This section is devoted to the determination of the zero-state response of an LTIC system. This is the system responsey(t)to an inputx(t)when the system is in the zero state, that is, when all initial conditions are zero.We shall assume that the systems discussed in this section are in the zero state unless mentioned otherwise.Under these conditions, the zero-state response will be the total response of the system.
We shall use the superposition property for finding the system response to an arbitrary input x(t). Let us define a basic pulsep(t)of unit height and widthτ, starting att=0 as illustrated in Fig. 2.3a. Figure 2.3b shows an inputx(t)as a sum of narrow rectangular pulses. The pulse starting att=nτ in Fig. 2.3b has a heightx(nτ)and can be expressed asx(nτ)p(t−nτ). Now,x(t) is the sum of all such pulses. Hence,
x(t)= lim
τ→0
"
τ
x(nτ)p(t−nτ)= lim
τ→0
"
τ
x(nτ) τ
!
p(t−nτ)τ
The term[x(nτ)/τ]p(t−nτ)represents a pulsep(t−nτ)with height x(nτ)/τ . As τ→0, the height of this strip→ ∞, but its area remainsx(nτ). Hence, this strip approaches an impulsex(nτ)δ(t−nτ)asτ→0 (Fig. 2.3e). Therefore,
x(t)= lim
τ→0
"
τ
x(nτ)δ(t−nτ)τ (2.22)
To find the response for this inputx(t), we consider the input and the corresponding output pairs, as shown in Figs. 2.3c–2.3f and also shown by directed arrow notation as follows:
input⇒output δ(t)⇒h(t) δ(t−nτ)⇒h(t−nτ)
[x(nτ)τ]δ(t−nτ)⇒ [x(nτ)τ]h(t−nτ)
τ→0lim
"
τ
x(nτ)δ(t−nτ)τ
x(t) [see Eq.(2.22)]
⇒ lim
τ→0
"
τ
x(nτ)h(t−nτ)τ
y(t)
(c)
(d)
(e)
nt t 0 y(t)
(f )
(b) t
t
t nt x(nt)
x(t)
t
0 d(t)
t
0
d(t nt)
nt t
0 t p(t) 1
(a)
t 0
[x(nt)t]d(t nt)
nt
0 h(t)
t
nt 0
h(t nt)
t
t
x(nt)h(t nt)t
nt 0 y(t)
Figure 2.3 Finding the system response to an arbitrary inputx(t).
Therefore,†
y(t)= lim
τ→0
"
τ
x(nτ)h(t−nτ)τ
=
# ∞
−∞
x(τ)h(t−τ)dτ (2.23)
This is the result we seek. We have obtained the system responsey(t)to an arbitrary inputx(t) in terms of the unit impulse responseh(t). Knowingh(t), we can determine the responsey(t)to any input.Observe once again the all-pervasive nature of the system’s characteristic modes. The system response to any input is determined by the impulse response, which, in turn, is made up of characteristic modes of the system.
It is important to keep in mind the assumptions used in deriving Eq. (2.23). We assumed a linear time-invariant (LTI) system. Linearity allowed us to use the principle of superposition, and time invariance made it possible to express the system’s response toδ(t−nτ)ash(t−nτ).
2.4-1 The Convolution Integral
The zero-state responsey(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: theconvolution integral. The convolution integral of two functionsx1(t)andx2(t)is denoted symbolically byx1(t)∗x2(t)and is defined as
x1(t)∗x2(t)≡
# ∞
−∞
x1(τ)x2(t−τ)dτ (2.24) Some important properties of the convolution integral follow.
THECOMMUTATIVEPROPERTY
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 letz=t−τ so thatτ =t−zanddτ = −dz, we obtain
x1(t)∗x2(t)= −
# −∞
∞
x2(z)x1(t−z)dz
=
# ∞
−∞
x2(z)x1(t−z)dz
=x2(t)∗x1(t) (2.25)
†In deriving this result we have assumed a time-invariant system. If the system is time-varying, then the system response to the inputδ(t−nΔτ)cannot be expressed ash(t−nΔτ)but instead has the formh(t,nΔτ). Use of this form modifies Eq. (2.23) to
y(t)=
# ∞
−∞x(τ)h(t,τ)dτ
whereh(t,τ)is the system response at instanttto a unit impulse input located atτ.
THEDISTRIBUTIVE PROPERTY According to the distributive property,
x1(t)∗ [x2(t)+x3(t)] =x1(t)∗x2(t)+x1(t)∗x3(t) (2.26)
THEASSOCIATIVEPROPERTY According to the associative property,
x1(t)∗ [x2(t)∗x3(t)] = [x1(t)∗x2(t)] ∗x3(t) (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.
THESHIFT PROPERTY If
x1(t)∗x2(t)=c(t) then
x1(t)∗x2(t−T)=x1(t−T)∗x2(t)=c(t−T) More generally, we see that
x1(t−T1)∗x2(t−T2)=c(t−T1−T2) (2.28) Proof. We are given
x1(t)∗x2(t)=
# ∞
−∞
x1(τ)x2(t−τ)dτ=c(t) Therefore,
x1(t)∗x2(t−T)=
# ∞
−∞
x1(τ)x2(t−T−τ)dτ
=c(t−T)
The equally simple proof of Eq. (2.28) follows a similar approach.
CONVOLUTION WITH AN IMPULSE
Convolution of a functionx(t)with a unit impulse results in the functionx(t)itself. By definition of convolution,
x(t)∗δ(t)=
# ∞
−∞
x(τ)δ(t−τ)dτ
Becauseδ(t−τ)is an impulse located atτ=t, according to the sampling property of the impulse [Eq. (1.11)], the integral here is just the value ofx(τ)atτ=t, that is,x(t). Therefore,
x(t)∗δ(t)=x(t) Actually this result was derived earlier [Eq. (2.22)].
THEWIDTH PROPERTY
If the durations (widths) ofx1(t)andx2(t)are finite, given byT1 andT2, respectively, then the duration (width) ofx1(t)∗x2(t)isT1+T2(Fig. 2.4). The proof of this property follows readily from the graphical considerations discussed later in Sec. 2.4-2.
T1
*
x1(t) x2(t)
t t t
x1(t) * x2(t)
T2 T1 T2
Figure 2.4 Width property of convolution.
ZERO-STATERESPONSE ANDCAUSALITY The (zero-state) responsey(t)of an LTIC system is
y(t)=x(t)∗h(t)=
# ∞
−∞
x(τ)h(t−τ)dτ (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 signalx(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 att=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 impulseδ(t)(which is located att=0) cannot begin beforet=0. Therefore, acausal 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 toτ (nott). If the inputx(t)is causal,x(τ)=0 forτ <0. Therefore,x(τ)=0 forτ <0, as illustrated in Fig. 2.5a. Similarly, ifh(t)is causal,h(t−τ)=0 fort−τ <0; that is, forτ >t, as depicted in Fig. 2.5a. Therefore, the productx(τ)h(t−τ)=0 everywhere except over the nonshaded interval 0≤τ ≤tshown in Fig. 2.5a (assumingt≥0). Observe that iftis negative,x(τ)h(t−τ)=0 for allτ, as shown in Fig. 2.5b. Therefore, Eq. (2.29) reduces to
y(t)=x(t)∗h(t)=$t
0−x(τ)h(t−τ)dτ t≥0
0 t<0 (2.30)
The lower limit of integration in Eq. (2.30) is taken as 0−to avoid the difficulty in integration that can arise ifx(t)contains an impulse at the origin. This result shows that ifx(t)andh(t)are both causal, the responsey(t)is also causal.
h(t t) 0
0 t t 0
(a) x(t) 0
t
t 0
t 0
(b) t
h(t t) 0 x(t)
Figure 2.5 Limits of the convolution integral.
Because of the convolution’s commutative property [Eq. (2.25)], we can also express Eq. (2.30) as [assuming causalx(t)andh(t)]
y(t)=
⎧⎨
⎩
# t
0−
h(τ)x(t−τ)dτ t≥0
0 t<0
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.
E X A M P L E 2.8 Computing the Zero-State Response
For an LTIC system with the unit impulse responseh(t)=e−2tu(t), determine the responsey(t) for the input
x(t)=e−tu(t)
Here bothx(t)andh(t)are causal (Fig. 2.6). Hence, from Eq. (2.30), we obtain y(t)=
# t
0
x(τ)h(t−τ)dτ t≥0 Becausex(t)=e−tu(t)andh(t)=e−2tu(t),
x(τ)=e−τu(τ) and h(t−τ)=e−2(t−τ)u(t−τ)
Remember that the integration is performed with respect to τ (not t), and the region of integration is 0≤τ ≤t. Hence,τ ≥0 andt−τ ≥0. Therefore,u(τ)=1 andu(t−τ)=1;
consequently,
y(t)=
# t
0
e−τe−2(t−τ)dτ t≥0
e2t 1
x(t)
et
0 t
(a)
1 h(t)
e2t
0 t
(b)
(c) et e2t y(t) 1
1
t 0
et
Figure 2.6 Convolution ofx(t)andh(t).
Because this integration is with respect toτ, we can pulle−2toutside the integral, giving us
y(t)=e−2t
# t
0
eτdτ=e−2t(et−1)=e−t−e−2t t≥0 Moreover,y(t)=0 whent<0 [see Eq. (2.30)]. Therefore,
y(t)=(e−t−e−2t)u(t) The response is depicted in Fig. 2.6c.
D R I L L 2.5 Computing the Zero-State Response
For an LTIC system with the impulse responseh(t)=6e−tu(t), determine the system response to the input:(a)2u(t)and(b)3e−3tu(t).
ANSWERS
(a) 12(1−e−t)u(t) (b) 9(e−t−e−3t)u(t)
D R I L L 2.6 Zero-State Response with Resonance
Repeat Drill 2.5 for the inputx(t)=e−tu(t).
ANSWER 6te−tu(t)
THECONVOLUTIONTABLE
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 inputx(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 (withλ1= −1 andλ2= −2) to be(e−t−e−2t)u(t). The following example demonstrates the utility of this table.
E X A M P L E 2.9 Convolution by Tables
Use Table 2.1 to compute the loop current y(t) of theRLC circuit in Ex. 2.4 for the input x(t)=10e−3tu(t)when all the initial conditions are zero.
The loop equation for this circuit [see Ex. 1.16 or Eq. (1.29)] is (D2+3D+2)y(t)=Dx(t) The impulse responseh(t)for this system, as obtained in Ex. 2.6, is
h(t)=(2e−2t−e−t)u(t) The input isx(t)=10e−3tu(t), and the responsey(t)is
y(t)=x(t)∗h(t)=10e−3tu(t)∗ [2e−2t−e−t]u(t) Using the distributive property of the convolution [Eq. (2.26)], we obtain
y(t)=10e−3tu(t)∗2e−2tu(t)−10e−3tu(t)∗e−tu(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)= 20
−3−(−2)[e−3t−e−2t]u(t)− 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)
TABLE 2.1 Select Convolution Integrals
No. x1(t) x2(t) x1(t)∗x2(t)=x2(t)∗x1(t)
1 x(t) δ(t−T) x(t−T)
2 eλtu(t) u(t) 1−eλt
−λ u(t)
3 u(t) u(t) tu(t)
4 eλ1tu(t) eλ2tu(t) eλ1t−eλ2t
λ1−λ2
u(t) λ1=λ2
5 eλtu(t) eλtu(t) teλtu(t)
6 teλtu(t) eλtu(t) 1
2t2eλtu(t)
7 tNu(t) eλtu(t) N!eλt
λN+1u(t)−
"N k=0
N!tN−k λk+1(N−k)!u(t)
8 tMu(t) tNu(t) M!N!
(M+N+1)!tM+N+1u(t) 9 teλ1tu(t) eλ2tu(t) eλ2t−eλ1t+(λ1−λ2)teλ1t
(λ1−λ2)2 u(t)
10 tMeλtu(t) tNeλtu(t) M!N!
(N+M+1)!tM+N+1eλtu(t)
11 tMeλ1tu(t) tNeλ2tu(t)
"M k=0
(−1)kM!(N+k)!tM−keλ1t k!(M−k)!(λ1−λ2)N+k+1u(t)
λ1=λ2 +
"N k=0
(−1)kN!(M+k)!tN−keλ2t k!(N−k)!(λ2−λ1)M+k+1u(t)
12 e−αtcos(βt+θ)u(t) eλtu(t) cos(θ−φ)eλt−e−αtcos(βt+θ−φ) (α+λ)2+β2 u(t) φ=tan−1[−β/(α+λ)]
13 eλ1tu(t) eλ2tu(−t) eλ1tu(t)+eλ2tu(−t) λ2−λ1
Reλ2>Reλ1
14 eλ1tu(−t) eλ2tu(−t) eλ1t−eλ2t λ2−λ1
u(−t)
D R I L L 2.7 Convolution by Tables
Use Table 2.1 to showe−2tu(t)∗(1−e−t)u(t)=1
2−e−t+12e−2t u(t).
D R I L L 2.8 Zero-State Response by Convolution Table
Rework Drills 2.5 and 2.6 using Table 2.1.
D R I L L 2.9 Another Zero-State Response by Convolution Table
For an LTIC system with the unit impulse responseh(t)=e−2tu(t), determine the zero-state responsey(t)if the inputx(t)=sin 3t u(t). [Hint:Use pair 12 from Table 2.1.]
ANSWER
1
13[3e−2t+√
13 cos(3t−146.32◦)]u(t)or 131[3e−2t−√
13 cos(3t+33.68◦)]u(t)
RESPONSE TOCOMPLEXINPUTS
The LTIC system response discussed so far applies to general input signals, real or complex.
However, if the system is real, that is, ifh(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), wherexr(t)andxi(t)are the real and imaginary parts of x(t), then for realh(t)
y(t)=h(t)∗ [xr(t)+jxi(t)] =h(t)∗xr(t)+jh(t)∗xi(t)=yr(t)+jyi(t)
where yr(t)andyi(t)are the real and the imaginary parts ofy(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)=xr(t)+jxi(t) ⇒ y(t)=yr(t)+jyi(t) then
xr(t) ⇒ yr(t) and xi(t) ⇒ yi(t) (2.31)
MULTIPLEINPUTS
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. Ifc(t)is the convolution ofx(t)withg(t), then
c(t)=
# ∞
−∞
x(τ)g(t−τ)dτ
One of the crucial points to remember here is that this integration is performed with respect toτ so thattis just a parameter (like a constant). This consideration is especially important when we sketch the graphical representations of the functionsx(τ)andg(t−τ). Both these functions should be sketched as functions ofτ, not oft.
The functionx(τ)is identical tox(t), withτ replacingt(Fig. 2.7c). Therefore,x(t)andx(τ) will have the same graphical representations. Similar remarks apply tog(t)andg(τ)(Fig. 2.7d).
To appreciate whatg(t−τ)looks like, let us start with the functiong(τ)(Fig. 2.7d). Time reversal of this function (reflection about the vertical axisτ =0) yieldsg(−τ)(Fig. 2.7e). Let us denote this function byφ(τ):
φ(τ)=g(−τ) Nowφ(τ)shifted bytseconds isφ(τ−t), given by
φ(τ−t)=g[−(τ−t)] =g(t−τ)
Therefore, we first time-reverseg(τ) to obtaing(−τ)and then time-shiftg(−τ)bytto obtain g(t−τ). For positivet, 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 functionsx(τ)andg(t−τ).
The convolutionc(t)is the area under the product of these two functions. Thus, to computec(t) at some positive instantt=t1, we first obtaing(−τ)by inverting g(τ) about the vertical axis.
Next, we right-shift or delayg(−τ)byt1to obtaing(t1−τ)(Fig. 2.7f), and then we multiply this function byx(τ), giving us the productx(τ)g(t1−τ)(shaded portion in Fig. 2.7f). The areaA1 under this product isc(t1), the value ofc(t)att=t1. We can therefore plotc(t1)=A1on a curve describingc(t), as shown in Fig. 2.7i. The area under the productx(τ)g(−τ)in Fig. 2.7e isc(0), the value of the convolution fort=0 (at the origin).
x(t)
(a)
g(t)
t t
0
(b) 0
(c) (d)
(e)
t
t t
x(t) g(t)
0 0
0
g(t) x(t)
(f )
(g)
(i)
0 c(t)
t1 t2
A2 A1
1 2
2 1
1
2
2 1
1 1
2
0 g(t t)
t t1 0
g(t t) t t2 0
g(t t) t t3 3
2 t1
x(t)
x(t)
x(t) 1
0 2
(h)
1 0
t1
2 t2 2 t2
t3
t3
2 t3
3
A1
A2
t
t
t
t Figure 2.7 Graphical explanation of the convolution operation.