EE571 Solution to HW#26
b)
Use semi-log paper to construct the Bode Plots (both magnitude and phase) for the following transfer functions:
i) G(s)=10(s-1)/[s2(s/10 + 1)]
10(s-1)
ii) G(s)= _______
2
s (s/10
+1 )
G(jw) deg
|G(jw)| dB
(s-1)
(s-1)
40
90
10
.1
1
10
100
1000 ω r/s
10
0
.1
0
1
10
1000 ω r/s
100
Ans.
-40
Ans.
-90
1/[(s/10)+1]
1/s
1/[(s/10)+1]
2
1/s 2
-180
ii) G(s)=10(s+1)/[s(s/10-1)2]
10(s+1)
ii) G(s)= _______
s(s/10 -1 )
(s+1)
0
.1
40
10
.1
1
(s+1)
G(jw) deg
|G(jw)| dB
2
10
1000
100
Ans.
ω r/s
10
1
10
100
1000
ω r/s
1/s
-90
Ans.
-40
1/s
-180
1/[(s/10)-1]
1/[(s/10)-1]
2
2
-360
-450
2
iii) G(s)=10(s+1)/[s((s/10) -1)]
10(s+1)
ii) G(s)= _______2
s[(s/10) -1 ]
(s+1)
0
.1
40
10
.1
-40
1
10
100
1000
1/s
ω r/s
Ans.
1/[(s/10) 2 -1]
b)
(s+1)
G(jw) deg
|G(jw)| dB
10
1
10
100
1000
ω r/s
-90
-180
1/s
Ans.
1/[(s/10)
2
+1]
-270
The magnitude plot is the same as the plot in HW#25. Thus, our minimum phase solution is:
G(s)=100(s/10 +1)/[s1(s/1+1)(s/100+1)]. However, the corresponding phase response of this G(s) should level off at -180
degrees at high frequencies. Our phase plot continues to decrease at high frequencies thereby implying that transportation
lag is present! Thus, our revised guess at G(s) is G(s)= 100(s/10 +1)e-sT /[s1(s/1+1)(s/100+1)]. To find T, we note that from
the Bode plot, the slope at high frequencies (i.e., frequencies well above the last break point at 100) is -225-(-180)
deg/(10,000-1,000 rad/sec) = -0.005 deg/rad/sec. Note that the slope of the theoretical transfer function G(s) at frequencies
well above the highest breakpoint is (-T)(180 deg)/(3.14159 rad/sec). Equating the two expressions, we obtain T=8.727 x 10-
5
and, therefore, G(s)= 100(s/10 +1)e-0.00008727s /[s1(s/1+1)(s/100+1)]. (Note: we will see that for such a system with such small
transportation delay, we can effectively ignore this delay when considering stability)
2. a)
Use your answers from HW#25 to sketch the ploar plots of:
i) G(s)=10(s+1)/[s2(s/10 + 1)]
i)
ii) G(s)=10(s+1)/[s(s/10+1)2]
ii)
ImG
iii) G(s)=10(s+1)/[s((s/10)2+1)]
iii)
ImG
ReG
ReG
ImG
ReG
b)
For the open-loop transfer function, G(s)=10/[s(s+1)(s/10 + 1)], draw the appropriate Nyquist Path then make a Nyquist
plot.
First, draw the Bode plot:
Bode Diagrams
Gm=0.82785 dB (at 3.1623 rad/sec), Pm=1.5763 deg. (at 3.0145 rad/sec)
100
Phase (deg); Magnitude (dB)
50
0
-50
-100
-50
-100
-150
-200
-250
-300
10-2
10-1
100
101
102
Frequency (rad/sec)
Note that the magnitude of the Bode plot when the phase = -180 degrees is -0.82785 dB which is slightly less than 0 dB. Therefore,
the polar plot will cross the negative real axis just slightly to the right of the point –1+j0 and won’t encircle it (see below)
Nyquist Path
Nyquist Plot
ω=0−
j ω
Im(GH)
ωcp = 3.16 r/s
-10 -1
σ
-1
ω=0+
Re(GH)
c)
Use your Nyquist plot to determine what would happen if we formed a closed-loop system out of G(s), would the closedloop system be stable (i.e., does the Nyquist plot encircle the point –1+j0)? Hint: You may want to have Matlab make a
Bode plot of G(s) then look at the magnitude of G(jω) when the phase plot crosses –180 degrees. If this magnitude is greater
than 0 dB (i.e., greater than 1), then the Nyquist plot will encircle the point –1+j0.
The closed-loop system will have 0 closed-loop poles in the RHP because the Nyquist Plot does NOT encircle the –1+j0
point.