3.8 SYSTEM [RESPONSE TO](#page-9-0) EXTERNAL INPUT: THE ZERO-STATE RESPONSE

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

The zero-state response y[n] is the system response to an input x[n] when the system is in the zero state. In this section we shall assume that systems are in the zero state unless mentioned otherwise, so that the zero-state response will be the total response of the system. Here we follow the procedure parallel to that used in the continuous-time case by expressing an arbitrary input x[n] as a sum of impulse components. A signal x[n] in Fig. 3.20a can be expressed as a sum of impulse components, such as those depicted in Figs. 3.20b–3.20f. The component of x[n] at n = m is x[m]δ[n−m], and x[n] is the sum of all these components summed from m = −∞ to ∞.

Therefore,

$$x[n] = x[0]\delta[n] + x[1]\delta[n-1] + x[2]\delta[n-2] + \cdots$$

• x[-1]\delta[n+1] + x[-2]\delta[n+2] + \cdots
  =

$$\sum_{m=-\infty}^{\infty} x[m]\delta[n-m]$$

(3.30)

282 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS

For a linear system, if we know the system response to impulse δ[n], we can obtain the system response to any arbitrary input by summing the system response to various impulse components. Let h[n] be the system response to impulse input δ[n]. We shall use the notation

$$x[n] \Longrightarrow y[n]$$

to indicate the input and the corresponding response of the system. Thus, if

$$\delta[n] \Longrightarrow h[n]$$

then because of time invariance

$$\delta[n-m] \Longrightarrow h[n-m]$$

and because of linearity

$$x[m]\delta[n-m] \Longrightarrow x[m]h[n-m]$$

and again because of linearity

$$\underbrace{\sum_{m=-\infty}^{\infty} x[m]\delta[n-m]}_{x[n]} \quad \Longrightarrow \quad \underbrace{\sum_{m=-\infty}^{\infty} x[m]h[n-m]}_{y[n]}$$

The left-hand side is x[n] [see Eq. (3.30)], and the right-hand side is the system response y[n] to input x[n]. Therefore,†

$$y[n] = \sum_{m = -\infty}^{\infty} x[m]h[n-m]$$

\n(3.31)

The summation on the right-hand side is known as the convolution sum of x[n] and h[n], and is represented symbolically by x[n] ∗ h[n]

$$x[n] * h[n] = \sum_{m=-\infty}^{\infty} x[m]h[n-m]$$

PROPERTIES OF THE CONVOLUTION SUM

The structure of the convolution sum is similar to that of the convolution integral. Moreover, the properties of the convolution sum are similar to those of the convolution integral. We shall enumerate these properties here without proof. The proofs are similar to those for the convolution integral and may be derived by the reader.

$$y[n] = \sum_{m=-\infty}^{\infty} x[m]h[n,m]$$

^† In deriving this result, we have assumed a time-invariant system. The system response to input ^δ[^{n − m}] for a time-varying system cannot be expressed as h[n − m]; instead, it has the form h[n, m]. Using this form, Eq. (3.31) is modified as follows:

The Commutative Property.

$$x_1[n] * x_2[n] = x_2[n] * x_1[n]$$

The Distributive Property.

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

The Associative Property.

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

The Shifting Property. If

$$x_1[n] * x_2[n] = c[n]$$

then

$$x_1[n-m]*x_2[n-p] = c[n-m-p]$$

\n(3.32)

The Convolution with an Impulse.

$$x[n] * \delta[n] = x[n]$$

The Width Property. If x1[n] and x2[n] have finite widths of W¹ and W2, respectively, then the width of x1[n] ∗ x2[n] is W¹ + W2. The width of a signal is 1 less than the number of its elements (length). Thus the signal in Fig. 3.22h has six elements (length of 6) but a width of only 5. Alternately, the property may be stated in terms of lengths as follows: if x1[n] and x2[n] have finite lengths of L¹ and L² elements, respectively, then the length of x1[n] ∗ x2[n] is L¹ + L² − 1 elements.

CAUSALITY AND ZERO-STATE RESPONSE

In deriving Eq. (3.31), we assumed the system to be linear and time-invariant. There were no other restrictions on either the input signal or the system. In our applications, almost all the input signals are causal, and a majority of the systems are also causal. These restrictions further simplify the limits of the sum in Eq. (3.31). If the input x[n] is causal, x[m] = 0 for m < 0. Similarly, if the system is causal (i.e., if h[n] is causal), then h[x] = 0 for negative x so that h[n − m] = 0 when m > n. Therefore, if x[n] and h[n] are both causal, the product x[m]h[n−m] = 0 for m < 0 and for m > n, and it is nonzero only for the range 0 ≤ m ≤ n. Therefore, Eq. (3.31) in this case reduces to

$$y[n] = \sum_{m=0}^{n} x[m]h[n-m]$$

\n(3.33)

We shall evaluate the convolution sum first by an analytical method and later with graphical aid.

EXAMPLE 3.20 Convolution of Causal Signals

Determine c[n] = x[n] ∗ g[n] for

$$x[n] = (0.8)^n u[n]$$

and $g[n] = (0.3)^n u[n]$

We have

$$c[n] = \sum_{m=-\infty}^{\infty} x[m]g[n-m]$$

Note that

$$x[m] = (0.8)^m u[m]$$

and $g[n-m] = (0.3)^{n-m} u[n-m]$

Both x[n] and g[n] are causal. Therefore [see Eq. (3.33)],

$$c[n] = \sum_{m=0}^{n} x[m]g[n-m] = \sum_{m=0}^{n} (0.8)^m u[m] (0.3)^{n-m} u[n-m]$$

In this summation, m lies between 0 and n (0 ≤ m ≤ n). Therefore, if n ≥ 0, then both m and n−m ≥ 0 so that u[m] = u[n−m] = 1. If n < 0, m is negative because m lies between 0 and n, and u[m] = 0. Therefore,

$$c[n] = \begin{cases} \sum_{m=0}^{n} (0.8)^m (0.3)^{n-m} & n \ge 0\\ 0 & n < 0 \end{cases}$$

or

$$c[n] = (0.3)^n \sum_{m=0}^n \left(\frac{0.8}{0.3}\right)^m u[n]$$

This is a geometric progression with common ratio (0.8/0.3). From Sec. B.8-3 we have

$$c[n] = (0.3)^n \frac{(0.8)^{n+1} - (0.3)^{n+1}}{(0.3)^n (0.8 - 0.3)} u[n]$$

= 2[(0.8)^{n+1} - (0.3)^{n+1}]u[n]

DR ILL 3.15 Convolution of Causal Signals

Show that (0.8)nu[n] ∗ u[n] = 5[1−(0.8)ⁿ+¹]u[n].

CONVOLUTION SUM FROM A TABLE

Just as in the continuous-time case, we have prepared a table (Table 3.1) from which convolution sums may be determined directly for a variety of signal pairs. For example, the convolution in Ex. 3.20 can be read directly from this table (pair 4) as

$$(0.8)^n u[n] * (0.3)^n u[n] = \frac{(0.8)^{n+1} - (0.3)^{n+1}}{0.8 - 0.3} u[n] = 2[(0.8)^{n+1} - (0.3)^{n+1}]u[n]$$

We shall demonstrate the use of the convolution table in the following example.

No.	x1[n]	x2[n]	x1[n] ∗ x2[n] = x2[n] ∗ x1[n]
1	δ[n−k]	x[n]	x[n−k]
2	γ nu[n]	u[n]	1−γ n+1 ! u[n] 1−γ
3	u[n]	u[n]	(n+1)u[n]
4	γ n 1 u[n]	γ n 2 u[n]	γ n+1 −γ n+1 1 2 u[n] γ1 = γ2 γ1 −γ2
5	u[n]	nu[n]	n(n+1) u[n] 2
6	γ nu[n]	nu[n]	γ (γ n −1) ! +n(1−γ ) u[n] (1−γ )2
7	nu[n]	nu[n]	1 6 n(n−1)(n+1)u[n]
8	γ nu[n]	γ nu[n]	(n+1)γ nu[n]
9	nγ n 1 u[n]	γ n 2 u[n]	! γ1γ2 γ1 −γ2 γ n 2 −γ n nγ n u[n] γ1 = γ2 1 + 1 −γ2)2 (γ1 γ2
10	n cos(βn+θ γ1 )u[n]	nu[n] γ2	1 n+1 cos[β(n+1)+θ−φ]− γ2 n+1 cos(θ−φ)]u[n] R[ γ1
			1/2 R = 2 + 2 −2 γ1 γ2 γ1 γ2 cosβ
			! ( γ1 sinβ) φ = tan−1 ( γ1 cosβ − γ2 )
11	γ n 1 u[−(n+1)]	γ n 2 u[n]	γ2 γ1 γ n γ n 2 u[n] + 1 u[−(n+1)] γ1 > γ2 γ1 −γ2 γ1 −γ2

TABLE 3.1 Select Convolution Sums

EXAMPLE 3.21 Convolution by Tables

Using Table 3.1, find the (zero-state) response y[n] of an LTID system described by the equation

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

if the input x[n] = 4−nu[n].

The input can be expressed as x[n] = 4−nu[n] = (1/4)nu[n] = (0.25)nu[n]. The unit impulse response of this system, obtained in Ex. 3.18, is

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

Therefore,

$$y[n] = x[n] * h[n]$$

= (0.25)ⁿu[n] * [(-0.2)ⁿu[n] + 4(0.8)ⁿu[n]
= (0.25)ⁿu[n] * (-0.2)ⁿu[n] + (0.25)ⁿu[n] * 4(0.8)ⁿu[n]

We use pair 4 (Table 3.1) to find the foregoing convolution sums.

$$y[n] = \left[\frac{(0.25)^{n+1} - (-0.2)^{n+1}}{0.25 - (-0.2)} + 4 \frac{(0.25)^{n+1} - (0.8)^{n+1}}{0.25 - 0.8}\right] u[n]$$

= $(2.22[(0.25)^{n+1} - (-0.2)^{n+1}] - 7.27[(0.25)^{n+1} - (0.8)^{n+1}])u[n]$
= $[-5.05(0.25)^{n+1} - 2.22(-0.2)^{n+1} + 7.27(0.8)^{n+1}]u[n]$

Recognizing that

$$\gamma^{n+1} = \gamma(\gamma)^n$$

we can express y[n] as

$$y[n] = [-1.26(0.25)^{n} + 0.444(-0.2)^{n} + 5.81(0.8)^{n}]u[n]$$

= [-1.26(4)^-n + 0.444(-0.2)ⁿ + 5.81(0.8)ⁿ]u[n]

DR ILL 3.16 Convolution by Tables

Use Table 3.1 to show that

(a)

$$(0.8)^{n+1}u[n] * u[n] = 4[1 - 0.8(0.8)^n]u[n]$$

(b) ⁿ3−nu[n] ∗ (0.2)nu[n] = ¹⁵ 4 (0.2)^{n − 1− 2} 3 n 3−ⁿ u[n]

(c)

$$e^{-n}u[n] * 2^{-n}u[n] = \frac{2}{2-e} \left[ e^{-n} - \frac{e}{2}2^{-n} \right] u[n]$$

EXAMPLE 3.22 Filtering Perspective of the Zero-State Response

Use the MATLAB filter command to compute and sketch the zero-state response for the system described by (E² +0.5E −1)y[n] = (2E² +6E)x[n] and the input x[n] = 4−nu[n].

We solve this problem using the same approach as Ex. 3.19. Although the input is bounded and quickly decays to zero, the system itself is unstable and an unbounded output results.

n = (0:11); x = @(n) 4.^(-n).*(n>=0); >> a = [1 0.5 -1]; b = [2 6 0]; y = filter(b,a,x(n)); >> clf; stem(n,y,‘k’); xlabel(‘n’); ylabel(‘y[n]’); axis([-0.5 11.5 -20 25]);

RESPONSE TO COMPLEX INPUTS

As in the case of real continuous-time systems, we can show that for an LTID system with real h[n], if the input and the output are expressed in terms of their real and imaginary parts, then the real part of the input generates the real part of the response and the imaginary part of the input generates the imaginary part. Thus, if

$$x[n] = x_r[n] + jx_i[n]$$

and $y[n] = y_r[n] + jy_i[n]$

using the right-directed arrow to indicate the input–output pair, we can show that

$$x_r[n] \Longrightarrow y_r[n]$$

and $x_i[n] \Longrightarrow y_i[n]$ (3.34)

The proof is similar to that used to derive Eq. (2.31) for LTIC systems.

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.

DR ILL 3.17 Response to Multiple Inputs

Show that the system described by y[n] − 0.6y[n − 1] − 0.16y[n − 2] = 5x[n] responds to input x[n] = δ[n] +4−nu[n] with output y[n] = [−1.26(4)−ⁿ +1.444(−0.2)ⁿ +9.81(0.8)ⁿ]u[n]. [Hint: Use the results of Exs. 3.18 and 3.21.]

3.8-1 Graphical Procedure for the Convolution Sum

The steps in evaluating the convolution sum are parallel to those followed in evaluating the convolution integral. The convolution sum of causal signals x[n] and g[n] is given by

$$c[n] = \sum_{m=0}^{n} x[m]g[n-m]$$

We first plot x[m] and g[n − m] as functions of m (not n), because the summation is over m. Functions x[m] and g[m] are the same as x[n] and g[n], plotted, respectively, as functions of m (see Fig. 3.22). The convolution operation can be performed as follows:

• 1. Invert g[m] about the vertical axis (m = 0) to obtain g[−m] (Fig. 3.22d). Figure 3.22e shows both x[m] and g[−m].
• 2. Shift g[−m] by n units to obtain g[n − m]. For n > 0, the shift is to the right (delay); for n < 0, the shift is to the left (advance). Figure 3.22f shows g[n − m] for n > 0; for n < 0, see Fig. 3.22g.
• 3. Next we multiply x[m] and g[n−m] and add all the products to obtain c[n]. The procedure is repeated for each value of n over the range −∞ to ∞.

We shall demonstrate by an example the graphical procedure for finding the convolution sum. Although both the functions in this example are causal, this procedure is applicable to the general case.

EXAMPLE 3.23 Graphical Procedure for the Convolution Sum

Find c[n] = x[n] ∗g[n], where x[n] and g[n] are depicted in Figs. 3.22a and 3.22b, respectively.

We are given

x[n] = (0.8) ⁿ and ^g[n] = (0.3) n

Therefore,

x[m] = (0.8) ^m and ^g[n−m] = (0.3) n−m

Figure 3.22 Graphical procedure to convolve x[n] and g[n].

290 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS

Figure 3.22f shows the general situation for n ≥ 0. The two functions x[m] and g[n−m] overlap over the interval 0 ≤ m ≤ n. Therefore,

$$c[n] = \sum_{m=0}^{n} x[m]g[n-m]$$

= $\sum_{m=0}^{n} (0.8)^m (0.3)^{n-m}$
= $(0.3)^n \sum_{m=0}^{n} \left(\frac{0.8}{0.3}\right)^m$
= $2[(0.8)^{n+1} - (0.3)^{n+1}]$ $n \ge 0$ (see Sec. B.8-3)

For n < 0, there is no overlap between x[m] and g[n−m], as shown in Fig. 3.22g, so that

c[n] = 0 n < 0

Combining pieces, we see that

$$c[n] = 2[(0.8)^{n+1} - (0.3)^{n+1}]u[n]$$

which agrees with the result found earlier in Ex. 3.20.

DR ILL 3.18 Graphical Procedure for the Convolution Sum

Find (0.8)nu[n] ∗ u[n] graphically and sketch the result.

ANSWER

5(1−(0.8)ⁿ+1)u[n]

AN ALTERNATIVE FORM OF GRAPHICAL PROCEDURE: THE SLIDING-TAPE METHOD

This algorithm is convenient when the sequences x[n] and g[n] are short or when they are available only in graphical form. The algorithm is basically the same as the graphical procedure in Fig. 3.22. The only difference is that instead of presenting the data as graphical plots, we display it as a sequence of numbers on tapes. Otherwise the procedure is the same, as will become clear in the following example.

EXAMPLE 3.24 Sliding-Tape Method for the Convolution Sum

Use the sliding-tape method to convolve the two sequences x[n] and g[n] depicted in Figs. 3.23a and 3.23b, respectively.

In this procedure we write the sequences x[n] andg[n] in the slots of two tapes: x tape and g tape (Fig. 3.23c). Now leave the x tape stationary (to correspond to x[m]). The g[−m] tape is obtained by inverting the g[m] tape about the origin (m = 0) so that the slots corresponding to x[0] and g[0] remain aligned (Fig. 3.23d). We now shift the inverted tape by n slots, multiply values on two tapes in adjacent slots, and add all the products to find c[n]. Figures 3.23d–3.23i show the cases for n = 0–5. Figures 3.23j, 3.23k, and 3.23l show the cases for n = −1,−2, and −3, respectively.

For the case of n = 0, for example (Fig. 3.23d),

$$c[0] = (-2 \times 1) + (-1 \times 1) + (0 \times 1) = -3$$

For n = 1 (Fig. 3.23e),

$$c[1] = (-2 \times 1) + (-1 \times 1) + (0 \times 1) + (1 \times 1) = -2$$

Similarly,

$$c[2] = (-2 \times 1) + (-1 \times 1) + (0 \times 1) + (1 \times 1) + (2 \times 1) = 0$$

\n

$$c[3] = (-2 \times 1) + (-1 \times 1) + (0 \times 1) + (1 \times 1) + (2 \times 1) + (3 \times 1) = 3$$

\n

$$c[4] = (-2 \times 1) + (-1 \times 1) + (0 \times 1) + (1 \times 1) + (2 \times 1) + (3 \times 1) + (4 \times 1) = 7$$

\n

$$c[5] = (-2 \times 1) + (-1 \times 1) + (0 \times 1) + (1 \times 1) + (2 \times 1) + (3 \times 1) + (4 \times 1) = 7$$

Figure 3.23i shows that c[n] = 7 for n ≥ 4.

Similarly, we compute c[n] for negative n by sliding the tape backward, one slot at a time, as shown in the plots corresponding to n = −1, −2, and −3, respectively (Figs. 3.23j, 3.23k, and 3.23l).

$$c[-1] = (-2 \times 1) + (-1 \times 1) = -3$$

\n

$$c[-2] = (-2 \times 1) = -2$$

\n

$$c[-3] = 0$$

Figure 3.23l shows that c[n] = 0 for n ≤ 3. Figure 3.23m shows the plot of c[n].

Rotate the g tape about the vertical axis as shown in (d)

DR ILL 3.19 Sliding-Tape Method for the Convolution Sum

Use the graphical procedure of Ex. 3.24 (sliding-tape technique) to show that x[n] ∗ g[n] = c[n] in Fig. 3.24. Verify the width property of convolution.

EXAMPLE 3.25 Convolution of Two Finite-Duration Signals Using MATLAB

For the signals x[n] and g[n] depicted in Fig. 3.24, use MATLAB to compute and plot c[n] = x[n] ∗ g[n].

3.8-2 Interconnected Systems

As with continuous-time case, we can determine the impulse response of systems connected in parallel (Fig. 3.26a) and cascade (Figs. 3.26b, 3.26c). We can use arguments identical to those used for the continuous-time systems in Sec. 2.4-3 to show that if two LTID systems S¹ and S² with impulse responses h1[n] and h2[n], respectively, are connected in parallel, the composite parallel system impulse response is h1[n] + h2[n]. Similarly, if these systems are connected in cascade, the impulse response of the composite system is h1[n] ∗ h2[n]. Moreover, because h1[n] ∗ h2[n] = h2[n] ∗ h1[n], linear systems commute. Their orders can be interchanged without affecting the composite system behavior.

Figure 3.26 Interconnected systems.

INVERSE SYSTEMS

If the two systems in cascade are the inverse of each other, with impulse responses h[n] and hi[n], respectively, then the impulse response of the cascade of these systems is h[n] ∗ hi[n]. But, the cascade of a system with its inverse is an identity system, whose output is the same as the input. Hence, the unit impulse response of an identity system is δ[n]. Consequently,

$$h[n] * h_i[n] = \delta[n]$$

As an example, we show that an accumulator system and a backward difference system are the inverse of each other. An accumulator system is specified by†

$$y[n] = \sum_{k=-\infty}^{n} x[k] \tag{3.35}$$

The backward difference system is specified by

$$y[n] = x[n] - x[n-1] \tag{3.36}$$

From Eq. (3.35), we find hacc[n], the impulse response of the accumulator, as

$$h_{\text{acc}}[n] = \sum_{k=-\infty}^{n} \delta[k] = u[n]$$

Similarly, from Eq. (3.36), hbdf[n], the impulse response of the backward difference system is given by

$$h_{\text{bdf}}[n] = \delta[n] - \delta[n-1]$$

We can verify that

$$h_{\text{acc}} * h_{\text{bdf}} = u[n] * {\delta[n] - \delta[n-1]} = u[n] - u[n-1] = \delta[n]$$

Roughly speaking, a discrete-time accumulator is analogous to a continuous-time integrator, and a backward difference system is analogous to a differentiator. We have already encountered examples of these systems in Exs. 3.8 and 3.9 (digital differentiator and integrator).

SYSTEM RESPONSE TO %^{n k}=−∞ ^x[k]

Figure 3.26d shows a cascade of two LTID systems: a system S with impulse response h[n], followed by an accumulator. Figure 3.26e shows a cascade of the same two systems in reverse order: an accumulator followed by S. In Fig. 3.26d, if the input x[n] to S results in the output ^y[n], then the output of the system in Fig. 3.26d is the %y[k]. In Fig. 3.26e, the output of the accumulator is the sum %x[k]. Because the output of the system in Fig. 3.26e is identical to that of system Fig. 3.26d, it follows that

if

$$x[n] \Longrightarrow y[n]$$

, then $\sum_{k=-\infty}^{n} x[k] \Longrightarrow \sum_{k=-\infty}^{n} y[k]$

^† Equations (3.35) and (3.36) are identical to Eqs. (3.10) and (3.8), respectively, with ^{T =} 1.

296 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS

If we let x[n] = δ[n] and y[n] = h[n], we find that g[n], the unit step response of an LTID system with impulse response h[n], is given by

$$g[n] = \sum_{k=-\infty}^{n} h[k] \tag{3.37}$$

The reader can readily prove the inverse relationship

$$h[n] = g[n] - g[n-1]$$

A VERY SPECIAL FUNCTION FOR LTID SYSTEMS: THE EVERLASTING EXPONENTIAL zⁿ

In Sec. 2.4-4, we showed that there exists one signal for which the response of an LTIC system is the same as the input within a multiplicative constant. The response of an LTIC system to an everlasting exponential input est is H(s)est, where H(s) is the system transfer function. We now show that for an LTID system, the same role is played by an everlasting exponential zⁿ. The system response y[n] in this case is given by

$$y[n] = h[n] * zn$$

=

$$\sum_{m=-\infty}^{\infty} h[m] z^{n-m}$$

=

$$zn \sum_{m=-\infty}^{\infty} h[m] z^{-m}$$

For causal h[n], the limits on the sum on the right-hand side would range from 0 to ∞. In any case, this sum is a function of z. Assuming that this sum converges, let us denote it by H[z]. Thus,

$$y[n] = H[z]z^n \tag{3.38}$$

where

$$H[z] = \sum_{m=-\infty}^{\infty} h[m]z^{-m}$$

\n(3.39)

Equation (3.38) is valid only for values of z for which the sum on the right-hand side of Eq. (3.39) exists (converges). Note that H[z] is a constant for a given z. Thus, the input and the output are the same (within a multiplicative constant) for the everlasting exponential input zn.

H[z], which is called the transfer function of the system, is a function of the complex variable z. An alternate definition of the transfer function H[z] of an LTID system from Eq. (3.38) is

$$H[z] = \frac{\text{output signal}}{\text{input signal}} \bigg|_{\text{input} = \text{everlasting exponential } z^n}$$

(3.40)

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

We repeat again that in this discussion we are talking of the everlasting exponential, which starts at n = −∞, not the causal exponential znu[n], which starts at n = 0.

For a system specified by Eq. (3.20), the transfer function is given by

$$H[z] = \frac{P[z]}{Q[z]} \tag{3.41}$$

This follows readily by considering an everlasting input x[n] = zn. According to Eq. (3.40), the output is y[n] = H[z]zⁿ. Substitution of this x[n] and y[n] in Eq. (3.20) yields

$$H[z]\{Q[E]z^n\} = P[E]z^n$$

Moreover,

$$E^k z^n = z^{n+k} = z^k z^n$$

Hence,

$$P[E]z^n = P[z]z^n \qquad \text{and} \qquad Q[E]z^n = Q[z]z^n$$

Consequently,

$$H[z] = \frac{P[z]}{Q[z]}$$

DR ILL 3.20 DT System Transfer Function

Show that the transfer function of the digital differentiator in Ex. 3.8 (big shaded block in Fig. 3.16b) is given by H[z] = (z−1)/Tz, and the transfer function of an unit delay, specified by y[n] = x[n−1], is given by 1/z.

3.8-3 Total Response

The total response of an LTID system can be expressed as a sum of the zero-input and zero-state responses:

total response =

$$\underbrace{\sum_{j=1}^{N} c_j \gamma_j^n}_{\text{ZIR}} + \underbrace{x[n] * h[n]}_{\text{ZSR}}$$

In this expression, the zero-input response should be appropriately modified for the case of repeated roots. We have developed procedures to determine these two components. From the system equation, we find the characteristic roots and characteristic modes. The zero-input response is a linear combination of the characteristic modes. From the system equation, we also determine h[n], the impulse response, as discussed in Sec. 3.7. Knowing h[n] and the input x[n], we find the zero-state response as the convolution of x[n] and h[n]. The arbitrary constants c1, c2,…, cn in the zero-input response are determined from the n initial conditions. For the system described by the equation

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

298 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS

with initial conditions y[−1] = 0, y[−2] = 25/4 and input x[n] = (4)−nu[n], we have determined the two components of the response in Exs. 3.13 and 3.21, respectively. From the results in these examples, the total response for n ≥ 0 is

total response =

$$\underbrace{0.2(-0.2)^n + 0.8(0.8)^n}_{\text{ZIR}} + \underbrace{0.444(-0.2)^n + 5.81(0.8)^n - 1.26(4)^{-n}}_{\text{ZSR}}$$

(3.42)

NATURAL AND FORCED RESPONSE

The characteristic modes of this system are (−0.2)ⁿ and (0.8)ⁿ. The zero-input response is made up of characteristic modes exclusively, as expected, but the characteristic modes also appear in the zero-state response. When all the characteristic mode terms in the total response are lumped together, the resulting component is the natural response. The remaining part of the total response that is made up of noncharacteristic modes is the forced response. For the present case, Eq. (3.42) yields

total response =

$$\underbrace{0.644(-0.2)^n + 6.61(0.8)^n}_{\text{natural response}} + \underbrace{-1.26(4)^{-n}}_{\text{forced response}} \qquad n \ge 0$$

Just like differential equations, the classical solution to difference equations includes the natural and forced responses, a decomposition that 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[n] as an explicit function of x[n]. 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 approach and its direct calculation of the forced and natural responses.