Step 1 of 16
(a)
Refer to Figure 3.54 (a) in the text book for the block diagram. Redraw the block diagram by reducing the feedback loops.
Figure 1 Here.
o ; =
Draw the signal flow graph.
Step 2 of 16
There are two fonvard paths and two loops.
Step 3 of 16 ^
Write the formula for the transfer function using Mason’s gain formula. y , P i+ j> .
Here.
p^ and are the forward path gains /^ and / j are the loop gains Determine the fon/vard path gains.
p ,= C ,< ^
f t = G,
Substitute — ---- for G *. G,G, P i ' Step 4 of 16Determine the loop gains. /, = -G|C^G,’ / 2 = -c,g;
Substitute — ^ — for ( j! and — —*— for G f . I - G j f f , ’ 1 - G , f f , ’ ' ■ [ I - G . W J I . I - G . W J _____________________ ( l - G , / f , ) ( l - G . f f , ) l2 = -G,G", . g,g. \- G ,H , Step 5 of 16
Determine the transfer function.
l- G ,H , ' 1
+
{ l- G ,H ,) ( \- G ,H ,) l-G ,H , rG ,G , + G , ( l - G , W Q l ( l - G . f f , )'( l-G ,W ,)( l-G /f ,) + G,G,G. + G ,G ,(l-G ,//0
G ,G .( I- G .W ,)^ ■ G ,( 1 - G .W .) ( I- G ,W ,)■(i-
g,« ,) (i-g,w,) +g,g.G3+g,g,(i-g,//,) Thus, the transfer function — using Mason’s gain formula isG ,G 3(l-G ,//.)+ G ,(l-G ,tf3)(l-G ,g,)
( l- G 3 g :) ( l- G , g .) + G ,G ,G , - t - G , G , ( l- G . W , )Step 6 of 16
(b)
Refer to Figure 3.54 (b) in the text book for the block diagram. Construct signal flow graph for the block diagram.
Figure 3
There are three forward paths and three loops.
Step 7 of 16 ^
Write the formula for the transfer function using Mason’s gain formula. Y p , * p , * p .
Here,
P\ ’ P i P i^^^ ^be fonvard path gains /j, /^and ly are the loop gains Determine the fonvard path gains.
a = - A s 1 . f t = -!■ * ! f t - ^ i s Step 8 of 16
Determine the loop gains.
, 1
, 1
I , - ----vO,
, 1
Determine the transfer function. y . f t + f t + p . R
• i 1 1 1 1 " , 1 1 1
j +fl,5 +ajj+<ij
Thus, the transfer function — of the system is byS ‘¥ b ^ -¥ b j
Step 9 of 16
(c)
Refer to Figure 3.54 (c) in the text book for the block diagram. Construct signal flow graph for the block diagram.
There are three fonvard paths and three loops.
Step 10 of 16 ^
Write the formula for the transfer function using Mason’s gain formula. y ^ f t + f t + f t
Pi ’ P i P i^^^ ^be fonvard path gains /|, /^and ly are the loop gains
Step 11 of 16
Determine the fonvard path gains.
f t - 4 ^
Determine the loop gains.
, 1
, 1
I , - ----vO,
S te p 1 2 o f1 6
, 1
Determine the transfer function. y _ Pl± £i± £l R ~ l - l , - l , - l , b,[^s* + a,s+ay)'*'by(s'*-a^)+ by s ^ + O fS ^ + a ^ + a ^ * { a ^ l\+ b y ) s + a ^ + a ^ b 2 + l^ +O yS+ai
Thus, the transfer function — of the system is + (a,6^ + 4^) j + ajAj + + 4^ s +a^s +ayS+ay
Step 13 of 16 ^
(d)
Refer to Figure 3.54 (d) in the text book for the block diagram. Construct signal flow graph for the block diagram.
5 ' * - \ + BH
There are two fonvard paths and two loops.
Step 14 of 16 ^
Write the formula for the transfer function using Mason’s gain formula.
y ft+ft
R ~ l - l , - l , Here.
p, and py are the fonvard path gains /i and iy are the loop gains Determine the fonvard path gains.
p, = AB' P i = D
\ + BHfor
A * U B HAB
Determine the loop gains. l,= - A B ’G l^= -D G Substitute U B H AB*G ABC for l + B ff Step 15 of 16 Step 16 of 16
Determine the transfer function.
y ft+ft
R ~ l - l , - l ^ AB J + B H * D \-¥BH AB-^BHD + D ^ \- ^ B H ^ A B G ^ D 0 - ¥ B H D GThus, the transfer function — of the system is AB+BHD+D [ + BH + ABG + DG+BHDG
UseMason’s rule to determine the transfer function between R(s) and Y(s) in Fig. Figure Block diagram
Step-by-step solution
Step-by-step solution
step 1 of 10
(b )
Construct signal flow graph from Figure 3.52 in textbook.
- H j
- Ha
Figure 1
Step 2 of 10
From Figure 1, the fon/vard path gains are 2 and loop path gain is € Consider Mason’s rule for Figure 1.
Determine the first Fonvard path gains from Figure 1. G 2 _________________G4
G i G e
Figure 2
Step 3 of 10 ^
Refer Figure 2 and write the equation for first fonvard path gain. Pi = ... (1)
Thus, the first fonvard path gain is G f i2G^G^ Determine the Second Fonvard path gains from Figure 1.
G i G e
G i
Gs
Figures
Step 4 of 10
From Figure 3, write the second fonvard path gain. P2 = G iG jG jG j... (2)
Thus, the second fonvard path gain is G,G)G5G^. Determine the first loop path gain from Figure 1.
- H i
Figure
4Step 5 of 10
From Figure 4, write the first loop path gain. / ,= - G ,G jG ,G * //, (3)
Thus, the first loop path gain is
Determine the second loop path gain from Figure 1.
- Ha
Figure 5
Step 6 of 10
From Figure 5, write the second loop path gain. I2 =-G ,G ,G ^G ,i¥, (4)
Thus, the second loop path gain is -G jG jG ^G j/f^ Determine the third loop path gain from Figure 1.
- H i
Gi
Gs
Figure 6
Step 7 of 10
From Figure 6, write the third loop path gain. /, = -G ,G ,G ,G * /f, (5)
Thus, the third loop path gain is -G jG jG jG ^Z/j Determine the fourth loop path gain from Figure 1.
G\
Step 8 of 10
From Figure 7, write the third loop path gain. U = -G ,G ,G jG */f^ (6)
Thus, the fourth loop path gain is -G fijG ^G JH ^ Determine fifth loop path gain from Figure 1.
Step 9 of 10
From Figure 8, write the fifth loop path gain. ... (7)
Thus, the fifth loop path gain is —G4H2 Determine the sixth loop path gain from Figure 1.
- H
2
Step 10 Of 10 ^
From Figure 9, write the fifth loop path gain. G , / 7 j... (8)
Thus, the sixth loop path gain is —GjJVj Consider mason’s gain formula.
P ,* P l (9)
Where,
p^,P2 is the fonvard path gains. is the loop path gains.
Substitute Equation {1),(2),(3),(4),{5),(6),(7) and (8) in Equation (9).
y (» )_____________________ g,G,G.G.+G,G,G.G.
I+ G1G2G4G4/ / } ^-G iG j^tG ^/f^+ G |G )G jG 4 /f) + G|G}G5G4/ f4 -\-G^H2 -¥G^H2 _______________ G |G ,(G ,G , + G ,G ,)_____________
1 + / / , ( G , + G ,) + G ,G ,( G ,G , + G , G , ) ( / / , + f f .
Problem 3.25PP
For the electric circuit shown in Fig., find the following: (a) The time-domain equation relating i(t) and v\ {t) ;
(b) The time-domain equation relating i(t) and v2 {t) ;
(c) Assuming all initial conditions are zero, the transfer function V2 {s) 1^1 (s^ and the damping ratio ^ and undamped natural frequency ojn of the system;
(d) The values of R that will result in v2 (fj having an overshoot of no more than 25%, assuming v1 {t) is a unit step. L = 10 mH, and C = 4 fJF.
(d) The values of R that will result in v2 {t) having an overshoot of no more than 25%, assuming v1 {t) is a unit step. L = 10 mH, and C = 4 fJF.
Figure Block diagrams
Step-by-step solution
step 1 of 10
The following is the given electric circuit:
L R
Step 2 of 10
(a)
Apply KirchhofTs current law to the input loop.
v . ( o = i ^ + « < ( o 4 r ' ( ' > *
Thus, the time domain equation relating and V |(r)is ,
step 3 of 10
(b)
From the circuit in Figure 1, the voltage across the capacitor is,
Thus, the time domain equation relating and v ^ {t) is.
Step 4 of 10
(c)
Apply KirchhofTs current law to the input loop.
Apply Laplace transform with zero initial conditions. y,{s) = s U ( s ) + R I(s )+ -^ I( s )
The capacitor voltage is.
Apply Laplace transform with zero initial conditions.
/ ( i ) = C t F , ( i ) (2)
step 5 of 10
Substitute equation (2) in equation (1).
Rearrange the terms to obtain the transfer function.
s*L+sR+— C 7t~2 r n L s^ + s— + — L I C ) J £ _ ^ R * V (5)
Thus, the transfer function. ( of the system is f ' . M
J £ _
step 6 of 10
R 1
Compare the characteristic equation + s — h— with the standard second-order L LC
characteristic expression. 5* •
2^i». - j
— = o f LC "
Thus, the undamped natural frequency, of the system is 1 T i e
Step 7 of 10
Substitute j for a>^ in 2^e>,
s L C * 1lC ~ L RyfZC 2L _ R [ C ' i ' l l
Thus, the damping ratio, ^ of the system is R C
2 V l
Step 8 of 10
(d)
The maximum peak overshoot is.
Where,
^ is the damping ratio which lies between 0 and 1. For maximum overshoot of 25%,
0.2s = e ~ ^
—13863 = —
Step 9 of 10
1.3863 s
Determine the value of ^ .
C 1.3863
V > - « " * - S — = 0.441
f = 0.441^1
Square the equation on both sides, < -* = 0 .4 4 l’ ( l - { - * )
f ’ = 0 . 1 9 5 ( l - f “) = 0 .1 9 5 - 0 .1 9 5 f ’ 1 .1 9 5 f’ = 0.195
Step 10 of 10
Simplify further to obtain ^ . 019 5
^ 1.195 -0 .1 6 3 2 ^• = 0.4
Write the expression for damping factor, ^ .
f-lJ!
The condition to be satisfied is,
* J ^ > 0 . 4 2>Il
Substitute 4 x 1 0 “*^°*^ C 3^d 1 0 x 1 0 “*^°*^ X . ^be equation.
- [ ■ 2V1 4x10-* 10x10-“> 0 .4 ^ ( 0 . 0 2 ) > 0 .4 j, > ( 0 . 4 ) ( 2 ) R > 4 0 Q
For the unity feedback system shown in Fig., specify the gain K of the proportional controller so that the output y(t) has an overshoot of no more than 10% in response to a unit step. Figure Unity feedback system
« *)o
Step-by-step solution
step 1 of 3
Step 1 of 3
T h e fbUowiiig is tfie b lo ck diag ram a u n ity feed b ack system :
Step 2 of 3
T h e tra n sfe r fu n ctio n o f th e sy stem is. K r ( j ) s(s+ 2 ) R (s) = - i ( i + 2 ) K ~ s (s + 2 )+ K K ^ + 2 s + K T h e tnaxiim im p e a k o v ersh o o t is»
W h e re ^ is th e d a n ^ n n g ratio v rfu d i lies b etw een 0 1. F o r m ax im u m o v e rsh o o t o f 10%, M . 0 . 1 = 8 ^ S o lv e fo r ^ 2 .3 0 2 6 = Q 2 3 0 2 6 - r^ — = 0.7 3 2 9 ^ = 0 . T 3 2 9 4 \ - C Square o n b o th sides, ^ = 0 . 7 3 2 9 ' ( l - ^ ) ^ ’ = 0 . 5 3 7 ( 1 - ^ “) ^ = 0 . 5 3 7 - 0 .5 3 7 ^ ^ 1.537^^ = 0.537 Sim p lify fiirth er to o b tam ^
^ 0-537 ^ 1.537 ^ = 0 . 3 4 9 4 ^ = 0 . 5 9
Step 3 of 3 ^
C o tn p a te ^ + 2 s + K w ife th e stan d ard second-order aqw essioii. 2 ^ < n = 2 K = a t S ubstitute 0 .5 9 fo r ^ in 2 ^ 0 1 = 2 2 ( 0 .5 9 ) 0 ,. = 2 1 0 .5 9 ^ 1 .6 9 5 T h e co n stan t K is, K = e i = 1 .6 9 5 ' = 2 .87
T h u s, fee gatn K o f f e e pro p o rtio n al con tro ller to h a v e a n o v ersh o o t o f m o re than 10% is Io<j: <z8 ^ .
Problem 3.27PP
For the unity feedback system shown in Fig., specify the gain and pole location of the compensator so that the overall closed-loop response to a unitstep input has an overshoot of no more than 25%, and a 1% settling time of no more than 0.1 sec. Verily your design using Matlab. Figure Unity feedback system
C om pensator P h o t
Step-by-step solution
Step-by-step solution
s te p 1 of 14 ^
Consider the unity feedback system shown in Figure 1. C o m p e n s a to r P la n t K 5 + a 100 *+ 2 S r y ( j ) Figure 1 Step 2 of 14
Find the transfer function for the system shown in Figure 1.
________ lo o y '( j+ o ) ( s + 2 5 ) + 1 0 0 X ___________ lOOK^__________ % ’ + » (o + 2 5 )+ (2 5 a+ 1 0 0 A :)
Step 3 of 14 ^
Write the standard form of transfer function for the second order system.
^ ---=■ s* + 2 ( ;6 } ^ + e i
Compare the denominator term: ^*+ j(a+25)+ (25fl+ 100A T ) with the denominator in the standard form: +2^Q>^+es^
Compare the coefficient of s terms. 2 ^< » .= (a+ 2 5 )
Compare the coefficient of constant terms. « ^ = (2 5 o + 1 0 0 X )
Step 4 of 14
Consider the following data: The peak overshoot, £ 0.25 The settling time, ^0 .1 sec Write the formula for settling time.
Re-write this expression. _ 4.6
f ® .= -
Substitute Q.l sec for t, in this expression. . 4.6
b46
Step 5 of 14 ^
Substitute 46 for in the expression; 2^(8), = ( a + 2 5 ) 2 (4 6 ) = (a + 2 S )
0 + 2 5 = 92 0 = 67
Write the formuia for peak overshoot.
4 / , = e ^
Take natural logarithm on both sides of this expression.
b > M = 7 ’^
Square on both sides of this expression.
1 r r ' .
Simplify further.
1 « ‘ + ( \ a M , 'f
, I
Substitute OJ25 for in this expression.
I (l"0-25)’ ’ ~ \ ) r ’ +(ln0.25)’
= 0 .4
Step 6 of 14
Substitute 0.4 for ^ in the expression: <i«i|.=46 0.4a>.=46
® '° 0 . 4 ®, = 115
Substitute 115 for or^ and 67 for o in the expression: =(2So+100/C )
115= = 25(67)+100Jf 100/: = 13225-1675
11550 100 K = 115.5
Thus, the gain of the system, is [ i i l s l
Step 7 of 14
Determine the pole location of the compensator.
Clearly, from Figure 1, the compensator has a pole aX s ^ - a Substitute for a in the expression: s ^ - a
s = -6 7
Thus, the pole location of the compensator is [5 a - ^7| .
Step 8 of 14
Consider the transfer function:
iooa: + s ( n + 2 5 ) + ( 2 5 a + 1 0 0 /:)
Substitute 115.5 for J f,a n d 67 for a in this transfer function.
» W = :t
100(115.5)
i ’ + s(6 7 + 2 5 )+ (2 5 (67)+100(115.5)) 11550
” i * + 9 2 j+ 1 3 2 2 5
Write the expression for transfer function.
y ( » ) 11550 « ( s ) “ * '+ 9 2 8 + 1 3 2 2 5
Step 9 of 14 ^
The input is unit step. That is r ( , ) = « ( /) Find the Laplace transfonn of unit step input.
M * ) = i
1 . X y(s) 11550
Substitute —for in the transferfunction: —r 4 = “ s---
* ' ’ R(s) *'+928+13225 r { s ) 11550 1 “ * '+ 9 2 i+ 1 3 2 2 5 y ( s ) = _____ 11550______ ^ ' * (* '+ 9 2 * + 1 3 2 2 5 ) Step 10 of 14 A
Consider the following function: " 5 5 0 ' ' *(*'+928+13225)
Use partial fraction approach to detennine the response H O
11550 J Bs+C * (* '+ 9 2 * + 1 3 2 2 5 )” * * '+ 9 2 * + 13225 11550 = X (* '+ 9 2 * + 1 3 2 2 5 )+ * (« * + C ) 11550 = .4 * '+ 9 2 .4 * +13 2 2 5 . 1 + & '+ C8 11550 = 4 * ' + B * '+ 9 2 .4 * + C 8 + 1 3 2 2 5 4 Simplify further. 11550 = ( 4 + f l ) * ' + ( 9 2 4 + C ) * +13 2 2 5 4 Compare the coefficients of 5*
Compare the coefficients of s 924< + C = 0
Compare the coefficients of constant tenris. 1322544 = 11550
v4 = 0.873
Substitute 0.873 for in the expression: 92*4 + C = 0 9 2 (0 .8 7 3 )+ C = 0
C = -80.316
Substitute 0.873 for ^ in the expression: *4<(>SsO 0.873+B = 0 5 = -0.873 Step 11 of 14 Simplify further. , , 4 Bs+C ^ * '° * ^ * '+ 9 2 * + 13225
Substitute 0.873 for -0.873 for and -$0,316 for C >n this expression. w V - 0-873 ^ (-0 .8 7 3 )^ -8 0 .3 1 6 s ^ j*-h92j-Fl3225 0.873 0.873s 80.316 s s*+ 92j + 13225 s*+92j + 13225 0.873 0.873s 80.316 s ( s + 4 6 ) +(105.4) ( s + 4 6 ) +(105.4) Step 12 of 14
Consider the following Laplace transform pairs.
------ " »c~* QO&bt ( s + a ) ^ + 6 *
b LT ^
« "* s in 6 / ( s + a ) ^ + d *
Apply inverse Laplace transform on both sides of the following expression.
y / j V _ f 0-873^ f 0-873S
\
r 80.316Y 105.4 * J [(*+ 4 6 )*+ (1 0 5 .4 )'J I 105.4 J [(* + 4 6 )’ +(105.4)' ,-if0.873'| 0.873* 1 ' I * J [(* + 4 6 )'+ (1 0 5 .4 )'J 80.316) / 105.4 1 . 105.4J
[(* + 4 6 )'+ (1 0 5 .4 )'J i - { y ( * ) } = 0.873tt(»)-0.873e-“ cos(105.4»)' s te p 13 of 14 ^ Simplify further.>'(/) = 0.873i((/)-0.873e-“ cos(105.4»)-0.762e'“ sin(105.4») Write a MATLAB code to get the plot of H O
» t = 0:0.01:0.14;
» y = 0.873 - 0.873.*exp(-46.*t).*cos(105.4.*t)- 0.762.*exp(-46*t).*sin(105.4.*t);
» plot(t,y);
The plot of output ^ ( / ) is shown in Figure 2.
Figure 2
Verify this design using MATLAB. Consider the following function:
" 5 5 0 ' ' * ( * '+ 9 2 8 + 1 3 2 2 5 ) Write the MATLAB code for this function. » numH = [11550]:
» denH = [1 92 13225 Oj; » sysH = tf(numH,denH); » impulse(sysH):
The result of this MATALB code is a plot shown in Figure 3.
Step 14 of 14
Hence, the plot obtained in MATLAB is similar to the plot obtained by theoretical calculations. Thus, the design is verified using MATLAB.
Suppose you desire the peak time of a given second-order system to be less than tp.. Draw the region in the s-plane that corresponds to values of the poles that meet the specification tp < tp.
Step-by-step solution
step 1 of 2 Second order sjrstem.
G=- j . 2 s = - ^ o ^ ± c v 7 i7 _ n _ n '* o c i V r ? S=-5o)„±0>4 => — < v *p *p step 2 of 2 - I X + -
Problem 3.29PP
A certain servomechanism system has dynamics dominated by a pair of complex poles and no finite zeros. The time-domain specifications on the rise time
(tr), percent overshoot {Mp). and settling time (fs) are given by tr< 0.6 sec,
M p< 17%. ts < 9.2 sec.
(a) Sketch the region in the s-plane where the poles could be placed so that the system will meet all three specifications.
(b)
Indicate on your sketch the specific locations (denoted by that will have the smallest rise time and also meet the settling time specification exactly.(b)
Indicate on your sketch the specific locations (denoted by that will have the smallest rise time and also meet the settling time specification exactly.Step-by-step solution
step 1 of 7
A certain servo mechanism system has dynamics dominated by a pair of complex poles and no finite zeros.
The time domain specifications on the rise time (fr). percent overshoot (Mp), and the settling time (fs) are given by t, 5 0.6 s, ^ 17%, /, £ 9.2 s.
(a)
Consider the percentage peak overshoot is ^ 17%. Write the expression for percentage peak overshoot.
xl00%
Substitute
17%
for %PO tha equation.17% *
0 M = exl00%
K . . 1 7 ) = - ^-1.7719*-
Step 2 of 7Apply squaring on both sides of the above equation. (9 .8 6 % fM R a ^ V
3.1398*^
3.1398-13.009f’
=0.24135
f = 0.5
Step 3 of 7Determine the damping angle.
= cos"' (0.5)
= 60®
Determine the real part of dominant pole of complex pole.