c (s) _ ~s + b • Doterml

turictlon R(s) - s2+as+ b .

s • mp Input s given
Consider a unfty feedback sy,tem with closed tran fer
with u111t ra
loop tran,t.r fu nction G(s). Show that the steady state error

l .
Ans: Refer Q 19.
.111 a forward gain of G(a),
016. Explain err.ct of Pl controller on a unity feedback control system wi

Ana: Rdi!r Q21.

. n,blnation with proportional c~'ll
017. What a.-. derivative controllers and why are they used 111 co , Syllabus • ~• I

Ans: R..-fcr Q22. , FREQUENCY IU!SPONSE.ANALYSIS : In troduction to fttquency domain opec:i&,,tiom - Bode wagram,, ~ Tramlt-r ~l<Wf1 &om_.
. rformance of feedback control the Bod~ d iagram - ~ ~ a nd gain margin - Stability analy•is &om Bode p lot, - Polar pkJls - Nyquist stabilit y ~
Q18. llllStratatheeff9ctsofpropor tionaldanvat1Wcontrolontra nSlen1pe ;

• l

p.
Ans: Refer Q23 .
Q19. Sh.ow that a derivative feedback has the effKt of Increasing the damping ratio without affe EAR N 'ING OBJECTIVE!~J ;
undamped natural freq uency of oscillations. ;,.
• •t-..
Ans: Refer Q25. C? Frequency domqln ,speclflcatio~
, ;;

,, (? Pold; p lot •) ( ~:

t
";.;~
I (? Bo~~ pl? ' '
For a lint onler systam, find out the output of the system when the input applied to the systemis unit ramp input. Sketch the' r(t)

'
Ql.

.
~.,,11.111I~,. show the study state error.
,/
C? Nyquist stab_l~~ ~rlt~~l~

~IDflt
,
. :
02. Acnty fldlack cantrohysllm has tll! open loop transferhltction given by Gisi· slsl~ 41 . Detennine .damping ratio, natural undamped

frlqlancy, ian:atlllll, peat RI an expression for error rasponse for a unit srep input function .
: :~~·,,\·-~ . . (
J~ • ,. ·<
INTRODUCTION
_ _ _ _ _ _ _ _ __ _ __ __ _
..
J.
03. A · f1ec1>ack . 10 . . • : • I
11111V nd se
control symm has a loop trwf1!r f111ctJon, Gisi · sis + 21. Find the nse time, percentage overshoot, peak time a_ The analysis .o f controi•~ystern ln ti~ domain hai several drawbacks which-can be overa,me by th. ·u,e.,of frequency response
. . . , ' ,.. .: ' . ' ,,..
I fflll ~ Df 12 llits. analysis. Frequenq response is,'!- •steady state response of a system for musoldal input. It ls usuayy, cflargcte rlze<;i b y. ,rariaTions
,\ . . .
of magnitude In dB and phase angle lri degrees wllh variatlans of frequency In radians. The analysis provides useful insights into

stability and p.erfo_nnarice d,aracteristia of the control system.

Frequency response analy;l, can be carried out using Bode plot,, Polar plot,, Nyquist plots or Nldlols plots. This unit a,ve r> the

conce pts of Bode •plat, Polar plot arid Nichols plot. These plot, are usually drawn for open loop synms. By using M and N circles

and Nichols d,art, the fre quencr rinponse of dose d loop system can be determined from the knowledge of open loop frequency

response . Thttse plot, are u~ _to ~ !ermine various ,_,.quency domain speciflcotl~, lo analyse synm stability and lo adJust the

gain in order to 'satlsfy the de!lred _specifkattons.

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:
~
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 coNTROL SYSTEMS IJNTU-K-A~,·
UNIT-3 (Frequency ResponseAnalysls )

' . NS (Aug./lep,;OI, l,1➔: \15!•1I N>fll/lil1y-Ol,llet◄, 05(1) I Denomimitor is ih .rectangular- form i.e .•. a + b. Now
INTRODUCTION, fREQUENC Y DOMAIN 5PECIFICATIO tt,'l Aprll/Moy-07, Sot.J, 05(1)) convert ing it to polar form i.e., ML.8
f linear c ontrol syetem~.
01 .
. I analysis o ,1, 11 , sonnnt l'~• k (M,) We have,
Explain why It is Important to c onduct frequen c y doma n
I. Ans:
. "
•1,, r ILis dcfincd,os the maximum 'ot peak value of magni tude
. . (5)
" f c1,,sc<l loo p transfer function.
Fn.•(t urnc v Rl•sponsr
,· . . . . ~ l)\\ll ll ' frl'tl lll' \H.' ) rr,,wn~c . ,t, r or II second order' syptcin, t'he; clo 9 ed loop trnnsfe r 291
Stoolly stotC:' o ulpul of o !--ystcm. \,,hen ~mus01dnl . . 1gnlll 1~ ll-" en'"' n . ·1 ri v the stan<lurd tcst al f'un l 11tl ll in sLan~urd form is,, ., 0 = 1an- 1( )
• . I 1I'll IL',t ~1gna l. :-.11111 u . 1
.., 1-u 2
In the lime- do nmm where Lhe unit ,1~r ,;:agnal p,, con!--1drn..---d a, ,l:lm '
1
lr l·1.1ucncy domuin 1s unit sinus oidal input ,ignsl.
... ( I)
lmporlunce of frcq ucnc) domnin ana.lys1s o~. , _ f'lltc ,ystcm ·mus i•
I t, 111d\\ 1J th Cllll lH) 1 l) ' • t ~ \\!he re ,
A dl.!,1gn of o system to the- freq uency donrnin pro, 1Jcs 1.hc drs1gnL•f" 111 •
, en popular OJTH.'ng,t engmeers...
• _ • •
' :
• mnJcb wh ic h ore renl1st1 0 nnd
· w, = ~ot~~ frequency \mdamped We know that, mathematically the ma,<im um value of
❖ This freq uency domn1n approoth mal..cs dcs1gnc1 ~,nd rcscat'\'has h) 1.:'.01l'ilT1.ll 1
• •' ~ = D qriiplng ratio. · any equati on can Ix: obtained by differentiating it resictwiU
appropriate nnd sotisfoctory simrl1fications. b · d fi to any variable in it and eq uating it to zero.
. . . . '
❖
. • I ·.1r cont n,1 :-) stems nrc o tame
rmn The ratio of actual damp ing to the critical damping is ·
Simple- models whtch g t, c eno ugh undrrstandahlc m1<m11at11.1n about llll \Ill ' Now in order to get the maximum val ue of magnitude of
do ma in analys b . nothin g but, the damP,ing _ratio. , ··
· · · closed loop ITansfer function difTc renliating equati on (5) w.r.t u
. . ,J thus fre qu ency domain analysis IS b
❖ \\11th the frequc-nc) domain. the tran~for funcnllO in cit.her s (1r: 1s obtallh.: · Now, substituting's ~ Jw in ~quatfon ( I), and eq uating it 10 ze ro we get,
evaluating transfer function of s or : domain. We get, '
. . .
dM d
. . · v•,rinnt system are usefu l.
❖ In the analys is o fstab1hty the frequency domain descnptH).11 ot lme-ar. tnnr Ill '
C(Jro) = - .....,.__w:.,:;_ __
du = du '( -:
J(=l=-=u=,=
) 2=+=4=~=, u=
. ,=)
Modr.-1 properties are e..,prc~sed dtrcclly. in S) stcm transfer tUJ1c1ion_ ... (2)
(jro) + 2~ro•.CJiJ +co~
2
The steady ~late response to ~lo\\ and fast oscillatof) inputs are chara..:tc rizcd easil).
R(jro)
· ·
The- computntm nally dt.fficuJt con\'olution integralsol-the li•me J omam
·
•
. , I ced b) equi valent multiplications of dM = .!.[(I - 11 2 )2 + 4'; 211 2 J3'' .(2(1- u2)(-2u ) +~ 1u)
an.: rep O • , 1 du 2
funcuon in the frcquenc) domain. (·: P= - Il
-_- ro' + 2~ro.roJ + ro~ 312
The data can be obtained from the measurement on the phisical system without de rivi ng its mathematical m'o de],
Frequency response analysis is most powerful in co nventional control iheory. They are ind ispensabk to ro~ust.
Now dividing with w , · to both the numerator and - [(1-u~/+ 4~
2112
1 x (-(4u(I -u~ J)+8~ 2u)
denominator in tHe above fractio n, we get, ·
2
~~
. I
❖ Frequency domain anal) s i.s is very simple and can be made accurate by the use qf readily ava ilable sinusoidal :r• can be wrillen as :r" we get,
genera tors and precise measurement eq uipm ents.
B ) the use of a frequenc) response. a system may be designed so that the effecls of undesirable noise
4u (l-u 2 )-8fu
are n~gligi
such analysis and design can be extended to certain non-linear control systems.
Substituting real and \m~ginary parts in eq uation (2), we
TI1e transfer fu nction of such co mp licated systems can be determined experimentall y by the use of frequency test. Substituted u, in place of u nnd is made to equate it 10
gel.
From the kn owledge of the open loop response and frequency response, the estlmation of abso lute and relative zero.
of a closed loop system can be done. Where.
02. Define about frequency domain specifications. (Mod1l P1por~. CM I Aprtl/May-19,
OR
Define frequency doma in specifications .
4u, [1 -u; ]-8~ 1 u,
OR ⇒ =O
Explain frequency domain specifications . ... (3) 2((1-u,1 ) 2 +4'; 2 u;)' ' '

=> 411, (1- u,') - 811, ~• = 0

OR
=> 411, - 411,'- 811, ~• = O
Explain the frequency domain specifications of a second order system.
The ra tio of w, and w. is called as nofmalizcd frequency (11) ⇒ 411,= 4u/ + 811,~2
OR .. w, 411,= 411, [11,' + 2~')
1.e .,11 = - ... (4)
Derive the expressions for frequency domain specifications of a second order system. ro,
⇒ I = u,' + 2~'
, 'I Substitutin g equat ion (4) in equation (3), we get, ⇒ 11,' =I - 2~'
OR
Derive th~ expressions for resonant peak and resonant frequency and hence establish ·the co
between time response and frequency response.
C(Jro) = - -- -
R(jw) I - 11 2 + }2~11
=> l 11, =.i1-2~~ I
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 CONTROL SYSTEMS IJNTU-KAl<INA u Nl1 -3 (P'niqVIJhcv Rll9J>OllM Allal-~•s) t, I a.tr

r •~inllmf!,. i; • 0 m M, a.od M we ~
- (j)~ l(c pl3.in~ ' •' ,-a!ue, Wt get,
Lrt. th< N'{1n>ltr«l t,mdv-1Jth. N, W,

r: [I w, •,f,if,'"i(,1' l
,, '\\'htn. . • .
ti,< •onnu.k
-..-,.. of 1hc do•c-.1 lo"I' f)'

Thus M, xhz",r, I whidi u fh< I IJfJ"/4 1JY.-nbool If liK

muunum v1J1.1< of M lrt ~OJUni ·i, • 0 ll'I M.

\( J I I
M ~ - ---r==.; ~ - = ~

QJ. Oorive the con.lafion ~ n time domain and · 2.c,,/1- c,1 2ro>-.1 - o· o
froquency domain speclncatfons.
M , " ID "'P!"oach ll> -
Ans: ~ -1f, ll4t•l , Q54•J
H.mce., boo> M, a.,d Al_ :tr< undna:ahk f:am the <),.mi
( orrtl• lloo ~ u n the Timt and Pr~utncy Rtspome
= \l ·(~n·<(~r -Ji
1
Inc peak ovenboot, M, and the rct00an1 peak. .1( zr• the
point of """·
Case (?)
'·"" domain and frequency domain speciliatiom of a second
1•rdt:r , ) stem respectively. When E, increasa be)ond =o.

= l1 -(::H-~=(::r=2 As known earlier

r~'\'-~]
Hae in this condinca, both (kt, M 3<ld M, dec:'casc. b"1
M, deer= at Wll:1' r= compa.-ing /,I_
P.:ak o,•cnboot, M , = ,!- ... II )
Case3

= [1 ~(:J-2(:J]- 4~'( :: J= Re,onant peak, M =

, ~
I
J1-E,' ·- (2) When. C= o ;117 in equaoon .... _grt,
I l
~;'(I-:;" I
From equations ( lj and (2). we can sa} that the tams
peak o,crshoot, M and ttSODaDI peak. M,, are related 1,0 each
J( = 21,,J1 -c; = 2x0.707x..,1 - ro7>1 ~!
Th<- .-al1>1: of M obumcd by su!lsbwDng a, ,. Mb)
cquaaoo ( 11 ,::.l mluccg M ">' M, ""°"""'
pe2k t<, other ..nd they bolharethefu.octio.n.s 10 -~;. wlrich means for a
The rCSOll2Jll peak l-I, gets = ~ a n d ~ ·1 ·
g" en value of .I( of its frequency response, there should be a
corr,-sponding value of M, existing if the sy~m is subjected which i, the lowest vahr of it.
M. C ~
:.;,1-~- l t1 a ~lep input.
Substituting I; =0.7117 m cquanoo t I}.. we §=1.
From equations (I) and (2), it is clear that the peak
Rttonrat Fn<jumc, (~J o, m hool, .If, is expooentially ,decreasing function in i;, and M, M, =0.2018 =20\I, ( oearlr)
1, " ddmal a. tbc ~ a: MllCh 1hc m.gmwd,, of
the d,b<:d loop =~fer ftmcuoo,.. = = r-20-20<-1=0 h-.:ing 1m ersely rcl&cd to 'E,'.
The variati.on of M, and M, can be explained b<,low,
By ,.ilicb ils clear llw.. 31_ gOl redl2ccd ID gr= ni..-ni
whereas M, still holds~ tmall valce.
lb<, above evaluation IS quadratic in nature.
V.c know um the ,aluc of u. is. Case I Case 4
soh'l!lg for the roots we got,
_ (6)
9.·ncrc u. bnng tbc oornw rCIOD.IXII frcqucncy i-c.., the ,=
2
2(1 - 2~ ) ± J4<1- 2~
2 2
) -4(1X-I) When,
I
i, =O
. When I; > 0.707 -..ith tl-.c UlCrc:asc in <; clwige,i to
0.707, DO more RSOD.1.Dl peak!~ md eJ<iso and tht c<><nl:won
rat,~ c f r=;u mq,.w::DCJ, 1,0 the uru!:unpcd mLC"iil frequc,cy.

- (7) z = 2(1 - 2~
2
20)

)± ✓4(1 - 4(, 2
+4l;') +4
r -,01,T

"1' \
. . r, breakdown. Al this positi!,n of<, thctt occuo co corrcl:ruon
b<etwtt11 M, and 'M; and 1hr stq, respoosc oscillirtioru
damped for lhi< v::Ju,, of; hena ils DO! • 1ut>b!rn1 IO notice.
=
well

[.qllllmg ~ (3J end (4), we get,

2
i
!"1 cases
I \. When I;= I thtte is co ovcnhoot produced in the 5' 51...-m
, = 2(1 - U, )=4/(1 -
2

2
,v.;2 ,v,' + I
T

~J----\..,,,
I O& Ollr l
any more as Af, geis decreased to a VCT)' low ,atw:e. Th..;,
be no scope of .If, as it grts decn:uod already 1,0 law nlu,: .u
<, = 0.707 itsd(
.,.,u
Rcsooanl frequcn..":.
(b) Thus. for correlat iag the transient response wa b
■ -- n. .._'"l)Ul ,_.,,. . ~flli<Alllt~•---·
M, 1.:11! M• .,, c.ar:ril.f &IM! lq,c,W ~
frcqU<'llcy rcspomc:of higher order sy!ltttru.. All th.osc alo™"
.r s (I - ~')± ✓~· - 4(; + 2 lh:QX\'---~ l correlation, arc useful. MoreoYcr ~ set of SJ>"cific.i1,0o.s
£.rpnssion fl>r Ba■dwldtb (WJ Fia- fonn time domain can b<, trans.lateil u freq11cncy dom:uo
From equ:n,on (5 ). we bavc, Ntglffling the negative ilgn, we get, l rom the figure (a) and figun: (b) its clear truit the) bavc specificauons from aU ~ c = of couelatioo.
ma,i mum values at initial potol i.e., at E, = 0. At c, = 0 both the Hence from , ba,,·e anal} ;is it is dear th.I the corTela.1100
4u.i.nt111es M, and M, ha>e h1g.b va)Lll!S i.t ., ,If has, =imum between time dom.un >1lJ.! frcqll<,"OC)' domam c,ust for the range
, alu,· where as M, lt nds . Infinite value, which is not desirable. of0 < I; < 0.707.

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 CONTROL SYSTEMS (JNTU -K~KIN UNIT-3 (m-iquency Response,Analysls) 3 .'7
L' I
Comparing equations ( l) and (2), we get, · • To.determine,
Q4. Denve the relation between phHe marvln and
damping ratio. Re1onaot peak. M, = 1
Ans: (1'odol ~
Relatio n S.twttn Phas~ Margin an d Damping Ratio
ConsulN • =ood order systm, whose open loop
functioo is given by.
. 01 I ~ y-11, Set~ 05<•ll

tranSm-
Phase an~lc al <•,.

¢11'- = ~900- un
"·

I
,l.,l
~
(J),
=> OJ.'• fs =-2:236
w. - 2.2l6 rad/sec

!;w= I
.Resonant frequency, ID, =?

Bandwidth. co, = ?

We know that,

.,.
•
= - 90 - tan
1 l,-:!_t,'+~1 2E,
!; = ..!.. ~_[_
w. 2.236
Peak ovcnhoot is given as..

G(s) = s(s ~ ~a,.) i; = 0.447

pt,ase m~m is g iven by,
\\ "here.
ru. = Unda:mp<!d natural frequcn..C) y = 1so 0 + <>,.. Reso nant Frequency

; = Dampmg ratio
0 - 1[ y - 2E,' + 4;◄ + The resonant fre/luency is given by, ⇒ 0.12 ~ ~ ,1t~l l
Subsntutmg F J W. we get. y = 1so0 - 90 - tan 2E,
Ol, = co, ✓i - 2E;2 Applying log. on both sides we get.

.,[ /-2;'+ R+l]

(i)"

G(/m) = J<il(jm:. U;m.) :;,, i 236 ../(-2 (0.447) 2 log,"-" = log;

y = 90° - tan 2f,
=2.236 >< 0.775
= -2. 1202 = -h1o..,
l v l -(, 2

✓-2e : ~ °'·.= 1.1l2 g,

'
~Jiz' ,
., ..
y = tan·' [
M:igrutude, Resonant Peak 2.1202= (,:t _ (I}
~
G(oi) = 1G{fel) 1 = ( ( , r- \2t"·. .. The above equation gives the relation
phase margin '/ and damping ration '<;'.
The rtso!UIIlt ~IS ~CD by,

"'· 1
~ ~ 2.1202 = {,:t
\ "'· ./ M= --J_ _ ,,; · I
~
I as. Detennine the resonant frequency m, r ' ~!;/1- E,
.,
2
2(0.447) ✓1 -(0.447) 2
·peak M and bandwidth for the system
2.1202 ( / 1- t:,' ) = (,:t
~
"'• \i 4t.'-l..!!!..r
. m,
p
transfer function is, GUo,) = .
5
U ,·
- - L -125
- 0.7997 .- ·
5 +J 2o,+ (i) Squaring DD bolh sides we get.
At gain aossover frequency a.., Band width
I
Ans: (Apri l-18, Sot-3, Q5( b) I April/May-1&, S.t
Band\\~dtb is given ~Y,
t
(2.1202)'( ✓1 r
-t:,' = r:.' tr .
Given that,
2 -+-4f,
w, = w, [ I - 21; 2 + ✓r2---4-E,' --.4 f ⇒ 4.4952 ( 1-(,1 = 9-8696 t:,2

IA, G(iw) = 5 +j2oi + (iw) 2 . ,---~,----~ p'

= 2.236 li - 2 <o.441) 2 + h-4(0.447) ' + 4(0.447) '
= 4.4952 - 4.4952 t:,2 = 9.8696 c,2
To determine, 9.8696 (,' + 4.49521:;' = 4.4952
G(w,) = z / 4f,' + r' Resonant frequency (w,) =? . I r=--:=-==~~•
= 2.236 I -p.3996 + /2- 0.1992 +o.159 f
⇒

⇒ 14.3648 I:;' = 4.4952

Magnitude of G{m,) at gain crossover lmjuency is I Resonant peak (M) = ? = 2.236 (0.6004 + 1.1661]"'
i.e .• G(o,,.) = I Bandwidth = ? ⇒ t:,' -- 143648
4.4952
= 2.236 (1.7665)"'
Now, Let, = 2.236 x I .329 ⇒ I:;' = 0.3129
j w =s = 2.972
G(o,.,) = z ,/4;' Tr' ⇒ (, = / 0.3 129
06. A unit step fHPOJIH test conducted on a
= ;r;/4f,' +x' = I G(iw)=G(s) = - --
5
5+1s +s2 second order system yjeldecl peak overshoot = IS= 0.31291
= x'(4; ' +z')= I M, = 0.12, and peak time'.~ 0.2a. Obtain the
= .r'+4-;'r -1 =O G(s) = 5 corresponding frequency response Indices
And also,

= (r')' + 4f,Y- I =O We know that,

,2 + 2s + 5
(M, ro, oiJ for1he·system. Peak time_ ,, = ;d
2 Ans: ~ -17, Sol,2, QS(b)
-4f,' ± ~ - 4f, ±2 ~ ⇒ 0.2= ...!....
x'= 2 = 2 The closed loop transfer function of second.order Given that, Old
is,
= - 2c;' ± ./4i.:+I C(s) w; Peak o vers.h oot, M, = 0.12 ⇒ m, = 02
z = ✓-2E,2± /4E,' + I R(s) = s 2 + 2E, w~s+ w; Peak lime, r, = 0.2 sec . ⇒ "', = 15.707

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 UNIT-3 •ft.iequilncy ReajsohsJiAnalysls)
3.9
OoiJt margin, ·.
' I
3.2 BoDE DIAGRAMS - TRANSFER FoMc . Step5
-
FROM THE
BoDE D1AGRAM - PHASI! MA,
· ·-· • 1 Starting from poirit M, draw a straight line for the fim
= ru,~ ru, ~ 1
= 15.707
ii\ND G Al"'
.. MARGIN - STABILITY ANA&.:
,
'
K =~
• 1GUwp,)1
·
fac1or with-its corresponding change of slope till the fi rst comer
15.707 frequency.
=> w .. = Fit0"1 BODE PLOTS
✓1 - r,i Where, IG(/<0,)lis ihemagnfrudeofG(/w)at Ci> ~ ci> . ,,, Strp 6
Explain procedure of bode plot and deh?_ n nln•
⇒ ru,. =
15.707 Q7.
of ain margin and phase margin from ~
The gain margin, K in decipe ls
1
= 20 log 10
K,
. From this point, <haw another straight line- fo r the next
/ 1- (0.56)2 gt Ap rll/May-19, Sot-1, . 1
po
I . factor in the table with its corresponding change of slope till
15 .707 OR = 201og,o-,o,.j = -201og,olG()co,.~ the next comer frequency.
⇒ ro. = JWp, )
✓ 1 - 0 .3136
Explain gain margin and phase margin. 7
_ 15.707 The gain margin of the system indicates the amount of Step
⇒ 00
•- / 0 .6864 the sySlem gain which can.tfe increased till system reaches on ,~ u e the above step till all the factors are considered.
_ 15 .707 (Refer Only Topic: G11i 11 Margin 'and Phast'Ma the ve rge of instability. · · Now this completes the magnitude plot.
0 _8285 = 18.95 rad / sec .
⇒ (I). -
Phase-Margin
OR Step 8
Why bode plots are cof!lmonly used.In The phase f!!argin (y), is the amount of additional phase
in order to draw the phase plot, frame the equation for
lag which can be introduced into the system till the system
Resonant peak. M = -=:--h frequency domain design? phase angle of the given transfer function . Th.is equation can be
' 2/;vl -r,· reaches the v.7rge ~finst~bilify_at gain crossover frequency.
framed by algebraically adding the angle contnl>utions of the
I Phase m3:Tgin'.· y = _180° + Ci>,, individual factors of the transfer function. This equation will
2 OR
2(0.56) / 1 -(0.56) be in tenns of ro.
Explain the significance of Bode Pio Where, Cl>;➔ Phase angle of the system at ga,in crossover
1
stability studies of linear control systems. frequency. Stepll
1.12 ✓1 - 0.3 136
Ans: __ (April/May-09,Set-1, QS(a) I Aprll/May-09,~et.J,
l QB. Explain the g;neral·procedure for constructing . . ·.. By taking arbitrary value of ro calculate the phase angle
1.12 ✓o . 6864 The Bode plot is a frequency response plo\ !!fthe
Bode.pli:,ts. :' · Aprtl/May-H, Set- , QS(II) until the value of phase angle crosses - J80°.
function of the system which can be used to analyze both 1
l
1.12 (0.8284)
loop and ~losed loop systems. The Bode plot consists ·o
OR St" 10
plots,
l. Magnitude plot drawn between magnitude of a s Enllst tlr!t :steps for the construction of Bode · In the same semitC:S
paper, just below the magnitude plot
= 0.9178
transfer function in dB and frequency w (log _w): plots. made the angles corresponding lo different values of ru. Now
join these points smootbly_with a free hand. This completes the
= l.0778 2. Phase plot dra wn between the phase angle of as Ans: AugJSop.-07, Set-4, QSlb) phase plot. . ,
transfer function in degrees and frequency W (I 0

Reso nant fr~quency, Ol, = wn J1- 2(, 2 Wh_e n compared to other plots, the Bode plot The following. are the sicps·invo!ved for the consin:iclion QI. Ellpl■ln the procedure for determination of
drawn easily and quickly as the locus can be repr of bode plots. '
= l 8.95 d-1-:...-2-(0-.5-6_) _2 tr■ n■t.r function from Efode plots.
by straight line asyn~ptotes. ·
= 18.95 /1- 2 (0.3136) I Slep I Ana: AugJSep.-07, Set~. Q54cl
Significance or Bode Plot

w.[/ 1-2(,
= 18.95 ✓1 - 0.6272
= l 8.95 / 0.3728
= 18.95(0.6105)
= 11 .568 {ad/sec
4
I.

2.

3.
Using bode plot the stability o f closed loop sys
be determined.
The bode plots are used to ca lculate freq ue ~cy r
specifications.
Bode pl ots provide th e necessary infom1athi
i
t
lo rn1 .·

Step 2
Represent the given t;ansfer function in time constant

Calcu)ate the comer frequency slope and change of slope

corresponding to factor of the given transfer function. Make a
Step I
From the given bode plots, calculate the value of k using
the formula;

(i)
•

20 log[:.] = Mro,, , When the in itial slope

t'
2 2 designing the control syste ms.
Bandwidth, ro, = + /2 - 4(, + 4(, ] tab ul ar colwnn of the above data such that the factors are placed OH>,,
4. The phase margin and gain margin of the system in the increasing order of the comer frequency. is - 20 < 11 dB/dee.
= 18.95 [ / 1 - 2(0.56) 2 + h- 4 (0.56) 2 + 4 (0,56)
4
] determ~ed by ~ ing bode plots, which helps th iil
ofrc!allve stab,ltty of the syste m. • ,f. \ Step 3 (ii) 201og[ Kw"]....._., = Mru,,, When the initial s lope
= 18.95 [ J, _2co .J 136) + ✓2 - 4(0.3 136) + 4 (0.0983) l
= 18.95[ ✓1 - 0 6272 + /17388]
5,

6.
. Both the low arid high frequency regiol)s can be bro
1mto focus simultaneously.
Transfer function of the system can be ~btained
bode plot. · . '
·1
j Calculate the magnitude of first factor using the foni,u!a,
.' .
M= 20 !og[G~·co)]w• OI
Where,
is +20 x n dB/dee.
0>, is the first comer frequency,
1

M 0>01 is the corresponding magni tude at We, .

)
Where, G(jw) is the first factor.
= 18.95[ ✓1 - 0.6272 + 1.067 1]
Step 1
Gain Margin
Step 4
~ 18.95 [ /1.43 99 l . The gai
. n ma rg m
· is
· d e. fi ncd as the reci'p ro ca."1 of
Mark the value M on the semi!og paper and start the
If the first line is having a slope of - 20 • 11 dB /J ee,
it indicates the presence , of pole at origin g iven by K/s' . On
= I 8.95 ( l.1 999) magnitude of open loop tra nsfer func tion at phase ·cr()s the other hand if the slope is +20 x n dB/dee, it indicates the
magnitude plot from this point.
frequency. ' • '.· presence of ze~o at origin given by Ks'. ·
= 22.738

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 CONTROL SYSTEMS tJNTU · KAl<I~. tJ NIT-3 (litilq uency ResponaeAnal~sls)
., 1 3 . 11
~ nllu dr l'lof · C ..
l~r'mt , Corner F re1ruucy ,81oJ!e Change In slope
'·
'1:-IQ"' <:ncck the chani;c io 811/rt' from first hno m n«t Hnc. (rad/~r•) (db/tee}. (db/sec)

(i)
.
u- the CM.lll)C "' s lope.- " . . . ' ' . h1' ' l\'Cllh)
O. th"" ti md1('Btes the presence ,,f • ttl\' "hie
(t +;-·' 1 ,r"
l t .'l',fc - .o -
f ,
,,.
; (ii) If tbe chnnge tP slope •~ - 40 ~ 11. then it ilidici,tcs lhe prcsc1wc of J"'k gl\'l'n b)
I

(1 +~'. .)'
,,

- ' I·
I +4J'!>
,JO ,, = 2..4 ,= 0,.2~ - 20 0 - 20 = - 20'

I
1'+2/o>' we,= 2 =0._s· 20 - 20 + 20= 0
I ,_.,•. I
II lhe change ,m slope 1s
• ,.20 ~
~ ,.nf7ero eL' l\'CO b,·
m. then •11 ind10Btes th e prcsm~- l + - s )'"
O\ '
liii) (
J I
Step J
-~ , w,, = 0.25 = 4 - 20 Q•".",20 = -20
Continue 1.tw above step till !Ill I.be comer frequencies•~ oons1dcrcd.
Step S . Let, Lower frequen~y, ro1< ·~,,' (ro, = o..J rad/sec)
• . . 3 11
d !es in the denom111a1or to get the co
Combine the factors o bt:uned ubo,>e by placm.g zeros m the numcrotor po , Higher frC<!uenc~, w>.'ro; (w = ~10 rad/sec)
transltr function . ' :I I l 1 , If 1

Q10. Write short notes on stability analysis from Bode Plots. Lower frequ~ncy,?J ;,,w;,',A".',IG(i~)l =70 Jog 5= 14 db
\t' ' ~I j 1. I : ,'-'.,,.,.,

OR At w =w, ;'A= IG(lro)1,=2qJog Sr 14 db

Write a note on determination of range of 'K' for stability using Bode plots.
II \ .- . ,..::' •i1"",'. •,, , 1•/: • .
Ans: (,t.ugJSep.~7. Set-1, QS(a) I AugJSep.-47, Sot At W = w,, , A=; r~-l~tefr~m w ~~.-~ · x lo~ r·wc, )j + A .

Tiie smbilit)' of a closed loop system using Bode plots can re assessed using two frequen cy domain specification!
· __.: ::,: ', 'j:i.,o:~:·(~ : co<i.. _ ., ......,,.
an, as fo llows. =~,20~log l-0:25!· +,14 =8 db

(1) TI1e gain margin At .:ii= (l); , ;

J
= [~lop~ ~Jm (l) .<- to -~ .
• •: ' ~ •. C.J
Xlog [ID·,]]+ A
• (l) L'z UIQ) ,. 11)4:i
.
(it) The phase mo.rgm
·Th, g 3 in margin is defined as the margin in gain •K' allowable' by which gain can be increased till the syst~.m rea ' ' = 0. x log '. i4 -·,.
' j_'Q3' :+,8
the ,,er~ of insubllity. · '
= ,8 db ' ~, '

r~i
(On lhc verge of instability. the value of K is K,_).
The range of val ues of K fo r a system be siabl, can be de1ennined from the value of gain margin. = co,,, A=
10 At w [slope from ,ro:
• L\
.;::co. x log
ff ro + A~~ - m
A system is said to be stable if the gain margin is +ve and is said to be unstable if gain margin is - ve. • :· -~·- C3 . "J

A.s the system gain K is increased from zero to K_, the gain lllll(gin is +ve and system is stable. ' =-20 log l~k~ ,
At K = K~.,. system is marginaUy stable. But as lhc value of K is inc reased beyond K_ (K > K= ) , the ga•i.? marg' =0db
and hence, the system is unslJlble. , ,I
Phase Plot
• I , • l • If
Hence, the range of val ues of K for a system LO be stable are, ♦ =, GLjw ·
0 < K < K_ ~ iart, 1.(2p>)- tan-1 (4w)- tan- 1 (0.25 ro)

5(1+2s) (l) tan 1 (2w) tan-1 (4w) tan-• (0.2Sco)

Q11 . Draw bode pl.o t for transfer function G(s] = ( + ) +( + _ ) . ♦
Ans:
1 45 1 0 255
1 0.1-
0.25
11.3
.. . 26.56
2 1.8
45
1.43
J .57
· - H .93. _; .12°
: 22.01 - 220.
Given that,
0.5 4~ 63.4) 7. 12 -25.55-
5( 1 +2s) 26"
G(s) = (I + 4s)( l + 0.25s) I 63.4) 75 .96 14.03 · - 26:56 - 270
4 82.87 ,. 86.42 · 45 - 48,55' - 4~•
Puts ; j<o, >,
10 ,87 .13 88.56 68 .1 9 -69.62 - 70°
. 5(1 + 2/m)
50 89.42 89.71
G(/ro) = ( I + 4Joi) (l + 0.25jw) - 85.42 85 ,7 1 - -86°

,'l"·f\
.. .
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 1,NIT-3 (Preq uency ResponieAMelylifs)
3.12 3.13
_, ~~-0... ~ 16
. 2ro, zq 0) • 0.8
IS(HS) 15x 5[J +.:'.]
G(.,) ~
2
, ., 5
8
\J' + lfu + I 00) s(s 2 + 16s +100)

7S(l+0,2.r) 0 ,75(l+0.2.r)
sxio L +~+I 2
s( l+0.01r + 0.1 _6 s)
100 100

Sinusoidal transfer function

, '
G(jw) = 0.75(l+o2jw) 0.75(1 +0.2jco)
(jw) [I +O.Ol(j<0) 2 + 0. l 6jw) (j(J})[I-O.Olco2 + jO.llia>J
Magnitude ,Plot
Comer frequencies,
•1 .
(J}q = - = 5 rad/sec
,0.2 ,
• l 00e, =co,= IOrad/sec

The various terms'of G(ico) \UC l!sted below in the form of table.

Corner
Term Frequency Slope Cll1■1elallope
(rad/sec) (db/dee) (db/dee)

0.75
--;
jro - 20 , - 20

I+ j 0.2 (J} 0),, = 5 20

.
"'-20+20 =O
---- __ . . . ,, ·", _,-·
,
I $
0> '2 = 10 -1() 0-40=-40
----~
1- 0.010> 2 +J0,16 w
' -
lir■ph Le t,
15(s+5) .
Q12. Sketch the bode plot for a system G(s) = • Hence.determine the stablllty of th w, = 0.1 rad/sec (·: w, < ru,, )
s(s 2 + 16s + 100)
Ans: AprlUMay-11, w, = 20 rad/sec ( ·: w,> roe, )
! '
Given that,
Let, A = I G(jro) I in db,
. I S(s+5)
G(s)"' s(? + 16s + I 00)
· On comparing the quadratic factor of G(s)with the standard form of quadratic factor, we have,
At,w = w, A =20log
.
lo.:751"' ~O,log
[Tool
~
1o.t7
s'-t2~w. s + w;= s' + 16s + 100 = 17.5 db '
ro; = 100
w. = ✓100 · = 10
Al, ro = roe, , A = 20 log 1-¾t~ ; 0 log 1°·; 51.
2~w.=_16 = - 16.5 db

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 UNIT -3. (Fre9uency Response Analysis)
3.15
>, .. _ _c. •o

AL(\) "' m..., , ·'

A = \slope from <O tow ' log "'' _' ] , !

'I •; U)
'I
1
-~
_ l0 x log 101 +(-16.5) =- l6.5 dt,
- r '
5
At.w = w,.
A = \ Slope from "-' 0 , to w"x log : ,h,1 , A'" 0 _ ''. 1

[
= - 40x \og 201 +(-16.5) =-.28.Sdb
10
In the scmilog graph sheet. choose a sulc of I imit = 20 db on y-axis and frequencies on x-ax is and draw the
plot '
Ph:t!ie Plot: Phase angle of G{fw) as a funcno n of'w· 1s given. by.
o = \ G(jro)
= tan 1 0.2w - 90° - tan·' [ ~ ] (for w :S w )
1-0.0 lw' •
q, = I G(jro)

= tan 102w - 90"-[tan'"'(~

1-o.01w·
) -1 80'1 (for w > W)"
The phase angle ofGt;w) are calculated fo r various values ofw and is listed below.

I 0. 1600
w rod/sec ta n· ' 0.2Ul ta o- - - - 2
$
1 - 0.0l co t" ; ~ .• J .t

O.l 1.14• 0.916° -89.77 = - 90° G11ph

0.5 5.7° 4.6 - 88.9 = - 89° 013. Sketch the bode plot'~nd determine the foll9wlng (i) gain cross over frequency (II) phase:cross over
I 11.30 9.2° -87.9 = -88°
frequenci (Iii) gain margin (Iv) phase margin for the transfer function Is given by G(s) = K
11
5 45° 46.8° - 9 1.8 = - 92° and detennlne· the K for stability. · ;1( + 1) (s+ 2)
10 63.4" 90° - 11 6.6= - 1°16°
Ans: April-11, Sot-2. 05
20 75.9" - 46.8° + 180° = I 33.2° - 147.3 = - 14 8
Give n that,
50 84 .3° - I 8.4° + 180° = 16 1.6° ~ 167 .3 = - 168°

100 87. 1° - 9.2° + 180° = 170.8° - 173.7 =- 174°

G(s)- K
' - s(s + l )(s +2)
150 88.09° - 6. 11° + 180° = 173.89° - 175.79= 176°
Let, K = I
200 88.56° - 4.58 + 180° = 175.42° - 176.95- 177°
G(s) - I
On the same se m 1Jog graph sheet, choose a scale of l umt = 20° on the y -axis on the ri ht hand ·d O [ h h ·' - s(s + I ) (s+ 2)
points-and j oin them . . g SI e I e S eet.
Put, s,; jw
Fro m the bode p lot, we hsve phase an gle at gain crossover frequency,

$,,=- 88' G( 'w) - I

J - Jw(jw + l)(Jw + 2)
Phase margin, y = 180° + $,., = 180° - 88° = 92°
0.5
Here, ga in margin =+= Jw( I + Jw) (I + 0.5Jw)
H ence, the gi ven system is stable since phase margin. and gain margin are positi ve.

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3 .16 CONTRO L SYSTEM S (JNTU·KAKJNADA
I
uNll'. J (Frequency Res~n~e f\n11lys1s) 3.17
tlldt l'to1 -
O • n~r Ill , lopr clb/,lcr : ""'t ~ F7:.
. ~flll<

05
Cornrr l' n,quenri 1,dl«•t Slopr d h'd('('
- :

.• ;v+:
~ ... I rt;•
:o ~~ .. . • !. '
/<U

I 1
·- -..
. .. 2() - 10 , ,40
}8 I I:
.I 1· ' I.'
' - .:-
'
I . !
•
' .
.,
'
.
: !•
•

- I
~
- 20 1
(i)
',
~

! :!_'., ·t :_
l + .1ro I
--. ..._ =. . •4n: .- i."1~";t ~.:f"i2'7P•-~-Ar
1 ': :_ ..1
r~r-:•·-;:-tr;. .r"';
t - ·!

·-
I
..
(i) :
I
-::! - 20
~t 20 = - 60 ~~ ' . t~,.·
7!± ,1.
f . I•
t!J 1 ~
f::4-- ~1,,_:-!.. 'll~f-~ -~ l-;~!l! ..,) .J"... -f.~.J.0.
.,.._!.......,._ __ ,
J.
t ,
-~•.:. .41 -~. .
r ~ -~~ '
•jlj,:.i..;..;.,h• . . , - I-~. ""r•
:. . ~l~:.➔ :":;c-'i
1
05
I t ;O ~"'
~~
~ lf
1
•. {......r"
t', . · 1•.
! , 1
·1,, .. t:.J~, . !":-w?.-=:--
·••.; t••I
•t -r,ct-.,+,1,F
. ,•:·,:_L '. .•; -'--•'•-~J
• · .

! I ': : ,· ! . .,
~ •_._ I - •~ -J ~~ • ~--: . ~ • ,- • !;.~ ,. '

Low.,,- cut-0fl frcqllfflC) , o, ~ 0 I rad :;e,

. i~;.!;,' ~.
.fi4 ;
-+· .~ '£
_-z-_ ~
·t •..,
~
',~ - ~ =~~-=,~
~ . ·: 1-:i.::i
Lppc.'f cut--Qff ~ucnc~. oo, - 10 rad'SC<" --_,_ = ~ ! -

·r l;., __i. ~I , .. r ·:L.::;J

No\\ , ' --♦ !ir' ~
. . . ."4-,..,.,
-
-·
- : ~ ...... 4 --
r ··7· - ,. .
+ rr •
.-l
A.I o>• co ~ 0.1 rad sec

Cr.Im.A• 20 loge I~~ I~ ~O log 5 ,. 13.979::: 14 db

'>=;:-~'T'-H'----:-7'"~t-·~;~•=--
,;.:'· -~. f~\~:; ;1, -:;~~
Al m•m< 1 rad ~
I- -.c.;__.ts.::G:.~~J
.' i ·-~ -;-- '
.
!o:1:
; .

Gain. A a 20 i<'>g 20 log [0.51= - 6.0205 ::: - 6 db y

Ar ru ~ o>. - ~ r:,d,scc

Gam. A~ - 40 log Ifj- (- 6) =- 40 log 2 -6 =·- l~.O-ll l-6 = - 18.041 1 ::: - 18 db

-i
·'

At ro= ro, ~ 10 rad.'scc

:.,.;
~

Gam.A :-60 log I

1
; l-'- (-18)=-60 1og \51- (- l8)= - 41.938- 18=-59.938 ::: - 60 db ::,~.1
'.· - 1
The magn,rude plol ts shown in graph.
Ph11s, Plo1
The pbBse angle for a gi,en S}·,aem is,
. ,.,f
~ ~
J O

~ (Jru) - - 90 - tan 1 (0.5 wJ-um·• (<:J)

(J) ta n·' (0.5 lil) tan·• (w) 4>(jw)

0.1 2.8624 5.7105 - 98.5729 ::: - 98.6
0.3 8.5307 16.6992 - 115.2299 ::: - 11 5.3
.. O
~ ,.0 0 0

0.6 16.6992 30.9637 . - 137.6629 ::: - 137.7

Gr■ ph
I 26.565 45 - 161.565::: - 16 1.6
1.3 33.0236 52.4314 - 175-.45 5 ::: ,- 175.5 Q1 4. Draw the bode plot and fi nd gain margin an d ph aH margin for the given, transfer fu nction
1.6 38.6598 57.9946 - 186.6544 ::: - 186.7 K
2 45 63.4349 - 198.4349 ::: - 198. 5 -G(s)"' s(1 + 0.1s)(1 + 0.5s)'
4 63 .4349 75 .9637 - 229.3986 ::: - 229.4 Ans: (Model Paper-N, Qi I Nov.-211. S.t-1, QS(bl)
6 71.565 80.5376 - 242.1026 ::: - 242 .1 Given'that,
8 75.9637 82.8749 - 248,8386 ~ - 24 8.8
JO 78.69 84.2894 Trartsfer function, G(s) ~ s( I + O. ls)( I + 0.5s)
- 252 .9794 - - 253
The phase plol is shown in graph. Assume K = !00
From graph,
Gain crossover frequency, ro,, ~ 0.5 rad/sec JOO
Phase crossover frequency, w,..= 1.5 rad/sec : . G(s) ~ s ( I + 0. ls)( I + 0.5s)
Gain margin. G.M = 12 db
Phase margin. P.M = 180 + 4>,. Put s= Jro
~- = 180 - 132 = 48° JOO
G(iro)- U~)( I + 0. I jcu) ( I + 0.5 }Ol)
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'
l \i
tJ NIT-t l(l!fequilnoy Rospohso Analysis) " :,' : , 3.19

.l\1.•gnlt,udt Plot
Rf'su1tan1
-
C'ornrr Slopr
~t"m Slopr
Frtqurnc~ -
100
--,- - - 20 -
) 0)

I 0) =...L= , - ~0 ~0 - :o = -40
T+o.V<o ,, 05 •
__. 0 _ :,1 = ~o
I
I + O l_1ro ro,, = 0'. 1 = 10 -:o
l. Coosidcnhe tmn K = 100
Its dB mng.nitudc is :!O 10!! K. = 40 dB . . . •• 1 3 distunce 40 dB nbovc the refi
'
S-0 the mngmtud~ plot corresponding 10 gain ' K ,s strnight hne parallel 10 .\ -<1xt. n
line.
Co<lSld~ pole ot origin ( ¾]
Toe slope of magnitude plot for I. pole at lhe origin is -20 dB/dee.
· · · 1r.11'ghl line whose slope is -20 dBi
Therefore. the resu.ltanl mag:nitu<k plot due to g:un K = I 00 :uid pole nt ongm ts 3 s -

IP> and wbosc magn itude at m = I is 40 dB.

J. . .
Consider simple pole l 1
~ 01.Sjro l
1
The above term is of the form - -
1 - TS

~
where comer firequcncv (J) = .!._ = ...!..._ = 1 radi'sec
, · ·• T 0.5 -

Therefore. the resultant slope from ro = 2 rad sec will be (-20 dB/dee as starting slope) + (- 20 dB/dee due to simple po
i.e re,,-ultWlt slope of -40 dB 'dee.

~ 4.
1
Consider simple pole(-.-- - )
I.,. 0 . 1JW
t,:~
I
The above 1erm is of the fom1 TTS
1

Where comer frequency w

' '>
=· .!._
T
= _!__ = 10 rad/sec
0.1
Therefore, the resultant slope from w = 10 rad/sec will be (-40 dB/dee as previou.s slope)+ (-20 dB/dee due to sun
pole) i.e., resullllJlt slope is -60 dB/dee. This will continue upto w - oo as there is no other factor present is G(S)
The magnitude plot is shown in graph.

Phase Plot
The phase angle of the system is given as,
q, = -90- tan·' 0. ](J) - tan·' 0.5 w
(D

0.2
tan· ' O.lw 13n· • 0.5 (J)
1.14' 5.7 1' -96.85°
~.
I 5.71° 26.56° - 122 .27°
5 26.56° 68 .19" -i84.75°
10 45• 78.69° -2 13.69°
'., 100 84.28' 88.85° -263 .14°
'.--,::it
,1 ~!~W.,.~iJ: .
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-11 .·
I \ '

CONTROL SYSTEMS iJNTU-KA~lt,I UN IT-~, (f raqu~ncy Rflsponse Analysis);

3 .21
3.lO
nlc11udc JJl()t-ls shown in grnp'b,
Q15· Sketch the bode plot and determine the gain margin and ph ase n,Argln.
From lhc bode plot,
10
for the tninah!r function Is glVt!n, G(s)" s( 1 + 0.4s) (f + 0.1s) <.l oin cfossover frcc1ueocy,
Ana:
w,. ~ 5 rod/sec
Grv('n that.
Phase crossover frequency,
G(s) ~ - 10
<\I -t- 0 4.,l\l • O l ,) co,~"' S.4 rad/sc_c
l'ut.. o · /W
Gain margin,,G.M = I db
G(/w) = - JO - Phase margin, P.t-,f = 180° + ~...
j ~>\l • 04 i m)(l + OJ _1wl

l\taj!nllutlc plot ·= ,180 + (- 175°)

Cbingc In Slope (db/sec) =S~
Terms Corner F'n'qucncy (radlsK) Slope (db/sec)

10 -cO - 20
/ (l)
-
l - 20 - 20-20 = -40
'" = _l_-7.5
I :0.4jro " 0.4 -
I -40-20 =-<iO
l + 0 .4jw
w,., = "efT =10 - 20

Let.
Lower c-omcr frcqucnc). (:) = 0. 1 rad/sec
Upper comer frequency w, = 50 rad,scc
No,,.,
TI1e gatn IS given as.

A= 20 log D~]
At 6) = Cil 1 = 0.1 radtsec
. JO
Garn A = 20 log D.1 = 40 db
At ro =Ol,1 =2.5 rad/sec
10
A = 20 log . = 12.04 :: 12 db
25
At w = w.., = IO rad/sec
A= l-40 x log l~ ]+ 12 =-J2 .08::-J2db

At w = (J), = 50 rad/sec

A= l-oo x log ~~ ] - 12 = - 53.94 :: - 54 db

Phase Plot
$ = - 90 - tan· 1 0.4 w - tan·• 0.1 w

tan· 1 0.4 w tan· 1(0. J w ) $

"' -
0,1 2.29 0.5 7 - 92 .86 =-93 °

I 2 1.8 5.7 J - 11 7.5 1 :: - 118°

2.5 45 14.04 - 149 .04 ::-149°

10 75 .96 45 - 210.96 =-211 °

78 .69 - 255.82 :: ..:256° Graph: Boda Plot for Gls) = II IO

50 87. 13 s + 0.4s)(I + O. ls)

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. .
 SYSTEMS (JNTU -KAKIN ' j)I ,'
, coNTRO L .
1000 _ ..::; UNIT •3 (Frequency Respdnne Analysis) 3 .23
3 ,22
. 1sg1v90t,yG(1)• , {oT1 + 1110.001 1 ~~• , 1
Q,16. Th&open loop.....,.runctlonof ■ lalllyfNdbeekconll'olays1Bm d hRte margin. 'onln, A '"'~ 40log l r&2-1+ 40
Bode plot. and from u,.,. plot. determine galn margin an P (Model Poi-r-11, QS J '-pt!IIM•r•17, lltt,2,
• - 40log 102+40
~:
= -- 40 i< 21og l0 -t 40
Gi,'Cll lhal.
1000 : - 40 " 2(1) + 40
G\S) - ·.~ )(O.llOh + I)
~ - 40 " 1 + 40
H{s) c 1
.. - 804 40
To d~e.nnin,,.
(,) Sketch l:lode plol ~ -40db
(u) Gain m3q,i,, Al oo • w4 ,., 2000 raJ/scc
(1i1) PhL~ rnru&in. Ooin, A .. •-.60logljgg8 l+(-40)
Let,
s cjro
=- 60 '1og 2 '- 40
=- 18.0617 ~ 40
1000 -= -58.0617 db
G(l') c jm(O.ljo.> ,.. l)(OOOIIQ)+!)
The magnitude plot in , hown in graph.
!00() ·-
~ G{;Ol) • j ro(l+j).lmHl+jO.OOlu>) Phose Plot
l\l~ltud t Pl 0 t The pb~e angle of system is given as,
Corner Change lo cj, = - 90 - tan·'.(O. lw) - tan-'(0.001 w)
Term Slope Slope
F'requtncy
(D tan- 1(0.lro) 1an- 1(0.00lru) ♦
~ - - 20 -
J 0) I 5 ,11 0.0572 - 95 .7672" ::: - 96°
I - 20 - 20 = -40
I + JO.l ro ro" = o'.i = JO - 20 3, 16.699 0.1 718 - 106.8708" =- 107"

I -40 - 20 = -{50 6 30,963 03437 - 12 LJ 067" ::: - 121.5°

ro,,,, = 0.601 = 1000 -20
I + j0.001 oo
10 45 0.5279 - 135.5729" ::: - 135.5°
Let, 30 71.565 1.7183 - 163.2833°::: - 163.3°
ro, = I rad/s.:c
60 80537 3.4336 - 173 .9706° ::: - 174°
oo, = 2000 rad.lsec
JOO 84,289 5.7 105 - 179.9995° ::: - 180"
Now,
At (J) =001 = I rad/sec 300 88.0908 16.699 - 194.7898" ::: - 195°

1 600 89.045 30.9637 - 2 10.0087° ::: - 210°

Gain.A= 201ogl ~ ,
1000 89,427 45 - 244.427" == - 224.5°
= 20 log I(}'= 20 x 3 log l O= 20 x 3())
2000 89.713 63.4349 - 243.1479" :: - 243.2°
= 20 , 3
The phase plot is shown in graph'.
= 60db
From ~ph,
co = ro = I Orad/sec . ro., = wP< = JOO rad/sec
1
Gain.A = ~logl ?8°1
And, ci,,<:= 0°
= 291ogl0' .
Hence,
= 20 x 2 log JO
Gain margin, G.M = 0 db
= 20 X 2(1)
Phase margin, P.H = 180° + ci,,,.
=20 x 2=4-0db
= 180"+0"
Al co = l 000 rad/sec
'l
= 180"
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. ~. _·; .:.i / ~-SPECTRUM All-lN·D~EJOURNAL FOR ENGINEERING STUDENTS ;',.~~ t'· \, ,• I 7
\1 '•l,

CONTRO L SYSTEM S IJNTU -K~~ UNIT -3 (Ftequenoy Re1poft11Milysl9 )

·<f':.3•;24'' 3 .25
C1lcul1tlon of Gat,i Cron Over f'rtq Uency
Consider the polar plut ~own in f1 gun! ( l ).

I•,.

a17 .. Expl~ln the procedure for con■tructlng the polar

plots . Apr11-1 a,9et.-3,Q5(1) \
\
\
OR \
""~ •J
Doflno polar plot' AugJS.p,-09 , S.M,06(1) figure
~ ....++--,-H-C-4--,....;"4..;--4-H ;......+..;..,;.+,;,,..-4c;"'-'.=; OR Now. in order to determine the gai n cross ovtr frequ;:ncy
of this polar plot, dr aw a circle with origin as center and radiu~
Explaln 'the _polar P!~ts. "
I cm. The point at which the polar plot inter.ccts the circle is
,o (AugJS.p,-Ot, 1141-2, 0 7(1) I Ap~I/May-4', S.t-3, 07(1) I
A . At point A , JC(/w~ = I and the corresponding frequenC) is
Ans: Nov./Ooc.-04, Slt-1 , 07(1))
gain cross over frequency (0>,. ).
It is a plot in which both magn itude and phase are drawn
0 11 a single plot. The polar plot of a sinusqidal transfer function
GV(I)) is a plot ofmagnitude,of G(iw) versus the phase angle of
(j(jo>) on polar coordinates as 'w' is varied from zero to infinity.
Pol,1r plot is useful in determining the stability of a closed loop
C ' sys tem fro m its open loop frequency response.
;.- J- Procedure

1. Let G(s) be the given transfer function . Figi,rt 12)

2. Subst itutes= jw in the transfer function G(s) to obtain Calculation of Phase Cross Over Frequency
2 G(jw). Ln order to calculate phase cross over frequency, we have
to locate a point of polar plot which intersects with th e negative
3. Eva luate \G(iw)I and L G(iw).
real axis (i.e., - 180"). In the polar plot shown in figure (3 ), point
4. At w = 0, Calculate IG(iw)I and L G(iw). B intersects with negative real nis . The freq uency a1 point B is
At w = 00, Cal~ulate \G(iw)I and L G(iw). the phase cross over frequency (wpc).

5. From the above obtained data, sketch the corresponding.

po lar plot. ·
The major advantage of the polar plot lies in stability
study of the sy~tem. Another advantage is that it depicts the
frequency-response characteristics of a system over the entire
freq uency range in a single plot.
The only disadvantage using polar plot is that it does
not clearly indicate the contributions of each individual factor
of the open loop transfer functi_on. Figura 13)
The point I L- 180" is same as point - I + j O. which
018 . Discuss the calculation of gain crossover is called as critical point in polar plots. The gain cross over
frequency and phase crossover frequency with frequency and phase cross over freq uency are often used to find
respective to the polar plots. stability of systems.
Q19 . Explain how the type of a system determines
Ans: (Mod1I P1p1r-ll, Qe J Ap~UM1y-16, S11 ◄, QS(a) the shape of polar plot.
Ans: llug./Sop.:07, S.t-2, 06(•)
Gain Cross Over Frequency Polar plots are one o f the tools used in frequ ency
domain analysis, wherein both magnitude and phase angle are
The frequency at which magnitude ofGUw) is I , is called
ple t1ed on the sam e paper fo r different val ues of w. The basic
as gain cross over frequency (w,). requirement to draw a polar plot is to have~ transfer function .
Type of a system is defined as m1mber of poles I~ing at
Ph~se C ro;is Over Frequency
origin or number of roots of the characteristic equation lying
The frequency at which the phase of GUw) is 180', is at orig in. The ty pe of a system has a great impact on th e shape
of the polar plot. This ca n be shown by drawing the polar plot
Graph: Boda Plot for Gisi = (O I 1000 ca lieu as phase cross over frequency (0>1,J
s , s + I) (0.00ls + l ) for different types of systems.

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.,-: 3.26 UNIT~!t (l=requenoy Response Analysis) 3 .27
11 • magnit1l(lc is infinity al au angle -1,
\\1,cn 111 - IJNpllry •u(jW!
C,ut I: 'fy~0 Sys.tcm, No Pole 111 Ortgln
·I line ,houl,1 he drnwn nt - 90". When 10' - 'l1rl' <JfJ'
LeL G\l) b<, 1M transfer function oftypc,-0. system" hich llrnn· n par,111<
• .." phnw nngk - 180°, hence !1\c ~)I
is glven_by, mn11.n1tud~ \"- 7 1.1 01 1 co • O
• oo•· nnd g<'r, townrds 1.cro nt - 1SO• Qs M(ti\l'"' -
w• -
1 no,, mo,·c~ th""' 111 ♦IIJ!) • ;- lll(f
G(s) ~ - - M/u>)•O
1+ j r.oT ,n figure( ~) ~CO)• 21(]' - Wf'
Rc'llli..0141)
linnga\JI) ;1,t5lJffi)
O,-<t l611'
G(iw) = - 1- 210·) ._ ~:r
1+ jwT
I
IG(iwJI~ M,w) = ~
vi+ or r- m:.s O
M1,u) = I I'/11' - 90"

L G(iw)= ~ru ) = ~ lMl ... OJT ... (2) wui ~ o

• ISO Figure (3~ Polar Plot ID< Typt--2 Syrtem
From equstions ( I) and (2), we ha,•e. I Hence, from the three cases considered, we can say that every time as the type of the system increases by I, the polar plot
- ISO
= I, O(l~) = 0°

\
(i) When OJ = 0, .H(w)
ro = roo shifts by 90° in clockwise direction. The tabular column shows polar plots for differen.t type and order of sys~m.
(1i) \Vhffi Ol = oo, ,11(0J) = 0, 41(1~) = - 90° MtCO\ 0
Polar Plo ll
Q(CO) = 0(/' S.No. Type r er TnnsferFunctlon
From above .datn we can sec that when w = 0. magnitude

F
is I and phase angle is zero. Therefore.Che polar plo15 strutS from 1
I. 0 I G(s)= t+sT 180
0° 1.c .• fowth qundr.un at • point I. When w =oo, magmtudr is 270 -9(!'
creasing
zero and phase an_gle is - 90°. Hence. the polar plots no" mo, e I
from I and go to zero at 900. The polar plot for typc-0 S) stem G(jW) = \+ jwT
Figure (2): Polar Plot for Type-I System
is sho\\11 in figure (I).

~
Case 3: Type-2 System, 1\vo Poles 31 Origin 1
2. 0 2 180
G(s) = {l+sTr)(l+sT,)
mg
I
1
'
G(j<il) = (I+ jwT1)(1+ j wT2 )
I

·~~·
C (s) = - , - --
' s· (l+sT) 1
3. I 2 G(s) = s(t+sT)
-~ ~
..,.,
M(0>)-0 1 .
G,(iw) = -- -ro-2 (-1 +
Q(o,) - - 90" I - 1-·w_T_)
G(jOl) = jw{ I+ jw n ~ 90

I 270
IC,(Jw)l= w = , ~
w- v l +w T 4. 0 3 G(s) = (l+sT,)(l+sT2 Xl+sT3 )
'-90" 180 -
270" 0
Figure (11: Polar Plot for Type-0 Syrum LC2(jw) = w = - 180° - tan-'wT G(jOl)
I
(I+ jmT1Xl + jwT2 )(I+ jo,T3 )
'----V~~;,""
90 .;;.-'"'
Case 2: Type-I System, One Pole at Origin
From equations (5) and (6), we have,
Le.t. G 1(s) be the transfe r function of a type- I system I 270
given by, (i) When w= 0 I 3 G(s) = 180 , ,-.,
s( l + sT,)( l +sT, ) 0
I M(w) = oo
I !II iocreas'rlg\\
G,(s) = s( I + sT)
<!>(W) = - 180° G(j<ll) = jw{ I+ joi'I;)(I + jwT,) 90
I
(ii) When w = = I 270
G,(iw)= j ro(l + jwT) 2 4 G{s) = s2 (l+sT,)(l+sT2 ) OJ:?
IG,(jw)I= M(w) = __ ~ ... (3)
,lf(w) = 0

<l>(w) =- 270°
I 180 Do
co.,,I +c,:i2 7' 2 G(jOJ) = (ioi)' (I+ jw Ti)( I+ jw T2 ) 90
Whenw~ o, magnitude is Zero at an angle - 1~0°. 'ft

~
L.G,(jOJ) = $(w) = - 90° - tan- 1(w7) ... (4) I .
a parallel line representing infinite magnitude is drawn at-I 7. 2 ',. G{s) = s'( l+sI;)(l+sT,)( l+sT,)
From e<juations (3) and (4), we have, . '
Whcnw =~. magnitude is zero, hence the polar plots noW • .
1 0
(i) When w = 0, M(w) = oo, $(w) = - 90° G{jW) 180
to zero magnitude point (i.e., origin) at - 270° as showrl in' U"')'(1~Jh_x1t j<i,T,)(I+ jwT3) 0
(ii) Whenw ,. oo_ M(w) = 0°,il>,(oi)=- l80° (3).
•
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 UNl'f'j (Froquonoy Reaporim, Analysis)
' ""' 111 fn.'qt1C1><:)' domAln 1! g1vcr~br, 3.29
20 1~ unn'-ft, I , 101.
Q. · : h• t 1• th • en,ct
of adding • r.ero at ortgll'I to
JOI 021. d1, cuu the efftct of adding one pole and one
• open loop tnnehlr functlol'I on polar plot? 1 zero (almultaneousty and uparat4ty) 101 given
(;/ "'' (I i J <llfi} tran1fer function on tti. polar plot.
Aftsl 4p,11,'lllay..a Sot ... Ol(b)
t . 81111 ph~~r onl:!,k mc l'.', lvc11 by, Ans:
"\,'(, 'CC the em.,, <•I "'1d,oi, a 7etf' " " lh~ f'('IBJ r iot. l,t 11lC OlA~lll I\ll \

IIJ C(ilhldcr 11 S ) ~t= "·•th 1,,., """'-

Cl!Ji' I "'(o>l - \/ I t w' fi'
<•)'
-
I o.12
I +<u'7i2
1~· _l! l \l"II
Lot u, comidcr II open loop truns for function orn1y,1.cm
hy,

(J(s) • _ I_ ... (I )
(1T + IJ
n lP
Let, • - fro then, Flpro t21
From figure (2). it ii clew lhal, the additwo or a pole ID a
1
11.., tran<fe, tuncti<'n Ill ~'10 .domain .,, l''"" b) . GUro) "" - -- tr.Wfcr function iJ1 polar plot •!hi.tis the plol by 9<f ill c;,,,,fcwlSO
1 (1+ JroT) direction. Similarly, when a rtJO-UfO pole lS added ID a b1Sfu
v'\J<ll)e L G(J<O) o(rn) {Lr&:tJon. ii ro<ate:, the plol by <JO" m dockw,:,c dm,ctJm rs c:, -, - .
() + / (1) 7;)
Now, the polar plot for the transfer function is sbown 1n £fft<t or Adding a :Uro to a Tran, ftr Function i• lb Polar
l ne n1agni1udc" l!'' en a.'- Plot .
figure ( I ) . ,_
1
•Guw> m1w1 -;== -2 9()> Wbrn a , ero ,s addc,l to a ltanSfer funcuoo rq,r=l<d
,,1+ r, ol in equation ( I ). thm tqllation ( I) becomes.
1k phase aoi:le " i;"= .,,., Then. G(s) = - -
.r + z
l+ sT
~ G(lto) \'l(W)
Let, •= Jw
-r.in ' roT,
G(iro) ~ j oH z - · /3)
=:::> ()(0) • o. m(O ) I+ jwT
O(oo ) 00. m(cc) = O The polar plot for a tl'1lllSfer function represmt.ed ,n
t equation (3), will be as shown 1n figun: (3). No"; tl:e magnitwk

.
,..~~·,
·•;/'
~
m figure (11
Q) r

(I)
0. m(W)

oo , m{W) :
I, O(W) ~

0. O( ClJ) = -9{r
O'

polar p lol for \he ·g,vo, L-.msfer funcllon ,s sbo"n

- ~70C'
=»=*
,p(0) = 90', ¢( oo) = O',
w =0, ro = 0, ro = 90'
w= oo
270•

Figuro (1)
IS g,vm by,

IGJWJ

9[(w)]
= J/;;;
= 1an-•
2] and the phase angl< is gi, eo by.

( ; ) - 1an- 1(w 7)
t:rtt, j I Effect of Adding a Pole to a Transfer Funcflon lo 115 Polar
ro = r. • ro = O'
I
Plot 9()

The polar plot of the system is shown in figure (2), Let tpe open loop transfer function be,
9<f'
w=- w=O 1
G(s)-= - -
( 1, 0)
oo (1 +sT)
180° (0. - o0° I W=O 0,

J
Clockw ise
Now, ifwe add a pole at origin then tile IIUnsfer funcuon (z. 0)

becomes,

180" (0. 9<f') I

G(s} = s(l+ sT)
Figure (11 270
Let, s =Jw
ca~2 Figure'l3i
0
z1~ ... (2)
Now, if a zero is added a1 origon ·s ' to the tramfer G(jro) = jro(I + JwT) From figure (3), it is clear th.:u, wb,,n • zuo is aJdod to
Figure (21
funouon. the 1ransfer funcwoa become, n transfer function, the high frequency portion of polllr plot gets
. Therefor.• addition of zero rotates the plot by 90' in The pol"' plot for the trunsftr function in equation (2)
s anuclockwtse dorccllon as the phase angle at zero is reduced ronuod by 90' in coumer-doctc...i se dinection.
G(s) ~ (l + s T,) 90° m antic lockwise direction. is shown in figure (2).
>. ,_t': i- '" •• -l

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 OL svsr EMS iJNl U·KAKINA l)Nlf-3 (M'8(]u9ncy ResponalJ A~l'fS19)
3,30 CONTA a.at
( 1+_1).~ (1+0.251)_ _ ■II I he J)Olor plol using fC(.1uOguJar coordinate• ls ,ke1ched us \ hown in following graph
CliU. ''fheopen looptran1h!rfunctio'1 of unltfecdbllc~ system ta gtvon by G(s) " 5 1 (1 + o.005s) ('1 + 0.001 ■) '

' polar plot 1.nd determine phase margin.

AM1

<.i(., l , . (I " O 3:!l_( ~+ 0.25 q

., 't 1 • ooos,·i 11. o ooi.)
PuL J ' /l\.l

G\ iml _ __l) • 02'<lll_( l • 0 t.'f(O)

V"')' {I• 0 OOS1wl( I • 0Wl1<,,)

'The m:agmrude 111,d rho><: •n@le an, given b).

"I '![lliI udc,

And al"'.

Phase angle. Q(w) - ~G{,<:l)

⇒ LG(JCl)~ un ' 02 w - um 025 w - 2"0' - lllll 1 0.005<:l - l:lll ' 0.001 w

The magrurude nod phas,, ulue< of G{,w) 1s.

"'(n d/s,c) IG(iwll LG(jw)

0.8 2.0( 7 - 249.87°
0.9 1.42 - 247.42'
1.0 1.05 - 244.99°
I.I 0.79 - :?42.59°
J.:? 0.62 - 240.21°
).5 0.33 - 233.26°
1.7 0.23 - 228.78°
2.0 0.15 - 222.32° Graph

From the grap)l, the required phase margin is,

The real and imaginary values of.GUw)
y c J80° _: ♦,..

w Real part lmuginaty Part "' 180.- 245°

(rad/sec) of G(jw) ofG(j w)
': -65°
0.8 - 0.69 1.89
Q23, Sketch the polar plot for a given open loop transfer function.
0.9 - 0.54 1.3 1
1.0 - 0.44 0.95 10
G(s) ., ii(•+ 1l(i; + 3)
I.I - 0.36 0.7 Aprfl.tlay,1 5, Sc1•2, QS(b)
Ans:
1.2 - 0.3 0.53
Given °tb.1t,
l .5 - 0. 19 0.26
G( ) 10 10/3
1.7 - 0.15 0.17 s ; s\s + l) (,+.3) - s( l + s) ( I + 0.33s)
2.0 - 0. 11 0.1 3.33
Ci(,) ~ ,·( I + s) cI + Q.33s)

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. ace proceedings.
 uNIT-3 (F1,9<1uency /'leopon1e An11Iy,I0 1 a.a3
3.3:1
l'u1, • t Jw
3 " + oJl ;w)
G(l,o) - -,ID( l◄ j;,lti°
Th( n1agn1tudo and phlis<, angle or< ~t\'t'n h) ,
M agni1udc.

3.33
0.33w .',o' ◄ 10 183-•1' + 9183
IG(!w)1~
10
- 09__ ·-
,o,'Gl' • 10.J8J m1 - 9183

And also,

Phase, angk, ♦( Ol) ~ L G{jM)

- 9Q• - tnn" 1<i) - tnO I Q.33 0)

w(ndl•••l IGQ1ull L GQw)

0. 1 33.11 33 - 97.6
0.2 16.2895 - 105.09
0.5 5.8768 - 125.93
2.2358 - 153.26
1.5 I 1035 - 172.64
2 0.0214 - I 86.86
5 0.0677 - 227.47

Now,
Eqwnion (I) Clln be wriucn as,
GU ) 3.33 3.33
"' • Jw ( I +/w) ( I ➔ 0.33jco) - l.33w 2 + J(w - 0.33oi3)

3.33 ( 1.33 w 2 - j(o, - 0.33 .,i))

2
'! - 1.33 w2 + /((/) - 0.33 (/)J) ] x (- 1.33 w - J(w - 0.33 wl))
Graph
-4.4289 w 2 ) J.33 (({> - 0.33 WJ)
1.768 w' + (w - 0.33 w1) 2 1.768 w' + (o> - 0.3Jo,l) 2
Th• reol and imaginory v~lues ofG(/w) Rre,
State the NY9_!'11t 1t11blllty criterion.
(Ap,1"111ty•17, Stl·2. 01fdJ I AuvJStp,-41, S.t,2, 00(•1 I
OJ(rad/1«) Rul Part ofG(Jw) Imaginary Part or G(jw)
Aug./S1p,.07, Sol-I, Ql(1) I AufJSlp..07, S.t~Qll•I I Ap{~y-47; Sot-4, 09{1)1
0.1 -4.38 - 32.83
-4.24 - 15.729
OR .
0.2 ' :.\ ~

0.5 - 3.499 -4.759 Moy/Juoo-15, s.t-2, 01(•1

State ti)• Ny.qui•! 1tablit1y th~orem.
- 1,997 - 1.005 OR
.,. 1.5 - 1.0.95 ---0.)41 State end-expl1(n Nyqul1\ 1t1qJllty crtterl~n• Ap,11•11 , 111-<I, Ol(bi

2 ---0.6)7 0.074 OR
s ---0.046 0.049 Expl1ln Nyqulat lilabUlty..~ffl•,rlQn,

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3 .34 tJ JIII r -3 ' (Freq[J11noy Ro~ponieA,i111y-,l1)
3 ,35
1•prll/loUr-Ot, s., 1. ~•I I "PIIIIMty-Ot,
Ans: - 11,r J111
. l·n\ 1r,,11, ,11- f1pcn~l0l'I' tf\-'t.lucnl!)' rcspon,c
"'l~-qu1,l l!l'ab1hry 1i.T tltt'\(''U} dr~mun~(. tbc "1Rhl hh C'1 lH ll\1-\."\i 11,"r . ' c;,t
loop !>"ll."'

C'(d G(,)
R{~) = I -+ G(,)H(,)
I the kll hall of.,-rlott<. The Nyqt~J,I
for ,rah1lit), all roo1s ofthe dl111'8Cl c,r1'1t~ equDtl<>n I • G<.,)H(.• )-O mu-I ,c m I f I , G(,)/.J(s) UU\I ·
(.nlcnon nd.'1\c-, the "fX'TI klop freql.£llC') ,-n;~ 'O....~. (r\,o,)H(,ol) 1l' the- numtx-1 ~,1 zcr\X""' :rn~t pv C'~ u · - ?(I'

~l h>lt ot •-rlJnc 026. Explain the procedure to con•truct the Nyquist Plot. '-"'11, s.w, Ql(• i
r(Wh.h.·, tJ1i: information nbout the
l\vqu1,t ~ntenon , .. base-Jon tbt- lJk.--orem of the- comrh:, , anablt"" 11 3 ~l, p
1 d
t, 11ti, ._, f O ,, , ti:m nnd t.!11.~ cgrcc o ms
r· I OR ,
-
stubtl1t, t'l a (Vnt:rol ~""'1=1C1Tl lt do- indicate \he dey~ of -.tab11lt, 1.c.. re-lame ,t.1 • ~. • • h . . Enlist the step-by-step procedure for the construction of Nyquist plots.
· . · p Lhe 1n.-qucncy d0111a111 c aructensli
:m unjlJtilr ' ~~tcrn 1·n e ~uctk."') do.mom plots ofGvW)Hvu>) ~ \ C'S mlom\Jlll' " l ' . Ans: tA"OJhl>.-GI, Sel-2, 09\<l I A"9,Jlop.47. 5"1-1, Q6(<1 I~ , 0 1. Sol"'- 06(0)1
d<1~d ln,Jp\~'-llml . ,· . The following are the steps involved in lhe construction ofNyquLS! plol
. f comrlcx van obles which 1s refe Step I
.\, ,tJtt'd S~qm.st ~r.ahilit} oih::non 1s based o n Ca-uch~ ·s m:1Juc-thC\.lt\'Ol O ·
Represent lhe given transfer func1ion in lime constant fonn i1., ( I + fl).
··pnn('tplc ,,1 3(}:.~menC.
Step 2
h finll< number of the poles in the
PntK rple o( rup1men1 is s,,ucd a,, "k t {_1(s) be a ,mgk ,·aluoJ funcnoo thJI as 3 Frame the magnilude an<l phase nngle equation of the given trarufer function.
th d ot 00 through :111y one of lite pol~, Slep 3
' UJlJ'OS< that an orb1tnlr) clo5"<1 p.1th T0 " chosen m the .<--J>t.:u,, 5" that the pa oes n - _ ,
Detennine the values of magni!Ude nod phase angle for w = 0 and w = - .
,,t' Q(s): the com-spondms T locus mapped in the Q(, \ plane ,11II encircle the ongm as many times as the difference be Step -I . .
P I T"
nwnb<.-r ol zeroes a.nJ the number of poll-, of Q(s\ !h.ll ore cnmckd b) the s-plllle ocus p Draw the polar plot from the values obtained in step 3.
Step 5
Draw the Nyquist plot by following lhe steps given below,
\\"here. (i) Take the mirror image o f the polar plot obtained.
\' •- :Sumb~ of ennrdcments of the on gin made b) the Q(s)-plane locus T0 (ii) Close the plols from w = 0- (mirror image) 10 w = o· (acrual plol) lhrough an angle mt tn anticlockwiu direction_ when:
11 is the type of the system.
Z = ;-..umb<.-r of zeros of Ql•) in RHS of s-plane enmcled by the s-plane contour.
However the plot can also be closed in clockwise direction but the numbe:rof encirclements N, to the point (- 1 + ; 0) must
P ; Number of poles of Qts) m RHS of s-plane encircled by the s-plane conlour. be taken as negative.
025. Write short notes on comparison of Polar and Nyquist plots. Slep 6
Mark the point of intersection of the plol with negative real axis, say (A) and dotennine the djsiance of this point from
OR origin (OA), using formula,

Distinguish between Polar plots and Nyquist plots. OA= m\ _.,,,,

Ans:
Where, m is the magnitude an<l°'ii>,,_is the phase crossover frequency.
Polar Plots Nyquist Plots w can be obtained by substituting <j, = - 180° in the phase angle equation obtained in step 2.
Step 7
"' '
.
I. It is a plol of magnilllde and phase angle of open I. h is a plot of magnitude and phase angle of open Now check whelher the point (- 1 +J O) lies inside or outside !he region 0.-1. lfi( lte:s inside then go 10 step 8. clse N = 0 go
loop lransfrr funcuon as a function of w (frequency) loop transfer function as a function of w and - Ol. to step 9.
Slcp 8 _
2. It can not he directly used exclusively without any 2. It can be used exclusively without any computing
Determine the number of encirclemenls (N) mnde by Nyquist plo110 lhe point - I + ; O,
powerful computing Lechniqul!S. lechruquc.
Step 9
Using Nyquist crilcrion dc1ennine the s1abili1y of boih open loop nnd closed loop syslem.
3. It does not d,rectl) yield the frequency charac- ). It gil'es complele infonna1ion about the frequency
Nyquist criterion is given by,
1enstics of n system. characteristics of a system. ... ( I l
N= P- Z
... ( 2)
4. It is plotted on complex plane. 4. It is also ploucd on complex plane. P =O
Where, pis the number of poles in RHS of s-plnne and Z is the number of zeroes of churacterisric eqlllltion will, p0si1tvc
5. Polar plot of the II1ln5fer func1,on, 5. Nyquis1 plot of the lransfer function, rcJ I parts. ·
From equation ( I) dclenniilc th< vnlue of Z. lf Z ~ O'. S)'Slt!lll is stable else unstable. _ _ .
I
G(s) = is shown as, G(s) = I If equation (2) is satisfied then open loop sysiem ''. s111bk else unstable. P c:111 be de1cnnm<"1 from the gl\en transtw
(l+sTt) ( l+sT, ) ( 1+ sT,) (I + sT ) is sho""' as,.
2
function.
r •
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~-36
. d nu tonu,tic con1 m l nnd signr1 l procrlslna UNIT-3 (Frequency Response Anal ys fs)
It 1S ~j+ 3 .37
027 . ' In addition to providing absolute stat>lllty, the 3. USC' Ul

Nyquist criterion also gives lnfonn1tlon on the nssr ssing t.he stnh1lity.
Pha..ea ngle , ¢, =- ta n- • ( ~ ) - u n ' <J<f!)
relative stablllty. Justify. Ft,:qucnt') rcsp0nsc ofbolh ,e,ics and pnrn llcl comr · 1-w'
4,
.Ans:
• One o f the nd\'Ontnges of Nyquist cri terion is that il gi,•cs
sysh~m~ can be l,MninL~d using Nyq uist plot. •
=- tan - • (2 x~
I -~'
)-tan
infommtton o frdaltw stability in additi on to nbso lute stnbi lity s.
of n control S)'Stcm without finding the rool5 ofthechamcteristic = - 1an · 1r-)-lan - 1(=)
usmg 11.
equati on 1.e .. locnti on of closed-loop poles. = - 90, - 90°
,s widel) used in the design o r ckctrica l sys1cn1.
R ,lativc Sru bUi~· 6. 11 = · 180"
is u..sed 1,, detcm1inc th e stnbi lil y o f closed loop S)•S
The rdati\'c sta bil ity indica tes the closeness of the 7, 11 From the above, it is c lear that the Nyquist co ntour.st.arts
system m sw bk fl'gion. lt is an iodication of the strength or without computing thl' i.' loscd loop poles . at ~ with magnitude 5 and ends at¢, ~ - 1S0° witb magn 1lu1.k
de.,-re,, of stah ihl) . Figure 11): Nyquist Contour z.cro. The mapping of action C, fro m s- planc to G(s) plan: 15
It is u,,clu l for sta bilit) analysis of system s with
8.
In the lime d om a in. relative stat,;lity is measured by delay which is not possible wi tli ,1 ther frequency anl\l Mnpping of section C, shown in figure (2)
pnramctcrs , uch as maximum overshoot and damping ratio. -Z? rf'
Fro m equation (2), the m agnit ude and phase ang le is
techniques. )ID
In fn.>quency domain. the relative stability ofa system can give n as,
1-., stuJicd from Nyquist plot. The rclati\''C.' stability of the system 9. Using Nyqui st crilcrion ii is possible to get informa ·
Magni tude, M = /G(jw)/
is !c'iven by closeness of polar plot to (-1 .,. jO) pomL As th e polar about closed loop poles location by plon.i ng , the
pk,1 gets closer lo (- 1 + j 0) point, the sys;em moves towards loop frequency response data.
0

lfl SfJbil it~ - = 1((1 - w' )+ j:w)(I+ j3oo)I

The relatwe stabtl it, in fr'C.'quency domain is quanrillltively 10.
interactions in mul tivariablc system s. 5 M~ 5
mcas un-d m te rm s of phase m3.rgin and gain margin.
Consider a G(/w)H(jw) Locus as shov.'11 in figure, whicb 11 . It C'!fl also be used for sta bility ana lys is of loop tr
.Jo- w2 ) 2 + (2 w) 2 •.j, + (3 w) 2
2
- 180"
cros ) the real ax is at A and a unit c ircle drawn ·with origin as function contai nin g exponential te rms.
cc11tc r cuts this locus al point --8. Let a be the angle between ... (3)
n ~_gJU \e real n.xi.s and OB. Q29. Given the open lpop transfer functl
.Jo-w')' + 4oo2 .✓1+ 9w2
5 3
G(s) = ) • Sketch the Nyq Ph~se a~glc, (I) = - , an· • ( ~:, ) - ta n· 1 ( ~) ... (4)
(1+2s+s 2 )(1+3s .. 1
plot and investigate the open loop and clolj For OJ = 0, we have, -91:1'
loop systems stability.
5 Fj1ure (21: Mappin, al soctiOJ1 C, in Gisi pl■ ne
Magni tude, M=
Ans: .j(l-w 2 ) 2 + 4w
1
.J,+9w 2 Mapping of Section C,
Figure Given tha~ 5
ff the gai n o f the system is increased then the locus Jci-o'i' +0)1+9(0) 2 In section C,. the variable ' w' vari~s from ~ to -TC . The
. 2 2
"',U shi ll upwards and it m ay c ross the real a.xis at (-1 + / 0) G(s) = 52
. 5
poin t. When th e locus passes thro ugh (-1 + J O) poin1 A ➔ l (I + 2s + s )(1 + 3s) ma~ping o f section C, fro m s-plane to G(s) plane is obtai ned
= ✓1../i = 5
and u ➔ 0 . Hence, the closeness o f G(jw)H(jw) locus lO the
Put s=jw , by letting s = . ~ R . e1• and by appro:<.imating ( I+s 7) rerm
critical point (- 1 + JO) can be measured in terms of intercept A
and a ngle a. The val ue o f A and a are quant itative ind.icalions Phase angle, cp =-tan· •( ~ 2) - tan · 1 (3w)
l- w to sT.
of re lative stability. These values arc used to de fine gain margin G(joi) = S
and phase margin as practical measures of relative stability. (I + 2j w+ (jw)' )( I + Jjw)
From equation ( I). we ha ve,
The concep15 of gain m argin and phase margin are defined for =- tan -I ( ~) - Inn · I (3(0))
1- (0) 2 5
open loop system but from the va lue of gain margin and phase
2 G(s)= (l+Js)( l +2s + s')
margin. the stab ilHy of c lose<! loop syste m can be j udged . (I+ 2J w-w )( 1+ Jjw) =- tan · (0) - tan·'(O)
1

Q28. Write short notes on appllcatlons of Nyquist =- 0 = 0 5

criterion. ((1-w')+ j2 w)( l +3jw)
For (l) = oo, we have, (3s)(s + l)(s + I)
Ans: Ap,tllllay-OI, kt-1 , Ql(b)
By inspecting equation (I ), it is clear that, there is no pO Magnitude, M = ~
The following are the applications of Nyquist criteri on.
at origin. Hence, th• enlire right half of s-plane is considec
Jo - ~' )' +4(~) 2
.y l +9(~J-
Js.(s )(s )
I. It _is used for st.ability analys is of linear tim e invariant
system. as NyquiS t contour. The Nyquist contour is shown in figure~(J. 5 5
2. ll is also used for determining the relative sta bility of a From figure()) , ii is clear that, th ere arc three sections C·' = .,r.;,..r.; = 3,.J
5

system . and C, '' =0

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SPECTRUM All-IN-ONE JOUR~AL f(IR .ENGINEERING STUDENTS ,' ,; ,-!l~ ._,
 coNiROL SYSTEMS [JNTU•KAltii -UNIT·,~, ([reqliel'ICYc,Reepor,u·
~l•l'Jlln 1tc>r Hellon
Analyels}
,.,,. 3.39
. ' .

Put In section C,, the vnrlabJc 'w' v■rlcJ f,om _.., to·O .ft , . _ .
c,1rrcspo11<llng contou~ ii aho~ in figure ( ). ' cnce, we can say !hat it IS a muror image; o( J«ti<lol 'C,'. 1he
r-/!_v 4
j, - 270"
G\s) = Lt __J_,__3
«- l(R rt"')
- LJ __s_.- ,I
• --- 3.Jtl _;,J30

5 Lt ...!.....,-,,.
.= 3 •-- R3
u Cl'
= ~.,,-,;e. LI ...!.._ -:180'
3 1c-- k1 /

- ~.,,-pe _..!._
3

S ~ .,,-,111 ,(0) C o.e~J3e -~ ~ 'r.,

3
!I
Flg~ra 141_: Ma,pin1 af .-tin c, It llal ,,_
When 0 =
2 , we have. To locate cro11ln1 point on real ■xii
G(,)1,. ll R.l!'<l ~ o.,,-I~ .. , ·una~-...,..
When the, locus crosses th~ ~al ,axis. • the .. .-.· will.. be· zero
· and thc co1TCSpOndmg
· frequtncy will be pbuc
R....-
crossover frequency. , .• · \ ' · .
: 0,e-;,rn From equatjpn (!), we have , ,
;",'c:
G(s)= . 5 ,;·.'•. .S ,
(l + 3s)(I + 2s +s 2) (I+ 3s)(s 2 2s+ I) ,+
When, 0 = -!I , we have,
2 , ·= 5 5
= 3sJ +7s 2 +Ss'l-1
_.,s 2, + 2s + I +3s J +6s 2 +3s
G(s )\,. Lt R.e16 =
R--
o.,,-{f) Puts =Jo>

G(jro)~ '. 5 . 5
1 2
· '3UCll) 3 +7(jro) 2 +5Uro)+I 3/oo +7fro +5J ro+i
From cqW1tions (5) and (6), ii is clcartha~ the section C, mapped from s-plane to G(s) plane in a circular are of zefl! 5
arowid origin with phase varying from
3
-)!I to n .
= _:3Jro3 - 7ro 2 +5Jro+I [ ·.- / =- 1]
2 2
• 5
The mapping of section C, from s-planc to G(s) plane is shown in figure (3).
· · - ~1-7oi2 ) + JOl(5-3tlj2 )

Rationalizi~g the denominator, we ge~

2
G(Jro) = 5 x(l-7ro ) - JOl(5-Joh
2 2
, • (I- 7ro2 ) + Jo(5-3ro2 ) (I - _7ro )- jo(5-Joo )

2
s[c1-70>2 ) -Jo(5-:3ro )] 5(1-7oh-5Jo(5-3ro1 )
2
= (I- 7ro2 )2-UOl(5-3ot)) 2 = (1-7002 / +oi2(5-3ro )'

2
G = 5(1-70)2 ) : • . • 5o(5-3ro ) .. . (7 )
1
U00 ) (l-7ro2 ) 2 .+or(S-'3ro2 ) 1 (l-7or)1·+oo2(5-Jro2 ) 2
When th~ locus irosses tlio ~ aitjs the imaginap' part will be zero and the com:sponding frequency ~ Ube phase crossover
frequency.
I
' Fl1ura !31 : M•Pl!int of uction c, In 6(1) pl■na.

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1
• - • : ,...,, _
1
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coNTAO L - ~ . UNIT ·3 .(Frequency Respon!Je Analysi9 ) ·' 3-.41
~ ow equoting imogintuy part
I fence, the part (- 1 +JO) is not encircled
10 n-ro <,f c,qunlion (7). we get,
N =, O
Also P~ 0
We hove,
N P-Z2

Z r P- N
. =0:- 0
Z=O
Wher\ Z = 0 it implies closed loop ,ystcm is s1able.
P = 0 it implies.open loop system is stable.

Q30. Construct Nyquist plot for a fHdbaek control aystllm whose open loop transfllf function is giv en by
. 5
G(s) H(s) ,. - ( - - , comment on stability of open loop a.nd eloHd loop system.
1 ! -~ .
Ans: ( - , -.u, QIS I Aprilll&ay,19, Sot◄ . 051bl)
= 1.29 rad / sec Given that,
Substituting w,., value m real part of equation (7), 5.
G(s) H(s)= s(l - s) .. ( I)
,.c ., w = w,,..= 1.29 rad / sec ' • I \

It is cle~ that; we have a pole at origin. Hence, choose the Nyquist contour as shown in figure ( I ). The Nyquist comour
We have.
consists of four section C1, C2 , CJ and C,. Therefore, the Nyquist plot is oblaincd by combining all the se_ction,.
2
5(1-7oi2) 5(1- 7(1 .29) )
jro
{I- 7cii )2 + ro'(5-3w2)' (1- 7(1.29) 2 )2 + 2' (5-3(1.29)2)2

5(1- 11.65) 5(-10.65)

= (1 -11.65)2+ 4(5-4.99)2 - (-10,6S/ +o.o4

=- 0.4693
The complete Nyquisl plot is as shown in figure (5).
- 270"

rlJ -Se,tion

Figure Ill

Mapping of Section C,

Put s =jw in equation (I).

- 0.4693
I- Section 5 5
G(iw) H(jw ) = jro(l - jw) = wL 9O° It+ ro 2 L-tan-1w
- 90'
= __5__ L-9O0 + 1an- 1©
. s w ✓l+w 2
Figure (5) : Nyquist plot for Gl1) =
2
(1+3s)(s + 2s+ I) 5
M(_w) = \G(jw) H(iw)\ = wJI + wl ... (2 )
The -Nyquist plot is not encircling the critical point (- l+JO). Th'is satisfies the Nyquist criterion or stability anc:then~
system is stable.
~(w) = LG(jw) H(j©) = -90 + 1an· 1w . (3)

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SYSTEMS [JNTU -KA~INA uNIT-3 (ffrAlquenoy Response M11lyala)
coNTA OL 3.43
3 .42
Joo
f'roru.cqu.-1ion (2) ond (3). "-'' hnvc.
When m c O : ,MI_Ol) •"" : +(co) ~ - 90'

(ii) When oo ~ I : Ml_ru) ~ 35~ : +(ml ~ 45•

\ili\ \Vbcn w = 00: ,l(o.>) • O : ~(tll) K 0

'
Using~ above- d.,ta. the polru- plOI is dra\\'11 such \Nil vv•O)'•v'"
r.."
U<i,n) I '<'LL' ,1,1rt.< Bl
' .
mec1 -0rigin :ilong (l uis when. m • "'·
- 90°

s-planc
)- .
figure 14~ Ml,.... ef SIC11en C,
~lopping of Seetlo~ C, •

pul, s -_ ~-"'
Lt R.eJO ·to G(s)
· H(s) In order to find the mapping of section C._wbc:re 8 vane$ from -~ to I·
5
G(s) H(s) = - - "' ~ [ ·: 1 - sT := I]
. s(l - s) s
fil"re (2): Mapping of Section C,
. ' . 5
G(s)H(s) = ,,- -_-., .,,~.
Mapping of Sec.lion C , Lt Re'"
n n R-0
Put. s = • ~~ R.e • in G(s) H(s) in order m fmd lhe mapping of section C, wh<r< 0 varies from 2 to - 2·
When9 =- f , G(!)H(s) = oo/I
[ ·.· J +sT :=s ll
7t . -1¾
[·:- 1 =•"I When 8 =
1, G(s) H(s) = ooe •
5
G(s) H(s) = - - -- - =
LI (R,"'f el'
O.e...,._.'
i Therefore. the- infinite section_ C, in s-plane is mapped as circular arc of radius oo "arying from ~ ro _ ~ as sbri \~-n ,n
•-~ figure (4).
2 2
jro
When0 = !!_
2•
G(s) H(s ) = 0 ;~'T" •l = 0.e"'"

When 8 = - - G(s) H(s) = 0 e+~··l

'I. • =O.e""
Therefore. lhe infmite secuon C, 1s s-plane is mapped as circular arc of zero radius around origin as shown i~I fi1 :
jw
- 270°

.oo

- 90°

figure 131: Mapping of Section c, Flgu11141: Mappin1 of Section C,

Mapping of Section c, . Now the mopping of cuch individ wil section C, . C,.
_ C1and C-4 me combined io order 10 obt·a m
· ti,e ..:011·re .Nyqu1st plot 1s
Herc, u, varies frQm -a, to 0 and locus of G(fw) H(.iw) gives the mapping of section C . The lo . • .
15 •' G(s) //(s} plane uod is shown In figure (S).
G(.im) H(.iw) and 1s g,ven by lhe mirror image of polar plot with respect to re•I axis. ' cus mvcrse polazc

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 f
L SYSTEMS uNIT-3 ,)frequency Responde Analysis )
coNTROJ~:.:_:.:..:..(JNTU- t~~
,_------ 3.45
111e' Nyqui~t contour ha., four scttion.< C,, c,, C, and C,.
j<l> ~tuppln~ or section C,
- 270° Section C, varies from Oto <» ,
Substituting s =Jo> in equation (I), we get,

G;) I
(I) = Jw(l+J 0.lw) ()+ j(I)) = j w(l+jw + ~ O.lw+ /0.1<1>2)

--+ -+ -- -r -0°
- 1soo • -1
/co(l+j l.lro-Q .lro2)
I
joo(l-0 .lco 2 + jl.lro)
1
= (1- 0.16>2) /ro+ /l.lw 2 = - l.lco2 + jw(I-O.lro2)
When the locus of GUw), crosses real axis, the imaginary value
will be zero and the con:espoodmg fn,qumcy will be phase
crossove r frequeocy.

Atw = wpc

,,
- 90"
oo( I - 0.lw')l. - •,c =Q
figure {51 · b h
. le the - 1 + JO pomt ut t e w,.(1-0 . J w!c) =0 ,
. ur in G(s) H(s) doesnot encirc
Hence, from the above :malysis. !he !-.')~I co:us the system
is unstable. I - 0.1 ro!c = I
tr:1I1Sferfunc1ion has one pole ngbi.half ofibcs-p ane.
. t lot for the open loop
• ·terlon Draw the Nyquis P w2 = J_ = 10
Q31 . State and explain Nyqul 5 I st•biloty en ·
· d loop system and dete pc 0.1
1 and discuss the stability of the c 1ose
functi on G(s) c s(1 + 0.1sl(1 + s) w"' = v'w = 3.1622 rad/sec
~ ,
relative stability.
(Model Paper-N, Qe IApt1t-'I At w = w"' = 3.1622 rad/sec
Ana:
I I
Nyquist StJ1bllity Criterion G(/(1_)}-=' -1.tw!, = -l.1(3.1622) 2
For answer re for Unit-ID, Q24. 1
=;- - I0.999 = - 0.0909
Problem
The open loop system is type- I and third order system. Hence,
Given that, the polar plot starts at - 90° axis at infinity, c:ros.scs real
axi, al -0.0909 and ends at origin in second quadrant. The mapping of
section C, is shown in figure (2).
G(s) = s( I + O. ls) (1 + s)
jv
To determine,
(i)
(ii)
Nyquist plot
Comment on stability. --=L__ --------u.
~
. •
By mspecun g th e S'J stem, it is clear that there is pole al origin. Hence, Nyquist contour will be as shown in fl
jw

Figure 121
c,
Mapping ofSectio n ·c,

Mapping of section C, from s-plane to G(s) pl ane is obtained

by lettings ,= R~'« R.e /0 in G(s) and vacyiog
Ofrom , lo - I I
and approximating the term (J+sT) to sT. i.e.,
c; ( l+sT):: sT

G(s) = s( J +O.ls)( l+s)

I I·
Fi~ure 111 = s(O.ls)(s) ~? ',
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 SYSTEMS \JNTU -KM IN
coNTRO L . UNIT•3 (Frequeocy Ae1pon~e An,lya ie]
,. 3.46 ,3 .47

0 f" ,-,o G(,) - 1

,_..
f .-,. • - - = O.r'"
, LI R oo .,/11

\\'hcnO " Whcno •·- f

G(s) ~ 0 r
·,
G(1) • O.e
+tl = O.,
1t
n
When 0 - - When .9 = '.!!.
2 2

G( , l - 0,
+,·\ f·\ ~ O"
-.....4 o, circular .u" of ,ero rn
d'
.
nround origin in G(s) plal\6;
I

G(,) = 1,1
o·.e- 1 = 0./
~
2
' I

I rom the obo, ~ ii II cleanha.1, the sectton C " m•r,~v HI!' •

'
The mapping of section C, is •fuiwn in figun, (S).
argume111 vnrying from - 2 3n 3rr . " ~•~ (3)
10 T ILS sho" -n in,.,;- - .
jro
ja,
jv,

~
00 C.

~=r
= \
u
a ⇒

~/

Figure Il l
I C

Mn pp.lo~ of Section C _, Figure (5]

..., ,an·es from_ a. o The mopping of section C, ,s given by the locus of The compl!'fe )'.lyqui_s1 plot is shown in figun, (6).
ln this section, th c val ue O I " ' ' 10 G(jw) as <ll
from - "' to O and " ill be the inverse polar• plo1 of G(/<iJ).• llie mapping
of sccuon C, ,s shown ·m fl, gure (4) .
· · JV
Cl(1)H(s~plano .
jv

J IU )i
+ w cc
o ~ -_d}-►u
Figure 141
I

Mapping of Section C,
The mappin g of section c, from s-planc lo G(s) plane '
can be obtained by lcning s = ~'~
R R.e"' in G(r) plane an~
O from - .';_ to _I;_ and approximating ( l +rT) 10 fleurafll
I. i.e., Stablllty
2 2 •
On inspecting the Nyqui~tcqplour, ii is clear lhat the
( l +sT) ::: I point - I+ JOnot er1circled. Hc11cc, Ilic given system
is stable.
I l I Q32. Theopen loop transfer function oh unity fffdba ck contro
G(., ) ~ s ( l + 0. ls)(I +s) = = -:;: . l systam Is G(s) • (
s(l ) (l )
stability to detarmlne tM, critica l value of gain s+1)(K2s+1 ) u . . Nyqui
• st
'K' for stablll ty.
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i
k,.,.______________ '
 l',I STEMS (JNTU -KAKl
coNTROLSV_:_:_::.:...- - - -.... .. UNIT ,J (Frequ ohcy Respbrnie Analysis)
~.4~8!...___ ___ ____ _---:---::-::-;::;;il
33· 3.49
- --;~
1,kpJc, 1111 ~ " '~
. Jli, c vnluc. we gel. - - . - - .
Ana: ~lllloy -17. Sal-3. 051•1 I freq uen tly Ash ed & Imp orta nt Que
l t1 t) ) ~ ~
stio ns ,
Gh·cnl hat , -- - - • -- - - -
- - - -----
, I c "' rq1101i,m (2). we gel, - --- - --- .;!

K S uh,111\1t ul g. ,,, ' rt ll 01 . Define about freque ncy domai n specJncatlo

(.I) ~ "{i+l)(2s7"TI . ns.
/..
jAprll/liloy-19, Sot-2, QS(•l I Aprll -11, hi◄, 0
,,t,.-! ( o. 27l1-I 1~
To dctcm,ine. 5(•) I Ap,tVM• y-18. Sot-1 , Q~(•l I """y/Jun6-1S,
.l fi\OI) 4(0 .2~l
An•: Refer Q2,
S.t-1 , QSl.t) I
M1y/Jun e-1', Sel•3, 05(•) I AugJSop Alt.
K vlllU; using N) quisl stability crilerion . !lot-4, OSi•J I Ap,WMa-,.01, s.t-4, 05(1) 1""'11/Mrt
K .(T! , s.t--3. ~ )!
a2. Determ ine the ·, ..on ant freque ncy co, rHona
nt peak M. and bandw idth for the system
1..cl, \ ~
i.; who•• tran1fe~ functio n Is, GOco)'" 5
s =jro + "; + (J<.0)2 ·
= ~ /1 + 0.3 136 5 12
K Ans: Refer Q5.
G(iro) = (jro + \) (2jro.,. \ ) i.; (April-11 , S.,.;J , QS(b) \ Aprll/Ma y-16, Set-4 , ~ (b))
: I J.0784 )iJ \ ) 6 Q3. Explai n proced ure of bode plot and determ
K ... (I) ination of gain margi n and phase margi n
= ( I + jro)(l +j 2@) K bode plot.
from
= ~
Now. fAptll/May-19, Sot-1 , 05(1) I ApttllMay-19, s.t-3,
Q5(b) I Aptll/May-16, Set,1, 01(e) I
K Ans: Refer Q7.
K = 1]90iT Ap,11/Moy-4t, s.1-1. 054•1 IApril/May-OIi , Set-3, 051•ll
Magniru dc, A11;ro) " J, + roiJi + 4<112 ... (2)
Q4. Explai n the genera l proced ure for constr ucting
Bode plots.
= 0.84025 K

1W
Ph:ise angle, «ro) = - tan 4 (ro)-w ,-'(2o,) ... (3)

For stability, Ans: Refer Q8,- (Apri!M ay-11, S.1-1 , Q5(b) f Au;JSop..07,
We know that. S.t-4, QS(b))
QS. Write short notes on stabili ty analys is from
0.84025 K" 1 Bode Plots.
Al phase crossover frequency, we have.
I
«ro) = -1 80' K = 0.84025 Ans: Refer QIO. ., (AugJllo p..01, S.1·1, Q5(b) IAugJs.p ..08, Set-3,
~ (b) f Aug.&p.-07, Set-1, 05(1) IAug./Sep
. .07. S.t-2, QS(a))
Substituting ~ro) value in equation (4). we get.. Q6. Explai n the proced ure for constr ucting the
⇒ K = l.1902 polar plots.

⇒ tan-'(ID) + un-'(2ro) = 180

Ans: Refer Q 17.

Q7 . State the Nyqui st stabili ty criterio n.

(April-18, S.t..l, QS(-a) I Au9.JSop.-OI, s.t-1.
06(•1 IAugJSep..06, Set-2., 07(11I
. A p r i ~ S.t-3, Q7(•) f No../Oec..-04, Sol-I
, 07(•)) 1W
(Aprtl/May-11; Sot-2, 01(d) I May/June-15, S.t-2,
Applyi ng Tao on both sides. we gel, 01(•) I April-11, Set-4. 06(b) f Apri~ 9. s.~1. 06(• 1I Aug./Sep..01. S11-2.
06(•) I
Ans: Refer Q24. · Aug.ts.p..07, S.M, Q6(i) I AugJS.p.-117, Set-4,
tan[tan- '(ro) + un-'(2ro )] = tan(\80") 06i•I f April/May.07, Set◄, Qlia) I Ap,Mlay .oJ , S..-3, Ql(b))

tan(tan - I(@)] + tan[tan - l (2ro)] QB. Derive the correlation between time domai n
0 and frequency domai n specif ication s.:
⇒ l - tan(tan - '(ro) . tan (rao- '(2ro)]
Ans: Refer Q3.
ro + 2Cil =o lmportJn t Quesdon
⇒ 1-(Cil) (2ro) Q9. Explal n the pr,ocedure for determ ination of
transf er functi on from Bode plots.
Ans: Refer Q9.
~ =2O
⇒ I - 2ro Important av.soon

Jro = l - 2Cil'
Q10 . Draw t~e 'tiode ' plot and· fin~ gain margi n and phase margi n for the given
⇒ transf er functi on
K
⇒ 2ro' +3ro- 1 = O G(s) = s(1 + 0.111) (1 + 0.5•) •
Ans: ReferQ l _4. •
3± ✓(3) 2 - 2A(2)H l Important Question
⇒ Ol = 2(2)
Q11. Explai n how· the type of a system
determ ines the shape of polar plol
- 3±./9+ 8
⇒ co = 4 Ans: Refer Ql 9.
Importan t Qu~tJon
-3±./17 Q12. Write sl;lort notes on compa rison
⇒ ro= - -
4- of Polar and Nyqui st plots.

Ans: Refer Q25 .

⇒ @ = 0.28, - 1.78 Important Ouution

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':. { .....~ nyona found guilty is LIABLE 1o face LEGAL proceedings •. EERIN.GSTUDENTS;1 . J ,~,:_~-if ,·,
l r
,~'f'' , _
t • ir-w ~.-1:"
- c oNTROL svs
TEMS (JNTU -KAKIN
. .
• f :-.: -: .....

..:_3 . 55(0~------~~:lii~«-~-ml-'
•i3ri9t:a·i-
£-i:1'"'5
1 1
I UNlf
• -
JG
+ '
CLAss1cAL CoNYRoL oes1G~
. dback system w,10 Gisi· s ,; ' OJ •
01 . F'md ~•• n•nt 1requency, monant peak •n
d band widlh ., • 111111y fee

, stem with open loop trans

K
fer function . Gisi• sil • 0.2s} (1 + 0.01sl
• So thllt It
4 ......:-
TECHNIQUES

Determine the value of the gam constant K1or the sy

02. . •
I K. find the new gem margin. Syllabus
pbase margin of about 35". for tllis value o

C LASSICAL CONTROL DESIG N TECHNIQU ES : Lag , Load, i..:.g-1'e.,i coropa=to.-s - O.Slg,, o f <ompensaror, using Bode plots.

( LEARNING OBJECTIVES J
Concept of log, Lead, Log•lead compens a tor

Cr Design of c~mpensotors using Bode plots

( INTRODUCTION
---------------
J
The process of ol~ring the design procedure of a system to mee t desired specl tication Is known as compemation. Howe¥et,
this is achieved by introducing on addltlonol de..,lce coUe.d compensator. The various fXpes of compensaton e mployed o ·Nt lag
comp~ator; l ~od compe n'sa to~ and lead -lag compensator.

Compensatlon 'ln..,olve syst~m redesign by mod ifk:atlon of the structure or by emplo ylng suitable devkes s.o as to meet the
performance specificotiom: This ls o process whldi compensa tes for the deflclent performance of the orlginaJ system., Modlf'KOrion
of the system con b e done In any fonn.

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