bode cutler

7/29/2019 Bode Cutler

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ac circuitanalysis

with illustrativeproblems

Phillip CutleEducation ResearchAssociate

anOrange Coas

CollegCosta Mes

Californi

McGraw-Hill Book CompanNew YorSt. Lou

San FranciscDusseldoJohannesburKuala Lumpu

LondoMexic

MontreNew Del

PanamRio de Janeiri

SingaporSydneToron


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A need f r equen t l y arises to e x a m i n e th e f requency re -sponse (bo th a m p l i t u d e and phase) of a c i r cu i t or a m p l i -fier . This is p a r t i c u l a r l y t r u e in m o r e advanced coursesw h e r e a t t e n t i o n is focused on t h e behav io r of feedbackampl i f ie rs or c o n t r o l systems. Hence i t is advisable toi n t r oduce t he sub je c t at an e l e m e n t a r y level so t h a t wew i l l be on f i r m e r g r o u n d in t h e f u t u r e .

1 the Bode plot11-1 The transfer functionMost an y c i r c u i t or system t ha t we might consider hassome kind of i n p u t and o u t p u t . Gene ra l l y we are in -terested in o u t p u t v o l t a g e versus i n p u t vo l t age or vol tagegain as a f u n c t i o n of f r equency , b u t th is need no t be t h ecase. We co u l d j u s t as wel l be interes ted in cu r ren t ga in ,or o u t p u t vo l t age ve r sus i n p u t c u r r e n t , and so on. W h e ndea l ing w i th c o n t r o l sys tems we m i g h t be interes ted inm otor speed versus a r m a t u r e vol tage , or height of l iquidin a t ank ve r sus a p o t e n t i o me t e r setting, etc. In anyevent , th e o u t p u t versus i n p u t r e l a t i onsh ip as a f unc t ionof frequency is , for o u r purposes, cal led t he t ransferf u n c t i o n . The t rans fe r func t ions o f most conce rn to usare exc lus ive ly elect r ical in n a t u r e and genera l ly avo l t age gain.

There are a few dif fe rent m et ho ds of p lo t t i ng t ransferfunc t i ons versus f requency . One of t he s imples t and m ostuseful is called th e Bode p l o t , af ter a f a m o u s scientist w hodid some outs tanding research in feedback theory . I tt u r n s o u t t h a t to convenien t ly cons t r uc t a Bode p lo t , thet rans fe r func t ion is best expressed in cer tain algebraicf o r m s w h i c h are readily conveyed by example .11-2 The low-pass filterThe first c i rcu i t we will develop th e t ransfer func t ionfo r is th e low-pass fil ter of Fig. 11-1. We seek th e voltage

c?i E

jE! 1'"'-'i

2 ,1" ~ RC&Low-pass filter

FIGURE 11-1

gain A y =E;/Ei as a func t ion of f requency. Clearly wehave a vo l t age divider here and hence


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w e can t r an s f e r y f r o m n u m e r a t o r to d e n o m i -and vice versa s i m p l y b y ch an g in g it s s ign, s incecor r e s pond s to t h e s i m i l a r t r a n s f e r of an ang le inf o r m .F i n d i n g a c o m m o n d e n o m i n a t o r and s i m p l i f y i n g ( 1 ) ,o b t a i n

..,- I + JDRC 1 + jw,,

T , = RCw i t t u r n s o u t t h a t t h e p r o d u c t R C is cal led a t i m eand it h as t h e d im en s io n s of seconds , s in ce 1

V A/secA"RCw i t is c o n ve n i e n t to reason t h a t t h e t i m e c o n s t a n t T ^t h e p e r i o d c o r r e s p o n d i n g t o a f i c t i t i ous r ad ian fre-

c o , , so t h a t( =

T A =he us e of t h e h su bsc r ip t wi l l be clar i f ied l a t e r . S u b s t i t u t -

(6 ) i n t o (2 ) yields , for t h e l o w - p a s s f i l te r ,-E , ( 1 1 - l f l )1l

c - > h hh o w s i m p l e th is express ion is and, in pa r t i cu l a r .it s f o r m . I n p o l a r f o r m w e s h o u l d express (7 ) as

E,E > yl +(///J2/tan-///,

1 /-tan-1///, ( 1 1 - l f e )v ' l + ( / /A ) 2reasons to b ec o m e apparent la ter, the f requency /,,led th e comer or break f requency and it is e x t r e m e ly

in cons t ruc t i ng Bode plot s .t m ig h t be interes t ing to correlate the m a t h e m a t i c a l11-la (o r 11-lfc) w i t h a q u a l i t a t i ve analysis of t h eat lo w frequencies Xc is so much largerR t h a t t h e mag ni tud e of E^ s1. Correspondingly( 1 1 - l a ) for/approaching zero , or m o r e significant lyt h e y t e rm approaches zero and A,, s1/0.a s imi l a r m a n n e r , th e t i m e cons tant L / R in an induct iv e c ircui t hasdi m e n s i on s of seconds, since

VL H A/secJ"'a=~v7A=sec

A s/increases , w e w o u l d expect Ay t o decrease ae x h i b i t some p h a s e lag since E; is t a k e n off t h e capac iand c a p a c i t o r vo l tage lags capac i tor cu r r e n t by 9I n s p e c t i o n of E q . ( 1 1 - l f e ) [or E q. ( 1 1 - l a ) ] ind ica te s t hwhen/ t h e f r e q u e n c y of t h e i n p u t s ignal \, is t he sam e/,,, t h e q u a n t i t y ///,, = 1 and h e n c e t h e m a g n i t u d eA, = 1\ 2 = 0.707 and th e phase an g l e is 0 = -tan1 = -45s.

As/becomes ve ry large, an d in p a r t i c u l a r f o r / > /t h e y t e r m in E q . ( 1 1 - l a ) a p p r o a c h e s i n f i n i t y , so t hA ,,sI 'y 'x = -jO, w h i c h means th e gain magn i tua p p r o a c h e s zero an d t h e gain phase is 9 0 . T h i s is toexpected, s ince at high f r equencies Xc -^ R . so t ha t tcircui t i m p e da n c e to E , i s essential ly resi s t ive and hent h e c u r r e n t t e n ds to be in phase w i t h E i . H o w e v e r , tha p p r o x i m a t e l y zero-degree p ha s e angle of t h e cu r r e n tadded to t h e 90 phase la g (-90) of t h e capac i toryield t h e p ha se of E;. Hen c e E.; lags E i b y ap p r o x i m a t e90 at/ ? > /,.O f course , w i t h Xc a p p r o a c h i n g zero in th i g h - f r e q u e n c y s p e c t r u m ( / > / , ) , X c < R and hence tm a g n i t u d e of ^ approach es ze ro . T h us th e gain magnt u d e approaches zero.N ow w o r k t h r o u g h p r o b l e m s PS 11-1, PS 11-2, aPS 11-3.11-3 The high-pass filterA n o t h e r circuit c o m m o n l y encoun te r ed is t h e high-paf i l te r of Fig. 11-2. T h i s circuit is c o m m o n l y used asc o u p l i n g n e t w o r k in m a n y ampl i f ie r c i rcu i t s where

H -Tc: R e .

. f

E, . 'T2 El - 'T&

High-pass filterFIGURE 11-2

c o n ta in s a dc as well as an ac com ponen t . U suallyw a n t o n ly t he ac component to appear at the outpT he capacitor C, also called a blocking capacitoassures t h is , since only a t ime-varying vol tage can caucapacitor cur r en t to f l o w . 2W e can develop t h e vol tage t ransfer funct ion by tvo l t a ge d iv id e r re la t ionsh ip and wri t ing

R-E ,) A, x ^ _-j X c i - y R wCR2

Recall that i, = C- .at

ac circuit analysis


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we le tI

C ~ R^can r e w r i t e ( 1 ) asA , ( 1 1 - 2 ^ )

1-7( 3 ) is p e r f e c t l y v a l i d , b u t i t is m o r e d e s i r a b l e

f o r r e a so n s t o b e c o m e a p p a r e n t l a t e r ) t o ge t r i d ofn e g a t i v e d e n o m i n a t o r t e r m . T h i s m a y be a c c o m-by m u l t i p l y i n g t h e n u m e r a t o r an d d e n o m i n a t o r( 3 ) by j f f i . I f y o u do t h i s a n d r e a r r a n g e t e r m s , t h e r e^ ( i i - 2 f r )A, =

'7,m ay a l so b e w r i t t e n asA, = j ' 1 1 = 90-' -tan"' / /,v 1 + ( / ^ ) 2 /

N o w w o r k t h r o u g h p r o b l e m s PS 1 1 - 4 , PS 1 1 - 5 . aPS 1 1 - 6 .N o t e t h a t , in a l l t h e p r o b l e m s , w e h a v e t r i e d t op r e s s t h e t r a n s f e r f u n c t i o n so t h a t i t c o n t a i n s t e rof t h e f o r m j x a n d ' o r 1 + j x . T h i s w i l l b e p o ss iw h e n e v e r w e d e a l w i t h c i r c u i t s c o n t a i n i n g a s inge n e r g y s t o r a g e e l e m e n t s u ch as j u s t o n e c a p a c i t o r o r oi n d u c t o r .

11-4 The decibelT h e t r a n s f e r f u n c t i o n s e n c o u n t e r e d in p r a c t i c e w i l lg e n e r a l c o n t a i n f a c t o r s in t h e n u m e r a t o r an d d e n o mn a t o r . T h e r e f o r e e v a l u a t i o n of t h e t r a n s f e r f u n c t i o n ap a r t i c u l a r f r e q u e n c y r e q u i r e s t h e p ro ces s o f m u l t i p l it i o n a n d / o r d i v i s i o n , n e i t h e r of w h i c h is c o n v e n i e np e r f o r m e d b y g r a p h i c t e c h n i q u e s . Since a g r a p h i c d i s pof t h e t r a n s f e r f u n c t i o n is o u r g oa l . w e seek an a l t e r n a ta p p r o a c h .Y o u w i l l r eca l l t h a t w i t h t h e aid of l o g a r i t h m s ,o p e r a t i o n s of m u l t i p l i c a t i o n and d i v i s i o n ar e c o n v e rt o a d d i t i o n an d s u b t r ac t i on , s i nce1 ) log a f r = log a + log h ( l l -

( H - 2 c )( 4 ) a p p e a r s m o r e c o m p l i c a t e d t h a n ( 3 ) . i teas ie r t o Bo de p l o t , w h i c h a f t e r al l is o u r

c o r r e l a t e ( 5 ) w i t h a q u a l i t a t i v e ana l ys i s of Fig. 1 1 - 2 ,m a y r eason as f o l l o w s . In th e l o w - f r e q u e n c y s p e c t r u m ,t h e n u m e r a t o r m a g n i t u d e s0 and t h es1 . Hence f o r / < $ / ( , A ,, s0.w e s h o u l d e x p e c t t h i s . since A "c > R at l o wT h e q u e s t i o n is wha t c o n s t i t u t e s l o w f r e -T h e a n s w e r is t h a t /in th e high-pass o r /,,h e l o w - p a s s f i l t e r f o r m s a c o n v e n i e n t spot f r e q u e n c yt h e l o w and h i g h f r e q u e n c y s p e c t r u m .l o w f r e q u e n c i e s A'c > R, t h e c u r r e n t l ead sb y a l m o s t 90; henc e t h e o u t p u t voltage w h i c h is t a k e nt h e r e s i s t o r s h o u l d b e in phase w i t h t h e c u r r e n t .of t h e phase p o r t i o n of (5 ) indicates that f o rt h e t a n g e n t is s0 and hence t h e net angle is

-0 o r S 90.t /= f t w h e r e /is the corner or break f r e q u e n c y ,= l/v'2 and t h e phase is 90 -45 = 45.n t h e h i g h - f r e q u e n c y s p e c t r u m (/> j \) w e see f r o m( 4 ) t h a t t h e r e a c t i v e t e r m in t h e d e n o m i n a t o r isan d hence A,, approaches 1/0'''. This is

w i t h t h e phys ica l fac ts , f o r X,- < R. E^\ S. F u r t h e r m o r e , t h e c u r r e n t i n t h i s case is essen t ia l l yR and hence is almost in phase w i t h EpE ^ is t a k e n off the res i s to r w h i c h has n o phaseassociated w i t h i t, it f o l l o w s t h a t E; is essen t ia l l yw i t h Ep

log = log a log b ( 1 l - 3b2 )and r a i s i n g a n u m b e r t o a p o w e r c o n v e r t s to m u l t i p l it i o n3) log ax = x log a ( l l - 3

N o w g r a p h i c a d d i t i o n and s u b t r a c t i o n is n o p r o b ls ince w e need o n l y c o u n t d i v i s i o n s o r lay off d i s t a n cw i t h a p a i r of d i v i d e r s . For t h i s reason, a m o n g o theth e ve r t i c a l scale in o u r graphic d i s p l a y wi l l be exp ressin a l o g a r i t h m i c u n i t cal led t h e dec i be l , a b b r e v i a t e d dThe d e c i b e l is f o r m a l l y de f ined as a un i t o f p o w e r gag i v e n b y

A, = 10 log - f l d B'i4)E q u a t i o n ( l l - 4 a ) T is r ead as, "the p o w e r gain Ap ,p ressed in d B, is e q u a l to 10 t imes t he l og of t h e p o wra t i o P ^ / P I . " A more common f o r m of exp ress ing E( l l - 4 a ) is s h o w n in E q . ( H - 4 f c ) :

p(dB)= 10 log^) A ( 11 -4A l t h o u g h E q . (ll-4a) is in better m a t h e m a t i c a l f o rE q . (H -4 f c ) , w h i c h is read and i n t e r p r e t e d in th e saw a y , is more p r e v a l e n t and hen ce b o t h f o rm s wi l lused here.S y m b o l s o t h e r t h a n A are commonly used to des ignthe ga in or ratio b e t w e e n two quantities. For e xampt h e sy m b o l K o r G is o f t en used, and as a fu r the r r e f in

( l l - 4


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th e s y m b o l G is used i f P ^ > Pi or x i f P y , < P\-e w i l l , ho weve r , s t i ck to t h e s y m b o l A i r r e spec t ive ofP ^ o r Pi is t h e g r e a t e r n u m b e r .Since e lec t r i ca l p o w e r is given by P = 2//? o r I 2 R ,e xpr e s s Eq. (4 ) as

A,= 10108^(18= lOlogl^-^dB i c\ ll\OogdBE^R:

^ E , R,= 10 lo g -2 d B + l O l o g d BV l/ 2E* n= 20 log d B + 10 lo g d B

' i "^P I R,A,= 101og-dB= l O l o g d B' i 'i "i^10 lo g

2= 10 lo g d B + l O l o g d B/? ,^= 20 log -2- d B + 10 lo g dB/i R^

E q u a t i o n s (6 ) and (7 ) a re p a r t i c u l a r l y useful i f o neto d e t e r m i n e power ga in (or loss) by m e a s u r i n gand R , o r / a n d R,whi ch i s u su a l ly easier to do t h a n aof p o w e r . H oweve r , m a n y elec t ronic

a r e p r i m a r i l y designed f ro m a s t a n d p o i n t ofo r cu r r en t gain, a n d hence w e drop th e res is t ivein (5) and (6 ) so t h a t th e def in i t ion of vo l t age andgain in decibels are respect ively,E!1

A, = 20 lo g dB = 20 log |/1,| dB ( l l - 5 a )

/ 4 , = 20 log dB = 2 0 1 o g | - 4 i | d B ( l l - 5 f r )The ratio of t h e o u t p u t o v e r the i n p u t is correc t ly

gain when tw o quantities are of s imi la r units.it is common prac t ice to refer to th e o u t p u t -r a t io as gain or system ga in even w h e n t h ey aredimens ional ly . For exam ple , w e migh t cons iderof an opt ica l sys tem as being th e o u t p u t lightof a l amp ve r sus applied vol tage . It shouldbe c lea r tha t ga in may be a number m o r e or lessun i ty . If t h e gain is less t h a n u n i ty , it representsor loss w i t h i n th e n e t w o r k or sys tem.

Note t h a t if the ga in is less t h a n uni ty , th e gain ex-in d B wi l l be nega t ive , since th e logarithm of ais nega t ive . For ex amp le ,

A = 20 lo g\d B = 20 ( log 1 -lo g 2) dBSince t h e l o g a r i t h m of 1 to a n y base is zero, th e a bor e d u c e s to

A = 2 0 ( 0 -l o g 2 ) d B = -20 log 2 d B= -20 (0.301) d B = -6.02dB

N ow w o r k t h r o u g h p r o b l e m s PS 11-7, PS 11-8. aPS 11-9.11-5 The Bode plotA m o s t useful m e a n s of d i s p l ay i n g th e a m p l i t u d e aphase response of a c i r cu i t or sy s t e m is to plotm a g n i t u d e ( a m p l i t u d e response) in decibels versus fq u e n c y as one c u r v e a n d t h e phase ve r s us f r equencya s e p a r a t e c u r v e . The same f r equ en cy axis is usedb o t h c u r v e s , and semi log g r a p h pape r is used so t haw id e f r e q u e nc y range can b e disp layed . C u r v e s w i t h tt ype of disp lay are k n o w n as Bode plots, and t heym o s t usefu l in e l e c t ro n ic c i r cu i t and c o n t r o l sys ta na ly s i s .To fami l i a r ize ourse lves w i t h Bode plots , w e w i l l fc o n s t r u c t th e Bode plo t s of some bas ic t r ansfe r functf ac tors . For t h e m o m e n t w e are n o t conce rned w i t h hthese f un c t io n s are phys ica l ly realized or even i f thcan be . We j u s t wan t to learn th e mechan i c s of cos t r u c t i n g t hem. O n c e these basic f o r m s are masterw e can m o v e on t o m o r e com pl i ca t ed fo rms which e ssen t ia l ly combinations of th e basic fo rms .Bode a l so showed tha t the phase and a m pl i t uresponse fo r m o s t t rans fe r func t ions are unique ly relaand t h a t th e s lopes of th e a m p l i t u d e responseind ica t ive of the phase response . Since it is usuaeasier to m easure amplitude rather than phase , ithooves us to see h o w w e can pred ic t th e phase shiftobse r v i ng th e a m p l i t u d e response .11-6 Basic formsThe f i rs t t r an s f e r f un c t io n fo rm ( fac tor) w e wil l consiis t h a t of A = k where A is the genera l symbo l w eusing to denote a t rans fer func t ion and k is s o m e art r a ry constant, a number which is independent ofquency . Thus if A = 100, w e have A in dB givenA = 20 lo g 100 dB = 40 dB . Similar ly if A = 0.2,have A =10 lo g 0.2 dB s-14 dB . These functioBode plo t as s traight l ines , as s h o w n in Fig. 11-3. Nt h a t there is no phase shift when th e ampli tude responis flat {A =a constant).Next as sum e a t rans fer func t ion of the form1) A = f . / ~ f ( l l - 6\ "c/where a > c is s o m e a rb i t r a ry constant which w e will re


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B o d e plot of a constan tA = *F I G U R E 11-3

s th e comer f r e q u e n cy , -a n d n is an integral exponent4 , . . . , n .3 To B o d e plot th is f u n c t i o n , le t u s firstn = 1 . In th is case th e func t ion is simply. . coA =] CO,

th e j indicates a constan t phase shift of 90of th e ratio C O / C D , ; . Hence th e phase responseas a s tra ight line (dashed) at 90 as shown in Fig.fo r n = 1.h e o n l y w ay we ca n plot th e magni tude (in d B) ofto assume v a r i o u s values of C D relative to o u r arbi-co... C o n s t r u c t i n g a table of values we havec u / c u ;

= 2 0 lo g (cu/to,)6

0.1-20

0.5-6.02

c^f\o

10

26.02

513.98

1020

f u n c t i on of (h i s form is s o me t i me s called a d e r i v a t i v e f u n c t i o n .

These values are plotted for th e n = 1 curve in Fig11-4, w i t h th e d a s h e d l ine for th e phase an d solid linfor t h e a m p l i t u d e .Note an d memorize th e fol lowing, as evidenced iFig. 11-4. For a t rans fer function of th e form in E q. (1wi t h n = 1, th e slope of th e Bode plot is positive an dconstan t e q u a l to 20 d B / d e c a d e s6 dB/octave wheredecade is a frequency ratio of 10, and an octave isf requency span of 2. Th e plot crosses 0 d B at co = co ,Th e phase angle associated w ith th is slope is 90.Now le t n = = 2 in E q . (1 ) so th at th e specific fun ction ware deal ing w i t h is/ c o '7CO:.

I CO^ ' CO


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the Bode plot 155


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^\

^f

'3d8 ^ , . , or u > = 0.5 u > ^ th e er rorth e a c t u a l a nd a p p r o x i m a t e a m p l i t u d e curvesa b o u t 1 d B a nd the phase e r r o r is o n ly about 6,is e s s e n t i a l ly th e largest p h a se e r ro r a t an y poin t .t o c we h a v e th e largest a m p l i t u d e error, which isdB , a n d th e p h a se error is zero since w e d r a w th ea p p r o x i m a t i o n c u r v e (dashed line) so t h a t i t0 a t (Uc/10 a n d 90 a t 10 a c . A s t r a i g h t l inethese tw o p o i n t s will intersect t h e p o i n t= 1, < f > = 45).t 0 ) ^ th e a p p r o x i m a t i o n to t h e a m p l i t u d e responsea b r u p t l y f r o m 0 d B / d e c to 20 d B / d e c orA t a n octave a b o v e a > c , or 2 c u , . , th e a m p l i t u d eis such t h a t th e t r u e response is only a b o u t 1 d B

a nd th e phase e r ro r is about 6 low.

Fo r ( D ~ s > u > c t h e s t ra i g h t - l i n e a p p r o x i m a t i o n a nd a c t uaa m p l i t u d e c u r v e merge for a l l prac t i ca l purposes andth e p h a s e e r r o r is a g a i n a b o u t 6. A s y o u c a n see, thea m p l i t u d e a n d t o a s o m e w h a t lesser e x t e n t t h e phaseresponse can be rea d i l y a p p r o x i m a t e d by me re ly d rawi n g s t r a i g h t l ines wi t h well -def ined slopes. If greatea c c u r a c y is desired, w e c a n use t he easily remembereda m p l i t u d e correct ion f ac t o r s of th e fo l lowi ng t a b l e :

(U/CU

dB< f >

0.106

0.516

130

216

1006

The above correc t ion factors a n d th e general shape ofth e p h a s e a n d amplitude curves sho uld be co mmit t edto m e m o r y .N o w le t us cons ider other values of n in th e functionA- 1+7") ( l l - 8 f l )v w j

w h i c h h a s a n a m p l i t u d e given by2) A^ = n 20 log,/!+ (ftJ/Mc)2 (ll-W

ac circuit analysis


10/30

ase ang le4 > = n t a n Ct J&> , ( l l - 8 c )

You w i l l n o t i c e t h a t t h i s express ion is th e rec ip roca l ofE q . ( l l - 8 a ) , w h i c h we jus t s tud ied . The a m p l i t u d eresponse is given by( 2 ) w e see t h a t the a m p l i t u d e in d B an d phase in

w i l l be n t i m e s the v a l u e s w e j u s t ob t a i n ed forFig. 11-6. Therefore , th e cor rec t ion fac to rs inpr ev i o u s t a b l e m u s t also be m u l t i p l i ed b y n.Bode plo t o f s t r a i g h t - l i n e a p p r o x i m a t i o n s to ( 1 )va lues of n is s h o w n in Fig. 11-7. Since th eslope f or ( y < ^c w as zero, i t w i l l s t i l l be zeroan y v a l u e of n. H o w e v e r , for u > > o ) c tne s lope m u s tm u l t i p l i e d b y t h e a p p r o p r i a t e n. Th u s , for n = 3.s lope for a > > c ' j , . is 3 (2 0 d B / d e c ) = 60 d B / d e c o rd B / o c t . The c o r r e s p o n d i n g phase as y mp to t i c a l l y3 {90') = 270". For n = 3 a t o > = c u , . - eis n ( 4 5 - ) = 1 3 5 ' a n d t he a m p l i t ud e er ro r is nd B ) = 9 d B .o w w o r k p r o b l e m PS 11-12.c o m m o n l y e n c o u n t e r e d f o r m of t rans fe r

is

2) - 4 d B = -"20 lo g ^' 1 + ((y,oj , ) 2 ( l l - 9 f c )The phase response is g iven by

0 = n t a n \-9c)0),A Bode plo t fo r n = 1 is s h o w n i n Fig. 11-8 and, as y o ucan see, the on ly di f fe rence i n t h e a m p l i t u d e response int h i s case and t h a t of F ig . 11-6 is t h a t for C D > o . c sgaindecreases. The slopes f or c o > ( L > ^ in E q . ( l l - 9 a ) for a n yv a lu e of n a re s i m p ly the nega t ive of the slopes for E q( l l - 8 a ) . The cor rec t ion f a c t o r s are also iden t i ca l inm a g n i t u d e excep t t h a t n ow they li e be low the s t r a igh t -l i ne a p p r o x i m a t i o n s . The phase cha rac te r i s t i c and i tsco r re c t i on f a c to r s for an y v a lu e of n is also i den t i c aw i t h the s ingle except ion tha t w e n ow have phase la gr a t he r t ha n lead. T h u s for c o t > (D , . and n = 1, th e phaseangle ap p r o ac h e s 9 0 .

N ow w o r k t h r o u g h prob l em PS 11-13 .T ra n s f e r f u nc t i o ns occas iona l ly c o n t a i n fac to rs ofthe fo rm

0"( l e a d )3 1 5

27 0

225

18 013 590

4 5

0

d860

5 5

50

4 5

4 0

r 35

30

2 5

2 015

10

5

0

.-1 / \ ,, 1 N U W WUIK UllUUgll p lUUH-l l l I J 11-1J.n + j -"-) ( l l - 9 a ) T ra n s f e r f u nc t i o ns occas iona l ly c o n t a i n fa\ " the fo rm

Ae

A M

dn tn 1

+an0

A nPh. CJ-'log

n p l i t u d ease

"-y."c\/\ + (ul

f^0.1 0.5 1 2 10

tJ/OJcF I G U R E 11-7

^ > 2

^--

^.",'

.'

<

-

|

^ ,--//.///^

///A

ity-//

/:'

-

/

i

' f

\/4

>

1

^

'

^

/

)f

t

in-3f

r< - .7n=1

n = 1

y ^ f r7 =

the Bode plot 157


11/30

0

--5003OJ" -10C1-15-20

0

-15

-30

12-45S '

-75

A = 11

+/ C

\

0.

^ 16

11^

0

0.1 0.5 1 2 10fc>/CJc

FIGURE 11-8

N

^6

C

1

S

0.5/

d E

>-S

u^

sS

C

>.

^

1C

//

3

^

^ 2B?

1 d

k\6

"C

s1

s-^

\

^

s

k,

6

\

i"

^

^

11 C

q

0 t c

s^

A =/ a > ^[j +7 -2 .5 -" +I&>.

^ cy^ co,., th e real term being squaredwil l o v e r w he lm th e imaginary term, so tha t

A = / c y '73) A s

+ j2 S w- + 1CO.

(U\ .u>,

1f?{We.

I tO"c.

1 - C OI.C"r. +j28to-co, ( 1 1 - l O f c )Thus the phase angle approaches 180 and th e ampli tudeapproaches zero. In general, for c o > c o , ; , th e amplitude isgiven by

a re called quadra t ic factors and theyarise due to interaction between tw o energyin a circuit o r sys tem.5 If th e circuitso that some formresonance p he nome non occurs, w e can expect toquadra t ic factors in their analysis. Since th eenergy s to r age e lements can be tw o capaci tors , tw o inductors , o r

and capaci tor .

C O 40 log ) A.IB = = 20 logWw CO,I f w e le t co/co,. = 1 in Eq. (4), we have A^y = 0. andfor c o / c o c = 2 w e obtain A^y S 1 2 . Thus th e slope i nth e region of c o > co , ; tends to approach -12 dB/oct,which is essentially th e same as 4 0 dB/dec. Exactlythe same conclusions can be d r a w n fo r (1 ) except thatthere the ampli tude response rises and th e phase leads

ac circuit analysis


12/30

Frequency, radians per se c Frequency, radians pe r se cF I G U R E 1 1-9

c o increases, whereas in ( 2 ) th e a m p l i t u d e decreasesd the phase lags. In bo th cases th e phase a symp to t i ca l l y180, t h o u g h from di f fe rent direct ions.N o w w h a t h a p p e n s at C D = o > < . in ( 2 ) ? Clea r ly th e func-reduces toA = -1-= ^-/-90j20 20i

th e phase at w = co, is -90 in (2 ) or +90 in ( I ) .e co r r e spond ing a m p l i t u d e in (2) isA^ = -20 lo g 26 ( 11 - lOc )

ske tching d e n o m i n a t o r quad ra t i c factors, we have Fi11-9 . N o t e t h a t as 6 (sometimes cal led d a m p i n g factoincreases, th e peak or resonant poin t shifts below aThe same curves m ay be used for n u m e r a t o r quarat ic fac tors i f y ou r em em ber t ha t for o > o c tha m p l i t u d e curve slopes u p w a r d s at 40 dB/dec; ando c , instead of a resonant rise, we m ay see a resonant das s h o w n in Fig. 11- lOa. The phase response is als imi l a r except tha t it approaches 180 of phase lead fu > > c o c . Fig. 1 1 - l O f c i l lustrates th e phase responseth is case.N ow wor k t h rough problem PS 11-14.

of (6) indicates t ha t i f < $ = ^ , lo g 25 = lo g 1=so t h a t A^ =0. If S < , say 6 = j for example ,26 w i l l be the lo g of a number less than 1 which is aq uan t i t y , and in this caseds = - 2 0 1 o g 2 ( ^ ) = -20(-0.3979) s+ 7 . 9 6 d B

we see t h a t th e a m p l i t u d e response actual ly r isesth e 0 d B level for 6 ?, sa yth e lo g is posi t ive and hence the n u m b e r o f d B isw negative. Both th e phase and m a g n i t u d e o f ( l ) ands t rongly depend u p o n th e value of i5 in th e vicinity of= u > c , and it is p robab ly best to plot a few points forHowever, as an aid to

180

Slope40dB/dec 90

( a > WFIGURE 11 -10

the Bode plot 15


13/30

11-7 Constructing the Bode plotIn ge ne ra l , a t r an s f e r f u n c t i o n w i l l c o n t a i n v a r i o u s f a c t o r sin th e n u m e r a t o r a n d d e n o m i n a t o r w h i c h c o n s i s t of t h eba s i c fo rm s w e j u s t learned to pl o t . W e c an u se t h i sb a c k g r o u n d to c o n s t r u c t t r a n s f e r f u n c t i o n Bode p lo t s ofp r a c t i c a l n e t w o r k s . T h e r e a re v a r i o u s w a y s of d o i n gt h i s , a l l of w h i c h a re best i l l u s t r a t e d b y e xa m pl e .A s s u m e w e w i s h to d e v e l o p a n d Bode p l o t t he vo l t a gega i n t r a n s f e r f u n c t i o n of t h e so-ca l led l ag n e t w o r k ofFig. 1 1 -1 la . Fi r s t le t u s deve lop the t r a n s f e r f u n c t i o na n d i n s pec t i t fo r t i m e c o n s t a n t s {R C p r o d u c t s o r L / Rr a t i o s ) w h o s e re c i p r o c a l s d e n o t e c o r n e r f r e que nc ie s . Ifse ve r a l t i me c o n s t a n t s a r i s e in t h e a n a l y s i s , w e c anc o r r e s p o n d i n g l y e xpe c t to h a v e severa l c o r n e r fre-q u e n c i e s to c o n s i d e r .

S t a r t i n g w i t h t he vo l t a ge d iv id e r r e l at i o n s h i p , w e m a yd e r i veR. R^ + l / j c o C ic,

/?! 4- Ri-]Xc , PI +P2+I/.AOC;,

(JojC.R, + 1)(1// )[juC,(R^ + R,) + l ] ( l / y c y C , )__1 + jwC^R^1 +juC,{R^ + R,)

Le tT , = Ci + R^) = 1/cdi , T, = C^R^ = l / w ^ .

w ^ = l O c ^ iE^ _ 1 + j w / W j A/1 )E,,, 1 + J U J / 0 ) i ^

E q u a t i o n ( 1 ) m a y be r e so l ve d i n t o t w o fa c t o r s-E"E.

U)1 +71 +7 QJ ,

The f i rs t f a c t o r w as s t u d i e d in c o n j u n c t i o n w i t h Eq( 1 1 - 8 ) a n d t h e s e c o n d w i t h Eq. (11 -9 ) . A Bode pl o t of tha m p l i t u d e r e s p o n s e o f th e f i r s t f a c t o r ( n u m e r a t o r ) is h o w n as c u r v e .V in Fig. l l - l l f r . The a m p l i t u d e respons

1 60 ac c i rcu i t ana ly s i s


14/30

t h e second f ac to r is s h o w n as c u r v e D . Since t h escale is l o g a r i t h m i c (dB) . m u l t i p l i c a t i o n o f t h eof t hese f ac to rs is e q u i v a l e n t to a d d i t i o n .t h e c o m p o s i t e a m p l i t u d e response i s s i mpl y t h ea d d i t i o n of c u r v e s N and D as s h o w n in

. 4 ^ g = .V + D. A p a i r of d i v i d e r s is m o s t useful inr e g a r d . C o r r e c t i o n f a c t o r s ca n b e ins e r t ed so t h a t ac u r v e can b e d r a w n , b u t ge ne ra l l y t h e s t r a i g h t -a p p r o x i m a t i o n s a re a d e q u a t e .T h e p h a s e of t h e f i r s t and second f a c t o r s h as also beenw i t h s t r a i g h t - l i n e s e g m e n t s as 4 > \ and 4 > oT h e phase a n g l e s m a y also b e a l g e b r a i c a l l y

w i t h a p a i r of d i v i d e r s o r b y c o u n t i n g s q u a r e s in as i m i l a r to t h e a m p l i t u d e r e s p o n s e .W i t h a l i t t l e prac t ice y o u w i l l in most cases be a b l e t ot h e a m p l i t u d e r esponse of A w i t h o u t h a v i n gc o n s t r u c t t h e i n d i v i d u a l fac tors . For e x a m p l e , w i t h> c ' j , in ( 1 ) . w e k n o w t h e f irst b r e a k ( c o r n e r ) occurs atas we go f r o m t h e lo w to t h e h i g h - f r e q u e n c y range .01, is in t h e d e n o m i n a t o r , ,4^3 wil l b r e a k do w n w a r dc y , w i t h a slope of 20 d B / d e c and c o n t i n u e d o i n g s o

c'j;. w h e r e u p o n t h e n u m e r a t o r s t a r t s c l i m b i n g a tT h u s t h e p o s i t i v e n u m e r a t o r slope cancels th ed e n o m i n a t o r s l o p e so t h a t the s lope of A^yo ff at 0 dB/dec.l t h o u g h i t is s o m e w h a t t r i c k i e r , t h e p h a s e responsey be s i m i l a r l y reasoned. To a fi rst a p p r o x i m a t i o n t h ep h a s e b r e a k s at O . l c j i a nd decreases a t45' dec. T h u s a t w , t h e de n o m i n a t o r a n g l e isb u t n o t e t h e n u m e r a t o r p h a s e is c o m i n g i n t o p l a y ,

f c > , = O . l c u , a n d t h e n u m e r a t o r p h a s e is increas ing4 5" dec. T h u s a t o i th e n u m e r a t o r a n d d e n o m i n a t o rslopes cancel , so t h a t t h e phase r e m a in s constan tth e -45 level it a c qu i r e d a t a i . This h o l d s t r ue unt i la t w h i c h p o i n t th e d e n o m i n a t o r phase has gone

i ts m a x i m u m possible sh i f t of 9 0 , so t h a t th eof t h e d e n o m i n a t o r phase angle is zero. The n u m e r -phase slope, h o w e v e r , c o n t i n u e s to c l i m b at 45 /deca decade a b o v e a>t . This p u l l s th e ne t phase slopeu p at 45 /dec a b o v e c o ^ u n t i l 10a , at w h i c h p o i n tn u m e r a t o r h a s gone t h r o u g h i ts m a x i m u m possible

sh i f t of 90. T h u s fo r a j >w^ both n u m e r a t o rd e n o m i n a t o r phase slopes are zero and the nets lope leve ls of f at 0 d ue to t h e -90 in th ecanceling th e +90 due to t h e n u m e r a t o r .

a p h y s i c a l v i e w p o i n t we can reason in Fig.la tha t a t lo w f requenc ies C^ l o o k s l ike an opena nd h e n c e E g = E ,n o r A^ = 0. A s o increases.c o m e s i n t o p lay so t h a t it s ta r ts to load Ry, w h i c hE g. This a c c o u n ts fo r th e break at (a,. A s a >f u r t h e r , C^ e v e n t u a l l y looks l ike a s h o r t and= E , n / ? 2 / ( ^ i + ^ 2 ) - This accounts fo r co; , where C;

out of t h e p ic t u r e .o w w o r k t h r o u g h p r o b l e m s PS 11-15, PS 11-16,1 1 -1 7 .

PROBLEMS WITH SOLUTIONSPS 1 1 - 1 W h a t is th e c o r n e r f req u ency o f t he low-pas sf i l t er in Fig. PS l l - l ? W h a t is t h e v o l t a g e ga in t r a n s f ef u n c t i o n at t h e c o r n e r f r e q u e n c y and at 1 , 1 0 , 1 ,2 , twiceand ten t i m e s th e c o r n e r f r e q u e n c y ?

1 0 kn-VSAr

0.01 A F .100V

F I G U R E PS l l - lSOLUTION The t i m e constan t is1 ) r,, = RC = 10 kQ(0 .01 ^F) = 1 x 10- - 4 secThe c o r r e s p o n d in g radian c o r n e r f r e q u e n c y is

2) o),, = = 10 4 rad/sechThe c o r r e s p o n d in g c o r n e r f r eq u ency is

^h/ = - = 1 592 Hz)The genera l e x p r e s s i o n fo r th e t r ans fe r fu nc t ion b y

E q . ( 1 1 - 1 ) is .Eo 1,/,, 1 + 7AP l u g g i n g in the g iven v a lu es of///,, = 0.1, 0.5, 1, 2, a nd

10 i n t o (4) , we h a v e/ /> ,1 - 4 > - 1[Q

0.10.995-5.71

0.50.894-26.6

10.707-45

->0.447-63.4

100.0995-84.3

PS 11-2 Prove t h a t th e corner f r eq u ency of a low-passf i l te r such as shown in Fig. 11-1 is t h a t f requency atw h i c h Xc =R.SOLUTION A p p l y i n g th e voltage div ider re la t ionsh ipto Fig. 11-1 , w e h a v e

2 -JXcA,=) R -j X c,D i v i d i n g n u m e r a t o r and d e n o m i n a t o r b y j X c andr e a r r a n g i n g terms, w e o b t a i n

I 1A,. = R R1-- 1+Jv.~]AC ^Cthe Bode plot 161


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w e l e t / ? = . V e i n Eq. (2 ) .1) A,. = = 0.707/-45 4) A,. =

. f _fi

+!i s p r e c i s e l y w h a t w e s h o w e d in t h e p r e v i o u sw h e n w e set/=yi,.

1 1 - 3 W h a t w o u l d be t h e c o r n e r f r e q u e n c y of t h eF ig . PS l l - 3 9

S e t t i n g / , 7 , e q u a l to 0.1. 0.5, 1 , 2, a n d 10 yields t h ef o l l o w i n g v a l u e s

/ /,A -,L

0.10.99508 4 . 3 s

0.50.44763.4'

10.70745 -

20.89426.6 s

100.9955 . 7 1 -

PS 11 - 5 D e t e r m i n e t h e v o l t a g e t r a n s f e r f u n c t i o n o ft h e c o u p l i n g n e t w o r k s h o w n in Fig. PS 11-5. Resis torR^ is a n a l o g o u s to t h e o u t p u t r e s i s t a n c e of a v o l t a g es o u r c e d r i v i n g a h i g h - p a s s f i l t e r .

F I G U R E PS 11-3B y i n s p e c t i o n w e se e t h a t R^ a n d R ^ m aye t h e v e n i z e d w i t h t h e v o l t a g e s o u r c e t o s i m p l i f y th ec i r c u i t t h a t t h e c a p a c i t o r l ook s b a c k i n t o .b y a n a l o g y w i t h Fi g . 11-1, i n t u i t i o n , or b o t h ,, = R,|,C = ( / ? i | R^}C. T h e r a d i a n c o r n e r f r e q u e n c y is

u > i , = 1 / T , , .11-4 W h a t is t h e corner f r e q u e n c y o f t h e h i g h - p a s s

in Fig . PS 11-4? W h a t is t h e v o l t a g e gain t r a n s f e rat t h e c o r n e r f r e q u e n c y a n d a t I /10 , 1/2, t w i c e ,n d te n t i m e s t h e c o r n e r f r e q u e n c y ?25>iF

-1 C

^fl;.

F I G U R E PS 11-5

SOLUTION A p p l y i n g t h e vol t age d i v i d e r r e l a t i o n s h i pw e h a v e

R.laE,) A ,. = RI +R^ - j X cioov nj)E,n 1^ E '

F I G U R E PS 11-4T h e t i m e constant f r o m Eq. (11-2) is

) T , = RC = 1 k Q ( 2 5 ^F) = 2 5 x 10-3 se ch e c o r r e s p o n d i n g radian comer f r e q u e n c y is

Divid ing n u m e r a t o r a n d denominator by Ri + R; yieldsRIR+R___2) ^-v=n , D V/ ? > + Pi Xr-JR^ + R ^ J R^ + R

R.R, +R, \-j (oC{Ri + R i)

(o, = =40 rad/secT!

n d t h e c o r r e s p o n d i n g c o r n e r f r e q u e n c y is

I f w e le t th e corner f r eq uency be13 ) co, = C(Ri +Rz )

) y, = -"L = 6.37 Hz271T h e ge ne ra l e xpre s s ion for t h e vol t age t r a n s f e r f u n c -

by E q. (ll-2b) is

w e c a n r e w r i t e (2 ) as4) A,,= ^R , + R ^ 1-7,",C O

ac circuit analysis


16/30

n u m e r a t o r a n d d e n o m i n a t e )./

1 \ - x j l"' i.+R-",^N o t e t h a t fo r /'> /, w e j u s t see t h e

s ince X^ ^ 0.P r o v e t h a t t h e c o r n e r f rea t w h i c h \c = Ri + P^ ln FI f w e d i v i d e n u m e r a t o r an

f Eq . ( 1 ) by R [ + R^ in t h e s o l u t i o nw e o b t a i n^ 1R , + / ? , ' . . X

1 Ri 4.Y^ = R i + ^2 t h e a b o v e e q u a t i o

R 1 /^ 2 - 070" ' (1 0 .L 0 1 ;1 \ PKI + /< 2 1 7 1 \rPS 11-8 E x pFig . PS 11-8 in

v o l t a g e

is t h e11-5.m i n a t o rs r ev i o u s

c

SOLUTIONes to .4, = 20 log) / 4 5 - = 20 log. ,, = -7.95ained b yu s w h e n PS 11-9 Then e t w o r k in Figv o l t ag e g a i n r a ti r c u i t inu en c y .

'E

t0

SOLUTION1 )

2)

3) ^=lc 1Therefore w e seE ^ / E i and w h (log"' i m p l i e stables p o s i t i v e. lated, and hencand s u b t r a c t 1,4) 2- = log-i

O l o g .0(2.063ress t hdB .

i

3 FIG

.10dB''in(0.4) dl88 dBv o l t a gPS 11io E - i / lr"j

Fie

/

-5.67d

i-s [ek the rose logth e antm a n t i se to obso t h a t'(1-

dB = 20 log 1- i nD d B = 41.26 dl

e v o l t a g e g a i n o8 kP .^A / "

12 k^

, L ' R E PS 1

= 20 logD L H ^

e g a i n in c-9 is A, =r )-i

dBA, =-5.67

dSA, =-5.6SURE PS 1

4, = 20 lo

B = 20 lo-5.67\20 )n u m b e r wharithm isi log). Y ousa va lues5ta in a pos

0.2835 -

1-8

12 k00.3979)

1B o f t-5.67

71-9

t - 'l

-' ilog-1

lich re ]-0.28wil l ro n l y ai t i ve n

1 ) = lo

13

>

-

c

hc

?E

iE

iE

(\3r3 Jecr


17/30

Lo o k in g in a t a b l e of l o g a r i t h m s (o r on a slide r u l e ) forh e n u m b e r c o r r e s p o n d i n g to 0.7165, w e h a v e 5.206 a n di t h a cha r a c t e r i s t i c of -1.*

The phase a n g l e is a l w a y s zero, since A eq ua l s a c o n s tan t .R e m e m b e r t h a t a c o n s t a n t a m p l i t u d e respo n se i mp l i e sz ero ph ase shi f t .E!, = 0.5206

S 1 1 - 1 0 C o n s t r u c t a Bode plot for th e a m p l i t u d efor each of t h e f o l l o w i n g t r a n s f e r f u n c t i o n s :. . co50A = 3200 3)

A = 0.004 4) A= (U300;

phase response in each case by i ns pe c t i ont h e a b o v e t r a ns f e r f u n c t i o n s .) A = 20 lo g 3200 = 70.1 d B

m a n t i s s a s f o r n e g a t i v e l o g a r i t h m s m a y be c o n v e n i e n t l yon a sl ide r u l e w i t h th e L a n d Cl scales.

2) A = 20 lo g 0.004 -48 d BThe p h a s e an g le is again zero a t a ll < y .3) The c o r n e r f r e q u e n c y is u > c = 50 rad/sec. Since n = 1 ,t h i s Bode p l o t wil l h a v e a pos i t i ve s lope of 20 dB/dec(o r 6 dB/o ct ) a n d pass t h r o u g h 0 dB a t c u = a , . . Thephase sh i f t is a c o n s t a n t 90.4) T h e c o r n e r f r e q u e n c y is & > , ; = 300 rad/sec. Hence theBode p l o t wil l pass t h r o u g h 0 dB a t a > ^ = 300. Since thec u t e r m is in th e d e n o m i n a t o r a n d n = 2, th e slope will bene ga t i ve a n d e q u a l to 40 dB/dec (o r 12 dB/oct ) . Forn = 2, ( j ) = -n90 = -2(90) = -180.

The r e s u l t a n t curv es a re plo t t ed in Fig. PS 11-10.PS 1 1 - 1 1 W h a t is th e t r a ns f e r f u n c t i o n co r respo n din gto t h e a m p l i t u d e response curv es l abe led a , b, c. d i nFig. PS 1 1 - I I ?SOLUTION C u r v e a crosses 0 d B at a c o r n e r fre-q u e n c y ^ = 40 Hz and a posi t ive s lope s6 dB/oct.

F I G U R E PS 11-10

| ac circuit analysis


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60

40

20

0

0

-

--"

--.-

,

[

--^ ,


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0.5 /, ^F I G U R E PS 1 1 - 1 2

The appr ox i ma t ed a mp l i tud e responsebe flat for/> /< . unt i l / c . A t / ^ th e slope changes f romto 60 dB/dec. Correct ion factors for sketching theresponse are 3, 9, and 3 dB at 0.5 f ^ f c , and 2 f cand + 18 and -18 at 0.5/^ and 2 /c respec-The phase c u r ve wi l l go through 1 3 5 at f cd approach 0 and -270 asymptotically for /< f cd f ^ > f c respectively. The resultant plot is show n inPS 11 -13 .

11-14 Determine the voltage gain transfer func-and construct a Bode plot for th e circuit of Fig.First w e must develop the t ransfer func-T hu s

jwCR + j LC-1 )Now the t r ick is to algebraically w restle th e final form of(1 ) into the quadrat ic form of Eq. (11- lOfc) , repeatedhere as (2) .2, a )CD +j2S- co -if w e let3) co, 2 = = 83.333 x 106 (rad/sec)2 or c o = 9129.n-/rad/sec and therefore obtain

A ^ - " , R + j ( X ^ -Xc)l//coC

K +J \WL-wC)4) c-co^Land then subst i tute (3 ) and (4) into (1 ) , we have

15) A,,= O )Rco^L

COco -

6 ac circuit analysis


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F I G U R E PS 1 1 - 1 3

\0>,7 CO.L CD .

) e-^-n.65e m ay s ubs t i t u t e (6 ) i n t o (5 ) to obtain

1A,, =) , ,(o\2 1 u'-WQ(7 ) is in the same form as (2), w i t h

1) 25 =Q

26S = =0.0429 s 0.043In co n s t r u c t i n g th e Bode plot , w e k n o w w e will have

fo r c o < u > c a s t ra igh t line along 0 dB u n t i l co , ; , at whichpo in t it breaks d o w n w a r d with a slope of -40 dB/defo r o > u > c - The response in the vicini ty of ca , . is determined by the damping fac to r 6. T u r n i n g to Fig. 11-9 wsee that for 6 =0.043 we are of f the p lo t . Hence w e t u rnt o E q . ( l l - l O c ) .10) dB peak = -20 lo g 26 = 21 .3 dBwhich y ie lds th e height of the resonan t peak at co,For a more accurate plot, w e might as well evaluate thet r ans fe r func t i o n . Thus w e have11 ) A =

1-( 9 2(0.0429) "91 2 9(U-

83.33x10- +7(9-399 x l(r6)(oan d plugg ing in various values of < u yields th e followingtab le of data, which is plotted in Fig. PS 11 - 14&:

the Bode plot 167


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co rad/seci f>-1,18

0. 1 (o .

912.9-0.4970.087

^0.5o,4564

-3.272.48

0.8(0,

7303-10.8'

8.72

"C

9129-9021.3

1.5(u,

-174'-1.98

2oi,

18258-177"-9.56

4o)(.

36516-178.7-23.5

lOc.^

91290-179.5-39.9

R L470n 0.6 H

E,n C0.02 M?

WF I G U R E PS 11-14

8 | ac circuit analysis


22/30

1 1 - 1 5 Develop th e t r an s f e r f unc t i on an d c o n s t r u c tBode plot of t he am p l i t u d e response of th e n e t w o r kin Fig. PS l l - 1 5 a . A s s u m e Ri = R^.

c * J i C J2F I G U R E PS 11 -15

F i r s t develop t h e t r an s f e r f unc t i on .R :R ^ - j X c , )RI -7^c,

R.R. +

Now let R i C , = T I = l / f c > i a n d R^C^ [ / ( . ) - , :""""eq'- I ' 2_Eo_ _ ^2 1 + j ( ) | t i ) Ê,,, /?i +R^ 1 +;&V(U^2 )

2. A n a l y z e th e t r an s f e r f u n c t i o n to asce r ta in w h a t cont r i b u t i on s t he va r i ou s f a c t o r s w i l l m a k e in th e Bode plotE q u a t i o n (2), th e t r a n s f e r f u n c t i o n , is of th e fo rmE o b=a

w h e r e a = RI,{R\ + Ri). b = 1 4 - j u / o } ^ a n d c = 1 +j t o / u i ^ . There fo re ,dB = 20 ( log a + lo g b -log c)

T he log a te rm will c o n t r i b u t e a c o m p o n e n t w h i c h ii ndependent of f r equenc y , whi le th e b and c t e r m s a ref requency-dependent . We hav e s h o w n in previous examples t h a t t he b te rm will b r eak at u } ^ and slope up-w a r d a t 20 dB per decade, whi le th e c t e rm b re aks a t wân d slopes d o w n w a r d at 20 dB per decade. Inspec t ionof th e c i rcu i t i nd ica tes tha t at low frequencies , wherA'c, is very large c o m p a r e d w i t h R i, th e o u t p u t will equa^R^(R^ + / ? 2 ) ] m . At f requenc ie s where Xc, is smalcompared wi th R^, th e o u t p u t will app roach t he i npu t3. C o n s t r u c t th e Bode plot . For th e purpose of sketching, let us a s sume t h a t R ^ = Rf Then R^/(Ri + R ^ ) =^, or 6 dB . This f ixes th e r a t i o of T I to T ;:

= R,C,R ,R . + julC^ Ri 4- l/ ju>C^R.

R ,Ri + 7-cuC\ {ju)R^C^ + l)//cuCi^2

PIRl + 1 +jwRtCi

RZR ^ + j w R t R ^ C i + R i1 +jo}RiCi

- ^2(t +7'tt>îCi)-/? i + R; + JCDR^R^C In u m e r a t o r and d e n o m i n a t o r by R [ + R ^

R ^ 1 + 7 ' u / ? i C iR I + R ^ 1 4-yco^Ci

/ ? , / ?2R,. =eq 1+R;

R ^ RZ C, or R )C, =0.5^C\1 R,+R,^ RiCiT - 2

"2ti 0.5RlC^ (i)t

T he Bode plot is sketched in Fig. PS 11-15&.PS 1 1 - 1 6 Bode plot th e t r a n s fe r fu n c t i o n

o2 0^10 1 + jA = . cy V . w }l + J ^ loo,

SOLUTION In th e very low f requency spec trum or foo a p p r o a c h i n g zero, w e see t h a t A = 10, or 20 dB . Thimeans w e have a n e t w o r k capable of providing anampl i f i ca t ion of ten t imes dow n to dc or zero f requencyas shown in Fig. PS 11-16. Electronic ampli f iers can dothis . A s c u increases , the fi rst break occurs a t < u =rad/sec, an d s ince th is is in th e denominator, th e gain( am p l i t u d e ) rolls of f a t -20 dB/dec from th e 20-dBlevel. T he nex t break at C D = 20 rad/sec is in th e n u m e r ator , which tries to pul l A u p a t 4-20 dB/dec. Howeverth i s is canceled by th e previous 20 dB/dec slope, sotha t th e n et slope is n ow zero and it s tays tha t w ay un t i

the Bode plot 169


23/30

F I G U R E PS 11-16

C D = 100, at w h i c h it breaks down again at 2 0 dB/decand remains so. If correction fac tors are inserted orA d a is ac tual ly c o mp u t e d , t h e smoothed amplituderesponse curve resul t s .Th e phase response is approx imated by lines slopingplus or m in u s 45/dec, depending upon wh e t h e r or no tt h e term is in t h e n u mer a t o r or denominator. The phaseapproxima t ions start at 0 at 0.1 a > c , go through 45at a > , , and t e rmina te at +90 at 10 u > c . The resultant phaseapproxima t ion is i l lus t ra ted by the stepped phase curve.For more accuracy, the various correction fac tors can beinserted or t h e actual phase angle of A m ay be computedto yield t h e smoothed phase curve of Fig. PS 11-16.PS 11 -17 Bode plot t he t rans fer func t ion

(UCD,A =1 ) 0)

1 +J 0,

SOLUTION The above t ransfer func t ion m ay be rew r i t t e n asC OCO iA =) . d} . W1 +J 1+;(Ui COi

Inspec t ion of (2) indicates that t h e first fac tor is t h a t ofhigh-pass RC t ype f i l ter and the second fac tor is t h a t oflow-pass f i l ter wi th identical comer frequencies. Therefore, in t h e frequency range c o < < U i , w e have th e lowpass response cl imbing at 2 0 dB/dec while t h e high-pasresponse is f lat at 0 dB . A t < u = coi, t h e high-pass response breaks to zero slope whi le t h e low-pass responsbreaks downward at 2 0 dB/dec. This yields t h e approximated amplitude response of Fig. PS 11-17.Th e approximate phase response may be sketched byfirst d raw in g t h e high-pass f i l ter phase response as a lins tar t ing at co/coi = 0.1, sloping down 45/dec from +90

170 ac circuit analysis


24/30

to 0 a t a > / ' ( 0 t = 10' d then ske tch ing th e low-passf i l t e r phase with th e same slope and < u values b u t s ta r t ingat 0' and t e r m i n a t i n g at 9 0 . The a pp r ox i m a t e phasema y t hen be ske tched by algebra ical ly adding both th e

h igh - and low-pass phase curves . S h o u l d grea te r accu r acybe requ i red , co r rec t ion f ac to r s may be inser ted or ac t u a lva lues com pu t e d in ( 2 ) . The ac t u a l ph as e an d m a g n i t u d ea re also s h o w n in Fig. PS 11-17.

the Bode plot


25/30

WITH ANSWERS11-1 Devee a m p l i t u d e re

L

.( a )

11-2 Devere60 kn

30 kn

1 n F?( a )

11-3 Deve

R

0

-6.02-dB

See Fig. P

un c t ion and Bode p lo t Pc u i t in Fig. PA I I - l a . t

".Y .20dB\.dec '

?

Developtude re spo

/?^ c?

( a )

?

I ft>fl

C?

(a )

^

t h em se

E

t h e)nsei

bode cutler

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