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Problem 3.20PP
Find the transfer functions for the block diagrams in Fig.
Figure Block diagrams
Step-by-step solution
step 1 of 19
(fli) T he fbUowiiig is Ihe givefi blodc diagram:
Step 2 of 19
F ro m F ig m el.
T he negative unity feedback transfer fim ction is.
q
i + q ( i )
q
i+q
I l+ G W j
step 3 of 19
F ^ u re 2 is d ie reduction diagram o f F igure l.
Step 4 of 19 ^
From Figure 2.
T he transfer function is.
R 1+G i
q+q,(i+q)
i+q
^ q + q + q q
i + q
Therefore, d ie transfer function o f the given b lock diagram is
R ~ l+ O j
Step 5 of 19
(b) T he following is d ie given blodc diagram:
F ig u re s
Step 6 of 19
First fitMltfig feedbadc paths and detennine transfer functions. F igure 4 is d ie 
m odified blodc diagram o f F igure 3.
Step 7 of 19
Transfer fonctioa for first feedback is H, = — ^ _ and
i+qq
Transfer function for second feedback is H ^ = — — — .
* I+ G 4G 5
Step 8 of 19 ^
Figure 5 is the reduction diagram o f F igure 4.
Cascade path 
F igure 5
Step 9 of 19
W e sinyilify d ie b lock diagram b y reducing d ie parallel cofnbination o£ d ie 
controller path (cascade padi). The resultant is show n in Figure 6.
/Jo- G .G jG .G . + ( ^
( l + q C j X l + G . G , )
F ig u red
Step 10 of 19
From the above block diagram , the transfer function becom es
r qqq,G. +0,
R ( l+ G iG ,) ( l+ G ,G 5) ’
qqq.q r+q1+G^G2 +G 4G5 +G i020^0^
+ -K ^ (l+ G iG 3+G 4G5+ 6 0̂ ,0405) 
1+ Q G j +G 4GS+QG3G4GS
Therefore, the transfer function o f the given b lock diagram is
~r \ __________ 1 + G ^G a + G 4G ^ +G ^G aG 4Q 5___________
Step 11 of 19
(c) G iven b lock diagram is
F ig u re ?
Step 12 of 19 ^
First fitidtfig feedbadc padis and detennine transfer functions. F igure 8 is d ie 
m odified blodc diagram o f F igure 7.
Feedbackl Feedbaclc2
F ig u re s
Step 13 of 19
Transfer function for first feedback is Hi = - ^ - and
1+G ,
Transfer fonctioa for second feedbadc is .
' l+ G ,
Step 14 of 19 ̂
F ^ u ie 9 is d ie reducdon diagram o f F igure 8.
Cascade pathl Cascade path2
F igure 9
Step 15 of 19 ^
W e sim plify d ie blodc diagram b y reducing d ie parallel com bination o£ d ie 
contrcdler path (cascade path).
Resuttant equation for cascade p a d il is 
Resultant equation for cascade p a d il is
OiOiOj
1 + G ,
G4GS
I+ G 4 ’
Step 16 of 19 ^ 
F ^ u re 10 is the reduction d ia ^ a m o f F igure 9.
F igure 10
Step 17 of 19 ^
From th e above block diagram , the sum m ing equation becom es
. G.GjG, ^
1+G. (1+G,)(1+G.)
 ̂ G .G ,G ,( l+ G ,)+ q C ^ q G ,G ,
(l+Q X l+G ,)
. g|GsG,(i+Gi)+qGiqq,Gi 
l-^G ^+G ^+G ^G t
Step 18 of 19 ^ 
F ^ u re 11 is the reduction d ia ^ a m o f F igure 10.
Step 19 of 19
From the above block diagram , d ie transfer function becom es
r G ,G ,G .(l+Q ,)+G i