CONNECTION BETWEEN THE LAPLACE TRANSFORM AND THE z-TRANSFORM

5.8 CONNECTION BETWEEN THE LAPLACE TRANSFORM AND THE z-TRANSFORM

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 discretetime 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. 5.25 . [ † ] Let the sampled signal be x (t), consisting of impulses spaced T seconds apart with the nth impulse of strength x[n]. Thus

Figure 5.25 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. (4.46) ]. 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 same as z being replaced by e sT . Therefore H(s)=H[e sT ]. Let us now apply x[n] to the input of H[z] and apply x (t) at the input of H[e sT ]. Whatever operations are of such a delay is e −sT [see Eq. (4.46) ]. 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 same as z being replaced by e sT . Therefore H(s)=H[e sT ]. Let us now apply x[n] to the input of H[z] and apply x (t) at the input of H[e sT ]. Whatever operations are

continuous-time system in Fig. 5.25b , would be a sequence of impulse whose nth impulse strength is y[n]. Thus

Figure 5.25: Connection between the Laplace transform and the z-transform. The system in Fig. 5.25b , being a continuous-time system, can be analyzed via the Laplace transform. If

then

Also

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

and

Substitution of Eqs. (5.82) and (5.83) in Eq. (5.81) yields

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

or or

It is clear from this discussion that the z-transform can be considered to be the Laplace transform with a change of variable z=e sT or s=(1/T) In z. Note that the transformation z=e sT transforms the imaginary axis in the s plane (s=j ω) into a unit circle in the z plane (z=e sT =e j ωT , 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.

[ † ] We can construct such x(t) from the sample values, as explained in Chapter 8 .