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Problem 6.34PP
A magnetic tape-drive speed-control system is shown in Fig. The speed sensor is slow enough 
that its dynamics must be included. The speedmeasurement time constant is rm = 0.5 sec; the 
reel time constant \sTr= J/b = A sec. where b = the output shaft damping constant = 1 N m- sec; 
and the motor time constant is r1 = 1 sec.
(a) Determine the gain K required to keep the steady-state speed error to less than 7% of the 
reference-speed setting.
(b) Determine the gain and phase margins of the system. Is this a good system design? 
Figure Magnetic tape-drive speed control
I Amplifier [ 7 ™ , ^
p and motor I ^ drive
Step-by-step solution
Step 1 of 10
Refer to the Figure 6.94 from the text book.
It is clear that the value of speed measurement time constant is 0.5 seconds, the value of 
reel time constant r^is 4 seconds, the value of output shaft damping constants is 1 N.nLsecand 
the value of motor time constant r . is l second.
Obtain the transfer function of the system.
Thus, the transfer function of the system for the Figure 6.94 is as follows.
T {s ).
f—If—1
1 -1-
( s r , + ! ) ( « , + l ) ( y s + i ) + iT
( 1)
Step 2 of 10
(a)
It is clear from equation (1) that the system is of Type 1.
It is clear that for Type 1 system, there exists positional error constant . 
The steady state error of the system is.
1
1+*. (2 )
It is clear that it^in equation (2) represents a positional error constant. 
Calculate the steady state error.
sR{s)
l + G{s)H{s)
= lim-
1+
i+ ( i ) ( j r )
n
Step 3 of 10
It is clear that the value of b is 1 N.m.sec 
The steady state error of the system is.
I
Substitute 1 N.m.sec ^ in steady state emor.
1
Equate this expression to equation (2).
1 _ 1
1 + / : = ! + * ,
Thus, it is clear that the value of K is equal to k^ .
Step 4 of 10
7^0.07 
z ^ \ + K
Calculate the value of gain such that the steady state error is less than . 
e ^£ 7 %
1
1 + ^ :"
1
0 .0 7 "
I4 .2 8 S 1 + /:
1 4 .2 8 -!£ a:
13.28£JC
Hence, approximately the value of K is.
Thus, for the emor less than 7% of the reference, the gain must be .
Step 5 of 10
(b)
Recall equation (1).
r(j)= Ar(tr, + 1)
{ « ’, + l ) ( i r , + l ) ( J s + 4 )+ ^ T
i ( s T , + l)(*r,+ + 1 j + X
Substitute 13 for/C, 1 N.m.sec for b. 0.5 secfor . and 1 sec for r,-
13(0.5j + 1)
'■ ' (0 .5 i + l ) ( j + l ) ( 4 j + l)+ 1 3
13(0 .5 i + l)
0.5*’ + 3 .5 * '+ 5 .5 * + 1 4
Step 6 of 10
The transfer function of the fonvard loop is.
1
 + 1
I
( l + 4