[5.9 CONNECTING THE](#page-11-0) LAPLACE AND z-TRANSFORMS

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5.9 CONNECTING THE LAPLACE AND z**-TRANSFORMS**

We now show that discrete-time systems also can be analyzed by means of the Laplace transform. In fact, we shall see that the z-transform is the Laplace transform in disguise and that discrete-time systems can be analyzed as if they were continuous-time systems.

So far we have considered the discrete-time signal as a sequence of numbers and not as an electrical signal (voltage or current). Similarly, we considered a discrete-time system as a mechanism that processes a sequence of numbers (input) to yield another sequence of numbers (output). The system was built by using delays (along with adders and multipliers) that delay sequences of numbers. A digital computer is a perfect example: every signal is a sequence of numbers, and the processing involves delaying sequences of numbers (along with addition and multiplication).

Now suppose we have a discrete-time system with transfer function H[z] and input x[n]. Consider a continuous-time signal x(t) such that its nth sample value is x[n], as shown in Fig. 5.30.† Let the sampled signal be x(t), consisting of impulses spaced T seconds apart with the nth impulse of strength x[n]. Thus,

$$\bar{x}(t) = \sum_{n=0}^{\infty} x[n]\delta(t - nT)$$

Figure 5.30 shows x[n] and the corresponding x(t). The signal x[n] is applied to the input of a discrete-time system with transfer function H[z], which is generally made up of delays, adders, and scalar multipliers. Hence, processing x[n] through H[z] amounts to operating on the sequence x[n] by means of delays, adders, and scalar multipliers. Suppose for x(t) samples, we perform operations identical to those performed on the samples of x[n] by H[z]. For this purpose, we need a continuous-time system with transfer function H(s) that is identical in structure to the discrete-time system H[z] except that the delays in H[z] are replaced by elements that delay continuous-time signals (such as voltages or currents). There is no other difference between realizations of H[z] and H(s). If a continuous-time impulse δ(t) is applied to such a delay of T seconds, the output will be δ(t −T). The continuous-time transfer function of such a delay is e−sT [see Eq. (4.30)]. Hence, the delay elements with transfer function 1/z in the realization of H[z] will be replaced by the delay elements with transfer function e−sT in the realization of the corresponding H(s). This is the same

^† We can construct such x(t) from the sample values, as will be explained in Ch. 8.

Figure 5.30 Connection between the Laplace transform and the z-transform.

as z being replaced by esT . Therefore, H(s) = H[esT ]. Let us now apply x[n] to the input of H[z] and apply x(t) at the input of H[esT ]. Whatever operations are performed by the discrete-time system H[z] on x[n] (Fig. 5.30a) are also performed by the corresponding continuous-time system H[esT ] on the impulse sequence x(t) (Fig. 5.30b). The delaying of a sequence in H[z] would amount to delaying of an impulse train in H[esT ]. Adding and multiplying operations are the same in both cases. In other words, one-to-one correspondence of the two systems is preserved in every aspect. Therefore if y[n] is the output of the discrete-time system in Fig. 5.30a, then y(t), the output of the continuous-time system in Fig. 5.30b, would be a sequence of impulse whose nth impulse strength is y[n]. Thus,

$$\bar{y}(t) = \sum_{n=0}^{\infty} y[n]\delta(t - nT)$$

The system in Fig. 5.30b, being a continuous-time system, can be analyzed via the Laplace transform. If

$$\overline{x}(t) \Longleftrightarrow \overline{X}(s) \quad \text{and} \quad \overline{y}(t) \Longleftrightarrow \overline{Y}(s)$$

then

$$\overline{Y}(s) = H[e^{sT}]\overline{X}(s)$$

\n(5.53)

Also,

$$\overline{X}(s) = \mathcal{L}\left[\sum_{n=0}^{\infty} x[n]\delta(t - nT)\right]$$

Now because the Laplace transform of δ(t −nT) is e−snT ,

$$\overline{X}(s) = \sum_{n=0}^{\infty} x[n]e^{-snT} \quad \text{and} \quad \overline{Y}(s) = \sum_{n=0}^{\infty} y[n]e^{-snT}$$

Substitution of these expressions into Eq. (5.53) yields

$$\sum_{n=0}^{\infty} y[n]e^{-snT} = H[e^{sT}]\left[\sum_{n=0}^{\infty} x[n]e^{-snT}\right]$$

By introducing a new variable z = esT , this equation can be expressed as

$$\sum_{n=0}^{\infty} y[n]z^{-n} = H[z] \sum_{n=0}^{\infty} x[n]z^{-n}$$

or

$$Y[z] = H[z]X[z]$$

where

$$X[z] = \sum_{n=0}^{\infty} x[n]z^{-n}$$

and $Y[z] = \sum_{n=0}^{\infty} y[n]z^{-n}$

It is clear from this discussion that the z-transform can be considered to be the Laplace transform with a change of variable z = esT or s = (1/T)lnz. Note that the transformation z = esT transforms the imaginary axis in the s plane (s = jω) into a unit circle in the z plane (z = esT = ejω*^T* , or |z| = 1). The LHP and RHP in the s-plane map into the inside and the outside, respectively, of the unit circle in the z plane.