1 (100)

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Problem 4.03PP
Compare the two structures shown in Fig. with respect to sensitivity to changes in the overall 
gain due to changes in the amplifier gain. Use the relation
as the measure. Select H1 and H2 so that the nominal system outputs satisfy 
F1 = F2, and assume /CH1 > 0.
Figure Block diagrams
I h p ' - — { ! ] — Q —
^ T— ----------1
Step-by-step solution
step 1 of 10
S = -
I t is given diat.
The eatpression to m easure se n sitiv i^ is, 
r fh iF
F d K
W here.
K , the feed forw ard gain 
, the ou^nit the system 
iS, fee m easure o f sensitivity
Step 2 of 10 ^
The U o d c d ia g rm o f lh e first structure is feow n in Figure 1.
F igure 1
F igure 2
Step 3 of 10 ^
The U o d c diagram o f the second structure is show n in F igure 2.
Step 4 of 10
A ssum ing fee intetm ediate outyut from fee felt b lock in F i^ n e 1. Calculate th e outyut 
the blodc diagram in F ^ u re 3.
F ig u re s
Step 5 of 10 ^
Calculate the gam o f fee b lock left blodc in F igure 3.
7 ^ = - ? ^ (1)
' \+K H ^
C a l c u l i the gain o f fee blodc r ^ i t U o d c in F igure 3.
( F , -F ^ , )K = F ,
F ^= F^{y+K H ^)
7 ? = - ^ C )
' 1 + ja r ,
Thus, com bine equation (1 ) a n d ^ ) to obtain fee iityut output relation.
p -____ K ( SK ^
.... ®
Step 6 of 10 ^
Calculate fee gam o f fee b lock d io w n in F ^ u re 2.
( 7 J -7 r^ ,) ( ^ : ) ( J i: )= F ,
7ac’ =7rj( i+J i:*77 j)
.......
step 7 of 10
n a I t i» i^vcu uiot, luc MMpuw wi »
th e r e la t io n b e tw e e n .? ] a n d ? 2 -
7!?77j =(l+JE H i)“ - I 
' ' K
(5)
Step 8 of 10
The m easure o f sensitivity due to changes in fee am plifier gain is,
' F [d K
d
f—[ l + K H j
( K
J » ]
( (l+ ^ ,)^ (7 i)"| 2K {l+ X H ,f-2 {l+ K H ,){H ,)(K ^)
[ J[ .
(i+Jsar,)’ (2A:+2is:’7r, -2 k ĥ )
" W(l+7SSf,)‘
Thus, the m easure o f sensitivity due to changes in fee an ^ lifie r gain for fee first blodc is,
S j = — 2— ....... (fi)
' l+TSK
Step 9 of 10
The m easure o f sensitivity due to changes in fee am plitW gain is, 
' 7̂ 1 l iK
{ (l+JC^7r,)(77)Y 2 K (1 + K ^ H ^ )- (2 K H ^ ){k ^ )
2 K
( K ) { l+ K ^ H ^ )
Thus, the m easure o f sensitivity due to c f a a n ^ in fee antylifier gain f iv fee first b lock is,
2
S f .
tu tefe
l+ K ^ H ^
Substitute fee value ? 2 from equation (̂ ) in feis equation to obtain the
2
s f =
_______ 2_______
i+ ( i :^ 7 r j‘ + 2 K H ;)
2
2
■ m
H en c^ from equations (6) and (7) observe feat, fee sensitivity g a in c h a n ^ fo r fee
second block is related to sensitivity o f first block by.
Step 10 Of 10 ^
Therefine, as J i? ] is greater fean zero, w e can o rach ide feat fee second blodc is less 
sensitive fee first blodc.