Question: It is shown in Example 6.6 that the system of Fig. 6.2-2 has the parameters = 0.250 , n = 0.9191 , and

It is shown in Example 6.6 that the system of Fig. 6.2-2 has the parameters ζ = 0.250 , ωn = 0.9191 , and τ = 4.36 s .(a) Find , , and for the sample period 7 = 0.5. (b) Repeat part (a) for T = 0.1. (c) Repeat part (a) for the

Example 6.6

For this example we will consider the system of Example 6.4. For this system the closed-loop transfer function was calculated to beG(z) 1 + G(z) Thus the system characteristic equation is The poles are then complex and occur at then Since,

Example 6.4

The system for this example is shown in Fig. 6-2. As in the first example, we will calculate the unit-step response. This system will also appear in many of the following examples. As was demonstrated in Chapter 5, the system output can be expressed asR(s) =  T=1 1-8-7's S G(s) 1 s(s+1) C(s)where, from the tables in Appendix VI, 1 Z = (  = ) - = [ 5 (5! S(S + 1)]T=1 G(z) Then Since then = 0.368z +

(a) Find , , and for the sample period 7 = 0.5. (b) Repeat part (a) for T = 0.1. (c) Repeat part (a) for the analog system, that is, the system with the sampler and data hold removed. Note that this case can be considered to be the limit as the sample period approaches zero. (d) Give a table listing the three parameters as a function of sampling frequency f = 1/T . State the result of decreasing the sampling frequency on the parameters.

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