EXAMPLE 3.17 Iterative Determination of the Impulse Response

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EXAMPLE 3.17 Iterative Determination of the Impulse Response

Iteratively compute the first two values of the impulse response h[n] of a system described by the equation

y[n] −0.6y[n−1] −0.16y[n−2] = 5x[n]

To determine the unit impulse response, we let the input x[n] = δ[n] and the output y[n] = h[n] in the system’s difference equation to obtain

$$h[n] - 0.6h[n-1] - 0.16h[n-2] = 5\delta[n]$$

subject to zero initial state; that is, h[−1] = h[−2] = 0. Setting n = 0 in this equation yields

h[0] −0.6(0)−0.16(0) = 5(1) ⇒ h[0] = 5

Setting n = 1 in the same equation and using h[0] = 5, we obtain

h[1] −0.6(5)−0.16(0) = 5(0) ⇒ h[1] = 3

Continuing this way, we can determine any number of terms of h[n]. Unfortunately, such a solution does not yield a closed-form expression for h[n]. Nevertheless, determining a few values of h[n] can be useful in determining the closed-form solution, as the following development shows.

3.7-1 The Closed-Form Solution of ^h[n]

Recall that h[n] is the system response to input δ[n], which is zero for n > 0. We know that when the input is zero, only the characteristic modes can be sustained by the system. Therefore, h[n] must be made up of characteristic modes for n > 0. At n = 0, it may have some nonzero value A⁰ so that a general form of h[n] can be expressed as†

$$h[n] = A_0 \delta[n] + y_c[n]u[n]$$

\n(3.27)

where yc[n] is a linear combination of the characteristic modes. We now substitute Eq. (3.27) in Eq. (3.26) to obtain Q[E](A0δ[n] +yc[n]u[n]) = P[E]δ[n]. Because yc[n] is made up of characteristic modes, Q[E]yc[n]u[n] = 0, and we obtain A0Q[E]δ[n] = P[E]δ[n], that is,

$$A_0 (\delta[n+N] + a_1 \delta[n+N-1] + \cdots + a_N \delta[n]) = b_0 \delta[n+N] + \cdots + b_N \delta[n]$$

Setting n = 0 in this equation and using the fact that δ[m] = 0 for all m = 0, and δ[0] = 1, we obtain

$$A_0 a_N = b_N \quad \Longrightarrow \quad A_0 = \frac{b_N}{a_N} \tag{3.28}$$

Hence,‡

$$h[n] = \frac{b_N}{a_N} \delta[n] + y_c[n]u[n] \tag{3.29}$$

The N unknown coefficients in yc[n] (on the right-hand side) can be determined from a knowledge of N values of h[n]. Fortunately, it is a straightforward task to determine values of h[n] iteratively, as demonstrated in Ex. 3.17. We compute N values h[0], h[1], h[2], … , h[N −1] iteratively. Now, setting n = 0, 1, 2, … , N −1 in Eq. (3.29), we can determine the N unknowns in yc[n]. This point will become clear in the following example.

EXAMPLE 3.18 Closed-Form Determination of the Impulse Response

Determine the unit impulse response h[n] for a system in Ex. 3.17 specified by the equation

$$y[n] - 0.6y[n-1] - 0.16y[n-2] = 5x[n]$$

^† We assume that the term yc[n] consists of characteristic modes for ^{n &}gt; 0 only. To reflect this behavior, the characteristic terms should be expressed in the form γ ^{n j} u[n − 1]. But because u[n − 1] = u[n] − δ[n], cjγ ^{n j u}[^{n − 1}] = cj^{γ n j u}[n] − cjδ[n], and yc[n] can be expressed in terms of exponentials ^{γ n j} u[n] (which start at n = 0), plus an impulse at n = 0.

^‡ If aN ⁼ 0, then ^A0 cannot be determined by Eq. (3.28). In such a case, we show in Sec. 3.12 that ^h[n] is of the form A0δ[n] + A1δ[n − 1] + yc[n]u[n]. We have here N + 2 unknowns, which can be determined from N +2 values h[0],h[1],…,h[N +1] found iteratively.

This equation can be expressed in the advance form as

$$y[n+2] - 0.6y[n+1] - 0.16y[n] = 5x[n+2]$$

or in advance operator form as

$$(E^2 - 0.6E - 0.16)y[n] = 5E^2x[n]$$

The characteristic polynomial is

$$\gamma^2 - 0.6\gamma - 0.16 = (\gamma + 0.2)(\gamma - 0.8)$$

The characteristic modes are (−0.2)ⁿ and (0.8)ⁿ. Therefore,

$$y_c[n] = c_1(-0.2)^n + c_2(0.8)^n$$

Inspecting the system difference equation, we see that aN = −0.16 and bN = 0. Therefore, according to Eq. (3.29),

$$h[n] = [c_1(-0.2)^n + c_2(0.8)^n]u[n]$$

To determine c¹ and c2, we need to find two values of h[n] iteratively. From Ex. 3.17, we know that h[0] = 5 and h[1] = 3. Setting n = 0 and 1 in our expression for h[n] and using the fact that h[0] = 5 and h[1] = 3, we obtain

\begin{array}{c}\n5 = c_1 + c_2 \\ 3 = -0.2c_1 + 0.8c_2\n\end{array}\n\right\} \implies c_1 = 1\n\begin{array}{c}\nc_1 = 1 \\ c_2 = 4\n\end{array}

Therefore,

$$h[n] = [(-0.2)^n + 4(0.8)^n]u[n]$$

DR ILL 3.14 Closed-Form Determination of the Impulse Response

Find h[n], the unit impulse response of the LTID systems specified by the following equations:

• (a) y[n+1] −y[n] = x[n]
• (b) y[n] −5y[n−1] +6y[n−2] = 8x[n−1] −19x[n−2]
• (c) y[n+2] −4y[n+1] +4y[n] = 2x[n+2] −2x[n+1]
• (d) y[n] = 2x[n] −2x[n−1]

ANSWERS

• (a) h[n] = u[n−1]
• (b) ^h[n]=−^{19 6 δ}[n] + ^{3 2} (2)^{n + 5 3} (3)ⁿ u[n]
• (c) h[n] = (2+n)2nu[n]
• (d) h[n] = 2δ[n] −2δ[n−1]

EXAMPLE 3.19 Filtering Perspective of the Unit Impulse Response

Use the MATLAB filter command to solve Ex. 3.18.

There are several ways to find the impulse response using MATLAB. In this method, we first specify the unit impulse function, which will serve as our input. Vectors a and b are created to specify the system. The filter command is then used to determine the impulse response. In fact, this method can be used to determine the zero-state response for any input.

Comment. Although it is relatively simple to determine the impulse response h[n] by using the procedure in this section, in Ch. 5 we shall discuss the much simpler method of the z-transform.