Question 13.6: The missile shown in Figure 13.14(a) can be aerodynamically ...
The missile shown in Figure 13.14(a) can be aerodynamically controlled by torques created by the deflection of control surfaces on the missile’s body. The commands to deflect these control surfaces come from a computer that uses tracking data along with programmed guidance equations to determine whether the missile is on track. The information from the guidance equations is used to develop flight-control commands for the missile. A simplified model is shown in Figure 13.14(b). Here the computer performs the function of controller by using tracking information to develop input commands to the missile. An accelerometer in the missile detects the actual acceleration, which is fed back to the computer. Find the closed-loop digital transfer function for this system and determine if the system is stable for K = 20 and K = 100 with a sampling interval of T = 0.1 second
The input to the control system is an acceleration command developed by the computer. The computer can be modeled by a sample-and-hold. The s-plane model is shown in Figure 13.14(c). The first step in finding the z-plane model is to find G(z), the forward-path transfer function. From Figure 13.14(c) or (d),
G (s) = \frac{1 − e^{−Ts}}{s} \frac{Ka}{s (s + a)} (13.48)
where a = 27. The z-transform, G(z), is (1 − z^{−1}) z {Ka/[s² (s + a)]}.
The term Ka/[s² (s + a)] is first expanded by partial fractions, after which we find the z-transform of each term from Table 13.1. Hence,
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 |
z \left\{\frac{Ka}{s² (s + a)} \right\} = Kz \left\{\frac{a}{s² (s + a)} \right\} = Kz \left\{\frac{1}{s²} – \frac{1/a}{s} + \frac{1/a}{s + a} \right\} (13.49)
= K \left\{\frac{Tz}{(z − 1)²} – \frac{z/a}{z – 1} + \frac{z/a}{z − e^{-aT} } \right\}
= K \left\{\frac{Tz}{(z − 1)²} – \frac{(1 − e^{-aT} ) z}{a (z − 1) (z − e^{-aT})} \right\}
Thus,
G (z)= K \left\{\frac{T (z − e^{-aT}) − (z − 1) \left(\frac{1 − e^{-aT}}{a} \right) }{(z − 1) (z − e^{-aT})} \right\} (13.50)
Letting T = 0.1 and a = 27, we have
G (z)= \frac{K (0.0655z + 0.02783)}{(z − 1) (z − 0.0672)} (13.51)
Finally, we find the closed-loop transfer function, T(z), for a unity-feedback system:
T (z) = \frac{G (z)}{1 + G (z)} = \frac{K (0.0655z + 0.02783)}{z² + (0.0655K − 1.0672) z + (0.02783K + 0.0672)} (13.52)
The stability of the system is found by finding the roots of the denominator. For K = 20, the roots of the denominator are 0.12 ±j 0.78. The system is thus stable for K = 20, since the poles are inside the unit circle. For K = 100, the poles are at −0.58 and −4.9. Since one of the poles is outside the unit circle, the system is unstable for K = 100.
Students who are using MATLAB should now run ch13apB5 in Appendix B. You will learn how to use MATLAB to determine the range of K for stability in a digital system. This exercise solves Example 13.6 using MATLAB.