[5.11 SUMMARY](#page-12-0)

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5.11 SUMMARY

In this chapter we discussed the analysis of linear, time-invariant, discrete-time (LTID) systems by means of the z-transform. The z-transform changes the difference equations of LTID systems into algebraic equations. Therefore, solving these difference equations reduces to solving algebraic equations.

The transfer function H[z] of an LTID system is equal to the ratio of the z-transform of the output to the z-transform of the input when all initial conditions are zero. Therefore, if X[z] is the z-transform of the input x[n] and Y[z] is the z-transform of the corresponding output y[n] (when all initial conditions are zero), then Y[z] = H[z]X[z]. For an LTID system specified by the difference equation Q[E]y[n] = P[E]x[n], the transfer function H[z] = P[z]/Q[z]. Moreover, H[z] is the z-transform of the system impulse response h[n]. We showed in Ch. 3 that the system response to an everlasting exponential zn is H[z]zⁿ.

We may also view the z-transform as a tool that expresses a signal x[n] as a sum of exponentials of the form zⁿ over a continuum of the values of z. Using the fact that an LTID system response to zⁿ is H[z]zⁿ, we find the system response to x[n] as a sum of the system’s responses to all the components of the form zⁿ over the continuum of values of z.

LTID systems can be realized by scalar multipliers, adders, and time delays. A given transfer function can be synthesized in many different ways. We discussed canonical, transposed canonical, cascade, and parallel forms of realization. The realization procedure is identical to that for continuous-time systems with 1/s (integrator) replaced by 1/z (unit delay).

The majority of the input signals and practical systems are causal. Consequently, we are required to deal with causal signals most of the time. Restricting all signals to the causal type greatly simplifies z-transform analysis; the ROC of a signal becomes irrelevant to the analysis process. This special case of z-transform (which is restricted to causal signals) is called the unilateral z-transform. Much of the chapter deals with this transform. Section 5.8 discusses the general variety of the z-transform (bilateral z-transform), which can handle causal and noncausal signals and systems. In the bilateral transform, the inverse transform of X[z] is not unique, but depends on the ROC of X[z]. Thus, the ROC plays a crucial role in the bilateral z-transform.

In Sec. 5.9, we showed that discrete-time systems can be analyzed by the Laplace transform as if they were continuous-time systems. In fact, we showed that the z-transform is the Laplace transform with a change in variable.

REFERENCES

• 1. Lyons, R. G. Understanding Digital Signal Processing. Addison-Wesley, Reading, MA, 1997.
• 2. Oppenheim, A. V., and R. W. Schafer. Discrete-Time Signal Processing, 2nd ed. Prentice-Hall, Upper Saddle River, NJ, 1999.
• 3. Mitra, S. K. Digital Signal Processing, 2nd ed. McGraw-Hill, New York, 2001.

PROBLEMS

• 5.1-1 Using the definition, compute the z-transform of ^x[n] = (−1)n(u[n] − ^u[^{n − 8}]). Sketch the poles and zeros of X[z] in the z plane. No calculator is needed to do this problem!
• 5.1-2 Determine the unilateral z-transform X[z] of the signal x[n] shown in Fig. P5.1-2. As the picture suggests, x[n]=−3 for all n ≥ 9 and x[n] = 0 for all n < 3.
• 5.1-3 (a) A causal signal has z-transform given by ^X[z] = ^z2 *^z*3−¹ . Determine the time-domain signal x[n] and sketch x[n] over −4 ≤ n ≤ 11. [Hint: No complex arithmetic is needed to solve this problem!]

Figure P5.1-2

576 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM

• (b) Consider the causal semiperiodic signal y[n] shown in Fig. P5.1-3. Notice, y[n] continually repeats the sequence [1, 2, 3] for n ≥ 0. Determine the unilateral z-transform Y[z] of this signal. If possible, express your result as a rational function in standard form.
• 5.1-4 Using the definition of the z-transform, find the z-transform and the ROC for each of the following signals.
  • (a) u[n− m]
  • (b) ^{γ n} sinπn u[n]
  • (c) ^{γ n} cosπn u[n]

Figure P5.1-3

(d)

$$\gamma^n \sin \frac{\pi n}{2} u[n]$$

\n(e) $\gamma^n \cos \frac{\pi n}{2} u[n]$
\n(f) $\sum_{k=0}^{\infty} 2^{2k} \delta[n-2k]$
\n(g) $\gamma^{n-1} u[n-1]$

(h) ^{nγ n u}[n]

$$(i) \; n \, u[n]$$

(j)

$$\frac{\gamma^n}{n!}u[n]$$

(k)