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Test signals Impulse
Step
Ramp
Sinusoidal (frequency response)
Impulse (t) 1
step u(t) 1/s
Ramp t u(t) 1/s2
Laplace Transforms
Chapter 5:Time Response Analysis:
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Examples of zeroth order systems :
Potentiometerdc amplifier
dc tachogenerator
zeroth order system :
C(s) / R(s) = K ; a constant
(algebraic equation)
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First order systems:C(s) / R(s) = K / (1+s )
K : steady state value of the Function : time constant,
smaller, faster response.
C(t) = K ( 1- e-t / ) ;for unit step
Step Response
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0
50
100
150
200
250
300
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
t1
t2
Time constant : t2 < t1
dc/dt = 1/ t=0K=1
Error e(t) = e-t /
ess = 0
c
t
If K 1 and input is not unitythen ess ?
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Ramp response of first order systems
r (t) = t u(t); R (s) = 1 / s2
1=KLet;ts1
K
R(s)
C(s)
+=
)s1(s
1)
2 +=c(s
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+= s1s1s1)s(C 2
)e1(-t=c(t) -t/
Error e(t) = (1-e-t /)
ess
=
If K 1 and input is not unity
then ess ?
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2
nn
2
2
n
+s2s)s(R)s(C
+
=
Second order systems
Second order example
position control
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For unit step input
R ( s ) =1
s
C(t) t =1.0
Applying final value theorem
Steady state error for step input = 0
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r LKp KA+
-
+
-
KTRa
1
s(Js+B)
sKb
position control Example
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Taking KP = 0.1 V / rad
KA = 10 V / VKT (Nm/A) = Kb (V/rad/sec. ) = 0.5
J = 1 Kg m2
B = 0.5 Nm / rad / sec
1+ss
1
)s(
)s(2
r
L
+=
n = 1 rad / sec and = 0. 5
s2 + s + 1 = (((( )))) (((( ))))s + 0.52 3
22++++
j 3 2
- j3
2
- 0.5
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In general
( ) ( )22n2n 1+s +=s2 + 2 n s + n 2
2
nn1,2 1js =
= - n j d ; d = n 1 - 2
d damped freq of oscillations damping factor
n undamped natural frequency
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roots of s are complex conjugate
= 0 : undamped
dampedunder:10
roots of s are real and distinct
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Unit step response
r (t) = u(t) ; R(s) =1/s
)s(s)s(C 2n2
2
n
+
=
c ( t ) = 1- cos nt (pure oscillation)
Case 1: =0
n2
C(s) =s (s2 + 2 ns + n2)
Case 2: 0 < < 1
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1 -n 2
- n
n
Where - n isreal part and d is the imaginary part of the roots
t)(coset)(sine
1
-1=(t)c dt-
d
t-
2
nn +
)+t(sin1
e-1=(t)c d
2
t- n
where cos
=
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C(t)
1.0
Time(t)
1+e- nt
1- 2
1-e-nt
1-2
Unit step response of 2nd order system for
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C(t)
1.0
tp
Mp
trTime
Tolerance band
Time response specification
ts
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sin (dt) = 0, dt =
Time at peak overshoot ( tp ) = /d
Peak overshoot (Mp):d c
d t==== 0
At t = tp ,
)+(sin
1
1)(2
.
= pn t
p
etc
21
eMp
=
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Settling time (ts):
Assume tolerance band 5%
0.05 = e
- nts
n
ssnn
3=or tt)05.0(l
=
n
s4isbandetoleranc2%fort
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Delay time (tr):
)+t(sin1
e
-1=0.5=)(tc dd2
t-
d
dn
So delay time (td) (1.1+.125+.469 2)/n
Rise time (tr):
)+t(sin1
e-1=1=)(tc rd
2
t-
r
rn
d
r -t =
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Second order system response
with ramp input
n2
s2 (s2 + 2 ns + n2)C(s) =
)+t(sin1
e2-t)t(c d
2
n
t-
n
n
+
=
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1 n2
E(s) = 1-s2 (s2 + 2 ns + n2)
Steady state error of a second order
system with ramp input:
2
nn
2
n
2
2 +s2s
s2s.
s
1
++
=
n
2
=
For the example taken,
= 0.5 & n = 1
ess = 1 unitess
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R(s) +
-
E(s) C(s)G(s)
Error coefficients
C (s) = R (s) . G (s) / ( 1+G (s) H (s) )
E (s) G (s) = C (s)
( )(s)H(s)G+1(s)R=(s)E
H (s)
(s)EsLt=)(eerrorstateSteady 0sss
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Step input
R (s) = 1 /s (constant position)
)s(H)s(G1
1.
s
1sLte
0sss +
=
coeff.)error(positionK)S(H)s(GLt p0s
=
p
ss
K
e
+
=
1
1So
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Ramp input
R (s) = 1 /s2 (constant Velocity)
)s(H)s(G11.
s1sLte2
0sss +
=
coeff.)error(VelocityK)S(H)s(sGLt v0s
=
vss K
1e =So
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Parabolic input
R (s) = 1 /s3 (constant Acceleration)
)s(H)s(G11.
s1sLte3
0sss +
=
coeff.)erroron(AccleratiK)S(H)s(GsLt a2
0s=
ass K
1e =So
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systemtheoftypen
)(....)ps)(ps(s)(....)zs)(zs()s(H)s(G21
n21
++++=
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Type 0 system
valuefiniteK1
1e
valuefinite)s(H)s(GLtK
p
ss
0sp
=+
=
==
===
ss0s
v e0)s(H)s(sGLtK
===
ss
2
0s
a e0)s(H)s(GsLtK
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Type 1 system
0e)s(H)s(GLtK ss0s
p ===
finiteK1e
finiteH(s))s(sGLtK
vss
0sv
==
==
===
ss
2
0s
a e0)s(H)s(GsLtK
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Type 2 system
0e)s(H)s(GLtK ss0s
p ===
0e)s(H)s(sGLtK ss0sv ===
finiteK1e
finite)s(H)s(GsLtK
ass
2
0sa
==
==
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)t(u2
t2
11++++Kp
1
Kv
aK
1
0
1
2
0
0 0
u(t) t u(t)InputType
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Compensation Techniques:
R +
-
K 1s(s+1)
C
Example
Kss
K
)s(R
)s(Cand
)1s(s
KG
2
++
=
+
=
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K2
1
,Kn ==
If K = 100n = 10 rad / sec, = 0.05
Mp= 85 %
ess for ramp input = 2/n = 0.01 unit
.sec84%)2( == nst
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While the steady state error is acceptable
Mp is not.Damping () to be increased to say, 0.5
10.521=1=2Then nn =
ess
for ramp input = 2/n
= 1unit
ess increases to 1 - not acceptable.
change theamplifierconfiguration.
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R +
-K+sKD
1
s(s+1)
C
Derivative error Compensation
Required Specifications:
= 0.5 and ess for ramp input = 0.01
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)1s(s
sKK
)s(H)s(G
D
+
+
=
KKsss
KsK
)s(R
)s(C
D
2 D ++++
=
n =K
= 10 rad/sec2n = 1 + KD
for = 0.5, KD = 9
sec.8.04
%)2(t
n
s =
= ess for ramp input
remains the same
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R +-
K 1s(s+1) C
sKt
+
-
Derivative output Compensation
Introduce another feed back loop
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K)K1(ss
K
)s(R)s(C
)K1s(s
K
)s(H)s(G
t
2
t
+++=
++=
Required Specifications:
= 0.5 and ess for ramp input = 0.01
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(1)-----100
K1
K
)s(H)s(sGLtK
t
0sv
=
+
=
=
)2(K1KorK12 ttn +=+=
99K
10000K100K
t =
==
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Comparing both methods
derivative output compensation:additional transducer required
gain reduced due to feedback sohigh gain amplifier required and will
have a higher n and hence lesser ts
I t l C t l
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+
-K
K
sA
I++++
1
1s s( )++++
Integral Control
)1s(s
KsK)s(H)s(G
2
IA
++=
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type 2 ess for step input = 0
ess for ramp input = 0ess for parabolic input is finite
However system becomes of 3 rd order -
stability problem arises and will bediscussed in later chapters
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Addition of Zeros and Poles to the system
2
nn
2
z
2
n
2
nn
2
2
n
2
nn
2
z
2
n
+s2s
sT
+s2s)s(R
)s(C
+s2s
)sT1(
)s(R
)s(C
+
+
+
=
+
+=
y(t)= y1(t)+ Tz dy1/dt
Zero added to closed loop Transfer function
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Unit step response when zero is added to closed loop Transfer
function at various mentioned positions
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Unit step response showing the effect of zero addition to closed
loop Transfer function.
Zero added to forward path
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Zero added to forward path
Unit step response when zero is added to Forward path at various positions
6s)T62(s3s
)sT1(6
)s(R
)s(C
2p,1p,6K,)ps)(ps(s
)sT1(K)s(G
z
23
z
21
21
z
+++++
=
===++ +=
Pole added to forward path
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Pole added to forward path
Unit step response when pole is added to Forward path at various mentioned positions
2nn
2pn
3p
2
n
pn
2
n
s2+)sT21(sT)s(R
)s(C
s)T)(12s(s)s(R)s(C
+++
=
++=
Pole added to closed loop Transfer Function
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Pole added to closed loop Transfer Function
Unit step response when pole is added to closed loop transfer function at
various mentioned positions.
s)T)(1s2(s)s(R
)s(C
p
2
nn
2n
2 +++=