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Problem 6.43PP
determine the value for K that will yield PM > 30° and the maximum possible closed-loop 
bandwidth. Use Matlab to find the bandwidth.
For the open-loop system
Step-by-step solution
step 1 of 10
Step 1 of 10
The open loop system is.
j= (s + 2 0 )
The phase margin of the system is.
PM = 180®+^^
The phase of the system is.
* -1 8 0 ® + t a n " '® - 2 ta n " '— 
20
Step 2 of 10
Hence, the phase margin of the system is.
PM = l8 0 * -1 8 0 » + t iin - '® -2 ta n - '
20
■ - I= tan *0 - 2 tan — 
20
Since the phase margin is calculated at gain cross over frequency
PM * Ian"' - 2 ta n " ‘ — 
20
Step 3 of 10
Since the phase margin. PM ^30®>
30“ S t a n " ' - 2 tan " '- £ •
” 20
tan"'«i - la n " '^ - t a n " '^ a 3 0 “ 
20 20
tan"'
a- 20
” 20 J
- t a n " '- 2 - a 30“ 
20
t a n " ' - t a n " ' ^ 2 30“1̂20+n.J, J 20
Step 4 of 10
Further simplify.
1 9 ® , - ^
20
379a.
2 3 0 “
2 0 + (2 0 + ® i) » ,
379® .
2 0 + ( 2 0 + « ^ . ) ® ^ ] ^ '^ 
656.447® ^ 2 20 + 20® ^ + aij, 
® ? - 6 3 6 .4 4 7 ® .+ 2 0 S 0
2 0 ;
2 tan 30“
1
The roots are,
0 ^ — 25.24 
 /® '+ i )
|A:c(a)|=
i*(V®“ +400)
Since the gain of the system at 0 ^ is unity or 0 dB,
L + 400T
^ ® ; . f e + 4 0 0 )
(fflj,+ 4 0 0) = a: (
( 26085.37
Thus, the value o f /25.21
The bandwidth is,
= 2 5 .2 1 -0 .0 3 1 4 
= 20.178 rad/s
Thus, the maximum possible bandwidth is 120.178 rad/sl •
Step 9 of 10
The MATLAB code to
k=26085.37;
num=[k k]:
den=[1 40 400 0 0];
sys=tf(num,den);
bodeplot(sys)
margin(sys)
grid
Step 10 of 10 ^
Thus, the plot of MATLAB is sown in Figure 1.
B ode D iagram
F reqacncy (rad /s)