d) This system could be called either a sample-data or a digital system—I don't
think our book is very clear on the distinction. Ogata's Discrete-Time Control.
EGES 523 Spring 2006
Homework 1 Solutions
Assigned 1/16/06 Due 1/23/05
1. (textbook Problem 1.1) 1 rev 1 min 60 sec 10 sec • • = Î Ts = 10 seconds (0.1 Hz) a) 1 sample 6 rev 1 min 1 sample 9 mi 1 hr 60 min 60 sec 60 sec Î Tc = 60 seconds. (0.0167 • • • = b) 1 sample 540 mi 1 hr 1 min 1 sample Hz) c) i. continuous ii. discrete iii. digital iv. continuous d) This system could be called either a sample-data or a digital system—I don’t think our book is very clear on the distinction. Ogata’s Discrete-Time Control Systems adds a little bit to the definition (2nd Ed., p. 3): “Loosely speaking, terminologies such as discret-time control systems, sampled-data control systems, and digital control systems imply the same type or very similar types of control systems. Precisely speaking, there are, of course, differences in these systems. For example, in a sampleddata control system both continuous-time and discrete-time signals exist in the system; the discrete-time signals are amplitude-modulated pulse signals. Digital control systems may include both continuous-time and discrete-time signals; here, the latter are in a numerically coded form. Both sampled-data control systems and digital control systems are discrete-time control systems.” In summary, discrete-time signals can have any value, but are only defined at specific times. Digital systems are both discrete-time and quantized. e) If the plane follows a sinusoidal path that crosses zero every 10 seconds (0.1 Hz), its zig-zag cannot be detected by the radar. In order for a signal to be recoverable, it must be sampled at a rate at least twice the highest frequency in the signal (the nyquist frequency), so a sinusoid with frequency half that of the radar, or 0.05 Hz, is the lowest frequency sinusoid. In the plot, sin(ωt) = sin(2π(0.05)t). 1
sin(wt)
0.5
0
-0.5
-1 0
10
20 Time (sec)
30
40
EGES 523 Spring 2006
Homework 1 Solutions
Assigned 1/16/06 Due 1/23/05
Y (s ) = D(s )G (s )E (s ) = D(s )G (s )[R(s ) − Y (s )] Y (s ) D(s )G (s ) = R(s ) 1 + D(s )G (s ) K (s + a ) = 3 s + (b + c )s 2 + (bc + K )s + Ka b) The closed-loop zero is at -1, and the closed-loop poles are {-0.3044, -2.3478±1.0289j} Since the closed-loop poles are all in the left half plane, the closed-loop system is stable. c) The Final Value Theorem is valid if the system is stable and has a final, constant value, which is true for this system.
2.
a)
1 2(s +1) R(s) = ⇒ Y (s) = 3 s s s + 5s2 + 8s + 2
lim y (t ) = lim sY (s ) t →∞
s →0
= lim s →0
(
)
2s (s + 1) s (s + 5s 2 + 8s + 2 ) 3
=1
r(t)
e(t)
s+a D(s) = K s+b
1 G(s) = s (s + c )
Figure 1: Continuous-time feedback system for Problem 1
y(t)
EGES 523 Spring 2006
Homework 1 Solutions
Assigned 1/16/06 Due 1/23/05
3. The root locus and bode plots should look like this: Open-Loop Bode Editor (C)
Root Locus Editor (C) 8
20 0
6
-20 4 -40 2
-60 G.M.: Inf Freq: Inf Stable loop -80
0
-45 -2 -90 -4 -135
-6
P.M.: 92.7 deg Freq: 0.345 rad/sec
-180
-8 -3
-2.5
-2
-1.5 -1 Real Axis
-0.5
0
-1
0
10
1
2
10 10 Frequency (rad/sec)
10
The bode plot shows that the system is stable, agreeing with the results in 2(b). The closed-loop poles should be identical to those in 2(b). The step response looks like this: Step Response 1 0.9 0.8 0.7
Amplitude
0.6 0.5 0.4 0.3 0.2 0.1 0
0
2
4
6
8
10
12
14
16
Time (sec)
The steady-state value is 1, which agrees with the answer to 2(c).
18
20
