EXAMPLE 2.5 Impulse Response via Impulse Matching

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EXAMPLE 2.5 Impulse Response via Impulse Matching

Find the impulse response h(t) for a system specified by

$$(D2 + 5D + 6)y(t) = (D + 1)x(t)$$

\n(2.14)

In this case, b⁰ = 0. Hence, h(t) consists of only the characteristic modes. The characteristic polynomial is λ² + 5λ + 6 = (λ + 2)(λ + 3). The roots are −2 and −3. Hence, the impulse

^† It might be possible for the derivatives of δ(t) to appear at the origin. However, if ^{M ≤ N}, it is impossible for h(t) to have any derivatives of δ(t). This conclusion follows from Eq. (2.11) with x(t) = δ(t) and y(t) = h(t). The coefficients of the impulse and all its derivatives must be matched on both sides of this equation. If h(t) contains δ(1) (t), the first derivative of δ(t), the left-hand side of Eq. (2.11) will contain a term δ(N+1) (t). But the highest-order derivative term on the right-hand side is δ(N) (t). Therefore, the two sides cannot match. Similar arguments can be made against the presence of the impulse’s higher-order derivatives in h(t).

response h(t) is

$$h(t) = (c_1 e^{-2t} + c_2 e^{-3t}) u(t)$$

\n(2.15)

Letting x(t) = δ(t) and y(t) = h(t) in Eq. (2.14), we obtain

$$\ddot{h}(t) + 5\dot{h}(t) + 6h(t) = \dot{\delta}(t) + \delta(t)$$

\n(2.16)

Recall that initial conditions h(0−) and h˙(0−) are both zero. But the application of an impulse at t = 0 creates new initial conditions at t = 0+. Let h(0+) = K¹ and h˙(0+) = K2. These jump discontinuities in h(t) and h˙(t) at t = 0 result in impulse terms h˙(0) = K1δ(t) and h¨(0) = K1δ(˙ t) + K2δ(t) on the left-hand side. Matching the coefficients of impulse terms on both sides of Eq. (2.16) yields

$$5K_1 + K_2 = 1$$

, $K_1 = 1$ $\implies$ $K_1 = 1, K_2 = -4$

We now use these values h(0+) = K¹ = 1 and h˙(0+) = K² = −4 in Eq. (2.15) to find c¹ and c2. Setting t = 0⁺ in Eq. (2.15), we obtain c¹ + c² = 1. Also setting t = 0⁺ in h˙(t), we obtain −2c¹ −3c¹ = −4. These two simultaneous equations yield c¹ = −1 and c² = 2. Therefore,

$$h(t) = (-e^{-2t} + 2e^{-3t})u(t)$$

Although the method used in this example is relatively simple, we can simplify it still further by using a modified version of impulse matching.

SIMPLIFIED IMPULSE MATCHING METHOD

The alternate technique we present now allows us to reduce the procedure to a simple routine to determine h(t). To avoid the needless distraction, the proof for this procedure is placed in Sec. 2.8. There, we show that for an LTIC system specified by Eq. (2.11), the unit impulse response h(t) is given by

$$h(t) = b_0 \delta(t) + [P(D)y_n(t)]u(t)$$

\n(2.17)

where yn(t) is a linear combination of the characteristic modes of the system subject to the following initial conditions:

$$y_n(0) = \dot{y}_n(0) = \ddot{y}_n(0) = \dots = y_n^{(N-2)}(0) = 0$$

and $y_n^{(N-1)}(0) = 1$ (2.18)

where y(k) ⁿ (0) is the value of the kth derivative of yn(t) at t = 0. We can express this set of conditions for various values of N (the system order) as follows:

$$N = 1 : y_n(0) = 1$$

\n

$$N = 2 : y_n(0) = 0, \dot{y}_n(0) = 1$$

\n

$$N = 3 : y_n(0) = \dot{y}_n(0) = 0, \ddot{y}_n(0) = 1$$

and so on.

As stated earlier, if the order of P(D) is less than the order of Q(D), that is, if M < N, then b⁰ = 0, and the impulse term b0δ(t) in h(t) is zero.

EXAMPLE 2.6 Impulse Response via Simplified Impulse Matching

Determine the unit impulse response h(t) for a system specified by the equation

$$(D2 + 3D + 2) y(t) = Dx(t)$$

(2.19)

This is a second-order system (N = 2) having the characteristic polynomial

$$(\lambda^2 + 3\lambda + 2) = (\lambda + 1)(\lambda + 2)$$

The characteristic roots of this system are λ = −1 and λ = −2. Therefore,

$$y_n(t) = c_1 e^{-t} + c_2 e^{-2t}$$

\n(2.20)

Differentiation of this equation yields

$$\dot{y}_n(t) = -c_1 e^{-t} - 2c_2 e^{-2t} \tag{2.21}$$

The initial conditions are [see Eq. (2.18)]

$$\dot{y}_n(0) = 1 \qquad \text{and} \qquad y_n(0) = 0$$

Setting t = 0 in Eqs. (2.20) and (2.21), and substituting the initial conditions just given, we obtain

$$0 = c_1 + c_2$$ $$1 = -c_1 - 2c_2$$

Solution of these two simultaneous equations yields

$$c_1 = 1 \qquad \text{and} \qquad c_2 = -1$$

Therefore,

$$y_n(t) = e^{-t} - e^{-2t}$$

Moreover, according to Eq. (2.19), P(D) = D so that

$$P(D)y_n(t) = Dy_n(t) = \dot{y}_n(t) = -e^{-t} + 2e^{-2t}$$

Also in this case, b⁰ = 0 [the second-order term is absent in P(D)]. Therefore,

$$h(t) = [P(D)y_n(t)]u(t) = (-e^{-t} + 2e^{-2t})u(t)$$

Comment. In the above discussion, we have assumed M ≤ N, as specified by Eq. (2.11). Section 2.8 shows that the expression for h(t) applicable to all possible values of M and N is given by

$$h(t) = P(D)[y_n(t)u(t)]$$

where yn(t) is a linear combination of the characteristic modes of the system subject to initial conditions [Eq. (2.18)]. This expression reduces to Eq. (2.17) when M ≤ N.

Determination of the impulse response h(t) using the procedures in this section is relatively simple. However, in Ch. 4 we shall discuss another, even simpler method using the Laplace transform. As the next example demonstrates, it is also possible to find h(t) using functions from MATLAB’s symbolic math toolbox.

EXAMPLE 2.7 Using MATLAB to Find the Impulse Response

Determine the impulse response h(t) for an LTIC system specified by the differential equation

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

This is a second-order system with b⁰ = 0. First we find the zero-input component for initial conditions y(0−) = 0, and y˙(0−) = 1. Since P(D) = D, the zero-input response is differentiated and the impulse response immediately follows as h(t) = 0δ(t)+ [Dyn(t)]u(t).

y_n = dsolve(‘D2y+3*Dy+2*y=0’,‘y(0)=0’,‘Dy(0)=1’,‘t’); h = diff(y_n) h = 2/exp(2*t) - 1/exp(t)

Therefore, h(t) = (2e−2*^t* −e−^t )u(t).

DR ILL 2.4 Finding the Impulse Response

Determine the unit impulse response of LTIC systems described by the following equations:

• (a) (D+2)y(t) = (3D+5)x(t)
• (b) D(D+2)y(t) = (D+4)x(t)
• (c) (D² +2D+1)y(t) = Dx(t)

ANSWERS

• (a) 3δ(t)−e−2*^t u*(t)
• (b) (2−e−2*^t* )u(t)
• (c) (1−t)e−^t u(t)

SYSTEM RESPONSE TO DELAYED IMPULSE

If h(t) is the response of an LTIC system to the input δ(t), then h(t−T) is the response of this same system to the input δ(t − T). This conclusion follows from the time-invariance property of LTIC systems. Thus, by knowing the unit impulse response h(t), we can determine the system response to a delayed impulse δ(t − T). Next, we put this result to good use in finding an LTIC system’s zero-state response.

2.4 SYSTEM RESPONSE TO EXTERNAL INPUT: THE ZERO-STATE RESPONSE

This section is devoted to the determination of the zero-state response of an LTIC system. This is the system response y(t) to an input x(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 pulse p(t) of unit height and width τ , starting at t = 0 as illustrated in Fig. 2.3a. Figure 2.3b shows an input x(t) as a sum of narrow rectangular pulses. The pulse starting at t = nτ in Fig. 2.3b has a height x(nτ ) and can be expressed as x(nτ )p(t−nτ ). Now, x(t) is the sum of all such pulses. Hence,

$$x(t) = \lim_{\Delta \tau \to 0} \sum_{\tau} x(n\Delta \tau) p(t - n\Delta \tau) = \lim_{\Delta \tau \to 0} \sum_{\tau} \left[ \frac{x(n\Delta \tau)}{\Delta \tau} \right] p(t - n\Delta \tau) \Delta \tau$$

The term [x(nτ )/τ ]p(^{t − n}τ ) represents a pulse ^p(^{t −} *ⁿ*τ ) with height x(nτ )/τ . As τ → 0, the height of this strip → ∞, but its area remains x(nτ ). Hence, this strip approaches an impulse x(nτ )δ(t −nτ ) as τ → 0 (Fig. 2.3e). Therefore,

$$x(t) = \lim_{\Delta \tau \to 0} \sum_{\tau} x(n\Delta \tau) \delta(t - n\Delta \tau) \Delta \tau$$

(2.22)

To find the response for this input x(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

$$\Rightarrow$$

output
\n

$$\delta(t) \Longrightarrow h(t)$$

\n

$$\delta(t - n\Delta\tau) \Longrightarrow h(t - n\Delta\tau)$$

\n

$$[x(n\Delta\tau)\Delta\tau]\delta(t - n\Delta\tau) \Longrightarrow [x(n\Delta\tau)\Delta\tau]h(t - n\Delta\tau)$$

\n

$$\lim_{\Delta\tau \to 0} \sum_{\tau} x(n\Delta\tau)\delta(t - n\Delta\tau) \Delta\tau \Longrightarrow \lim_{\Delta\tau \to 0} \sum_{\tau} x(n\Delta\tau)h(t - n\Delta\tau) \Delta\tau$$

\n

$$x(t) \quad [\text{see Eq. (2.22)}]$$

Figure 2.3 Finding the system response to an arbitrary input x(t).

Therefore,†

$$y(t) = \lim_{\Delta \tau \to 0} \sum_{\tau} x(n\Delta \tau)h(t - n\Delta \tau)\Delta \tau$$

=

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

(2.23)

This is the result we seek. We have obtained the system response y(t) to an arbitrary input x(t) in terms of the unit impulse response h(t). Knowing h(t), we can determine the response y(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 to δ(t −nτ ) as h(t −nτ ).