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SOLUTIONS MANUAL FOR SELECTEDSOLUTIONS MANUAL FOR SELECTEDSOLUTIONS MANUAL FOR SELECTEDSOLUTIONS MANUAL FOR SELECTED
PROBLEMS INPROBLEMS INPROBLEMS INPROBLEMS IN
PROCESS SYSTEMS ANALYSIS AND
CONTROL
DONALD R. COUGHANOWR
COMPILED BY
M.N. GOPINATH BTech.,(Chem)M.N. GOPINATH BTech.,(Chem)M.N. GOPINATH BTech.,(Chem)M.N. GOPINATH BTech.,(Chem)
CATCH ME AT [email protected]
Disclaimer: This work is just a compilation from various sources believed to be reliable and I am not responsible for any errors.
CONTENTS
PART 1: SOLUTIONS FOR SELECTED PROBLEMS
PART2: LIST OF USEFUL BOOKS
PART3: USEFUL WEBSITES
PART 1
1.1 Draw a block diagram for the control system generated when a human
being steers an automobile.
1.2 From the given figure specify the devices
Solution:
Inversion by partial fractions:
3.1(a) 0)0()0(1 '
2
2
===++ xxxdt
dx
dt
dx
)0()0()( '2
2
2
xsxsXsdt
dxL −−=
)0()( xsXsdt
dxL −=
L(x) = X(s)
L{1} = 1/s
+−− )0()0()( '2 xsxsXss
sXxsXs1
)()0()( =+−
ssXss
1)()1( 2 =++=
)1(
1)(
2 ++=
ssssX
Now, applying partial fractions splitting, we get
)1(
11)(
2 +++
−=ss
s
ssX
2222
2
3
2
1
2
3
3
2
2
1
2
3
2
1
11)(
+
+
−
+
+
+−=
ss
s
ssX
tetCosesXLtt
2
3sin
3
1
2
31))(( 2
1
2
1
1−−− −−=
+
−=
−tSintCosetX
t
2
3
3
1
2
31)( 2
1
b) 0)0()0(12 '
2
2
===++ xxxdt
dx
dt
dx
when the initial conditions are zero, the transformed equation is
s
sXss1
)()1( 2 =++
)1(
1)(
2 ++=
ssssX
12)1(
122 ++
++=
++ ss
CBs
s
A
sss
CsBsssA ++++= 22 )12(1
)(20
)(0 2
sofeffecientscotheequatingbyCA
sofeffecientcotheequatingbyBA
−+=
−+=
2,1,1
2
1
0
)(1
−=−==
−=
−=
=+
−=
CBA
AC
B
BA
constofeffecientscotheequatingbyA
12
21)(
2 +++
−=ss
s
ssX
( )( )
+
++−= −−
2
11
1
111)}({
s
s
sLsXL
( )
++
+−= −
2
1
1
1
1
11)}({
ssLtX
)1(1)}({ tetX t +−= −
3.1 C 0)0()0(13 '
2
2
===++ xxxdt
dx
dt
dx
by Applying laplace transforms, we get
ssXss
1)()13( 2 =++=
)13(
1)(
2 ++=
ssssX
13
)(2 ++
++=
ss
CBs
s
AsX
CsBsssA ++++= 22 )13(1
)(30
)(0 2
sofeffecientscotheequatingbyCA
sofeffecientcotheequatingbyBA
−+=
−+=
3,1,1
33
1
0
)(1
−=−==
−=−=
−=
=+
−=
CBA
AC
B
BA
constofeffecientscotheequatingbyA
+++
−= −−
13
31)}({
2
11
ss
s
sLsXL
−
+
+−= −−
22
11
2
5
2
3
31)}({
s
s
sLsXL
−
+
−
−
+
+−= −−
2222
11
2
5
2
3
2
5
5
2.
2
3
2
5
2
3
2
3
1)}({
ss
s
sLsXL
tt
CosetX
t
2
5sinh
5
3
2
5(1)( 2
3
+−=−
3.2(a)
1)0(
0)0()0()0(;
11
''''
3
3
4
4
=
====+
x
xxxtCosdt
xd
dt
dx
Applying Laplace transforms, we get
1)0()0()0()()0()0()0()0()(
2
'''23'''''1234
+=−−−+−−−−s
sxsxxssXsxsxxsxssXs
1)1()()(
2
34
+=+−+s
sssssX
34
2)1()1
1()( ssss
ssX +
+++
+=
= )1)(1(
12
)1)(1(
123
23
23
23
+++++
=++++++
sss
sss
sss
ssss
11)1)(1(
1223223
23
++
++
+++=+++++
s
FEs
s
D
s
C
s
B
s
A
sss
sss
)1()()1()1)(1()1)(1()1)(1(12 323222223 +++++++++++++=+++ ssFEssDssscssBsssAssss
A+B+E=0 equating the co-efficient of s5.
A+B+E+F=0 equating the co-efficient of s4.
A+B+C+D+F=0 equating the co-efficient of s3.
A+B+C=0 equating the co-efficient of s2.
B+C=2 equating the co-efficient of s.
A+B+E=0 equating the co-efficient of s2.
C=1equating the co-efficient constant.
C=1
-B=-C+2=1
A=1-B-C=-1
D+F=0
E+F=0D+E=1
D-E=0
2D=1
A=-1; B=1; C=1
D=1/2; E=1/2; F =-1/2
{ }
+−
++
+++−
= −−
1
)1(2/1
1
2/1111)(
232
11
s
s
ssssLsL
{ }
+−
++
+++−
= −−
1
)1(2/1
1
2/1111)(
232
11
s
s
ssssLsXL
{ } int2
1
2
1
2
1
21)(
2
StCoset
ttX t −++++−= −
2)0(;4)0(2 12
2
2
−==+=+ qqttdt
dq
dt
qd
applying laplace transforms,we get
23
'2 22)0()(()0()0()(
ssqssQqsqsQs +=−+−−
+=−+−+ 112
424))((2
2
ssssssQ
)(
)24()1(2
)(2
3
ss
ss
s
sQ+
+++
=
= )1(
24224
34
++++
ss
sss
)1(
3*2
)1(
2
1
14)(
4 ++
++
+
=sssss
sQ
31
3
1)1(24)())(( teetqsQL tt +−+== −−−
therefore tet
tq −++= 23
2)(3
3.3 a)
+−
+=
++ 4
1
1
1
3
3
)4)(1(
32222 ss
s
ss
s
+−
+=
2222 2
1
1
1
ss
tCosCostss
L 22
1
1
12222
1 −=
+
−+
−
b) [ ] 522)1(
1
)52(
12222 +−
++=
+−=
+− ss
CB
s
A
sssss
A+B=0
-2A+C=0
5A=1
A=1/5 ;B=-1/5;C=2/5
We get
+−
−+=
52
21
5
1)(
2 ss
s
ssX
Inverting,we get
=
−+ tCosetSine tt 222
11
5
1
=
−+ tCostSinet 222
11
5
1
c) 2222
22
)1(1)1(
233
−+
−++=
−+−−
s
D
s
C
s
B
s
A
ss
sss
233)1()1()1( 232222 +−−=+−+−+− sssDssCssBsAs
233)()12()2( 2322323 +−−=+−++−++− sssDsssCssBsssA
A+C=3
-2A+B-C+D=-1
A-2B=-3
B=2;
A=2(2)-3=1
C=3-1=2
D=2(1)-2+2-1=1
We get 22 )1(
1
1
221)(
−+
+++=
sssssX
By inverse L.T
[ ] tt teettXL +++=− 221)(1
[ ] )2(21)(1 tettXL t +++=−
3.4 Expand the following function by partial fraction expansion. Do not evaluate
co-efficient or invert expressions
)3()1)(1(
2)(
22 +++=
ssssX
3)1(11)(
222 ++
++
+++
++
=s
F
s
EDs
s
CBs
s
AsX
22222 )1)(1()3)(1)(()1)(3)(1)(()3()1( ++++++++++++++= ssFssEDssssCBsssA
)14)(1()34)(()1)(34)(()3)(12( 222224 ++++++++++++++++= sssFssEDssssCBssssA
2333)4
34()346()342()43()( 2345
=++++++
+++++++++++++++++=
FEACAFE
BCAsCBCAsBCBAsFBCAsFBAs
A+B+F=0
-3A+C+4B+F=0
2A+B+4C+3B=0
6A+C+4B+3C=0
A+4C+3B+3D+4E+F=0
3A+3C+3E+F=2
by solving above 6 equations, we can get the values of A,B,C,D,E and
33 )3()1)(1(
1)(
+++=
sssssX .
3232 )3()3(321)(
++
++
++
++
++++=
s
H
s
G
s
F
s
E
s
D
s
C
s
B
s
AsX
by comparing powers of s we can evaluate A,B,C,D,E,F,G and H.
c))4()3)(2(
1)(
+++=
sssssX
4321)(
++
++
++
+=
s
D
s
C
s
B
s
AsX
by comparing powers of s we can evaluate A,B,C,D
3.5 a) )15.0)(1(
1)(
++=
ssssX
)15.0(1)15.0)(1(
1
++
++=
++ s
C
s
B
s
A
sssLet
1)(2
12
3
2
222
=++
++
++= ssCs
sB
ssA
A=1
2
1
20
22−=+==++ C
BC
BA
2
30
2
3−=+==++ CBCB
A
B/2=1/2 *-3/2=-1;
B=-2;
C= -3/2+2=1/2
+
++
−=15.0
1
2
1
1
21)(
ssssX
tt eetxsXL 21 21)())(( −−− +−===
b) 0)0(;22 ==+ xxdt
dx
Applying laplace trafsorms
ssXxssX /2)(2)0()( =+−
)2(
2))((1
+=−
sssXL
+
= −−
)2(
22))(( 11
ssLsXL
=
+
−= −−
2
2/12/12))(( 11
ssLsXL
=te 21 −−
3.6 a) 52
1)(
2 +++
=ss
ssY
= 52
1)(
2 +++
=ss
ssY
4)1(
12 ++
+=
s
s
=
++
+= −−
4)1(
1))((
2
11
s
sLsYL
using the table,we get
tCosetY t 2)( −=
b) 4
2 2)(
s
sssY
+=
32
21)(
sssY +=
Y(t)= 21 ))(( ttsYL +=−
c) 3)1(
2)(
−=
s
ssY
= 3)1(
222
−+−
s
s
32 )1(
2
)1(
2
−+
−=
ss
−
+
−
= −−3
1
2
1
)1(
2
)1(
2)(
sL
sLtY
= )2(2
(2 22
tteet
te ttt +=+
3.7a) )1()1()1(
1)(
222 ++
+++
=+
=s
DCs
s
BAs
ssY
1)1)(()( 2 =++++ sDCsBAsthus
= 1)()(23 =+++++ DBsCADsCs
C=0,D=0
Also A=0;B=1
222222 )()()()()()(
1
)1(
1)(
is
D
is
C
is
B
is
A
isisssY
−+
−+
++
+=
−+=
+=
1)())(()())(( 2222 =+++−+−+−+ isDisisCisBisisA
1)()22()()( 23 =−+−−+++−++++−++ DCiBAiDiCBiAsDCiBAisCA
Thus,A+C=0
-Ai+B+Ci+2Di=0 ; B=D
A-2Bi+C+2Di=0
-Ai-B+Ci-D=1 Also D=-Ci;B=-Ci,
A=-C,C=-i/4
A=i/4 ; B=-1/4; D=-1/4
22 )(
4/1
)(
4/
)(
4/1
)(
4/)(
isis
i
isis
isY
−−
+−
−+
+−
++
=
22 )(
4/1
)(
4/
)(
4/1
)(
4/)(
isis
i
isis
itY
−−
+−
−+
+−
++
=
22 )(
4/1
)(
4/
)(
4/1
)(
4/)(
isis
i
isis
itY
−−
+−
−+
+−
++
=
itititit teeeeitY 4/14/14/14/)( −−−= −−
)(4/1)( itititit teieteietY −−−= −−
)()()()((4/1)( tSinitCosttiSintCositSiniCostttiSinCostitY +−+−−−−= )
)22(4/1)( tCosttSintY −=
)(2/1)( tCosttSintY −=
3.8 )1(
1)(
2 +=
sssf
= 1
)(2 +
++=s
C
s
B
s
Asf
1)1()1( 2 =++++= CssBssA
Let s=0 ; A=1
s=1; 2A+B+C=1
s=-1: C=1
B=-1
1
111)(
2 +++=sss
sf
tettf −+−= )1()(
PROPERTIES OF TRANSFORMS
4.1 If a forcing function f(t) has the laplace transforms
s
e
s
ee
ssf
sss 3
2
21)(
−−−
−−
+=
2
231
s
ee
s
e sss −−− −+
−=
)]2()2()1()1[()]3()([)}({)( 1 −−−−−+−−== − tuttuttutusfLtf
)3()2()2()1()1()( −−−−−−−+= tututtuttu
graph the function f(t)
4.2 Solve the following equation for y(t):
1)0()(
)(0
==∫ ydt
tdydy
t
ττ
Taking Laplace transforms on both sides
=∫ dt
tdyLdtyL
t)(
})({0
τ
)0()(.)(.1
ysyssys
−=
1)(.)(.1
−= syssys
1)(
2 −=s
ssy
)cosh(1
)}({)(2
11 ts
sLsyLty
−== −−
4.3 Express the function given in figure given below the t – domain and the
s-
domain
This graph can be expressed as
)}6()6()5()5()5({)}3()3()2()2({)}5()1({ −−+−−−−+−−−−−+−−−= tuttuttututtuttutu
)6()6()5()5()3()2()2()2()1()( −−+−−−−−−−−+−= tuttuttuttuttutf
2
6
2
53
2
3
2
2
)}({)(s
e
s
e
s
e
s
e
s
e
s
etfLsf
ssssss −−−−−−
+−−−+==
2
53623
s
eeee
s
ee ssssss −−−−−− −−++
−=
4.4 Sketch the following functions:
)3()1(2)()( −+−−= tutututf
)2()1(3)(3)( −+−−= tututtutf
4.5 The function f(t) has the Laplace transform
22 /)21()( seeSf ss −− +−=
obtain the function f(t) and graph f(t)
2
221)(
s
eesf
ss −− +−=
2
2
2
1
s
ee
s
e sss −−− −−
−=
)]2()2()1()1{()()1()1()}({)( 1 −−−−−−+−−−== − tuttutttututsfLtf
)2()2()1()1(2)( −−+−−−= tuttutttu
4.6 Determine f(t) at t = 1.5 and at t = 3 for following function:
)2()3()1(5.0)(5.0)( −−+−−= tuttututf
At t = 1.5
)2()3()1(5.0)(5.0)( −−+−−= tuttututf
)1(5.0)(5.0)5.1( −−= tutuf
05.05.0)5.1( =−=f
At t = 3
0)33(5.05.0)3( =−+−=f
RESPONSE OF A FIRST ORDER SYSTEMS
5.1 A thermometer having a time constant of 0.2 min is placed in a
temperature bath and after the thermometer comes to equilibrium with
the bath, the temperature of the bath is increased linearly with time at the
rate of I deg C / min what is the difference between the indicated
temperature and bath temperature
(a) 0.1 min
(b) 10. min
after the change in temperature begins.
© what is the maximum deviation between the indicated temperaturew
and bath temperature and when does it occurs.
(d) plot the forcing function and the response on the same graph. After the
long enough time buy how many minutes does the response lag the input.
Consider thermometer to be in equilibrium with temperature bath at
temperature Xs
0,)/1()( >°+= ttmXtX S
as it is given that the temperture varies linearly
X(t)-Xs = t
Let X(t) = X(t) - Xs = t
Y(s) = G(s).X(s)
22 1
1
1
1)(
s
C
s
B
s
A
sssY ++
+=
+=
ττ
A = 12 =−= CB ττ
2
2 1
1)(
ssssY +−
+=
ττ
τ
tetY t +−= − ττ τ/)(
(a) the difference between the indicated temperature and bath temperature
at t = 0.1 min = X(0.1)_ Y(0.1)
= 0.1 - (0.2e-0.1/0.2 - 0.2+0.1) since T = 0.2 given
= 0.0787 deg C
(b) t = 1.0 min
X(1) - Y(1) = 1- (0.2e-1/0.
2 - 0.2 +1) = 0.1986
(c) Deviation D = -Y(t) +X(t)
= -τe-t/T+T =τ (-e-t/T+1)
For maximum value dD/dT = τ (-e-t/T+(_-1/T) = 0
-e-t/ = 0
as t tend to infinitive
D = τ (-e-t/T+(_-1/T) = τ =0.2 deg C
5.2 A mercury thermometer bulb in ½ in . long by 1/8 in diameter. The
glass envelope is very thin. Calculate the time constant in water flowing
at 10 ft / sec at a temperature of 100 deg F. In your solution , give a
summary which includes
(a) Assumptions used.
(b) Source of data
(c) Results
T = mCp/hA = )(
)(
DLAh
CAL p
π
ρ
+
Calculation of
nm
ed CRK
hDNU (Pr)==
4.967710
10)3048.0*10)(10*54.2*8/1(Re
3
32
=== −
−
µρDv
d
KgKKJK
C p/2.4Pr ==
µ
Source data: Recently, Z hukauskas has given c,m ,ξ,n values.
For Re = 967704
C = 0.26 & m = 0.6
NuD = hD/K = 0.193 (9677.4)*(6.774X10-3) = 130
.h = 25380
5.3 Given a system with the transfer function Y(s)/X(s) = (T1s+1)/(T2s+1).
Find Y(t) if X(t) is a unit step function. If T1/T2 = s. Sktech Y(t) Versus
t/T2. Show the numerical values of minimum, maximum and ultimate values
that may occur during the transient. Check these using the initial value
and final value theorems of chapter 4.
1
1)(
2
1
++
=sT
sTsY
X(s) =unit step function = 1 X(s) = 1/s
siT
B
s
A
sTs
sTsY
22
1
)1(
1)( +=
++
=
A = 1 B = T1 - T2
sT
TT
ssY
2
21
1
1)(
+−
+=
2/
2
211)( TteT
TTtY −−
+=
If T1/T2 = s then
2/41)( TtetY −+=
Let t/T2 = x then x
etY−+= 41)(
Using the initial value theorem and final value theorem
)()(0
ssYLimTYLimST ∞→→
=
= 51
1
1
1
2
1
2
1
2
1 ==+
+=
++
∞→∞→ T
T
sT
sT
LimsT
sTLim
SS
)()(0
ssYLimTYLimST ∞→→
= = 11
1
2
1
0=
++
→ sT
sTLimS
Figure:
5.4 A thermometer having first order dynamics with a time constant of 1
min is placed in a temperature bath at 100 deg F. After the thermometer
reaches steady state, it is suddenly placed in bath at 100 deg F at t = 0 and
left there for 1 min after which it is immediately returned to the bath at
100 deg F.
(a) draw a sketch showing the variation of the thermometer reading with
time.
(b) calculate the thermometer reading at t = 0.5 min and at t = 2.0 min
min)1(1
1
)(
)(=
+= τ
ssX
sY
−=
−
s
e
ss
s110)(
−=
−
s
esY
s110)(
+−
+=
−
)1()1(
110)(
ss
e
sssY
s
1)1(10)( <−= − tetY t
( ) 1)1()1(10)( )1( ≥−−−= −−− teetY tt
At t = 0.5 T = 103.93
At = 2 T =102.325
5.5 Repeat problem 5.4 if the thermometer is in 110 deg F for only 10 sec.
If thermometer is in 110 deg F bath for only 10 sec
60/10110 teT −−=
sec60&sec100 =<< Tt
535.101sec)10( ==tT
sec10535.1100 60/)10( >+= −− teT t
T(t=30sec) = 101.099 deg F
T(t=120sec) = 100.245 deg F
5.6 A mercury thermometer which has been on a table for some time,is
registering the room temperature ,758 deg F. Suddenly, it is placed in a 400
deg F oil bath. The following data are obtained for response of the
thermometer
Time (sec) Temperature, Deg F
0 75
1 107
2.5 140
5 205
8 244
10 282
15 328
30 385
Give two independent estimates of the thermometer time constant.
−
=
T
t
400
325ln
τ
From the data , average of 9.647,11.2,9.788,10.9,9.87,9.95, and 9.75 is 10.16 sec.
5.7 Rewrite the sinusoidal response of first order system (eq 5.24) in
terms of a cosine wave. Re express the forcing function equation (eq 5.19)
as a cosine wave and compute the phase difference between input and
output cosine waves.
τ
τωω
τ 1
1
)(1
1)(
22
+
+=
+=
ss
As
ssY
splitting into partial fractions then converting to laplace transforms
)sin(11
)(22
/
22φω
ωτωταωτ τ +
++
+= − t
Ae
AtY t
where φ = tan-1 (ωτ)
As t →∝
)
−−+
=++
= φπ
ωωτ
φωωτ 2
cos(1
)sin(1
)(2222
tA
tA
stY
−=+= tAtAtY ωπ
φω2
cos)sin()(
−=2
cos)(π
ωtAtY
The phase difference = φππ
φ =
−−−22
5.8 The mercury thermometer of problem 5.6 is allowed to come to
equilibrium in the room temp at 75 deg F.Then it is immersed in a oil
bath for a length of time less than 1 sec and quickly removed from the
bath and re exposed to 75 deg F ambient condition. It may be estimated
that the heat transfer coefficient to the thermometer in air is 1/ 5th that in
oil bath.If 10 sec after the thermometer is removed from the bath it reads
98 Deg F. Estimate the length of time that the thermometer was in the bath.
t < 1 sec τ/1
1325400 teT −−=
Next it is removed and kept in 75 Deg F atmosphere
Heat transfer co-efficient in air = 1/5 heat transfer co-efficient in oil
hair = 1/5 hoil
hA
mC=τ sec10=oilτ
sec50=airτ
50/
12)75(75 t
F eTT −−+=
CTempFinalTF deg98==
50/1010/)325325(7598 1 −−−+= ee
t
91356.010/ =−te
t 1 = 0.904 sec.
5.9 A thermometer having a time constant of 1 min is initially at 50 deg C.
it is immersed in a bath maintained at 100 deg C at t = 0 . Determine the
temperature reading at 1.2 min.
τ = 1 min for a thermometer initially at 50 deg C.
Next it is immersed in bath maintained at 100 deg C at t = 0
At t = 1.2
)1()( /τteAtY −−=
50)1(50)2.1( 1/2.1 +−= −eY
Y(1.2) = 84.9 deg C
5.10 In Problem No 5.9 if at, t = 1.5 min thermometer having a time
constant of 1 minute is initially at 50 deg C.It is immersed in a bath
maintained at 100 deg C at t = 0.Determine the temperature reading at t =
1.2 min.
At t = 1.5
CY °= 843.88)5.1(
Max temperature indicated = 88.843 deg C
AT t = 20 min
)1(843.13843.88 1/8.18−−−= eT
T = 75 Deg C.
5.11 A process of unknown transfer function is subjected to a unit impulse
input. The output of the process is measured accurately and is found to be
represented by the function Y(t) = t e-t. Determine the unit step response in
this process.
X(s) = 1 Y(t) = te-t
2)1(
1)(
+=
ssY
2)1(
1
)(
)()(
+==
ssX
sYsG
For determining unit step response
2)1(
1)(
+=
ssY
22 )1(1)1(
1)(
++
++=
+=
s
C
s
B
s
A
ssY
A = 1 B = -1 C = -1
2)1(
1
1
11)(
+−
+−=
ssssY
tt teetY −− −−= 1)(
Response of first order system in series
7.1 Determine the transfer function H(s)/Q(s) for the liquid level shown in
figure P7-7. Resistance R1 and R2 are linear. The flow rate from tank 3 is
maintained constant at b by means of a pump ; the flow rate from tank
3 is independent of head h. The tanks are non interacting.
Solution :
A balance on tank 1 gives
dt
dhAqq 111 =−
where h1 = height of the liquid level in tank 1
similarly balance on the tank 2 gives
dt
dhAqq 2221 =−
and balance on tank 3 gives
dt
dhAqq 302 =−
here 1
11
R
hq =
2
22
R
hq = bq =0
So we get
dt
dhA
R
hq 1
1
1
1 =−
dt
dhA
R
h
R
h 22
2
2
1
1 =−
dt
dhAb
R
h3
2
2 =−
writing the steady state equation
011
1
1 ==−dt
dhA
R
hq SsS
dt
dhA
R
h
R
h SSS 22
2
2
1
1 =−
02
2 =−bR
h S
Subtracting and writing in terms of deviation
dt
dHA
R
HQ 1
1
1
=−
dt
dHA
R
H
R
H 21
2
2
1
1 =−
dt
dHA
R
H3
2
2 =
where Q = q –qS
H1= h1-h1S
H1= h2-h2S
H = h - hS
Taking Laplace transforms
)()(
)( 11
1
1 sHsAR
sHsQ =− ---------(1)
)()()(
22
2
2
1
1 sHsAR
sH
R
sH=− --------(2)
)()(
3
2
2 sHsAR
sH= ----------(3)
We have three equations and 4 unknowns(Q(s),H(s),H1(s) and H2(s). So we
can express one in terms of other.
From (3)
sAR
sHsH
31
22
)()( = -------------(4)
)1(
)()(
21
122 +
=sR
sHRsH
τ where 222 AR=τ ------------(5)
From (1)
)1(
)()(
1
11 +
=s
sQRsH
τ, 111 AR=τ ---------(6)
Combining equation 4,5,6
)1)(1)((
)()(
213 ++=
sssA
sQsH
ττ
)1)(1)((
1
)(
)(
213 ++=
sssAsQ
sH
ττ
Above equation can be written as
i.e, if non interacting first order system are there in series then there overall
transfer function is equal to the product of the individual transfer function in
series.
7.2 The mercury thermometer in chapter 5 was considered to have all its
resistance in the convective film surrounding the bulb and all its
capacitance in the mercury. A more detailed analysis would consider both
the convective resistance surrounding the bulb and that between the bulb
and mercury. In addition , the capacitance of the glass bulb would be
included.
Let
Ai = inside area of bulb for heat transfer to mercury.
Ao = outside area of bulb, for heat transfer from surrounding fluid.
.m = mass of the mercury in bulb.
mb = mass of glass bulb.
C = heat capacitance of mercury.
Cb = heat capacity of glass bulb.
.hi = convective co-efficient between the bulb and the surrounding fluid.
.ho = convective co-efficient between bulb and surrounding fluid.
T = temperature of mercury.
Tb = temperature of glass bulb.
Tf = temperature of surrounding fluid.
Determine the transfer function between Tf and T. what is the effect of bulb
resistance and capacitance on the thermometer response? Note that the
inclusion of the bulb results in a pair of interacting systems, which give an
overall transfer function different from that of Eq (7.24)
Writing the energy balance for change in term of a bulb and mercury
respectively
Input - output = accumulation
dt
dTCmTTAhTTAh b
bbbiibf =−−− )()(00
dt
dTCmTTAh bii =−− 0)(
Writing the steady state equation
0)()(00 ==−−−dt
dTCmTTAhTTAh bs
bbsbsiibsfs
0)( =− sbsii TTAh
Where subscript s denoted values at steady subtracting and writing these
equations in terms of deviation variables.
dt
dTCmTTAhTTAh b
bbmbiibf =−−− )()(00
dt
dTCmTTAh m
mbii =−− 0)(
Here TF = Tf - TfS
TB = Tb - TbS
Tm = T - TS
Taking laplace transforms
)()())()((00 sTCmTTAhsTsTAh BbbmBiiBF =−−− ----(1)
And )())()(( ssTmCsTsTAh BmBii =− ------(2)
= )()())()((00 ssTCmsmCSTsTsTAh BbbmBF =−−
From (2) we get
)1()(1)()( +=
+= ssTs
Ah
mCsTsT im
ii
mB τ
Where ii
iAh
mC=τ
Putting it into (1)
0)1))(1()()(00
0 =
+++− s
Ah
mCsssTsT imF ττ
=
+++= s
Ah
mCsssTsT imF
00
0 )1))(1()()( ττ
=
1)(
1
)(
)(
00
0
2
0 ++++=
sAh
mCs
sT
sT
iiF
m
ττττ
=
1)(
1
)(
)(
00
0
2
0 ++++=
sAh
mCs
sT
sT
iiF
m
ττττ
Or we can write
1)(
1
)(
)(
00
0
2
0 ++++=
sAh
mCs
sT
sT
iif ττττ
ii
iAh
mC=τ and
00
0Ah
Cm bb=τ
We see that a loading term mC/ hoAo is appearing in the transfer function.
The bulb resistance and capacitance is appear in 0τ and it increases the
delay i.e Transfer lag and response is slow down.
7.3 There are N storage tank of volume V Arranged so that when
water is fed into the first tank into the second tank and so on. Each tank
initially contains component A at some concentration Co and is equipped
with a perfect stirrer. A time zero, a stream of zero concentration is
fed into the first tank at volumetric rate q. Find the resulting
concentration in each tank as a function of time.
Solution:
. ith tank balance
dt
dCVqCqC i
ii =−−1
0)1( =−− issi qCqC
=
=−−
q
V
dt
dC
q
VCC i
ii
τ
)1(
Taking lapalce transformation
)()()()1( ssCisCsC ii τ=−−
)()1()()1( sCissC i τ+=−
ssC
sC
i
i
τ+=
− 1
1
)(
)(
1
Similarly
issC
sC
sC
sC
sC
sC
sC
sC
sCo
sC
i
i
i
ii
)1(
1
)(
)(
)(
)(
)(
)(
)(
)(
)(
)(
2
1
1
2
0
1
τ+=×−−−−−−−−−××=
−
−
Or
N
N
ssCo
sC
)1(
1
)(
)(
τ+=
NNss
CsC
)1()( 0
τ+
−=
+−−−−−−−
+−
+−−=
− ssssCsC
NNN ττ
ττ
ττ
1)1()1(
1)(
10
−−−−−
−−
−−−=
−−
−
−−
−
−
τττ
ττ
tN
N
t
N
N
t
N eN
te
N
teCtC
)!2(.
)!1(.1)(
2
2
1
10
+−−−−−−
+−
−−=
−−
−1
)!2(.
)!1(.1)(
21
0N
t
N
t
eCtC
NN
t
N
τττ
7.4 (a) Find the transfer functions H2/Q and H3/Q for the three tank system
shown in Fig P7-4 where H1,H3 and Q are deviation variables. Tank 1 and
Tank 2 are interacting.
7.4(b) For a unit step change in q (i.e Q = 1/s); determine H3(0) , H3(∞)
and sketch H3(t) vs t.
Solution :
Writing heat balance equation for tank 1 and tank 2
dt
dhAqq 111 =−
dt
dhAqq 2221 =−
1
211
R
hhq
−=
2
22
R
hq =
Writing the steady state equation
01 =− ss qq
021 =− ss qq
Writing the equations in terms of deviation variables
dt
dHAQQ 111 =−
dt
dHAQQ 2
221 =−
1
211
R
HHQ
−=
2
22
R
HQ =
Taking laplace transforms
)()()( 111 ssHAsQsQ =−
)()()( 2121 ssHAsQsQ =−
)()()( 2111 sHsHsQR −=
)()( 222 sHsQR =
Solving the above equations we get
( )[ ]1)(
)(
2121
2
21
22
++++=
sRAs
R
sQ
sH
ττττ
Here 111 AR=τ
222 AR=τ
Now writing the balance for third tank
dt
dhAqq 3332 =−
Steady state equation
032 =− SS qq
3
3
3R
hq =
dt
dhA
R
HQ 3
3
3
3
2 =−
Taking laplace transforms
)()(
)( 3
3
3
2 ssHAR
sHsQ =−
( )1)()( 3
3
3
2 += sR
sHsQ τ where 333 AR=τ
From equation 1,2,3,4 and 5 we got
[ ]1)(
1
)(
)(
2121
2
21 ++++=
sRAssQ
sQs
ττττ
Putting it in equation 6
[ ]( )11)()(
)(
32121
2
21
33
+++++=
ssRAs
R
sQ
sH
τττττ
Putting the numerical values of R1,R2 and R3 and A1,A2,A3
[ ]( )12164
4
)(
)(
2
3
+++=
ssssQ
sH
[ ]164
2
)(
)(2
2
++=
sssQ
sH
Solution (b)
ssQ
1)( =
[ ]( )12164
41)(
23 +++=
sssssH
From initial value theorem
)()0( 33 ssHLimHS ∞→
=
= )164)(12(
42 +++∞→ sss
LimS
=
)164()12(
4
23
ssss
LimS
+++∞→
H3 (0) = 0
From final value theorem
)()( 30
3 ssHLimHS→
=∞
= )164)(12(
420 +++→ sss
LimS
H3 (∞) = 4
7.5 Three identical tanks are operated in series in a non-interacting fashion
as shown in fig P7.5 . For each tank R=1, ττττ = 1. If the deviation in
flow rate to the first tank in an impulse function of magnitude 2,
determine
(a) an expression for H(s) where H is the deviation in level in the third
tank.
(b) sketch the response H(t)
(c) obtain an expression for H(t)
solution :
writing energy balance equation for all tanks
dt
dhAqq 1
1 =−
dt
dhAqq 2
21 =−
dt
dhAqq =− 32
R
hq 11 =
R
hq 22 =
R
hq =3
So we get
01 =− SS qq
021 =− SS qq
032 =− SS qq
writing in terms of deviation variables and taking laplace transforms
)()(
)( 11 sHAR
sHsQ S=−
)()()(
221 sHAR
sH
R
sQS=−
)()()(2 sHA
R
sH
R
sHS=−
solving we get
33 )1(
1
)1()(
)(
+=
+=
ss
R
sQ
sH
τ
33 )1(
2
)1(
)()(
+=
+=
ss
sQsH
τ
{ } tet
sHLtH −− ==2
2)()(2
1
tettH −= 2)(
02)(
=−= −− tt tetedt
tdH
22 tt ==
at t = 2 max will occur.
7.6 In the two- tank mixing process shown in fig P7.6 , x varies from 0 lb
salt/ft3 to 1 lb salt/ft3 according to step function. At what time does the
salt concentration in tank 2 reach 0.6 lb/ ft3 ? The hold up volume of each
tank is 6
ft3.
Solution
Writing heat balance equation for tank 1 and tank 2
dt
dyVqq yx =−
dt
dlVqq cy =−
steady state equation
0=− ysxs qq
0=− csys qq
writing in terms of deviation variables and taking laplace transforms
)()()( sYsq
VsYsX =−
q
V
ss
q
VsX
sY=
+=
+
= ττ
;1
1
1
1
)(
)(
2)1(
)(
)1(
)()(
+=
+=
s
sX
s
sYsC
ττ
2)1(
1
)(
)(
+=
ssX
sC
τ
ssX
1)( =
23
6===
q
Vτ
2)12(
)()(
+=
ss
sXsC
2)2
1(
)4/1()(
+=
ss
sC
+
=
2)2
1(
11
4
1)(
ss
sC
+−
+
−=
2
1
1
2
11
2
1
1)(
2
ss
sC
22
2
11)(
tt
etetC−−
−−=
3/61.0)( ftsaltlbtC =
t = 4.04 min
7.7 Starting from first principles, derive the transfer functions H1(s)/Q(s)
and H2(s)/Q(s) for the liquid level system shown in figure P7.7. The
resistance are linear and R1= R2 = 1. Note that two streams are flowing
from tank 1, one of which flows into tank 2. You are expected to give
numerical values of the parameters and in the transfer functions and to
show clearly how you derived the transfer functions.
Writing heat balance equation for tank 1
dt
dhAqqq a
111 =−−
1
11
R
hq =
a
aR
hq 1=
dt
dhA
R
h
R
hq
a
11
1
11 =−−=
writing the balance equation for tank 2
dt
dhAqq 2221 =−
dt
dhA
R
h
R
h 22
2
2
1
1 =−
writing steady state equations
01
1 =−−R
sh
R
hq
a
ss
02
2
1
1 =−R
sh
R
sh
writing the equation in terms of deviation variables
dt
dHA
RRHQ
a
11
1
1
11=
+−
dt
dHA
R
H
R
H 22
2
2
1
1 =−
taking laplace transforms
sHARR
RRsHsQ S
a
11
1
211 )()( =
+− -----------(1)
and )()()(
22
2
2
1
1 sHsAR
sH
R
sH=− -----------(2)
from (1) we get
++
=
a
a
RR
RRsA
sQ
sH
1
11
1 1
)(
)(
+
+
+=
1)(
)(
1
11
1
1
1
sRR
ARR
RR
RR
sQ
sH
a
a
a
a
[ ]1)(
)(
1
1
1
1
+
+=
s
RR
RR
sQ
sH a
a
τ ;
a
a
RR
ARR
+=
1
111τ
and from (2 ) we get
[ ]( )11)(
)(
21
1
2
1
2
1
++
+=
ss
R
R
RR
RR
sQ
sH a
a
ττ 222 AR=τ
putting the numerical values of parameters
+
=1
3
4
3
2
)(
)(1
ssQ
sH
( )113
4
3
2
)(
)(2
+
+
=ss
sQ
sH
8.1 A step change of magnitude 4 is introduced into a system having the
transfer
46.1
10
)(
)(2 ++
=sssX
sY
Determine (a) % overshoot
(b)Rise time
(c)Max value of Y(t)
(d)Ultimate value of Y(t)
(e) Period of Oscillation.
Given s
sX4
)( = )46.1(
40)(
2 ++=
ssssY
The transfer function is
)1)4
6.1()(2.0
25.010
)(
)(
2 ++
×=
sssX
sY =
)14.025.0
5.22 ++ ss
5.0;25.02 == ττ
4.02 =τξand )1(4.0)5.0(2
4.0dunderdampeissystem=<==ξ
we find ultimate value of Y(t)
104
40
)46.1(
40)()(
200
==++
==→→∞→ ss
sLtssYLttYLtSSt
thus B= 10
now, from laplace transform tables
+
−−=
−)sin(
1
1110)(
2φα
ξτξt
etY
where ξξ
φτξ
α22 1
tan,1 −
=−
= −
(a) Over shoot =
×−=
−
−=
84.0
4.0exp
1exp
2
π
ξ
πξB
A = 0.254
thus % overshoot = 25.4
c)thus, max value of Y(t) = A+B = B(0.254)+B
= 2.54+10 = 12.54
e) Period of oscillation = 21
2
ξ
πτ
−= 3.427
b) For rise time, we need to solve
r
t
ttforte ==
+
−−
−10)sin(
1
1110
2φα
ξτξ
= )sin( φατξτ
+−
rter
= 0
= 0)1589.1833.1sin(5.0
4.0
=+−
rterτ
solving we get tr = 1.082
thus
SOLUTION: % Overshoot = 25.4
Rise time = 1.0842
Max Y(t) = 12.54
U(t) Y(t) = 10
Period of oscillation = 3.427
Comment : we see that the Oscillation period is small and the decay ratio
also small = system is efficiently under damped.
8.2 The tank system operates at steady state. At t = 0, 10 ft3 of wateris
added to tank 1. Determine the maximum deviation in level in both tanks
from the ultimate steady state values, and the time at which each
maximum occurs.
A1 = A2 = 10 ft3
R1 = 0.1ft/cfm R2 = 0.35ft/cfm.
As the tanks are non interacting the transfer functions are
)1(
1.0
1)(
)(
1
1
+=
+=
ss
K
sQ
sH
τ
)15.3)(1(
35.0
)1)(1()(
)(
21
22
++=
++=
ssss
R
sQ
sH
ττ
Now, an impulse of providedisftt 310)( =∂
tes
sHsQ −=+
===1
1)(10)( 1
and 15.45.3
5.3
)15.3)(1(
5.3)(
22 ++=
++=
sssssH
Now 871.15.32 === ττ
202.12
5.45.42 ====
τξξτ
thus, this is an ovedamped system
Using fig8.5, for 2.1=ξ , we see that maximum is attained at
min776.1,95.0 == tt
τ
And the maximum value is around 325.02 =τ Y2 (t) = 0.174
= H2(t) = 0.174x3.5 = 0.16ft
thus max deviation is H1 will be at t = 0 = H1 = 1 ft
max deviation is H2 will be at t = 1.776 min = H2Max = 0.61 ft.
comment : the first tank gets the impulse and hence it max deviation turns out
to be higher than the deviations for the second tank. The second tank exhibits an
increase response ie the deviation increases, reaches the H2Max falls off to zero.
8.3 The tank liquid level shown operates at steady state when a step
change is made in the flow to tank 1.the transaient response in critically
damped, and it takes 1 min for level in second tank to reach 50 % of total
change. If A1/A2 = 2 ,find R1/R2 . calculate ττττ for each tank. How long does it take for level in first tank to reach 90% of total change?
For the first tank, transfer function 1
11
1)(
)(
+=
s
R
sQ
sH
τ
For the second tank )1)(1()(
)(
21
2
++=
ss
R
sQ
sH
ττ
= 1)()(
)(
21
2
21
22
+++=
ss
R
sQ
sH
ττττ
1)(
1)(;
1)(
21
2
21
22 +++
==ss
R
ssH
ssQ
ττττ
21)( τττ =parameter
For 21)( τττ =parameter
+−==
−21
21
22 11)(,1ττ
ττξ
t
et
RtHfor
given, t = 1 for 21)( τττ =parameter
( ) 222 )0(1)( RRtH =−=∞→
IR
eR −=
+−=
−
2
111 2
1
21
221ττ
ττ
also 212 ττξτ +=
5.02
11
2
2
12211
21 ======+
==A
A
R
RRARAτ
ττξ
from I
τ
τ
11
15.01−
+=− e
min372.1
min596.0
5.0
min372.1;1.0
)1(9.0
)1()(94.0
)1()(;)1(
)()3.8
%90
21
2
1
596.0
596.011
1
11
1
11
1
1
=
==
=
==
−=
−=∞→
−=+
=
−
−
−
−
t
R
R
thus
te
eRR
eRt
eRtHss
RsH
t
t
t
t
ττ
τ
τ
τ
Comment :
.tan
tansec,., 2121
kfirstthanchangestoslowlymoreresponds
ondtheRRasAlsoquicklystatesteadytheregainssystemtheindicateofvaluesSmall >ττ
8.4 Assuming the flow in the manometer to be laminar function between
applied pressure P1 and the manometer reading h. Calculate a) steady sate
gain ,b) τ ,c) ξ . Comment on the parameters and their relation to the physical nature of this problem.
Assumptions:
Cross-sectional area =a
Length of mercury in column = L
Friction factor = 16/Re (laminar flow)
Mass of mercury = mrg
Writing a force balance on the mercury
Mass X acceleration = pressure force - drag force - gravitational force
)(2
)(2
12
2
ghAu
AfApdt
hdAL ρ
ρρ −−=
g
ph
dt
dh
gDdt
hd
g
L
ρρµ 1
2
2 8=++
At Steady state, g
ph ss ρ
1=
= g
pH
dt
dH
gDdt
Hd
g
L
ρρµ 1
2
2 8=++
= [ ]g
spsHssH
gDsHs
g
L
ρρµ )(
)()(8
)( 12 =++
= [ ] )()(1 132
2
1 spksHsksk =++
= )1()(
)(
2
2
1
3
1
1
++=
sksk
k
sp
sH
Where ;1g
Lk = ;
82
gDk
ρµ
= ;1
3g
kρ
=
Thus )12()(
)(22
1
1
++=
ss
R
sp
sH
ξττ
Where ;1
gR
ρ= ;2
g
L=τ ;
82
gDρµ
ξτ =
Now ;)g
Lb =τ
1
4
2
1.
8)
−
==
g
L
gDgDc
ρµ
τρµ
ξ
Steady state gain
;1
)(0 g
RsGLtS ρ
==→
Comment : a) τ is the time period of a simple pendulum of Length L.
b) ξ is inversely proportional to τ , smaller the τ ,the system will
tend to move from under damped to over damped characteristics.
8.5 Design a mercury manometer that will measure pressure of upto 2
atm, and give responses that are slightly under damped with ξ = 0.7
Parameter to be decide upon :
.a) Length of column of mercury
.b) diameter of tube.
Considering hmax to be the maximum height difference to be used
;13600*81.9
10*01325.1*2 5
maxmax1 === hghp ρ
;51.1max mh =
Assuming the separation between the tubes to be 30 cm,
We get an additional length of 0.47 m;
Which gives us the total length L= 1.5176.47
L = 2 M
Now, ξ = 0.7 = 7.04
=
L
g
gDρµ
00015.0
10*5.181.9*13600*74.0
2
81.9*10*6.1*4
7.0
47
3
=
=== −
−
g
L
g
Dρ
µ
As can be seen, the values yielded are not proper, with too small a diameter
and too large a length. A smaller ξ value and lower measuring range of
pressure might be better.
8.6 verify that for a second order system subjected to a step response,
[ ]ξξ
τξ
ξτξ 2
12
2
1tan1sin
1
11)(
−+−
−−= −
− tetY
t
With ξ <1
)12(
11)(
22 ++=
ssssY
ξττ
baswhere
ssssss
+−=−
+−
=
−−=++
τξ
τξ
ξττ
1
))((12
2
1
21
22
bas −−=−
+−
=τ
ξτξ 12
2
))((
1
)(21
2
ssssssY
−−
=τ
−+
−+=
)()(
1)(
21
2 ss
C
ss
B
s
AsY
τ
−+
−+=
)()(
1)(
21
2 ss
C
ss
B
s
AsY
τ
0
1)()())(( 12211
2
=++
=−+−+++−
CBA
ssCsssBsssssssA s
1)( 121 =−−+− CsBsssA s
2121
1
11;1
ssCB
ssAsAs s −=+===
1
22
1
sCsBs −=+=
2121
2121
11
ssss
ssCsCs =−
+=+=
)(
11
)(
1
)(
1
12121122122 ssssssssB
sssC
−−=−
−−==
−==
( ) ( ) ( )
−−+
−=−
=
2122112121
2
1.)(
11.
11.
11)(
ssssssssssssssY
τ
8.6
−+
−−= tstS
esss
esssss
tY 21
)(
1
)(
111)(
12212121
2τ
21
2
1
ssτ= 1
−
−−= tStS e
se
ssstY 21
2112
2
11
)(
11)(
τ
[ ]tStS esesss
tY 21
12
12 )(
11)( −
−−=
[ ]tStSesestY 21
122 12
1)( −−
+=ξ
τ
[ ]btjbtjbabtjbtjbaje
tY
t
sin)(cos()sin)(cos(12
1)(2
−+−−+−−−
−=−
ξ
τ τξ
[ ])sincos(212
1)(2
btabtbjbje
tY
t
+−−
−=−
ξ
τ τξ
[ ])sin()cos(11
1)( 2
2tt
etY
t
αξαξξ
τξ
+−−
−=−
[ ]ξξ
α21−
=
[ ]
−= −
ξξ
φ2
1 1tan
verified
8.14 From the figure in your text Y(4) for the system response is
expressed
b) verify that for ,1=ξ and a step input
τ
τ
t
et
tY−
+−= 11)(
1
11)(
22 ++=
ssssY
ττ
22 )1()1(
1)(
++
+=+ ττ
CBs
B
A
sssY
1)12(222 =+++ CsBsssA ττ
02 =+ BAτ
02 =+CAτ
A=1; 2τ=B ; τ2=C
( )21
)1(1)(
+
++−=
s
s
ssY
ττττ
( ) 2)1(1
1)(
+−
+−=
ssssY
ττ
ττ
ττ
τ
tt
teetY−−
−−=1
1)(
τ
τ
t
tet
tY−
+−= )1(1)(
proved
c) for ,1>ξ prove that the step response is
[ ])sinh()cosh(1)( ttetY
t
αβατξ
+−=−
1
1
2
2
−=
−=
ξ
ξβ
τξ
α
Now ))((
/1)(
21
2
ssBsssY
−−=
τ
Where
τξ
τξ 12
1
−+−=s
τξ
τξ 12
2
−−−=s
from 8.6(a)
−−+
−−−=
)(
1
)(
1
)(
1
)(
1111)(
2122112121
2 ssssssssssssstY
τ
[ ]tStSeses
sstY 21
12
12 )(
11)( −
−−=
−+−−
−−−
−+=
−−−
−
−teeeetY
ttt
τξ
τξτ
ξ
τξ
τξξ
τξξ
ξ
τ 12
1
1
2
2
2
2
11
121)(
−−−−+−−
+=−
−−
−−
−−
tttt
t
eeeeee
tY τξ
τξ
τξτ
ξ
τξ
ξξξξξ
1
2
1
2
1
2
22
2
1112
1)(
+−
−
−−+=
−−−
2211)(
2
ttttteeee
etYαααα
τξ
ξ
ξ
[ ])sinh()cosh(1)( ttetY
t
αβατξ
+−=−
8.7 Verify that for a unit step-input
(1) overshoot =
−
−21
expξ
πξ
(2) Decay ratio =
−
−21
2exp
ξ
πξ
For a unit step input the response (ξξξξ<1):
−+−
−−= −
−
ξξ
τξ
ξ
τξ
2
12
2
1tan1
11)(
tSin
etY
t
(1) we have to find time t where the maxima occurs
= dY/dt = 0
−+−
−= −
−
ξξ
τξ
ξτ
ξ τξ
2
12
2
1tan1
1
) tSin
e
dt
dY
t
01
tan12
12 =
−+−− −
−
ξξ
τξ
τ
τξ
tCos
e
t
=
−+− −
ξξ
τξ
2
12 1tan1tan
t=
ξξ 21−
πξξ
nt=
− 21
for maxima
= πξξ
nt
21 2
=−
= 21 ξ
π
−=
tt
8.8 Verify that for X(t) =A sin ωωωωt, for a second order system,
( ) ( ))sin(
2)(1
)(222
φωξτω
++−
= t
t
AtY
2
1
)(1
2tan
ωτξωτ
φ−
−= −
)12(
1
)()(
2222 +++=
sss
AsY
ξττω
−+
−+
++
−=
)()()()(
2
1
1
111
2 ss
D
ss
C
js
B
js
AAsY
ωωτω
Now as tBtAtYt ωω sincos)(, 1111 +=∞→
Where 1111 BAA +=
)( 1111 BAjB −=
to determine ordertheinjjsputBA ωω −= ,, 11
))((2 21
1
sjsj
jA
−−−
=ωωω ))((2 21
1
sjsj
jB
++=
ωωω
−−−
++=
))((
1
))((
1
2 2121
11
sjsjjsjs
jA
ωωωωω
++
+++−−+−−−=
))((
)()(
2 22
2
22
1
2121
2
2121
211
ωωωωωωωω
ω ss
ssjsjsssjsjsjA
++
+=
))((
)(22
2
22
1
2111
ωω ss
ssA similarly
++
−=
))((
)(22
2
22
1
2
2111
ωωωωss
ssB
using 22121
12
ττξ
=−
=+ ssss
=2
2
22
22
2
2
1
)12(224
τξ
ττξ −
=−=+ ss
+−+
−=
42
2
2
4
2
11
)12(21
2
ωξτω
τ
τξ
τωA
A
22
2
2
3
21
2
+
−
−
=
τξω
τω
τωξA
= 222 )2())(1(
2
ξωτωτωξτ+−
− A
and
+
−
−=
22
2
2
2
211
21
1
τξω
ωτ
ω
ωτ
τϖA
B
222
2
)2())(1(
))(1(
ξωτωτωτ+−
−=
A
Thus 211
11
)(1
2tan
ωτωτξ
φ−−
==B
A
And,222 )2())(1(( ξωυωτ +−
=A
ANew (using NewABA =+22
1111
Thus, )()2())(1((
)(222
φωξωυωτ
++−
= tSinA
tY
proved
8.9) If a second- order system is over damped, it is more difficult to
determine the parameters τξ & experimentally. One method for determining
the parameters from a step response has been suggested by R.c Olderboung
and H.Sartarius (The dynamics of Automatic controls,ASME,P7.8,1948),as
described below.
(a) Show that the unit step response for the over damped case may
be written in the form.
21
2112
1)(rr
ererts
trtr −−=
Where r1 and r2 are the roots of
01222 =++ ss ξττ
(b) Show that s(t) has an inflection point at
)(
)/ln(
12
12
rr
rrti −
=
© Show that the slope of the step response at the inflection point
)()( 1
ittts
dt
sd
i
=−
Where, itrtr
i ererts 21
21
1 )( −=−=
)(
1
21
211 rrr
r
rr
−
−=
(d) Show that the value of step response at the inflection point is
)(1)( 1
21
211
ii tsrr
rrts += and that hence
21
1
11
)(
)(1
rrts
ts
i
i −−=−
(e) on a typical sketch of a unit step response show distances equal to
)(
1&
)(
)(1
11
ii
i
tsts
ts−
(f) Relate 21 && rrtoτξ
(a) ))((
1
12
1
)12
1)(
21
2
2
2
2
22 rsrsss
sssG
−−=
+
+
=++
= τ
ττξτ
ξττ
= ))((
1
)(21
2
rsrsssY
−−= τ
2121 ))((
1
rs
C
rs
B
s
A
rsrss −+
−+=
−−
)()())((1 1221 rscsrsBsrsrsA −+−+−−=
Put s= 0 = Ar1r2 =1 ; 2τ=A
Put )(
1;1)(
211
2111rrr
BrrBrrs−
==−==
)(
1;1)(
122
1222rrr
CrrCrrs−
==−==
−−+
−−+=
))((
1
))((
11)(
21221211
2
2 rsrrrrsrrrssY
ττ
−+
−+=
)()(
1)(
122211
2
2
21
rrr
e
rrr
etY
trtr
ττ
[ ]
−
−−= trtr erer
rrtY 12
21
21
11)(
)(1)(
21
2112
rr
erertY
trtr
−
−−=φ
(b) For inflection point , 0&02
3
2
2
==dt
sd
dt
sd
21
21 )( 22
rr
eerr
dt
dstrtr
−
−−=
0)(
21
1221
2
2 22
=−
−−=
rr
ererrr
dt
sdtrtr
itrrtrtr er
rereu
)(
1
212
2112 −====
21
1
2ln
rr
r
r
ti −
==
(c ) )()( '
ittts
dt
tds
i
==
]
−
−−=
−− 12
1
21
1
2
1
2
21
21rr
r
rr
r
r
r
r
r
r
rr
rr
−
−−=
−
−
− 12
1
21
1
2
21
2
1
1
2
21
21rr
r
rr
r
r
r
r
rr
r
r
r
rr
rr
=
−
−−
− 12
1
1
2
21
211 )( rr
r
r
r
r
rr
rrr
−=
−
=
12
1
1
21
)( rr
r
tt
r
i r
rr
dt
tds
Also )(
()(
21
)
2112
rr
eerr
dt
tdstrtr
tt i −
−−=
=
−
−
−−=
=1
)(
)(
2
1
21
211
r
r
rr
err
dt
tdstr
tt i
trtr
tterer
dt
tds
i
21
21
)(−=−=
=
(d) 21
1
2
2
11
21
21
)(
11)(12
rr
r
r
r
rts
rr
ererts
irttr
i
ii
−
−
+=−
−−=
= 21
1
2
2
11 )(
1)(rr
r
r
r
rts
ts
i
i −
−
+==
Now
+==−
21
1 11)(1)(
rrtsts ii
−−==
−
21
'
11
)(
)(1
rrts
ts i
21
21221
1;
11
rrrrrr ====+ τ
ττ
;2
21 =−
=+τξ
rr ξ2121 2 rrrr −=+
+−=
1
2
2
1
2
1
r
r
r
rξ
proved.
8.10 Y(0),Y(0.6),Y(∞ ) if
)12(
)1(251)(
2 +++
=ss
s
ssY
)125
2
25(
111)(
2
++
+=sss
sY
Y(s) impulse response + step response of G(s)
Where
)125
2
25(
1)(
2
++=
sssG
ξξ
τξ
τξτξ 2
12
2
1tan1sin
1
1)(
−+−
−= −− t
etY
t
Y(t) = 1+5.0.3e-t sin (4.899t)-1.02e
-t sin(4.899t+1.369)
Y(0)= 1-1=0
Y(0.6) = 1+0.561+0.515
Y(∞ ) =1
Comment : as we can see ,the system exhibits an inverse response by
increasing from zero to more than 1 and as t tend to ∞ ,will reach the steady state value of 1.
8.11 In the system shown the dev in flow to tank 1 is an impulse of
magnitude 5 . A1 = 1 ft2, A2 = A3 = 2 ft
2 , R1 = 1 ft/cfm R2 = 1.5 ft/cfm .
(a) Determine H1(s), H2(s),
H3(s)
Transfer function for tank 1 )1(
1
)(
)(
1
1
+=
ssQ
sH
τ
)1(
5)(1 +=
ssH
from tank 2, )13)(1(
5.1
)1)(1()(
)(
21
22
++=
++=
ssss
R
sQ
sH
ττ
for tank 3, dt
dhAqq c
332 =−
dt
dhAQconstqq c
3323 )( ===
dt
dhA
R
H 33
2
2 =
thus, ssH
sH
R
sHsSHA
3
1
)(
)()()(
2
3
2
233 ===
ssH
sH
R
sHsSHA
3
1
)(
)()()(
2
3
2
233 ===
)13)(1(
5.0
)(
)(3
++=
ssssQ
sH
8.11© )1(
5)(1 +=
ssH
tetH −= 5)(1
AH 155.0)46.3(1 =
143
5.1
)(
)(2
2
++=
sssQ
sH
143
5.7)(5)1(
22 ++===
sssHQ
3=τ
42 =ξτ
155.132
4
2
4===
τξ
from fig 8.5
τξ
tand155.1= = 2
3
46.3===
τξ
t
5.7265.0)(2 XtH =τ
147.15.7265.0
)(2 ==τX
tH
)143(
5.0
)(
)(2
3
++=
ssssQ
sH
)143(
5.2)(5)(
23 ++===
ssssHsQ
3=τ3
2=ξ
from fig 8.2 at 155.1,2 == ξτt
Y(t) =0.54
H3(t) =0.54*2.5 = 1.35
8.12 sketch the response Y(t) if )12.1(
)(2
2
++=
−
ss
esY
S
Determine Y(t) for t = 0,1,5,∞
22
2
22
2
2
2
)8.0()6.0(
)8.0(
8.0
1
)8.0()6.0()12.1()(
++=
++=
++=
−−−
s
e
s
e
ss
esY
SSS
0)(,
14.0)5(,5
0)1(,1
0)0(0
2))2(8.0sin(25.1)( )2(6.
=∞∞=
==
==
==
≥−= −−
Yt
Yt
Yt
Ytfor
ttetY t
Problem 8.13 The system shown is at steady state at t = 0, with q = 10
cfm
A1 = 1ft2,A2=1.25ft
2, R1= 1 ft/cfm, R2= 0.8 ft/cfm.
a) If flow changes fro 10 to 11 cfm, find H2(s).
b) Determine H2(1),H2(4),H2(∞ ) c) Determine the initial levels h1(0),h2(0) in the tanks.
d) obtain an expression for H1(s) for unit step change.
Writing mass balances,
( ))1tan(1
1
1
21 kfordt
dhA
R
hhq =
−−
At steady state ( )
SS
ss hhR
hhq 21
1
21
2 −=−
−
Also for tank 2
( )dt
dhA
R
h
R
hh 22
2
2
1
21 =−−
At steady state ( )
810*8.08.01
2
221 ====−
S
Sss hhhh
181 =Sh
C) 181 =Sh ft fth 8)0(2 =
The equations in terms of deviation variables
dt
dHAQQ 111 =− where
1
211
R
HHQ
−=
dt
dHAQQ 2
221 =−2
22
R
HQ =
18.2
8.0
1)()(
)(2
2121
2
21
22
++=
++++=
sssRAs
R
sQ
sH
ττττ
))(31.8()18.2(
8.0)(
22 aAnssss
sH++
=
Step response of a second order system
4.12
8.2;8.22
112
===
===
ξξτ
ττ
)(176.0)22.0(8.0)(;11) 2 figfromfttHt
ta =====τ
)(624.0)78.0(8.0)(;44) 2 figfromfttHt
tb =====τ
fttHtc 8.0)() 2 =∞→=∞→
Thus
ftH 176.0)1(2 =
ftH 624.0)4(2 =
ftH 8.0)(2 =∞
8.13(d) we have
)()()()( 111 sHsAsQsQ =
)()()()( 2221 sHsAsQsQ =−
)()()()()()( 22112 sHsAsHsAsQsQ +=−
+=− )()(
1)()()( 22
2
11 sHsAR
sHsAsQ
+= )(
12
2
2 sHR
sτ
−=
sH
ssHAsQRsH
2
1122
)()(()(
τ
We have Deg
R
sRAs
R
sQ
sH 2
2121
2
21
22
1)()(
)(=
++++=
ττττ
+
−=
)1(
)()(()(
2
11222
s
ssHAsQR
Deg
sHR
τ
−−=
+
)(
)(1
1 11
2
sQ
sHsA
Deg
sτ
−−=
−=
Deg
sDeg
sADeg
sH
sAsQ
sH2
1
2
1
11)(1
1
)(
)( ττ
−−++++=
Deg
ssRAs
sAsQ
sH22121
2
21
1
11)(1
)(
)( τττττ
=
++=
Deg
sRAs
sAsQ
sH )1
)(
)(21121
1
τττ
++++
++=
1)(
)1(
)(
)(
2121
2
21
212
sRAs
sRR
sQ
sH
τττττ
8.14
( ))18.04(
422)(
2 +++
=ss
s
ssY
( )
)18.04(
24)(
2 +++
=ss
s
ssY
)18.04(
1214)(
2 ++
+=sss
sY
)18.04(
8)(
2 ++=
ssssY +
)18.04(
42 ++ ss
= (step response) + (impulse response)
24, ==τNow ; 8.02 =ξτ
2.0=ξ
also, 22
4==
τt
impulse response τY(t) = 4*0.63 = 2.52 (from figure)
step response = 8*1.15 = 9.2 (from figure)
Y(4) = 1.26+9.2
Y(4) =10.46
Q 9.1. Two tank heating process shown in fig. consist of two identical,
well stirred tank in series. A flow of heat can enter tank2. At t = 0 , the flow rate of heat
to tank2 suddenly increased according to a step function to 1000 Btu/min. and the temp
of inlet Ti drops from 60oF to 52oF according to a step function. These changes in heat
flow and inlet temp occurs simultaneously.
(a) Develop a block diagram that relates the outlet temp of tank2 to inlet
temp of tank1 and flow rate to tank2.
(b) Obtain an expression for T2’(s)
(c) Determine T2(2) and T2(∞)
(d) Sketch the response T2’(t) Vs t.
Initially Ti = T1 = T2 = 60oF and q=0
W = 250 lb/min
Hold up volume of each tank = 5 ft3
Density of the fluid = 50 lb/ft3
Heat Capacity = 1 Btu/lb (oF)
Solution:
(a) For tank 1
w
Ti
T1
T2
w
q
Input – output = accumulation
WC(Ti – To) - WC(T1 – To) = ρ C V dt
dT1 -------------------------- (1)
At steady state
WC(Tis – To) - WC(T1s – To) = 0 ------------------------------------(2)
(1) – (2) gives
WC(Ti – Tis) - WC(T1 – T1s) = ρ C V dt
dT 1'
WTi’ - WT1
’ = ρ V
dt
dT 1'
Taking Laplace transform
WTi(s) = WT1(s) + ρ V s T1(s)
ssTi
sT
τ+=1
1
)(
)(1 , where τ = ρ V / W.
From tank 2
q + WC(T1 – To) - WC(T2 – To) = ρ C V dt
dT2 -------------------------- (3)
At steady state
qs + WC(T1s – To) - WC(T2s – To) = 0 ------------------------------------(4)
(3) – (4) gives
Q ‘ + WC(T1 – T1s) - WC(T2 – T2s) = ρ C V dt
dT 2'
Q ‘ + WCT1’ - WCT2
’ = ρ C V dt
dT 2'
Taking Laplace transform
Q (s) + WC(T1(s) - T2(s)) = ρ C V s T2(s)
++
= )()(
1
1)( 12 sT
WC
sQ
ssT
τ , where τ = ρ V / W.
(b) τ = 50*5/250 = 1 min
WC = 250*1 = 250
Ti(s) = -8/s and Q(s) = 1000/s
Now by using above two equations we relate T2 and Ti as below and after taking laplace
transform we will get T2(t)
( )
( )
( )4)84()(
1
1
1
118
)1(
114)(
1
8
)1(
4)(
)(1
1
250
)(
1
1)(
2
22
22
22
−+=
+−
+−−
+−=
+−
+=
++
+=
−t
i
ettT
ssssssT
sssT
sTs
sQ
ssT
ττ
(c) T2’(2) = -1.29
T2(2) = T2’(2) + T2s = 60 – 1.29 = 58.71 oF
T2’(∞) = -4
T2(∞) = T2’(∞) + T2s = 60 – 4 = 56 oF
Q – 9.2. The two tank heating process shown in fig. consist of two identical , well stirred
tanks in series. At steady state Ta = Tb = 60oF. At t = 0 , temp of each stream changes
according to a step function
Ta’(t) = 10 u(t) Tb’(t) = 20 u(t)
(a) Develop a block diagram that relates T2’ , the deviation in the temp of tank2,
to Ta’ and Tb’.
(b) Obtain an expression for T2’(s)
(c) Determine T2(2)
W1 = W2 = 250 lb/min
V1 = V2 = 10 ft3
ρ1 = ρ2 = 50 lb/ft3
C = 1 Btu/lb (oF)
0.5
0.85
-4
0
T2’(t)
t
Solution:
(a) For tank1
ssTa
sT
1
1
1
1
)(
)(
τ+= , where τ1= ρ V / W1.
For tank2
W1C(T1 – To) +W2C(Tb – To) – (W1+W2)C(T2 – To)= ρ C V dt
dT2 ------ (1)
At steady state
W1C(T1s – To) +W2C(Tbs – To) – (W1+W2)C(T2s – To)= 0 -----------------(2)
(1) – (2)
W1T1’ + W2Tb
’- W3T2
’ = ρ V
dt
dT 2'
Taking L.T
W1T1(s) + W2Tb(s)- W3T2(s) = ρVs T2(s)
W1
Ta
T1
T2
W3=W1+W2
Tb
W2
W1
[ ])(3
2)(3
11
1)( 12 ST
WWST
WW
ssT b+
+=
τ where τ= ρ V / W3.
(b) τ1 = 50*10/250 = 2 min
τ = 50*5/250 = 1 min
W1/W3 = 1/2 = W2/W3
Ta(s) = 10/s and Tb(s) = 0/s
Now by using above two equations we relate T1 and Ta as below and after taking laplace
transform we will get T2(t)
( )
( )
( )
( )2
2
2
2
2
2
1
2
10515)(
21
20
1
515)(
)21)(1(
2015)(
1
10
)21)(1(
5)(
1
)(2
1
)21)(1(
)(2
1
)(
1
)(2
1
)1(
)(2
1
)(
tt
ba
b
eetT
ssssT
sss
ssT
ssssssT
s
sT
ss
sTsT
s
sT
s
sTsT
−− −−=
+−
+−=
+++
=
+−
++=
+−
++=
+−
+=
(c) T2’(2) = 10.64 oF
T2(2) = T2’(2) + T2s = 60 + 10.64 = 70.64 oF
Q – 9.3. Heat transfer equipment shown in fig. consist of tow tanks, one nested inside the
other. Heat is transferred by convection through the wall of inner tank.
1. Hold up volume of each tank is 1 ft3
2. The cross sectional area for heat transfer is 1 ft2
3. The over all heat transfer coefficient for the flow of heat between the tanks is 10
Btu/(hr)(ft2)(oF)
4. Heat capacity of fluid in each tank is 2 Btu/(lb)(oF)
5. Density of each fluid is 50 lb/ft3
Initially the temp of feed stream to the outer tank and the contents of the outer tank are
equal to 100 oF. Contents of inner tank are initially at 100 oF. the flow of heat to the inner
tank (Q) changed according to a step change from 0 to 500 Btu/hr.
(a) Obtain an expression for the laplace transform of the temperature of inner
tank T(s).
(b) Invert T(s) and obtain T for t= 0,5,10, ∞
Solution:
(a) For outer tank
WC(Ti – To) + hA (T1 – T2)- WC(T2 – To) = ρ C V2 dt
dT2 -------------------------- (1)
At steady state
WC(Tis – To) + hA (T1s – T2s)- WC(T2s – To) = 0 ------------------------------------ (2)
(1) – (2) gives
WCTi’ + hA (T1’ – T2’)- WCT2’ = ρ C V2 dt
dT '2
Substituting numerical values
10 Ti’ + 10 ( T1’ – T2’) – 10 T2’ = 50dt
dT '2
Taking L.T.
Ti(s) + T1(s) – 2T2(s) = 5 s T2(s)
Now Ti(s) = 0, since there is no change in temp of feed stream to outer tank. Which gives
ssT
sT
52
1
)(
)(
1
2
+=
Q 10 lb/hr
T1
T2
For inner tank
Q - hA (T1 – T2) = ρ C V1 dt
dT1 --------------------- (3)
Qs - hA (T1s – T2s) = 0 ------------------------------- (4)
(3) – (4) gives
Q’ - hA (T1’– T2’ ) = ρ C V1 dt
dT '1
Taking L.T and putting numerical values
Q(s) – 10 T1(s) + 10 T2(s) = 50 s T1(s)
Q(s) = 500/s and T2(s) = T1(s) / (2+ 5s)
)(5052
)(10)(10
5001
11 ssT
s
sTsT
s=
++−
++
−= 152
15)(
501
sssT
s
)11525(
)52(50)(
21 +++
=sss
ssT
( )( )50
18.2650
82.3
)52(2)(1
++
+=
sss
ssT
( ) ( )50
18.26
29.5
5082.3
71.94100)(1
+−
+−=
ssssT
5018.26
5082.3'
1 e 5.29 -e 94.71 - 100 (t)Ttt −−
=
and
5018.26
5082.3
1 e 5.29 -e 94.71 - 200 (t)Ttt −−
=
For t=0,5,10 and ∞
T(0) = 100 oF
T(5) = 134.975 oF
T(10) = 155.856 oF
T(∞) = 200 oF
Q – 10.1. A pneumatic PI controller has an output pressure of 10 psi,
when the set point and pen point are together. The set point and pen point are suddenly
changed by 0.5 in (i.e. a step change in error is introduced) and the following data are
obtained.
Determine the actual gain (psig per inch displacement) and the integral time.
Soln:
e(s) = -0.5/s
for a PI controller
Y(s)/e(s) = Kc ( 1 + τI-1/s)
Y(s) = -0.5Kc ( 1/s + τI-1/s2)
Y(t) = -0.5Kc ( 1 + τI-1 t )
At t = 0+ y(t) = 8 � Y(t) = 8 – 10 = -2
2=0.5Kc
Kc = 4 psig/in
At t=20 y(t) = 7� Y(t) = 7-10 = -3
3 = 2 ( 1 + τI-1 20 )
τI = 40 sec
Q-10.2. a unit-step change in error is introduced into a PID controller. If KC = 10 , τI = 1
and τD = 0.5. plot the response of the controller P(t)
Soln:
P(s)/e(s) = KC ( 1 + τD s+ 1/ τIs)
For a step change in error
Time,sec Psig
0- 10
0+ 8
20 7
60 5
90 3.5
P(s) = (10/s)(1 + 0.5 s + 1/s )
P(s) = 10/s + 5 + 10/s2
P(t) = 10 + 5 δ(t) + 10 t
Q – 10.3. An ideal PD controller has the transfer function
P/e = KC ( τD s + 1)
An actual PD controller has the transfer function
P/e = KC ( τD s + 1) / (( τD/β) s + 1)
Where β is a large constant in an industrial controller
If a unit-step change in error is introduced into a controller having the second
transfer function, show that
P(t) = KC ( 1 + A exp(-βt/ τD))
Where A is a function of β which you are to determine. For β = 5 and KC =0.5,
plot P(t) Vs t/ τD. As show that β � ∞, show that the unit step response approaches that
for the ideal controller.
Soln:
P/e = KC ( τD s + 1) / (( τD/β) s + 1)
For a step change, e(s) = 1/s
P(s) = KC s( τD s + 1) / (( τD/β) s + 1)
10
15
10(1+t)
P(t)
t
=
+
−
+
βτ
βτ
ssK
D
D
C
1
111
P(t) =
−
+−
D
t
D
D
C eK τβ
βτ
βτ 11
1
=
−+
−D
t
C eK τβ
β )1(1
So, A = β – 1
P(t) = 0.5 ( 1 + 4 exp(-5t/ τD))
As β � ∞ then τD/β � 0 and
P/e = KC ( τD s + 1) / (( τD/β) s + 1) becomes
P/e = KC ( τD s + 1) that of ideal PD controller
Q – 10.4. a PID controller is at steady state with an output pressure of a psig. The set
point and pen point are initially together. At time t=0, the set point is moved away from
2.5
0.5
P(t)
t/τD
the pen point at a rate of 0.5 in/min. the motion of the set point is in the direction of lower
readings. If the knob settings are
KC = 2 psig/in of pen travel
τI = 1.25 min
τD = 0.4 min
plot output pressure Vs time
Soln:
Given de/dt = -0.5 in/min
s e(s) = -0.5
Y(s)/e(s) = KC ( 1 + τD s+ 1/ τIs)
Y(s) = -( 1/s + 1/ τIs2 + τD )
Y(t) = -( 1 + t/1.25 + 0.4 δ(t) )
Y(t) = y(t) – 9 = - ( 1 + t/1.25 + 0.4 δ(t) )
y(t) = 8 – 0.8 t – 0.4 δ(t)
Q – 10.5. The input (e) to a PI controller is shown in the fig. Plot the output of the
controller if KC = 2 and τI = 0.5 min
9
8
7.6
10
y(t)
t
e(t) = 0.5 ( u(t) - u(t-1) - u(t-2) + u(t-3) )
e(s) = (0.5/s) ( 1 – e-s - e
-2s + e
-3s )
P(s)/e(s) = KC ( 1 + (1 / τI s) ) = 2 ( 1+ 2/s )
P(s) = ( 1/s + 2/s2 ) (1 – e-s - e-2s + e-3s )
P(t) = 1 + 2t 0 ≤ t < 1
= 2 1 ≤ t < 2
= 5 – 2t 2 ≤ t < 3
= 0 3 ≤ t < ∞
Q – 12.1. Determine the transfer function Y(s)/X(s) for the block diagrams shown.
Wxpress the results in terms of Ga, Gb and Gc
0 1
2 3
4
t, min
0.5
e
-0.5
Soln.
(a) Balances at each node
(1) = GaX
(2) = (1) – Y = GaX – Y
(3) = Gb(2) = Gb(GaX – Y)
(4) = (3) + X = Gb(GaX – Y) + X
Y = Gc(4) = Gc (Gb(GaX – Y) + X)
= GaGbGcX – GbGcY + GcX
GbGc
GaGbGc
X
Y
++
=1
)1(
(b) Balances at each node
(1) = X – (4)
(2) = Gb(1) = Gb( X – (4))
(5) = GcX/Ga
(3) = Gc(2) = GbGc( X – (4))
(4) = (3) + (5) --------------------------- 5
= GbGc( X – (4)) + GcX/Ga
Y = Ga(4)
From the fifth equation
(4) = GbGcX – GbGc(4) + GcX/Ga ----------- 6
GaGbGc
XGcGaGbGc
)1(
)()4(
++
=
From the sixth equation
)1(
)1(
GbGc
GcGaGb
X
Y
++
=
Q – 12.2
Find the transfer function y(s)/X(s) of the system shown
Soln:
Balance at each node
(1) = X – Y ---------(a)
(2) = (1) + (3) ----------(b)
(3) = G1(2) where G1 = 1/(τ1s + 1) ----------(c)
(4) Y = G2(3) where G2 = 0.5/(τ1s/2 + 1) ----------(d)
From (d) and (c)
Y = (2)G1G2
= G1G2 (X – Y + (3) ) ----------(e)
Also from (b) and (c)
(3) = G1((1) + (3))
(3)(1 – 1/(τ1s + 1)) = 1/(τ1s + 1)
(3) τ1s = 1
(3) = 1/(τ1s ) = (X – Y) / (τ1s)
Substitute this in (e)
( )YXss
s
Y −
+
++=
111
11
)12
)(1(
5.0
τττ
12
1
1
22
1 ++=
ssX
Y
ττ
Q – 12.3. For the control system shown determine the transfer function C(s)/R(s)
Soln.
Balances at each node
(1) = R – C ------------------(a)
(2) = 2 (1) = 2(R – C) ------------------(b)
(3) = (2) – (4) = 2(R – C) – (4) -------------------(c)
(4) = (3)/s = (2(R – C) – (4))/s -------------------(d)
(5) = (4) – C -------------------(e)
C = 2(5) -------------------(f)
Solving for (4) using (d)
s (4) = 2(R – C) – (4)
(4) = 2(R – C) / (s +1)
Using (e)
(6) = 2(R – C) / (s +1) – C
( )
−−+
= CCRs
C1
22
( )
73
4
)1(24)1(4
+=
++++=
sR
C
ssCR
Q – 12.4. Derive the transfer function Y/X for the control system shown
Soln.
Balance at each node
(1) = (5) + X -----------------(a)
(2) = (1) – (4) -----------------(b)
(3) = (2)/s ------------------(c)
Y = (3)/s ------------------(d)
(5) = 2 (3) ------------------(e)
(4) = 25Y ------------------(f)
From (b)
(4) = (1) – (2)
= (1) – s (3) from (c)
= (1) – s2 Y from (d)
= (5) + X - s2 Y from (a)
= 2 (3) + X - s2 Y from (e)
= 2 s Y + X - s2 Y
From (f)
Y = (2 s Y + X - s2 Y)/25
X = Y( 25 – 2s + s2 )
252
12 +−
=ssX
Y
13.1 The set point of the control system in fig P13.1 given a step change
of 0.1 unit. Determine
(a) The maximum value of C and the time at which it occurs.
(b) the offset
(c) the period of oscillation.
Draw a sketch of C(t) as a function of time.
)12)(1(
51
)12)(1(
5
+++
++=
ssK
ss
R
C
932
82 ++
=ssR
C
sR
1.0=
b) 0889.09
8.0
932
8.0)(
20==
++=∞
→ ssLtCS
offset = 0.0111
c) 22
1
3
12;
3
2;
9
8.0=⇒=== ξξττK
overshoot = 305.01
exp2
=
−−
ξ
πξ
= Maximum vslue of C = 1.0305*0.0889=0.116
Maximum value of C = 0.116
−+−
−−= −−
ξξ
τξ
ξτξ 2
12
2
1tan1sin
1
11
9
8.0116.0
te
t
6.11
tan1
2
1
2=
−
−= −
ξξ
ξ
τt
Time at which Cmax occurs = 1.6
(c ) Period of ociullation is 166.31
2
2=
−=
ξ
πτT
T =3.166
Decay ratio = (overshoot)2 = 0.093
13.2 The control system shown in fig P 13.2 contains three-mode controller.
(a) For the closed loop, develop formulas for the natural period of oscillation
τ and the damping factor ξ in terms of the parameters K, Dτ , Iτ and 1τ . (b) Calculate ξ when K is 0.5 and when K is 2. (c ) Do ξ & τ approach limiting values as K increases, and if so, what are these values?
(d ) Determine the offset for a unit step change in load if K is 2.
(e ) Sktech the response curve (C vs t) for a unit-step change IN LOAD
WHEN k is 0.5 and when K is 2.
(f) In both cases of part (e) determine the max value of C and the time at
which it occurs.
++
++
++
+=
ss
s
k
ssk
s
R
Ca
I
D
I
D
ττ
τ
ττ
τ
11
11
11
1
1
)
1
1
( )
++
++
+=
ssk
s
s
U
C
I
D ττ
τ
τ
11
1
11
1
1
1
1
=
112 +
++
+ s
k
ks
k
k
s
III
DI
I
τττ
ττ
τ
++++
++
=
ssks
ssk
R
C
I
D
I
D
τττ
ττ
111
11
1
( )ksksk
ssk
R
C
IIDI
IID
++++
++=
ττττττττ
)1()(
12
1
2
k
k
k
k IIDI ττξ
ττττ
)1(2;
)(2 +=
+=
k
k
k
k IIDI τξ
τττ )1()(2
+=
+×=
)(2
)1(
1τττ
ξ+
+==
D
I
kk
k
)(2
)1()(4
)(2
1
2
1
2
1
1
2
ττ
τττ
τττπ
ξ
τπ
+
+−+
+×
=−
==
D
DD
DI
kk
kkk
k
k
T
21
1
)1(4
)(4
+−
+
+=
kkk
kT
II
D
D
ττ
ττ
ττπ
B) Dτ = Iτ =1; 1τ .=2
For k = 0.5 ; ξ =0.75 671.0)5.2(5.0
1=
For k = 2 ; ξ =1.5 530.032
1=
×
C)
+
+=
+
+==
k
k
kk
k
ID
I
D
I
ττ
τ
τττ
ξ
2
1
2
11
2
1
)(
)1(
2
1
As 3535.02
1,
1
==∞→ττ
ξ Ik
21
1
)1(4
)(4
+−
+
+=
kkk
kT
II
D
D
ττ
ττ
ττπ
k
k IDI τττττ 1+
=
k
I
DI
1τττττ +=
DIK τττ =∞→ =1
2552.7
14
4, =
−
=∞→
I
D
DTkAs
ττ
πτ
(d)
++
++
+=
ss
s
k
s
U
C
I
D ττ
τ
τ
11
11
1
1
1
1
ksksk
s
U
C
IDI
I
++++=
2
1 )()1( τττττ
sU
1=
kskskC
IDI
I
++++=
2
1 )()1( τττττ
00
)(0
==→ k
ssfLtS
K=2
For a unit step change in U
0)( =∞C
Offset = 0
(e) k = 0.5, ξ =0.671 & τ =2.236
95.181
2
2=
−=
ξ
πτT
If k = 0.5
135
22 ++
=ss
s
U
C;
135 2 ++=
ss
sC
If k = 2
15.12
5.02 ++
=ss
s
U
C;
15.12
5.02 ++
=ss
C
In general τ
ξξτ
τξ
tetC
t
2
21sin
1
11)( −
−=
−
The maximum occurs at ξξ
ξ
τ 2
1
2
1tan
1
−
−= −t
If k = 0.5 tmax = 2.52 Cmax=0.42
If k = 2 tmax = 1.69 Cmax=0.19
13.3 The location of a load change in a control loop may affect the system
response. In the block diagram shown in fig P 13.3, a unit step chsange
in load enters at either location 1 and location 2.
(a) What is the frequency of the transient response when the load enters at
location Z?
(b) What is the offset when the load enters at 1 & when it enters at 2?
(c) Sketch the transient response to a step change in U1 and to a step
change in U2.
0;1
21 == Us
U
Css
UCRS =
+
+
+−12
1
12
2))(( 1
R = 0
1144
2
1)12(
25
)12(
2
2
2
2
1 ++=
++×
+=
ss
s
s
U
C
1144
22
1 ++=
ssU
C
11
1
11
42;
11
2;
11
2=⇒=== ξξττK
2516.02
10
2
11
2
112
==−
===πτ
ξπT
fFrequency
C(∞) = 2/11
Offset = 2/11 =0.182
U1=0;U2=1/s
Cs
UCRs
=
+
+−+
×⇒12
1)(
12
25 2
R=0
112
1
12
1012
1
2 ++
×+
+=⇒
ss
s
U
C
1144
122
2 +++
=ss
s
U
C
C(∞) = 1/11
Offset = 1/11=0.091
a) if ;1
1s
U = frequency = 0.2516
if ;1
2s
U = frequency = 0.2516
b)if ;1
1s
U = frequency = 0.182
if ;1
2s
U = frequency = 0.091
13.5A PD controller is used in a control system having first order process and a
measurement lag as shown in Fig P13.5.
(a) Find the expressions for ξ and τ for the closed –loop response.
(b) If τ1 = 1 min, τm = 10 sec, find KC so that ξ = 0.7 for the two cases: (1)
τD =0,(2) τD =3 sec,
(c) Compare the offset and period realized for both cases, and comment on the
advantage of adding derivative mode.
)1)(1(
)1(1
)1(
)1(
)
1
1
+++
+
++
=
ss
sK
s
sK
R
Ca
m
DC
DC
τττ
ττ
)1()(
)1)(1(
1
2
1 +++++
++=
CDCmm
mDC
KsKs
ssK
R
C
τττττττ
= 1
12
+=
C
m
K
τττ
1
1
+=
C
m
K
τττ
)1(2
1
1
1
+
++=
Cm
DCm
k
k
ττ
τττξ
b) 7.0;10min;11 === ξττ sm
1)
)1(600
70
2
17.0
0
+×=⇒
=
C
D
k
τ
kc=3.167
2)
255.5
)1(600
370
2
17.0
3
=
+
+×=⇒
=
C
C
C
D
k
k
k
sτ
c)for 1
)(;1
+=∞=
C
C
k
kc
sR
periodsperiodsfor
periodperiodfor
offsetCsfor
offsetCFor
D
D
D
D
==−
==
==−
×==
==∞=
==∞=
17.86)7.0(1
255.6
6002
;3
57.1051
167.4
6002
;0
16.0;84.0)(;3
24.0;76.0)(;0
2
2
πτ
ξ
πτ
τ
τ
Comments:
Advantage of adding derivative mode is lesser offset lesser period
13.6The thermal system shown in fig P 13.6 is controlled by PD controller.
Data ; w = 250 lb/min; ρ = 62.5 lb/ft3; V1 = 4 ft
3,V2=5 ft3; V3=6ft
3;
C = 1 Btu/(lb)(°F)
Change of 1 psi from the controller changes the flow rate of heat of by 500
Btu/min. the temperature of the inlet stream may vary. There is no lag in the
measuring element.
(a) Draw a block diagram of the control system with the appropriate transfer
function in each block.Each transfer function should contain a numerical values of
the parameters.
(b) From the block diagram, determine the overall transfer function relating the
temperature in tank 3 to a change in set point.
(c ) Find the offset for a unit steo change in inlet temperature if the controller
gain KC is 3psi/°F of temperature error and the derivative time is 0.5 min.
CWTTTCVqCWT 10110 )( +−=+ ρ
CWTTTCVCWT 21221 )( +−= ρ
CWTTTCVCWT 32332 )( +−= ρ
)()( 1110 CVCWTqCVWCT ρρ +=++
1
01CVWC
qTT
ρ++=
T1= T2 = T3
⇒+
+=1
03CVWC
qTT
ρ ( )sCVWC
sqsT
1
3
)()(
ρ+=
13.6 (b)
)15.1)(125.1)(1(
2)1(1
)15.1)(125.1)(1(
2)1(
)(
)('3
+++++
++++
=
ssssk
ssssk
sR
sT
DC
DC
τ
τ
= )1(2)175.2875.1)(1(
)1(2
)(
)(2
'
3
sksss
sk
sR
sT
DC
DC
ττ
+++++
+=
12)275.3(625.4875.1
)1(2
)(
)(23
'
3
+++++
+=
cDC
DC
kskss
sk
sR
sT
ττ
c) kC=3;s
soffsetD
1)(?,,5.0 '
0 === ττ
12)275.3(625.4875.1
1
)(
)(230'
'
3
+++++→CDC
si kskss
LtsT
sT
τ
= 143.07
1
12
1==
+Ck
Offset =0.143
13.7 (a) For the control system shown in fig P 13.7, obtain the closed loop
transfer function C/U.
(b) Find the value of KC for whgich thre closed loop response has a ξ of 2.3.
(c) find the offset for a unit-step change in U if KC = 4.
Cs
UCRs
sKC =
+−+
+×
1)(
125.0
1
=
125.0
1.1
1
++
+=
s
s
s
Ks
U
C
C
)1(25.0
125.02 +++
+=
sKss
s
U
C
C
CC KsKs
s
U
C
4)1(4
42 +++
+=
b) ξ=2.3
C
C
CK
K
K
12;
4
1 +== ξττ
= C
C
CK
K
K
13.22
4
1 +=×
3.21=
+=
C
C
K
K
KC=2.952
C) KC=4,U = 1/s
= 1620
412 ++
+×=
ss
s
sC
4
1)( =∞C
offset = 0.25.
13.8 For control system shown in Fig 13.8
(a) C(s)/R(s)
(b) C(∞∞∞∞) (c) Offset
(d) C(0.5)
(e) Whether the closed loop response is oscillatory.
(a)
)1(
41
)1(
4
++
+=
ss
ss
R
C
4
42 ++
=ssR
C
b) C(∞) =2*1=2
C(∞) =2
C) offset = 0
d) 4
1
4
12;
2
1=⇒== ξξττ
+
−
−= −−15tan
4
15sin
4
11
11
2
)( 12
2 τt
etC
t
=
+−= −−
15tan4
15sin
15
412)5.0( 14
1
eC
C(0.5)=0.786
.e) ξ<1, the response is oscillatory.
13.9 For the control system shown in fig P13.9,determine an expression for
C(t)
if a unit step change occurs in R. Sketch the response C(t) and compute
C(2).
1̀2
1
111
11
++
=
++
+=
s
s
R
C
s
s
R
C
2
2
11)(
12
11
)12(
1
1
t
etC
ssss
sC
sR
−−=
+−
+=+
+=
=
C(2) = 0.816
13.10 Compare the responses to a unit-step change in a set point for the
system shown in fig P13.10 for both negative feedback and positive feedback.Do
this for KC of 0.5 and 1.0. compare the responses by sketching C(t).
-ve feed back :
))1(( ++
=C
C
Kss
KC
+ve feed back
1
11
1
1
+−
+×
=
sK
sK
R
C
C
C
))1(( C
C
Kss
KC
−+=
For KC = 0.5 , response of -ve feed back is
32
3
2
3
1
)32(
1
+
−+=
+=
ssssC
)1(3
1
3
1
3
1)( 2
3
2
3 tt
eetC−−
−=−=
response of +v feed back is
21)(
12
21
)12(
1
t
etC
ssssC
−−=
+−
+=+
=
For KC = 1, response of -ve feed back is
tetC
ssssC
2
2
1
2
1)(
2
2
1
2
1
)2(
1
−−=
+
−+=
+=
response of +ve feed back is
ttC
sC
=
=
)(
12
14.1 Write the characteristics equation and construct Routh array for
the control system shown . it is stable for (1) Kc= 9.5,(ii) KC =11; (iii) Kc=
12
Characteristics equation
0)66(116
0)66(116(
06)3)(2)(1(
0)3)(2)(1(
61
23
2
=++++
=++++
=++++
=+++
+
Kcsss
Kcsss
Kcsssor
sss
Kc
Routh array
)1(6
)66(6
111
2
3
Kcs
Kcs
s
+
+
For Kc=9.5
= 10-(Kc)= 10-9.5=0.5>0 therefore stable.
For Kc=11
= 10-(Kc)= 10-11=-1<0 therefore unstable
For Kc=121
= 10-(Kc)= 10-12=--2<0 therefore unstable
14.2 By means of the routh test, determine the stability of the system shown
when KC = 2.
Characteristic equation
01042
102
3121
2=
++
++sss
060252
01205042
0120401042
010.2).3(2)1042(
23
23
23
2
=+++
=+++
=++++
=++++
sss
sss
ssss
ssss
Routh Array
1 25
2 60
-10/2
The system is unstable at Kc = 2.
14.4 Prove that if one or more of the co-efficient (a0,a1,….an) of the
characteristic equation are negative or zero, then there is necessarily an
unstable root
Characteristic equation :
0..................................1
10 =+++ −n
nn axaxa
0)/......................./( 0
1
010 =++ −aaxaaxa n
nn
0.............................,
0).(..........).........)((
21
210
<
=−−−
n
n
haveWe
xxxa
ααα
ααα
As we know the second co-efficient a1/a0 is sum of all the roots
2/)1(11
2
0
1
−= ∑∑
==
n
j
ji
n
ia
aαα
Therefore sum of all possible products of two roots will happen twice as
21αα dividing the total by 2.
And
00/
)00(0
202 >⇒>∴
<<>
aaa
jiji αααα
Similarly
),.......1(0
0/
00)1()1(
0/01)1(
)()1(
0
0
0
0
njforaso
aacasebothin
againisa
asoissumtheandisoddjif
aasoissumtheandisevenjif
rootsjofproductspossibleaollofsuma
a
j
j
jj
j
j
jj
=>
>
><−−=
>>−=
−=
14.5 Prove that the converse statement of the problem 14.4 that an
unstable root implies that one or more co-efficient will be negative or
zero is untrue for all co-efficient ,n>2.
Let the converse be true, always .Never if we give a counter example we
can contradict.
Routh array
0
2
3
23
01
31
21
32
s
s
s
s
sss
−
+++
System is unstable even when all the coefficient are greater than 0; hence a
contradiction,
14.6 Deduce an expression for Routh criterion that will detect the
Presence of roots with real parts greater than σ for any rectified σ >0
Characteristic equation
0.................................1
10 =+++ −n
nn axaxa
Routh criteria determines if for any root, real part > 0.
Now if we replace x by X such that
.x + σ =X
Characteristic equation becomes
0.................................)()( 1
10 =++−+− −n
nn aXaXa σσ
Hence if we apply Routh criteria,
We will actually be looking for roots with real part >σ rather than >0
0.................................2
2
1
10 =+++ −−n
nnn axaxaxa
Routh criterion detects if any root jα is greater than zero.
Is there any )1(0.,...............,,........., 21 −−−−−>= njx αααα
Now we want to detect any root
)1(
0
From
j
j
>
−>
ασα
σα
σασ
σασ
σασ
σασ
σασ
σ
α
α
α
α
αααα
−>
>+
+=
>+=+
>+=+
>+=+
>+=+
>=
>=
>=
>=
>=
j
j
n
j
n
j
nj
or
anyecttocriteriaRouthapplyand
xXLetso
x
x
x
x
anythereis
sidesbothonadd
x
x
x
x
anythereisimplies
x
0det
,
0
.
.
.
0.
.
.
.
0
0
0
.
.
.
0
.
.
.
0
0
0,.................,............................,.........,
2
1
2
1
21
14.7 Show that any complex no S1 satisfying ,1<S yields a value of
s
sZ
−+
=1
1that satisfies Re(Z)>0,
Let S=x+iy, 122 <+ yx
s
sZ
−+
=1
1
( )
+−
+=
>
<−++<
==
=−=
−++
<+>+−
<<−
<<−
<+
<+
>++−
>+−>
++−+−
=
++−++−
=
+−+
−−+=+−
+−+−
−−++
−
)5.05.0(
1
)5.05.0(
:
0)Re(
4)2)(1(0
00&1
40&1
2)(1
10)(1
int11
11
1
1
0)(21
0)(10)Re(
)(21
)(1)Re(
)(21
2)(1
21
))11(1(
)1(
)1(
)1(
1
1
22
22
2222
22
22
22
22
22
22
22
22
22
22
isL
partrealthetodueunstableissystemtheisif
example
z
xyx
isitthenyxif
isitthenyxif
xyx
Now
yxthereforetrueisyx
circleunittheinsPoy
xareRanges
yx
yxhavewe
yxx
andyxthenzif
yxx
yxZ
yxx
iyyx
yxx
yiyxxx
iyx
iyx
iyx
iyx
)5.05.0()5.05.0(
1 5.01 tSintCoseis
L t +=
+−−
14.8 For the output C to be stable, we analyze the characteristic
equation of the system
0)1()1)(1(
111 3
21
=+×++
+ ssssI
ττττ
01)()(
01)1(
3
2
21
3
21
3212
2
1
=+++++
=+++++
sss
sssss
III
I
ττττττττ
ττττττ
Routh Array
1
0
1)(
0
21
2
321
3
s
s
s
s
I
II
α
τττ
τττττ
+
+
)(
))((
21
21321
τττττττττττ
α+
−++=
I
III
Now
0)1( 12 >τττ I
Since 21 &ττ are process time constant they are definitely +ve
0;0 21 >> ττ
(2) 0)( 21 >+τττ I
(3) 21321 ))((0 ττττττττα III >++⇒>
0
)()(
0
3
21
21
2132121
21322311
>
−+
>
+−>+
>−+++
I
I
I
II
also τ
τττ
τττ
ττττττττ
ττττττττττ
14.9 In the control system shown in fig find the value of Kc for which
the system is on the verge of the instability. The controller is replaced by a
PD controller, for which the transfer function is Kc(1+s). if Kc = 10,
determine the range for which the system is stable.
Characteristics equation
0)66(116
06)3)(23(
06)3)(2)(1(
0)3)(2)(1(
61
23
2
=++++
=++++
=++++
=+++
+
Kcsss
Kcsss
Kcsssor
sss
Kc
Routh array
+−
+
)3
13
13
31
2
3
Kcs
Kcs
s
For verge of instability
8
)3
13
=
+=
Kc
Kc
Characteristics equation
03)1(
)1(101 =
++
+s
kcs
011)103(3 23 =++++ Kcssss
Routh Array
30/2
230
11)103(3
113
1031
2
3
>
>
>+
+
D
S
s
D
vegefor
s
s
s
τ
τ
τ
τ
14.10 (a) Write the characteristics equation for the central system shown
(b) Use the routh criteria to determine if the system is stable for Kc=4
© Determine the ultimate value of Kc for which the system is unstable
(a) characteristics equation
02)1(32
0)2()12)((
0)1(
1
12
1
3
21
23
2
=++++
=++++
=+
=
++
kcskcss
skcsss
ss
skc
0)1(33 23 =++++ kcsss
Kc=4Routh array
3;03
03
4)1(3
3/1
83
52
2
3
==−
=−+
−
KcKc
kckc
stablenot
s
s
s
For verge of instability
14.11 for the control shown, the characteristics equation is
0)1(464 234 =+++++ kssss
(a) determine value of k above which the system is unstable.
(b) Determine the value of k for which the two of the roots are on
the imaginary axis, and determine the values of these imaginary roots and
remaining roots are real.
0)1(464 234 =+++++ kssss
k
ks
ks
s
ks
+
+−
+
+
11
)1(5
44
15
44
)1(61
2
3
4
For the system to be unstable
1
1
01
4
5
11
05
114
−>
−<
<+
>
+<
<
+−
k
k
k
k
k
k
The system is stable at -1<k<4
(b) For two imaginary roots
4);1(5
44 =+= kk
Value of complex roots
is
s
±=
=+ 055 2
5454641 22342 +++++++ sssssss
24 0 ss ++
0
55
55
404
54
2
2
3
23
+
+
++
+
s
s
ss
ss
SOLUTION:
ii
s ±−=±−
=−±−
= 22
24
2
20164
PART 2
LIST OF USEFUL BOOKS FOR PROCESS
CONTROL
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PUBLICATIONS, INDIA ( WIDE VARIETY OF SOLVED PROBLEMS ARE
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2. ADVANCED CONTROL ENGINEERING BY RONALD.S .BURNS ,
BUTTERWORTH AND HIENEMANN.
3. PROCESS MODELLING SIMULATION AND CONTROL FOR
CHEMICAL ENGINEERS, WILLIAM.L.LUYBEN, MCGRAW HILL.
4. A MATHEMATICAL INTRODUCTION TO CONTROL THEORY BY
SCHOLOMO ENGELBERG, IMPERIAL COLLEGE PRESS
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