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Problem 6.32PP
For the system depicted in Fig{a), the transfer-function blocks are defined by
1 1
G ( s ) = ' H ( s ) =
(5-1-2)2(5-I-4 ) 5-1-1
(a) Using hocus and riocfind, determine the value of K at the stability boundary.
(b) Using hocus and riocfind, determine the value of K that will produce roots with damping 
corresponding to 0.707.
(c) What is the GM of the system if the gain is set to the value determined in part (b)? Answer 
this question without using any frequency-response methods.
(c) What is the GM of the system if the gain is set to the value determined in part (b)? Answer 
this question without using any frequency-response methods.
(d) Create the Bode plots for the system, and determine the GM that results for PM = 65®. What 
damping ratio would you expect for this PM?
(e) Sketch a root locus for the system shown in Fig (b). How does it differ from the one in part
(a)?
(f) For the systems in Figs, (a) and (b). how does the transfer function V2(s)/R(s) differ from 
Y1 {s)^(s)7 Would you expect the step response to r(t) to be different for the two cases?
Figure Block diagram; (a) unity feedback; (b) H(s) in feedback
S tep -by-s tep s o lu tio n
Step 1 of 25
(=0
Given
o ( s ) = ------------2------------
 ̂ ' ( s + 2 ) \s + A )
H (s ) — ! - 
 ̂ ' s+1
(s + 2 )^ (s+ 4 )(s + l)
Step 2 of 25
MATLAB program for root locus
» b = [1 4 ] ;
» c=[l 1];
» 8 1 system is unstable
Step 5 of 25
b)
^F=[14 4]; 
» b = [1 4 ] ;
» c=[l 1];
» ^conv(a,b); 
» dei^onv(c,d); 
» num=[l];
» ^£(num ,den)
Step 6 of 25
Transfer function:
1
sM + 9s^3 + 28 s^2 + 36s + 16
» rlocus(x)
» v=[.6 0 -3 3];
» sgrid([0.7a7],[])
» gtext('\2eta*0.707')
» axis(v);
» rlocfmd(x)
Select a point in the graphics window
Step 7 of 25
selected_point =
-0.9092 + 0.90861 
ans = 5.9288
Step 8 of 25
Root locus is
K = 5.9288 Produces roots with damping ratio corresponds to 0.707
Step 9 of 25
c)
Gain margin is
g = - ? L
5.928 
G = 13.66
Step 10 of 25
, = -115®
Step 13 of 25 ^
With 1 the db gain at 135® is -2 9 .6 d b . This gain shouldbe made zero to have 
to PM of 65®. Hence to every point of magnitude plot a db gain of 26.6db shouldbe 
added. The corrected magnitude plot is obtained by shifting the plot with R=1 by 29.6db 
iqswards. The magnitude correction is independent of frequency. Hence the magnitude of 
29.6db is contributed by the term K. The value is calculated by equating 2 0 1 o ^ to 
29.6db
201og£'=29.6
294
AT=10»
^ = 30.12
Step 14 of 25
With K = 30.12 open loop transfer function is
a {s )H [s ) -----------------------------
 ̂ ' (s+ 2 )’ ( s + 4 ) ( s + l)
Now the bode plot for above system is 
» a = [1 4 4];
» b = [1 4 ] ;
» c=[l 1];
» ^conv(a,b);
» den*conv(c,^;
» num=[30.12];
» ^tf(num,den)
Transfer function:
30.12
sM + 9 s^3 + 28 s^2 + 36 s + 16
» bode(x) 
» grid
Step 15 of 25
From the above bode plot 
Gain margin = 8.56db 
Gain margin in linear scale 
201ogK=8.56
8.S6
K = - [ 0 ^
K = 2.6B
So gain margin is 2.68
Step 16 of 25
Step 17 of 25
Approximate danq^ing ratio 
PM 
100 
65 
100 
^ '*0 .6 5
c = -
Step 18 of 25
Fig (a)
Step 19 of 25
Fig(b)
Step 20 of 25
Root locus is drawn for 0(^b)H(^s)
For figures
0 {s ) H {s ) = K O {s )H (s )
So Root locus for both the figures is same 
And root locus is
Step 21 of 25
£)
For figure (a) Transfer function is 
T,(s) K a (s )H {s )
R [s)~ l+ K O {s )H {s )
1 . . 1
yi{s) (s+ 2 )^ (s+ 4 ) s+1
(s+ 2 ) ( s+ 4 ) s+ 1 
1
J^(s) (s+ 2 )^ (s+ 4 )(s + l)
(ff+2) (ff+4)(ff+l)
W __________ f
R{s) J5T + (s+ 2)^(s+ 4)(s+ l)
K 1
Step 22 of 25
For figure (b) Transfer function is 
r , ( s ) _ K a js )
R (s )~ \+ K O [s )H [s )
jv 1
r^(s) (s+ 2 )^ (a+ 4 )
^ W " l + -
( s+ 2 ) (s+ 4 ) s+ 1
jv ]
^ (s+ 2 )^ (s+ 4 )
^ ------------
(s+ 2 ) ( s + 4 ) ( s + l)
r , ( s ) i : ( s + i )
5 (s ) J i:+ (s+ 2 )’ ( s + 4 )(s + l)
step 23 of 25
MAILAB program for step response
» a = [1 4 4];
» b = [1 4 ] ;
» c=[l l]i
» ^conv(a,b);
» dei^onv(c,d);
» n u m = [ l 1];
» ^f(num ,den)
Step 24 of 25
Step 25 of 25