that the time it takes for a signal of width (duration) T¹ to completely pass another signal of width (duration) T² so that they become non-overlapping is T1+T2. When the two signals become non-overlapping, the convolution goes to zero.

DR ILL 2.10 Interchanging Convolution Order

Rework Ex. 2.11 by evaluating g(t) ∗ x(t).

DR ILL 2.11 Showing Commutability Using Two Causal Signals

Use graphical convolution to show that x(t) ∗ g(t) = g(t) ∗ x(t) = c(t) in Fig. 2.11.

DR ILL 2.12 Showing Commutability Using a Causal Signal and an Anticausal Signal

DR ILL 2.13 Showing Commutability Using Shifted Signals

Repeat Drill 2.11 for the functions in Fig. 2.13.

THE PHANTOM OF THE SIGNALS AND SYSTEMS OPERA

In the study of signals and systems we often come across some signals such as an impulse, which cannot be generated in practice and have never been sighted by anyone.† One wonders why we even consider such idealized signals. The answer should be clear from our discussion so far in this chapter. Even if the impulse function has no physical existence, we can compute the system response h(t) to this phantom input according to the procedure in Sec. 2.3, and knowing h(t), we can compute the system response to any arbitrary input. The concept of impulse response, therefore, provides an effective intermediary for computing system response to an arbitrary input. In addition, the impulse response h(t) itself provides a great deal of information and insight about the system behavior. In Sec. 2.6 we show that the knowledge of impulse response provides much valuable information, such as the response time, pulse dispersion, and filtering properties of the system. Many other useful insights about the system behavior can be obtained by inspection of h(t).

Similarly, in frequency-domain analysis (discussed in later chapters), we use an everlasting exponential (or sinusoid) to determine system response. An everlasting exponential (or sinusoid), too, is a phantom, which nobody has ever seen and which has no physical existence. But it provides another effective intermediary for computing the system response to an arbitrary input. Moreover, the system response to everlasting exponential (or sinusoid) provides valuable information and insight regarding the system’s behavior. Clearly, idealized impulses and everlasting sinusoids are friendly and helpful spirits.

Interestingly, the unit impulse and the everlasting exponential (or sinusoid) are the dual of each other in the time-frequency duality, to be studied in Ch. 7. Actually, the time-domain and the frequency-domain methods of analysis are the dual of each other.

WHY CONVOLUTION? AN INTUITIVE EXPLANATION OF SYSTEM RESPONSE

On the surface, it appears rather strange that the response of linear systems (those gentlest of the gentle systems) should be given by such a tortuous operation of convolution, where one signal is fixed and the other is inverted and shifted. To understand this odd behavior, consider a hypothetical impulse response h(t) that decays linearly with time (Fig. 2.14a). This response is strongest at t =0, the moment the impulse is applied, and it decays linearly at future instants so that one second later (at t = 1 and beyond), it ceases to exist. This means that the closer the impulse input is to an instant t, the stronger is its response at t.

Now consider the input x(t) shown in Fig. 2.14b. To compute the system response, we break the input into rectangular pulses and approximate these pulses with impulses. Generally, the response of a causal system at some instant t will be determined by all the impulse components of the input before t. Each of these impulse components will have different weight in determining the response at the instant t, depending on its proximity to t. As seen earlier, the closer the impulse is to t, the stronger is its influence at t. The impulse at t has the greatest weight (unity) in determining

^† The late Prof. S. J. Mason, the inventor of signal flow graph techniques, used to tell a story of a student frustrated with the impulse function. The student said, “The unit impulse is a thing that is so small you can’t see it, except at one place (the origin), where it is so big you can’t see it. In other words, you can’t see it at all; at least I can’t!” [2].

Figure 2.14 Intuitive explanation of convolution.

the response at t. The weight decreases linearly for all impulses before t until the instant t − 1. The input before t − 1 has no influence (zero weight). Thus, to determine the system response at t, we must assign a linearly decreasing weight to impulses occurring before t, as shown in Fig. 2.14b. This weighting function is precisely the function h(t − τ ). The system response at t is then determined not by the input x(τ ) but by the weighted input x(τ )h(t − τ ), and the summation of all these weighted inputs is the convolution integral.

2.4-3 Interconnected Systems

A larger, more complex system can often be viewed as the interconnection of several smaller subsystems, each of which is easier to characterize. Knowing the characterizations of these subsystems, it becomes simpler to analyze such large systems. We shall consider here two basic interconnections, cascade and parallel. Figure 2.15a shows S¹ and S2, two LTIC subsystems connected in parallel, and Fig. 2.15b shows the same two systems connected in cascade.

In Fig. 2.15a, the device depicted by the symbol inside a circle represents an adder, which adds signals at its inputs. Also the junction from which two (or more) branches radiate out is called the pickoff node. Every branch that radiates out from the pickoff node carries the same signal (the signal at the junction). In Fig. 2.15a, for instance, the junction at which the input is applied is a pickoff node from which two branches radiate out, each of which carries the input signal at the node.

Let the impulse response of S¹ and S² be h1(t) and h2(t), respectively. Further assume that interconnecting these systems, as shown in Fig. 2.15, does not load them. This means that the impulse response of either of these systems remains unchanged whether observed when these systems are unconnected or when they are interconnected.

To find hp(t), the impulse response of the parallel system S^p in Fig. 2.15a, we apply an impulse at the input of Sp. This results in the signal δ(t) at the inputs of S¹ and S2, leading to their outputs h1(t) and h2(t), respectively. These signals are added by the adder to yield h1(t) + h2(t) as the output of Sp:

$$h_p(t) = h_1(t) + h_2(t)$$

To find hc(t), the impulse response of the cascade system S^c in Fig. 2.15b, we apply the input δ(t) at the input of Sc, which is also the input to S1. Hence, the output of S¹ is h1(t), which now acts

Figure 2.15 Interconnected systems.

as the input to S2. The response of S² to input h1(t) is h1(t) ∗ h2(t). Therefore,

$$h_c(t) = h_1(t) * h_2(t)$$

Because of the commutative property of convolution, it follows that interchanging the systems S¹ and S2, as shown in Fig. 2.15c, results in the same impulse response h1(t) ∗ h2(t). This means that when several LTIC systems are cascaded, the order of systems does not affect the impulse response of the composite system. In other words, linear operations, performed in cascade, commute. The order in which they are performed is not important, at least theoretically.†

^†Change of order, however, could affect performance because of physical limitations and sensitivities to changes in the subsystems involved.

192 CHAPTER 2 TIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS

We shall give here another interesting application of the commutative property of LTIC systems. Figure 2.15d shows a cascade of two LTIC systems: a system S with impulse response h(t), followed by an ideal integrator. Figure 2.15e shows a cascade of the same two systems in reverse order; an ideal integrator followed by S. In Fig. 2.15d, if the input x(t) to S yields the output y(t), then the output of the system of Fig. 2.15d is the integral of y(t). In Fig. 2.15e, the output of the integrator is the integral of x(t). The output in Fig. 2.15e is identical to the output in Fig. 2.15d. Hence, it follows that if an LTIC system response to input x(t) is y(t), then the response of the same system to the integral of x(t) is the integral of y(t). In other words,

if

$$x(t) \Longrightarrow y(t)$$

then $\int_{-\infty}^{t} x(\tau) d\tau \Longrightarrow \int_{-\infty}^{t} y(\tau) d\tau$

Replacing the ideal integrator with an ideal differentiator in Figs. 2.15d and 2.15e, and following a similar argument, we conclude that

if

$$x(t) \Longrightarrow y(t)$$

then $\frac{dx(t)}{dt} \Longrightarrow \frac{dy(t)}{dt}$

If we let x(t) = δ(t) and y(t) = h(t) in Fig. 2.15e, we find that g(t), the unit step response of an LTIC system with impulse h(t), is given by

$$g(t) = \int_{-\infty}^{t} h(\tau) d\tau$$

\n(2.35)

We can also show that the system response to δ(˙ t) is dh(t)/dt. These results can be extended to other singularity functions. For example, the unit ramp response of an LTIC system is the integral of its unit step response, and so on.

INVERSE SYSTEMS

In Fig. 2.15b, if S¹ and S² are inverse systems with impulse response h(t) and hi(t), respectively, then the impulse response of the cascade of these systems is h(t) ∗ hi(t). But, the cascade of a system with its inverse is an identity system, whose output is the same as the input. In other words, the unit impulse response of the cascade of inverse systems is also an unit impulse δ(t). Hence,

$$h(t) * h_i(t) = \delta(t) \tag{2.36}$$

We shall give an interesting application of the commutative property. As seen from Eq. (2.36), a cascade of inverse systems is an identity system. Moreover, in a cascade of several LTIC subsystems, changing the order of the subsystems in any manner does not affect the impulse response of the cascade system. Using these facts, we observe that the two systems, shown in Fig. 2.15f, are equivalent. We can compute the response of the cascade system on the right-hand side, by computing the response of the system inside the dotted box to the input x˙(t). The impulse response of the dotted box is g(t), the integral of h(t), as given in Eq. (2.35). Hence, it follows that

$$y(t) = x(t) * h(t) = \dot{x}(t) * g(t)$$

\n(2.37)

Recall that g(t) is the unit step response of the system. Hence, an LTIC response can also be obtained as a convolution of x˙(t) (the derivative of the input) with the unit step response of the system. This result can be readily extended to higher derivatives of the input. An LTIC system response is the convolution of the nth derivative of the input with the nth integral of the impulse response.

2.4-4 A Very Special Function for LTIC Systems: The Everlasting Exponential est

There is a very special connection of LTIC systems with the everlasting exponential function est, where s is a complex variable, in general. We now show that the LTIC system’s (zero-state) response to everlasting exponential input est is also the same everlasting exponential (within a multiplicative constant). Moreover, no other function can make the same claim. Such an input for which the system response is also of the same form is called the characteristic function (also eigenfunction) of the system. Because a sinusoid is a form of exponential (s = ±jω), everlasting sinusoid is also a characteristic function of an LTIC system. Note that we are talking here of an everlasting exponential (or sinusoid), which starts at t = −∞.

If h(t) is the system’s unit impulse response, then system response y(t) to an everlasting exponential est is given by

$$y(t) = h(t) * e^{st} = \int_{-\infty}^{\infty} h(\tau) e^{s(t-\tau)} d\tau = e^{st} \int_{-\infty}^{\infty} h(\tau) e^{-s\tau} d\tau$$

The integral on the right-most side is a function of a complex variable s and a constant with respect to t. Let us denote this term by H(s), which is also complex, in general. Thus,

$$y(t) = H(s)e^{st}$$

\n^(2.38)

where

$$H(s) = \int_{-\infty}^{\infty} h(\tau) e^{-s\tau} d\tau$$

(2.39)

Equation (2.38) is valid only for the values of s for which H(s) exists, that is, if $ ^∞ −∞ ^h(τ )e−s^{τ dτ} exists (or converges). The region in the s plane for which this integral converges is called the region of convergence for H(s). Further elaboration of the region of convergence is presented in Ch. 4.

For a given s, note that H(s) is a constant. Thus, the input and the output are the same (within a multiplicative constant) for the everlasting exponential signal.

H(s), which is called the transfer function of the system, is a function of complex variable s. An alternate definition of the transfer function H(s) of an LTIC system, as seen from Eq. (2.38), is

$$H(s) = \frac{\text{output signal}}{\text{input signal}} \bigg|_{\text{input} = \text{everlasting exponential } e^{st}} \tag{2.40}$$

The transfer function is defined for, and is meaningful to, LTIC systems only. It does not exist for nonlinear or time-varying systems, in general.

We repeat again that this discussion is about the everlasting exponential, which starts at t = −∞, not the causal exponential estu(t), which starts at t = 0.

194 CHAPTER 2 TIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS

For a system specified by Eq. (2.2), the transfer function is given by

$$H(s) = \frac{P(s)}{Q(s)}\tag{2.41}$$

This follows readily by considering an everlasting input x(t) = est. According to Eq. (2.38), the output is y(t) = H(s)est. Substitution of this x(t) and y(t) in Eq. (2.2) yields

$$H(s)[Q(D)e^{st}] = P(D)e^{st}$$

Moreover,

$$D^r e^{st} = \frac{d^r e^{st}}{dt^r} = s^r e^{st}$$

Hence,

$$P(D)e^{st} = P(s)e^{st} \qquad \text{and} \qquad Q(D)e^{st} = Q(s)e^{st}$$

Consequently,

$$H(s) = \frac{P(s)}{Q(s)}$$

DR ILL 2.14 Ideal Integrator and Differentiator Transfer Functions

Show that the transfer function of an ideal integrator is H(s) = 1/s and that of an ideal differentiator is H(s) = s. Find the answer in two ways: using Eq. (2.39) and using Eq. (2.41). [Hint: Find h(t) for the ideal integrator and differentiator. You also may need to use the result in Prob. 1.4-12.]

A FUNDAMENTAL PROPERTY OF LTI SYSTEMS

We can show that Eq. (2.38) is a fundamental property of LTI systems and it follows directly as a consequence of linearity and time invariance. To show this let us assume that the response of an LTI system to an everlasting exponential est is y(s,t). If we define

$$H(s,t) = \frac{y(s,t)}{e^{st}}$$

then

$$y(s,t) = H(s,t) e^{st}$$

Because of the time-invariance property, the system response to input es(t−T) is H(s,t − T) es(t−T) , that is,

$$y(s, t - T) = H(s, t - T) e^{s(t - T)}$$

\n(2.42)

The delayed input es(t−T) represents the input est multiplied by a constant e−sT . Hence, according to the linearity property, the system response to es(t−T) must be y(s,t) e−sT . Hence,

$$y(s,t-T) = y(s,t) e^{-sT} = H(s,t) e^{s(t-T)}$$

Comparison of this result with Eq. (2.42) shows that

$$H(s,t) = H(s,t-T) \qquad \text{for all } T$$

This means H(s,t) is independent of t, and we can express H(s,t) = H(s). Hence,

$$y(s,t) = H(s) e^{st}$$

2.4-5 Total Response

Assuming distinct roots, the total response of a linear system can be expressed as the sum of its zero-input response (ZIR) and its zero-state response (ZSR):

total response =

$$\underbrace{\sum_{k=1}^{N} c_k e^{\lambda_k t}}_{\text{ZIR}} + \underbrace{x(t) * h(t)}_{\text{ZSR}}$$

For repeated roots, the zero-input component should be appropriately modified.

For the series RLC circuit in Ex. 2.4 with the input x(t) = 10e−3*^t u*(t) and the initial conditions y(0−) = 0, vC(0−) = 5, we determined the zero-input response in Ex. 2.1a [Eq. (2.9)]. We found the zero-state response in Ex. 2.9. From the results in Exs. 2.1a and 2.9, we obtain

total current =

$$\underbrace{(-5e^{-t} + 5e^{-2t})}_{\text{zero-input current}} + \underbrace{(-5e^{-t} + 20e^{-2t} - 15e^{-3t})}_{\text{zero-state current}} \qquad t \ge 0$$

(2.43)

Figure 2.16a shows the zero-input, zero-state, and total responses.

Figure 2.16 Total response and its components.

NATURAL AND FORCED RESPONSE

For the RLC circuit in Ex. 2.4, the characteristic modes were found to be e−^t and e−2*^t* . As we expected, the zero-input response is composed exclusively of characteristic modes. Note, however, that even the zero-state response [Eq. (2.43)] contains characteristic mode terms. This observation is generally true of LTIC systems. We can now lump together all the characteristic mode terms in the total response, giving us a component known as the natural response yn(t). The remainder, consisting entirely of noncharacteristic mode terms, is known as the forced response yφ(t). The total response of the RLC circuit in Ex. 2.4 can be expressed in terms of natural and forced components by regrouping the terms in Eq. (2.43) as

total current =

$$\underbrace{(-10e^{-t} + 25e^{-2t})}_{\text{natural response } y_n(t)} + \underbrace{(-15e^{-3t})}_{\text{forced response } y_\phi(t)} \qquad t \ge 0$$

(2.44)

Figure 2.16b shows the natural, forced, and total responses.

The classical solution to a differential equation includes the natural (also called the homogeneous or complementary) solution and the forced (also known as the particular) solution; traditional courses on differential equations provide simplified procedures to determine these components. Unfortunately, the classical solution lacks the engineering intuition and utility afforded by the zero-input and zero-state responses. The classical approach cannot separate the responses arising from internal conditions and external input. While the natural and forced solutions can be obtained from the zero-input and zero-state responses, the converse is not true. Further, the classical method is unable to express the system response to an input x(t) as an explicit function of x(t). In fact, the classical method is restricted to a certain class of inputs and cannot handle arbitrary inputs, as can the method to determine the zero-state response. For these (and other) reasons, we do not further detail the classical solution of differential equations.