mesb374 system modeling and analysis forced response
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MESB374 System Modeling and Analysis
Forced Response
Forced Responses of LTI Systems• Forced Responses of LTI Systems
– Superposition Principle
– Forced Responses to Specific Inputs
• Forced Response of 1st Order Systems– Transfer Function and Poles/Zeros
– Forced Response of Stable 1st Order Systems
• Forced Response of 2nd Order Systems– Transfer Function and Poles/Zeros
– Forced Response of Stable 2nd Order Systems
Forced Responses of LTI Systems
• Superposition Principle
forced responses of
simple inputs
y a y a y a y a y b u b u b unn
nm
m( ) ( ) ( )
11
2 1 0 1 0
Input Output
u1 (t)
u2 (t)
u(t)=k1 u1 (t) + k2 u2 (t)
y1 (t)
y2 (t)
y(t)=k1 y1 (t) + k2 y2 (t)
The forced response of a linear system to a complicated input can be obtained by studying how the system responds to simple inputs, such as unit impulse input, unit step input, and sinusoidal inputs with different input frequencies.
complicated input= simple inputs =forced response of complicated input
Linear System
Typical Forced Responses• Unit Impulse Response
– Forced response to unit impulse input
• Unit Step Response– Forced response to unit step input (u (t) = 1)
• Sinusoidal Response– Forced response to sinusoidal inputs at different input frequencies– The steady state response of sinusoidal response is call the Frequency
Response.
Time t
u(t)
Time t
u(t)
Time t
u(t)
1
Y s G s U s G s , 0
, 10, 0
tu t t t dt
t
1
1
U s
Y s G s U s G ss
0 0
1lim lim 0s s
y sY s s G s Gs
2 2
U s
Y s G s U s G ss
y 2 2
ssY s G s
s
If system is stable, SS is zero.
• TF and Poles/Zeros
• Unit Step Response – ( u=1 and zero ICs )
• Standard Form of Stable 1st Order System
KG s
s
Time t
y(t)
Forced Response of 1st Order Systems
y ay bu y y Ku
where : Time Constant
K : Static (Steady State, DC) Gain
1pole: 0p
zero: No zero
K
Y s G s U s K Ks s s s s s
1
1y t K e
yss(t) = KK
yT(t) = K e -t/
1
1y t K e
1 by y u
a a
1
a
bK
a
Stable system
Normalized Unit Step ResponseNormalized Unit Step Response (u = 1 & zero ICs)
(such that as , 1)
( ) (1 )
Normalized :
( )( ) 1
n
n
t
t
t y
y y Ku
y t K e
y ty t e
K
Time t
( 1 - e - t/ )
0 1 2 3 4 5 6Time [ t]
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Nor
mal
ized
Res
pon
se
0.6321 0.8647 0.9502 0.9817 0.9933
( ) ( )
( )( )
( )
( )
( )
y y Ku
y t K e
y ty t
Ke
d
dty t
d
dty
t
t
n
n
n
1
1
0
Unit Step Response of Stable 1st Order System
Effect of Time Constant
Normalized:
Initial Slope:
Q: What is your conclusion ?
1 t
e
0 2 4 6 8 10
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Time [sec]
Nor
mal
ized
Res
pons
e
1
Smallest
increases
Largest
The smaller is,
the steeper the initial slope is, and the faster the response approaches the steady state.
Q: How would you calculate the forced response of a 1st order system to a unit pulse (not unit impulse)?
Forced Responses of Stable 1st Order SystemQ: How would you calculate the unit
impulse response of a 1st order system?
Q: How would you calculate the sinusoidal response of a 1st order system?
Time t
u(t)1
1
s
dt u t
dt
Q (Hint: superposition principle ?!)
1s su t u t u t
1s sy t y t y t
-1Time t
u(t)1
1
s
dy t y t
dt
• I/O ModelStandard Form of 2nd Order Systems
0n a
1 02
1 0
b s bG s
s a s a
1 00, 0a a
y a y a y b u b u 1 0 1 0
where n : Natural Frequency [rad/s]
: Damping Ratio K : Static (Steady State, DC) Gain
• TF and Pole/Zeros
• Stability Condition
• Standard Form of Stable 2nd Order Systems without Zeros
21 1 0
1,2
4pole:
2
a a ap
0
1
zero: b
zb
1 0 0
2 21 0 0 2 n n n
a a b
y a y a y b u y y y K u
1 1
02 2n
a a
a
0 02
0n
b bK
a
• Stable 2nd Order Systems without Zeros
• Pole Locations
– Over-damped ( )
– Critically damped ( )
– Under damped ( )
1
1
1
Two distinct real poles
Two identical real poles at 1,2 np
Two complex poles at 2
1,2 1
d
n np j
Poles of Stable 2nd Order Systems2 2
2
222
2
2
( )1 22 1
n n n
n
n n
nn
y y y K u
K KG s
s s s s
Real
Img.
n
n
22
21,2
2 0
( 1)
n n
n n
s s
p
Under-damped 2nd Order System• Unit Step Response ( u=1 and zero ICs )
2 2
1,2221 2
1 11
1( ) ( ) ,
( )( )2
, 12
n nd
n n
d d d
K KY s G s p j
s s s p s ps s s
K A A KA j
s s j s j
1
1 1
2 Re
21
2
( )
1 cos sin
1sin , tan
1
d d
j td
j t j t
A e
t nd d
d
td
y t K A e A e
K e t t
KK e t
Unit Step Response of 2nd Order Systems
OS
yMAX
tP0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15Time [sec]
0
0.2K
0.4K
0.6K
0.8K
K
1.2K
1.4K
1.6K
Uni
t Ste
p R
espo
nse
Td
tS
Unit Step Response of 2nd Order System• Peak Time (tP)
Time when output y(t) reaches its maximum value yMAX.
• Percent Overshoot (%OS)At peak time tP the maximum output
The overshoot (OS) is:
The percent overshoot is:
21( ) 1MAX Py y t K e
21MAX SSOS y y Ke
21
% 100%(0)
100%
SS
OSOS
y y
e
2
2
1
2 2
1
d P
Pd n
ddn
t
t
T
Find such that ( ) 0P Pt y t
( ) 1 cos sint nd d
d
y t K e t t
2
sin1
n nd
dyKe t
dt
• Settling Time (ts)Time required for the response to be within a specific percent of the final (steady-state) value.
Some typical specifications for settling time are: 5%, 2% and 1%.
Look at the envelope of the response:
Q: Which parameters of a 2nd order system affect the peak time?
Unit Step Response of 2nd Order System
% 1% 2% 5%
tS
ln 100sxt
x% band settling time:
4.6 3.9 3
Damping ration and natural frequency
Q: Which parameters of a 2nd order system affect the % OS?
Damping ratio
Q: Which parameters of a 2nd order system affect the settling time?
Damping ratio and natural frequency
Q: Can you obtain the formula for a 3% settling time?
1
• Mass-Spring-Damper System
I/O Model:
In Class ExerciseQ: What is the static gain of the system ?
Q: How would the physical parameters (M, B, K) affect the step response of the system ?
(This is equivalent to asking you for the relationship between the physical parameters and the damping ratio, natural frequency and the static gain.)
xKs
M
B
f(t)
( )sM x Bx K x f t
2
2
2
2 2
1
1
2
s
s
n
n n
G sMs Bs K
MB K
s sM M
K
s s
2
1
sn
s
s
K
MB
MK
KK