Question 13.4: Given a zero-order hold in cascade with G_1(s) = (s + 2)/(s ...
Given a zero-order hold in cascade with G_{1} (s) = (s + 2)/(s + 1) or
G (s) = \frac{1 − e^{−Ts} }{s} \frac{(s + 2)}{(s + 1)} (13.38)
find the sampled-data transfer function, G(z), if the sampling time, T, is 0.5 second.
Equation (13.38) represents a common occurrence in digital control systems, namely a transfer function in cascade with a zero-order hold. Specifically, G_{1} (s) = (s + 2)/(s + 1) is in cascade with a zero-order hold, (1 − e^{−Ts})/s. We can formulate a general solution to this type of problem by moving the s in the denominator of the zero-order hold to G_{1} (s), yielding
G (s) = (1 − e^{−Ts} ) \frac{G_{1}(s)}{s} (13.39)
from which
G (z) = (1 − z^{−1} ) z \left\{\frac{G_{1}(s)}{s}\right\} = \frac{z – 1}{z} z \left\{\frac{G_{1}(s)}{s}\right\} (13.40)
Thus, begin the solution by finding the impulse response (inverse Laplace transform) of G_{1} (s)/s. Hence,
G_{2} (s) = \frac{G_{1} (s)}{s} = \frac{s + 2}{s (s + 1)} = \frac{A}{s} + \frac{B}{s + 1} = \frac{2}{s} – \frac{1}{s + 1} (13.41)
Taking the inverse Laplace transform, we get
g_{2} (t) = 2 – e^{-t} (13.42)
from which
g_{2} (kT) = 2 – e^{-kt} (13.43)
Using Table 13.1, we find
TABLE 13.1 Partial table of z- and s-transforms
| ƒ(t) | F(s) | F(z) | ƒ(kT) | |
| 1. | u(t) | \frac{1}{s} | \frac{z}{z – 1} | u(kT) |
| 2. | t | \frac{1}{s²} | \frac{T_{z} }{(z – 1)² } | kT |
| 3. | t^{n} | \frac{n!}{s^{n + 1} } | \underset{a\rightarrow 0}{lim} (-1)^{n} \frac{d^{n} }{d a^{n} } \left[\frac{z}{z – e^{-aT} } \right] | (kT)^{n} |
| 4. | e^{-at} | \frac{1}{s + a } | \frac{z}{z – e^{-aT} } | e^{-akT} |
| 5. | t^{n} e^{-at} | \frac{n!}{(s + a)^{n+1} } | (-1)^{n} \frac{d^{n} }{d a^{n} } \left[\frac{z}{z – e^{-aT} } \right] | (kT)^{n} e^{-akT} |
| 6. | sin ωt | \frac{ω}{s² + ω² } | \frac{z sin ωT}{z² − 2z cos ωT + 1} | sin ωkT |
| 7. | cos ωt | \frac{s}{s² + ω² } | \frac{z (z − cos ωT)}{z² − 2z cos ωT + 1} | cos ωkT |
| 8. | e^{-at} sin ωt | \frac{ω}{(s + a)² + ω² } | \frac{ze^{-aT} sin ωT }{z² − 2ze^{-aT} cos ωT + e^{-2aT}} | e^{-akT} sin ωkT |
| 9. | e^{-at} cos ωt | \frac{s + a}{(s + a)² + ω² } | \frac{z² – ze^{-aT} cos ωT }{z² − 2ze^{-aT} cos ωT + e^{-2aT}} | e^{-akT} cos ωkT |
G_{2} (z) = \frac{2z}{z − 1} – \frac{z}{z − e^{−T} } (13.44)
Substituting T = 0.5 yields
G_{2} (z) = z \left\{\frac{G_{1} (s)}{s}\right\} = \frac{2z}{z − 1 } – \frac{z}{z − 0.607} = \frac{z² − 0.213z}{(z − 1) (z − 0.607)} (13.45)
From Eq. (13.40),
G (z) = \frac{z – 1}{z} G_{2} (z) = \frac{z − 0.213}{z − 0.607} (13.46)
Students who are using MATLAB should now run ch13apB1 in Appendix B. You will learn how to use MATLAB to convert G1(s) in cascade with a zero-order hold to G(z). This exercise solves Example 13.4 using MATLAB.
Students who are performing the MATLAB exercises and want to explore the added capability of MATLAB’s Symbolic Math Toolbox should now run ch13apF3 in Appendix F located at www.wiley.com/go/Nise/ControlSystemsEngineering8e. MATLAB’s Symbolic Math Toolbox yields an alternative method of finding the z-transform of a transfer function in cascade with a zero-order hold. Example 13.4 will be solved using MATLAB and the Symbolic Math Toolbox with a method that follows closely the hand calculation shown in that example.