ME 3281 Spring 2013, University of Minnesota
Transform Solutions to LTI Systems – Part 4
April 2, 2013
Final Value Theorem
Given F(s), how can we find lim f(t)
t→∞
Final Value Theorem (FVT):
lim f(t) = lim sF(s)
t→∞
s→0
When is the FVT applicable?
1). F(s) should have no poles in the right half of the complex plane( Real
part should not be +v).
2). F(s) should have no poles on the imaginary axis, except at most one
pole at s=0.
Examples:
A
a) F(s) = , find lim f(t)
s
t→∞
lim sF(s) = lim s
s→0
b) F(s) =
A
s2
, FVT cannot be applied.
A
Note: ℒ −1 *𝐹(𝑠)+ = ℒ −1 { 2} = 𝐴𝑡.
s
Final value is not defined.
s→0
A
=A
s
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c) F(s) =
ω
s2 +ω2
s. ω
=0
s→0 s 2 + ω2
limF(s) = lim
s→0
ℒ −1 *𝐹(𝑠)+ = 𝑠𝑖𝑛ωt
No well-defined final value
The poles are s 2 + ω2 = 0, s 2 = −ω2 , s = ±jω
Real part=0
Hence the FVT cannot be applied.
d)
F(s) =
s
s 2 + ω2
The FVT cannot be applied.
Example on use of IVT and FVT
s 2 + 2s + 4
F(s) = 3
s + 3s 2 + 2s
IVT: Can it be applied? YES
2 4
1+ + 2
s 3 + 2s 2 + 4s
s s
f(0+ ) = lim sF(s) = lim 3
= lim
=1
2
3 2
s→∞
s→∞ s + 3s + 2s
s→∞
1+ + 2
s s
FVT: Can it be applied?
Poles are obtained from:
ME 3281 Spring 2013, University of Minnesota
s 3 + 3s 2 + 2s = 0
or s(s 2 + 3s + 2) = 0
or s(s + 2)(s + 1) = 0
Note: A second order polynomial with positive coefficients always has
roots with –ve real parts.
e.g. ms 2 + bs + k = 0 ⇒ the poles are always stable.
A higher order polynomial (3rd order or higher) need not be stable if all
coefficients are positive.
However, even if one coefficient is –ve, the system will be unstable.
(will have at least one pole with +ve real part)
The FVT is applicable in this example.
s 3 + 2s 2 + 4s
s 2 + 2s + 4 4
lim f(t) = lim sF(s) = lim 3
= lim 2
= =2
t→∞
s→0
s→0 s + 3s 2 + 2s
s→∞ s + 3s + 2
2
Example:
5s 2 + 8s − 5
F(s) = 2 2
s (s + 2s + 5)
Find f(t), feature: Repeated pole at s=0.
F(s) =
As + B
Cs + D
+
s2
s 2 + 2s + 5
After calculation, it turns out
A = −1, B = 2, C = −2, D = 2
Hence
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F(s) =
−s + 2
2s − 2
1 2
2(s − 1)
−
=
−
+
−
s2
s 2 + 2s + 5
s s 2 (s + 1)2 + 4
1 2 2(s + 1 − 2)
=− + 2−
s s
(s + 1)2 + 4
1 2
2(s + 1)
4
=− + 2−
+
s s
(s + 1)2 + 22 (s + 1)2 + 22
1 2
2(s + 1)
2
=− + 2−
+
2.
s s
(s + 1)2 + 22
(s + 1)2 + 22
f(t) = −1 + 2t − 2e−t cos2t + 2e−t sin2t
Example: To illustrate how to handle numerator and denominator of the
same order.
2s 2 + 7s + 8 2(s 2 + 3s + 2) + s + 4
F(s) = 2
=
s + 3s + 2
s 2 + 3s + 2
2(s 2 + 3s + 2)
s+4
s+4
= 2
+ 2
=2+
(s + 2)(s + 1)
s + 3s + 2
s + 3s + 2
=2+
A
B
2
3
+
=2−
+
s+2 s+1
s+2 s+1
Hence: f(t) = 2δ(t) − 2e−2t + 3e−t
ME 3281 Spring 2013, University of Minnesota
Standard Representation of a Second Order System
𝑚𝑥̈ + 𝑏𝑥̇ + 𝑘𝑥 = 𝐹(𝑡)
In the Laplace domain,
𝑚(𝑠 2 𝑋(𝑠) − 𝑠𝑥(0) − 𝑥̇ (0)) + 𝑏*𝑠𝑋(𝑠) − 𝑥(0)+ + 𝑘𝑋(𝑠) = 𝐹(𝑠)
(ms 2 + bs + k)X(s) = F(s) + (ms + b)x(0) = m𝑥̇ (0)
X(s) =
(ms + b)x0 + mv0
1
F(s)
+
ms 2 + bs + k
ms 2 + bs + k
1
→ Zero − state reponse
ms 2 + bs + k
(ms + b)x0 + mv0
→ Zero − input response
ms 2 + bs + k
For a stable system, the zero input response → 0 as t → ∞.
The zero-state response need not converge to zero as t → ∞.
It will have some terms that converge to zero and some terms that do not
converge to zero.
For example: if F(s) =
F0
s
lim f(t) =
t→∞
F0
( does not converge to zero)
k
The terms that do not converge to zero constitute the steady state
response of the system, and all the terms that converge to zero constitute
ME 3281 Spring 2013, University of Minnesota
the transient response.
If the force is a sinusoid, the steady state response will be a sinusoid.
If the force is some others bounded periodic function, the steady state
response will be a bounded periodic function.
The transfer function
1
ms2 +bs+k
can be written in the following standard
2nd order form:
ωn 2
s 2 + 2ζωn s + ωn 2
April 4, 2013
Continuing with the previous m, k, b system.
If the initial conditions are zero,
X(s) =
ms 2
1
F(s)
+ bs + k
To write in the standard second order form,
1). Divide numerator and denominator by m,
1
m
X(s) =
F(s)
b
k
2
s + s+
m
m
Coefficient of s 2 in the denominator is now 1.
2). Need to make the constant terms in the numerator and denominator
equal
ME 3281 Spring 2013, University of Minnesota
k
1
m
X(s) =
F(s)
k s2 + b s + k
m
m
Compare with
ωn 2
s 2 + 2ζωn s + ωn 2
We get
k
b
= ωn 2 , 2ζωn =
m
m
ωn : Undamped natural frequency of the system
ζ=
1 b
=
2ωn m
1
b
b
=
k m 2√km
2√
m
ζ is called the damping ratio.
If ζ > 1, the system id said to be over-damped. It has no overshoot and
no oscillations.
If ζ < 1, the system id said to be under-damped. It has oscillatory
behavior and it has overshoot.
To see why, note that the characteristic equation is:
s 2 + 2ζωn s + ωn 2 = 0
2
s 2 + 2ζωn s + ωn 2 ζ2 − ζ2 ωn + ωn 2 = 0
2
(s + ζωn )2 − ζ2 ωn + ωn 2 = 0
(s + ζωn )2 = ωn 2 (ζ2 − 1) … … … … … … ①
If ζ < 1, then:
(s + ζωn ) = ±jωn √1 − ζ2 , j = √−1
ME 3281 Spring 2013, University of Minnesota
Hence the poles are:
s = −ζωn ± jωn √1 − ζ2
Since
1
ωn 2
X(s) = . 2
F(s)
k s + 2ζωn s + ωn 2
1
If F(s) = (the response to a step input), then
s
1
ωn 2
1
X(s) = . 2
.
k s + 2ζωn s + ωn 2 s
After partial fraction expansion, the response will be of the type:
x(t) = xss + Ae−ζωn t sinωd t + Be−ζωn t cosωd t
(Likely xss =
F
k
from previous experience)
where ωd = ωn √1 − ζ2
(Imaginary part of the poles)
ωd is called the damped natural frequency. Thus the response of the
system is oscillatory.
On the other hand, consider ζ > 1,
The charact eqn gives:
(s + ζωn )2 = ωn 2 (ζ2 − 1)
Hence the poles are:
s = −ζωn ± √ωn 2 (ζ2 − 1)
s = −ζωn ± ωn √ζ2 − 1
Hence the response to
1
ωn 2
1
X(s) = . 2
.
k s + 2ζωn s + ωn 2 s
ME 3281 Spring 2013, University of Minnesota
is given by:
x(t) = xss + Ae−(ζωn±ωn √ζ
2 −1)t
+ Be−(ζωn ±ωn √ζ
2 −1)t
Thus the response is not oscillatory. No overshoot.
What about ζ = 1?
The value of ζ = 1 is called critical damping. It is the transition point
between no oscillations and oscillations.
The poles are given in this case by:
s = −ζωn ± ωn √ζ2 − 1
s = −ζωn = −ωn
Hence
1
ωn 2
1
X(s) = . 2
.
k s + 2ζωn s + ωn 2 s
1
ωn 2
1
1
ωn 2
1
= . 2
. (forζ = 1) = .
.
2
k s + 2ωn s + ωn s
k (s + ωn )2 s
After partial fraction expansion and inverse Laplace transforms, the
response is found to be :
x(t) = xss + Ae−ωn t + Bte−ωn t
(Why? ℒ −1 {
1
(𝑠+𝑎)2
} = 𝑡𝑒 −𝑎𝑡 )
Again the response is exponential ⇒ no oscillations.
ME 3281 Spring 2013, University of Minnesota
Frequency Response of Linear Time Invariant Systems
Complex Numbers: Recall that every complex number has a magnitude
and a phase.
Example: z = a + bj,
j = √−1
a is called the real part of z, a = Re(z)
b is called the imaginary part of z, b = Im(z)
Magnitude of z: |z| = √(a2 + b 2 ) = √Re(z)2 + Im(z)2
b
Im(z)
Phase of z: ∠z = tan −1 ( ) = tan −1 (
)
a
Re(z)
Both the magnitude and phase of a complex number are real.
What is the steady state response of any LTI system for a sinusoidal input
of frequency ω?
Assume that the system is stable : All its poles have negative real parts.
For example:
ME 3281 Spring 2013, University of Minnesota
1
F(s)
ms 2 + bs + k
ω
(sinusoid)
If F(s) = 2
s + ω2
1
ω
Then X(s) =
.
ms 2 + bs + k s 2 + ω2
X(s) =
After partial fraction expansion, and inverse Laplace transforms, we will
find:
x(t) = Ae−ζωnt sinωd t + Be−ζωn t cosωd t + Csinωt