Coefficient of Thermal Expansion and their Importance.pptx
lec04.pdf
1. Plan of the Lecture
I Review: dynamic response; transfer functions; transient
and steady-state response
I Today’s topic: dynamic response (transient and
steady-state) with arbitrary I.C.’s
2. Plan of the Lecture
I Review: dynamic response; transfer functions; transient
and steady-state response
I Today’s topic: dynamic response (transient and
steady-state) with arbitrary I.C.’s
Goal: develop a methodology for characterizing the output of a
given system for a given input.
3. Plan of the Lecture
I Review: dynamic response; transfer functions; transient
and steady-state response
I Today’s topic: dynamic response (transient and
steady-state) with arbitrary I.C.’s
Goal: develop a methodology for characterizing the output of a
given system for a given input.
Reading: FPE, Section 3.1, Appendix A
5. Dynamic Response
u y
h
Problem: compute the response y to a given input u
under a given set of initial conditions.
In particular, we wish to know both the transient response (due
to I.C.’s) and the steady-state response (once the effect of the
I.C.’s “washes away”).
6. Laplace Transforms Revisited
(see FPE, Appendix A)
One-sided (or unilateral) Laplace transform:
L {f(t)} ≡ F(s) =
Z ∞
0
f(t)e−st
dt (really, from 0−
)
7. Laplace Transforms Revisited
(see FPE, Appendix A)
One-sided (or unilateral) Laplace transform:
L {f(t)} ≡ F(s) =
Z ∞
0
f(t)e−st
dt (really, from 0−
)
— for simple functions f, can compute L f by hand.
8. Laplace Transforms Revisited
(see FPE, Appendix A)
One-sided (or unilateral) Laplace transform:
L {f(t)} ≡ F(s) =
Z ∞
0
f(t)e−st
dt (really, from 0−
)
— for simple functions f, can compute L f by hand.
Example: unit step
f(t) = 1(t) =
(
1, t ≥ 0
0, t < 0
9. Laplace Transforms Revisited
(see FPE, Appendix A)
One-sided (or unilateral) Laplace transform:
L {f(t)} ≡ F(s) =
Z ∞
0
f(t)e−st
dt (really, from 0−
)
— for simple functions f, can compute L f by hand.
Example: unit step
f(t) = 1(t) =
(
1, t ≥ 0
0, t < 0
L {1(t)} =
Z ∞
0
e−st
dt = −
1
s
e−st
∞
0
=
1
s
10. Laplace Transforms Revisited
(see FPE, Appendix A)
One-sided (or unilateral) Laplace transform:
L {f(t)} ≡ F(s) =
Z ∞
0
f(t)e−st
dt (really, from 0−
)
— for simple functions f, can compute L f by hand.
Example: unit step
f(t) = 1(t) =
(
1, t ≥ 0
0, t < 0
L {1(t)} =
Z ∞
0
e−st
dt = −
1
s
e−st
∞
0
=
1
s
(pole at s = 0)
11. Laplace Transforms Revisited
(see FPE, Appendix A)
One-sided (or unilateral) Laplace transform:
L {f(t)} ≡ F(s) =
Z ∞
0
f(t)e−st
dt (really, from 0−
)
— for simple functions f, can compute L f by hand.
Example: unit step
f(t) = 1(t) =
(
1, t ≥ 0
0, t < 0
L {1(t)} =
Z ∞
0
e−st
dt = −
1
s
e−st
∞
0
=
1
s
(pole at s = 0)
— this is valid provided Re(s) > 0, so that e−st t→+∞
−
−
−
−
→ 0.
15. Laplace Transforms Revisited
Example: f(t) = cos t
L {cos t} = L
1
2
ejt
+
1
2
e−jt
(Euler’s formula)
=
1
2
L {ejt
} +
1
2
L {e−jt
} (linearity)
L {ejt
} =
Z ∞
0
ejt
e−st
dt
16. Laplace Transforms Revisited
Example: f(t) = cos t
L {cos t} = L
1
2
ejt
+
1
2
e−jt
(Euler’s formula)
=
1
2
L {ejt
} +
1
2
L {e−jt
} (linearity)
L {ejt
} =
Z ∞
0
ejt
e−st
dt =
Z ∞
0
e(j−s)t
dt
17. Laplace Transforms Revisited
Example: f(t) = cos t
L {cos t} = L
1
2
ejt
+
1
2
e−jt
(Euler’s formula)
=
1
2
L {ejt
} +
1
2
L {e−jt
} (linearity)
L {ejt
} =
Z ∞
0
ejt
e−st
dt =
Z ∞
0
e(j−s)t
dt =
1
j − s
e(j−s)t
∞
0
18. Laplace Transforms Revisited
Example: f(t) = cos t
L {cos t} = L
1
2
ejt
+
1
2
e−jt
(Euler’s formula)
=
1
2
L {ejt
} +
1
2
L {e−jt
} (linearity)
L {ejt
} =
Z ∞
0
ejt
e−st
dt =
Z ∞
0
e(j−s)t
dt =
1
j − s
e(j−s)t
∞
0
= −
1
j − s
19. Laplace Transforms Revisited
Example: f(t) = cos t
L {cos t} = L
1
2
ejt
+
1
2
e−jt
(Euler’s formula)
=
1
2
L {ejt
} +
1
2
L {e−jt
} (linearity)
L {ejt
} =
Z ∞
0
ejt
e−st
dt =
Z ∞
0
e(j−s)t
dt =
1
j − s
e(j−s)t
∞
0
= −
1
j − s
(pole at s = j)
20. Laplace Transforms Revisited
Example: f(t) = cos t
L {cos t} = L
1
2
ejt
+
1
2
e−jt
(Euler’s formula)
=
1
2
L {ejt
} +
1
2
L {e−jt
} (linearity)
L {ejt
} =
Z ∞
0
ejt
e−st
dt =
Z ∞
0
e(j−s)t
dt =
1
j − s
e(j−s)t
∞
0
= −
1
j − s
(pole at s = j)
L {e−jt
} =
Z ∞
0
e−jt
e−st
dt
21. Laplace Transforms Revisited
Example: f(t) = cos t
L {cos t} = L
1
2
ejt
+
1
2
e−jt
(Euler’s formula)
=
1
2
L {ejt
} +
1
2
L {e−jt
} (linearity)
L {ejt
} =
Z ∞
0
ejt
e−st
dt =
Z ∞
0
e(j−s)t
dt =
1
j − s
e(j−s)t
∞
0
= −
1
j − s
(pole at s = j)
L {e−jt
} =
Z ∞
0
e−jt
e−st
dt =
Z ∞
0
e−(j+s)t
dt
22. Laplace Transforms Revisited
Example: f(t) = cos t
L {cos t} = L
1
2
ejt
+
1
2
e−jt
(Euler’s formula)
=
1
2
L {ejt
} +
1
2
L {e−jt
} (linearity)
L {ejt
} =
Z ∞
0
ejt
e−st
dt =
Z ∞
0
e(j−s)t
dt =
1
j − s
e(j−s)t
∞
0
= −
1
j − s
(pole at s = j)
L {e−jt
} =
Z ∞
0
e−jt
e−st
dt =
Z ∞
0
e−(j+s)t
dt = −
1
j + s
e−(j+s)t
∞
0
23. Laplace Transforms Revisited
Example: f(t) = cos t
L {cos t} = L
1
2
ejt
+
1
2
e−jt
(Euler’s formula)
=
1
2
L {ejt
} +
1
2
L {e−jt
} (linearity)
L {ejt
} =
Z ∞
0
ejt
e−st
dt =
Z ∞
0
e(j−s)t
dt =
1
j − s
e(j−s)t
∞
0
= −
1
j − s
(pole at s = j)
L {e−jt
} =
Z ∞
0
e−jt
e−st
dt =
Z ∞
0
e−(j+s)t
dt = −
1
j + s
e−(j+s)t
∞
0
=
1
j + s
24. Laplace Transforms Revisited
Example: f(t) = cos t
L {cos t} = L
1
2
ejt
+
1
2
e−jt
(Euler’s formula)
=
1
2
L {ejt
} +
1
2
L {e−jt
} (linearity)
L {ejt
} =
Z ∞
0
ejt
e−st
dt =
Z ∞
0
e(j−s)t
dt =
1
j − s
e(j−s)t
∞
0
= −
1
j − s
(pole at s = j)
L {e−jt
} =
Z ∞
0
e−jt
e−st
dt =
Z ∞
0
e−(j+s)t
dt = −
1
j + s
e−(j+s)t
∞
0
=
1
j + s
(pole at s = −j)
25. Laplace Transforms Revisited
Example: f(t) = cos t
L {cos t} = L
1
2
ejt
+
1
2
e−jt
(Euler’s formula)
=
1
2
L {ejt
} +
1
2
L {e−jt
} (linearity)
L {ejt
} =
Z ∞
0
ejt
e−st
dt =
Z ∞
0
e(j−s)t
dt =
1
j − s
e(j−s)t
∞
0
= −
1
j − s
(pole at s = j)
L {e−jt
} =
Z ∞
0
e−jt
e−st
dt =
Z ∞
0
e−(j+s)t
dt = −
1
j + s
e−(j+s)t
∞
0
=
1
j + s
(pole at s = −j)
— in both cases, require Re(s) 0, i.e., s must lie in the right
half-plane (RHP)
39. Laplace Transforms Revisited
Convolution: L {f ? g} = L {f}L {g}
(useful because Y (s) = H(s)U(s))
Example: ẏ = −y + u y(0) = 0
Compute the response for u(t) = cos t
40. Laplace Transforms Revisited
Convolution: L {f ? g} = L {f}L {g}
(useful because Y (s) = H(s)U(s))
Example: ẏ = −y + u y(0) = 0
Compute the response for u(t) = cos t
We already know
H(s) =
1
s + 1
(from earlier example)
U(s) =
s
s2 + 1
(just proved)
41. Laplace Transforms Revisited
Convolution: L {f ? g} = L {f}L {g}
(useful because Y (s) = H(s)U(s))
Example: ẏ = −y + u y(0) = 0
Compute the response for u(t) = cos t
We already know
H(s) =
1
s + 1
(from earlier example)
U(s) =
s
s2 + 1
(just proved)
=⇒ Y (s) = H(s)U(s) =
s
(s + 1)(s2 + 1)
y(t) = L −1
{Y }
42. Laplace Transforms Revisited
Convolution: L {f ? g} = L {f}L {g}
(useful because Y (s) = H(s)U(s))
Example: ẏ = −y + u y(0) = 0
Compute the response for u(t) = cos t
We already know
H(s) =
1
s + 1
(from earlier example)
U(s) =
s
s2 + 1
(just proved)
=⇒ Y (s) = H(s)U(s) =
s
(s + 1)(s2 + 1)
y(t) = L −1
{Y }
— can’t find Y (s) in the tables. So how do we compute y?
44. Method of Partial Fractions
Problem: compute L −1
s
(s + 1)(s2 + 1)
This Laplace transform is not in the tables, but let’s look at the
table anyway. What do we find?
45. Method of Partial Fractions
Problem: compute L −1
s
(s + 1)(s2 + 1)
This Laplace transform is not in the tables, but let’s look at the
table anyway. What do we find?
1
s + 1
L −1
1
s + 1
= e−t
(#7)
1
s2 + 1
L −1
1
s2 + 1
= sin t (#17)
s
s2 + 1
L −1
s
s2 + 1
= cos t (#18)
— so we see some things that are similar to Y (s), but not quite.
46. Method of Partial Fractions
Problem: compute L −1
s
(s + 1)(s2 + 1)
This Laplace transform is not in the tables, but let’s look at the
table anyway. What do we find?
1
s + 1
L −1
1
s + 1
= e−t
(#7)
1
s2 + 1
L −1
1
s2 + 1
= sin t (#17)
s
s2 + 1
L −1
s
s2 + 1
= cos t (#18)
— so we see some things that are similar to Y (s), but not quite.
This brings us to the method of partial fractions:
I boring (i.e., character-building), but very useful
I allows us to break up complicated fractions into sums of
simpler ones, for which we know L −1 from tables
47. Method of Partial Fractions
Problem: compute L −1{Y (s)}, where
Y (s) =
s
(s + 1)(s2 + 1)
48. Method of Partial Fractions
Problem: compute L −1{Y (s)}, where
Y (s) =
s
(s + 1)(s2 + 1)
We seek a, b, c, such that
Y (s) =
a
s + 1
+
bs + c
s2 + 1
(need bs + c so that deg(num) = deg(den) − 1)
49. Method of Partial Fractions
Problem: compute L −1{Y (s)}, where
Y (s) =
s
(s + 1)(s2 + 1)
We seek a, b, c, such that
Y (s) =
a
s + 1
+
bs + c
s2 + 1
(need bs + c so that deg(num) = deg(den) − 1)
I Find a: multiply by s + 1 to isolate a
50. Method of Partial Fractions
Problem: compute L −1{Y (s)}, where
Y (s) =
s
(s + 1)(s2 + 1)
We seek a, b, c, such that
Y (s) =
a
s + 1
+
bs + c
s2 + 1
(need bs + c so that deg(num) = deg(den) − 1)
I Find a: multiply by s + 1 to isolate a
(s + 1)Y (s) =
s
s2 + 1
= a +
(s + 1)(as + b)
(s2 + 1)
51. Method of Partial Fractions
Problem: compute L −1{Y (s)}, where
Y (s) =
s
(s + 1)(s2 + 1)
We seek a, b, c, such that
Y (s) =
a
s + 1
+
bs + c
s2 + 1
(need bs + c so that deg(num) = deg(den) − 1)
I Find a: multiply by s + 1 to isolate a
(s + 1)Y (s) =
s
s2 + 1
= a +
(s + 1)(as + b)
(s2 + 1)
— now let s = −1 to “kill” the second term on the RHS:
a = (s + 1)Y (s)
s=−1
= −
1
2
52. Method of Partial Fractions
Problem: compute L −1{Y (s)}, where
Y (s) =
s
(s + 1)(s2 + 1)
53. Method of Partial Fractions
Problem: compute L −1{Y (s)}, where
Y (s) =
s
(s + 1)(s2 + 1)
We seek a, b, c, such that
Y (s) =
a
s + 1
+
bs + c
s2 + 1
(need bs + c so that deg(num) = deg(den) − 1)
54. Method of Partial Fractions
Problem: compute L −1{Y (s)}, where
Y (s) =
s
(s + 1)(s2 + 1)
We seek a, b, c, such that
Y (s) =
a
s + 1
+
bs + c
s2 + 1
(need bs + c so that deg(num) = deg(den) − 1)
I Find b: multiply by s2 + 1 to isolate bs + c
55. Method of Partial Fractions
Problem: compute L −1{Y (s)}, where
Y (s) =
s
(s + 1)(s2 + 1)
We seek a, b, c, such that
Y (s) =
a
s + 1
+
bs + c
s2 + 1
(need bs + c so that deg(num) = deg(den) − 1)
I Find b: multiply by s2 + 1 to isolate bs + c
(s2
+ 1)Y (s) =
s
s + 1
=
a(s2 + 1)
s + 1
+ bs + c
56. Method of Partial Fractions
Problem: compute L −1{Y (s)}, where
Y (s) =
s
(s + 1)(s2 + 1)
We seek a, b, c, such that
Y (s) =
a
s + 1
+
bs + c
s2 + 1
(need bs + c so that deg(num) = deg(den) − 1)
I Find b: multiply by s2 + 1 to isolate bs + c
(s2
+ 1)Y (s) =
s
s + 1
=
a(s2 + 1)
s + 1
+ bs + c
— now let s = j to “kill” the first term on the RHS:
bj + c = (s2
+ 1)Y (s)
s=j
=
j
1 + j
57. Method of Partial Fractions
Problem: compute L −1{Y (s)}, where
Y (s) =
s
(s + 1)(s2 + 1)
We seek a, b, c, such that
Y (s) =
a
s + 1
+
bs + c
s2 + 1
(need bs + c so that deg(num) = deg(den) − 1)
I Find b: multiply by s2 + 1 to isolate bs + c
(s2
+ 1)Y (s) =
s
s + 1
=
a(s2 + 1)
s + 1
+ bs + c
— now let s = j to “kill” the first term on the RHS:
bj + c = (s2
+ 1)Y (s)
s=j
=
j
1 + j
Match Re(·) and Im(·) parts:
c + bj =
j
1 + j
=
j(1 − j)
(1 + j)(1 − j)
=
1
2
+
j
2
58. Method of Partial Fractions
Problem: compute L −1{Y (s)}, where
Y (s) =
s
(s + 1)(s2 + 1)
We seek a, b, c, such that
Y (s) =
a
s + 1
+
bs + c
s2 + 1
(need bs + c so that deg(num) = deg(den) − 1)
I Find b: multiply by s2 + 1 to isolate bs + c
(s2
+ 1)Y (s) =
s
s + 1
=
a(s2 + 1)
s + 1
+ bs + c
— now let s = j to “kill” the first term on the RHS:
bj + c = (s2
+ 1)Y (s)
s=j
=
j
1 + j
Match Re(·) and Im(·) parts:
c + bj =
j
1 + j
=
j(1 − j)
(1 + j)(1 − j)
=
1
2
+
j
2
=⇒ b = c = 1
2
59. Method of Partial Fractions
Problem: compute L −1{Y (s)}, where
Y (s) =
s
(s + 1)(s2 + 1)
60. Method of Partial Fractions
Problem: compute L −1{Y (s)}, where
Y (s) =
s
(s + 1)(s2 + 1)
We found that
Y (s) = −
1
2(s + 1)
+
s
2(s2 + 1)
+
1
2(s2 + 1)
61. Method of Partial Fractions
Problem: compute L −1{Y (s)}, where
Y (s) =
s
(s + 1)(s2 + 1)
We found that
Y (s) = −
1
2(s + 1)
+
s
2(s2 + 1)
+
1
2(s2 + 1)
Now we can use linearity and tables:
y(t) = L −1
−
1
2(s + 1)
+
s
2(s2 + 1)
+
1
2(s2 + 1)
62. Method of Partial Fractions
Problem: compute L −1{Y (s)}, where
Y (s) =
s
(s + 1)(s2 + 1)
We found that
Y (s) = −
1
2(s + 1)
+
s
2(s2 + 1)
+
1
2(s2 + 1)
Now we can use linearity and tables:
y(t) = L −1
−
1
2(s + 1)
+
s
2(s2 + 1)
+
1
2(s2 + 1)
= −
1
2
L −1
1
s + 1
+
1
2
L −1
s
s2 + 1
+
1
2
L −1
1
s2 + 1
63. Method of Partial Fractions
Problem: compute L −1{Y (s)}, where
Y (s) =
s
(s + 1)(s2 + 1)
We found that
Y (s) = −
1
2(s + 1)
+
s
2(s2 + 1)
+
1
2(s2 + 1)
Now we can use linearity and tables:
y(t) = L −1
−
1
2(s + 1)
+
s
2(s2 + 1)
+
1
2(s2 + 1)
= −
1
2
L −1
1
s + 1
+
1
2
L −1
s
s2 + 1
+
1
2
L −1
1
s2 + 1
= −
1
2
e−t
+
1
2
cos t +
1
2
sin t (from tables)
64. Method of Partial Fractions
Problem: compute L −1{Y (s)}, where
Y (s) =
s
(s + 1)(s2 + 1)
We found that
Y (s) = −
1
2(s + 1)
+
s
2(s2 + 1)
+
1
2(s2 + 1)
Now we can use linearity and tables:
y(t) = L −1
−
1
2(s + 1)
+
s
2(s2 + 1)
+
1
2(s2 + 1)
= −
1
2
L −1
1
s + 1
+
1
2
L −1
s
s2 + 1
+
1
2
L −1
1
s2 + 1
= −
1
2
e−t
+
1
2
cos t +
1
2
sin t (from tables)
= −
1
2
e−t
+
1
√
2
cos(t − π/4) (cos(a − b) = cos a cos b + sin a sin b)
66. Transient and Steady-State Response
Consider the system ẏ = −y + u y(0) = 0
u(t) = cos t −→ y(t) = −
1
2
e−t
| {z }
transient
response
+
1
√
2
cos(t − π/4)
| {z }
steady-state
response
67. Transient and Steady-State Response
Consider the system ẏ = −y + u y(0) = 0
u(t) = cos t −→ y(t) = −
1
2
e−t
| {z }
transient
response
+
1
√
2
cos(t − π/4)
| {z }
steady-state
response
— transient response vanishes as t → ∞ (we will see later why)
Let’s compare against the frequency response formula:
68. Transient and Steady-State Response
Consider the system ẏ = −y + u y(0) = 0
u(t) = cos t −→ y(t) = −
1
2
e−t
| {z }
transient
response
+
1
√
2
cos(t − π/4)
| {z }
steady-state
response
— transient response vanishes as t → ∞ (we will see later why)
Let’s compare against the frequency response formula:
H(s) =
1
s + 1
=⇒ H(jω) =
1
jω + 1
69. Transient and Steady-State Response
Consider the system ẏ = −y + u y(0) = 0
u(t) = cos t −→ y(t) = −
1
2
e−t
| {z }
transient
response
+
1
√
2
cos(t − π/4)
| {z }
steady-state
response
— transient response vanishes as t → ∞ (we will see later why)
Let’s compare against the frequency response formula:
H(s) =
1
s + 1
=⇒ H(jω) =
1
jω + 1
u(t) = cos t has A = 1 and ω = 1, so
y(t) = M(1) cos t + ϕ(1)
=
1
√
2
cos t − π/4
70. Transient and Steady-State Response
Consider the system ẏ = −y + u y(0) = 0
u(t) = cos t −→ y(t) = −
1
2
e−t
| {z }
transient
response
+
1
√
2
cos(t − π/4)
| {z }
steady-state
response
— transient response vanishes as t → ∞ (we will see later why)
Let’s compare against the frequency response formula:
H(s) =
1
s + 1
=⇒ H(jω) =
1
jω + 1
u(t) = cos t has A = 1 and ω = 1, so
y(t) = M(1) cos t + ϕ(1)
=
1
√
2
cos t − π/4
— the freq. response formula gives only the steady-state part!!
72. Transient and Steady-State Response
Consider the system ẏ = −y + u y(0) = 0
We computed the response to u(t) = cos t in two ways:
y(t) = −
1
2
e−t
+
1
√
2
cos(t − π/4)
— using the method of partial fractions;
y(t) =
1
√
2
cos t − π/4
— using the frequency response formula.
73. Transient and Steady-State Response
Consider the system ẏ = −y + u y(0) = 0
We computed the response to u(t) = cos t in two ways:
y(t) = −
1
2
e−t
+
1
√
2
cos(t − π/4)
— using the method of partial fractions;
y(t) =
1
√
2
cos t − π/4
— using the frequency response formula.
Q: Which answer is correct? And why?
74. Transient and Steady-State Response
Consider the system ẏ = −y + u y(0) = 0
We computed the response to u(t) = cos t in two ways:
y(t) = −
1
2
e−t
+
1
√
2
cos(t − π/4)
— using the method of partial fractions;
y(t) =
1
√
2
cos t − π/4
— using the frequency response formula.
Q: Which answer is correct? And why?
A: At t = 0,
1
√
2
cos(t − π/4) = 1
2 6= 0, which is inconsistent
with the initial condition y(0) = 0. The term −1
2e−t
t=0
= −1
2
cancels the steady-state term, so indeed y(0) = 0.
75. Transient and Steady-State Response
Consider the system ẏ = −y + u y(0) = 0
We computed the response to u(t) = cos t in two ways:
y(t) = −
1
2
e−t
+
1
√
2
cos(t − π/4)
— using the method of partial fractions;
y(t) =
1
√
2
cos t − π/4
— using the frequency response formula.
Q: Which answer is correct? And why?
A: At t = 0,
1
√
2
cos(t − π/4) = 1
2 6= 0, which is inconsistent
with the initial condition y(0) = 0. The term −1
2e−t
t=0
= −1
2
cancels the steady-state term, so indeed y(0) = 0.
Therefore, the first formula is correct.
76. Transient and Steady-State Response
Main message: the frequency response formula only gives the
steady-state part of the response, but the inverse Laplace
transform gives the whole response (including the transient
part).
77. Transient and Steady-State Response
Main message: the frequency response formula only gives the
steady-state part of the response, but the inverse Laplace
transform gives the whole response (including the transient
part).
— we will now see how to deal with nonzero I.C.’s ...
78. Laplace Transforms and Differentiation
Given a differentiable function f, what is the Laplace transform
L {f0(t)} of its time derivative?
79. Laplace Transforms and Differentiation
Given a differentiable function f, what is the Laplace transform
L {f0(t)} of its time derivative?
L {f0
(t)} =
Z ∞
0
f0
(t)e−st
dt
80. Laplace Transforms and Differentiation
Given a differentiable function f, what is the Laplace transform
L {f0(t)} of its time derivative?
L {f0
(t)} =
Z ∞
0
f0
(t)e−st
dt
= f(t)e−st
∞
0
+ s
Z ∞
0
e−st
f(t)dt
81. Laplace Transforms and Differentiation
Given a differentiable function f, what is the Laplace transform
L {f0(t)} of its time derivative?
L {f0
(t)} =
Z ∞
0
f0
(t)e−st
dt
= f(t)e−st
∞
0
+ s
Z ∞
0
e−st
f(t)dt (integrate by parts)
82. Laplace Transforms and Differentiation
Given a differentiable function f, what is the Laplace transform
L {f0(t)} of its time derivative?
L {f0
(t)} =
Z ∞
0
f0
(t)e−st
dt
= f(t)e−st
∞
0
+ s
Z ∞
0
e−st
f(t)dt (integrate by parts)
= −f(0) + sF(s)
83. Laplace Transforms and Differentiation
Given a differentiable function f, what is the Laplace transform
L {f0(t)} of its time derivative?
L {f0
(t)} =
Z ∞
0
f0
(t)e−st
dt
= f(t)e−st
∞
0
+ s
Z ∞
0
e−st
f(t)dt (integrate by parts)
= −f(0) + sF(s)
— provided f(t)e−st → 0 as t → ∞
84. Laplace Transforms and Differentiation
Given a differentiable function f, what is the Laplace transform
L {f0(t)} of its time derivative?
L {f0
(t)} =
Z ∞
0
f0
(t)e−st
dt
= f(t)e−st
∞
0
+ s
Z ∞
0
e−st
f(t)dt (integrate by parts)
= −f(0) + sF(s)
— provided f(t)e−st → 0 as t → ∞
L {f0
(t)} = sF(s) − f(0) — this is how we account for I.C.’s
85. Laplace Transforms and Differentiation
Given a differentiable function f, what is the Laplace transform
L {f0(t)} of its time derivative?
L {f0
(t)} =
Z ∞
0
f0
(t)e−st
dt
= f(t)e−st
∞
0
+ s
Z ∞
0
e−st
f(t)dt (integrate by parts)
= −f(0) + sF(s)
— provided f(t)e−st → 0 as t → ∞
L {f0
(t)} = sF(s) − f(0) — this is how we account for I.C.’s
Similarly:
L {f00
(t)} = L {(f0
(t))0
} = sL {f0
(t)} − f0
(0)
= s2
F(s) − sf(0) − f0
(0)
88. Example
Consider the system
ÿ + 3ẏ + 2y = u, y(0) = ẏ(0) = 0
(need two I.C.’s for 2nd-order ODE’s)
Let’s compute the transfer function: H(s) =
Y (s)
U(s)
89. Example
Consider the system
ÿ + 3ẏ + 2y = u, y(0) = ẏ(0) = 0
(need two I.C.’s for 2nd-order ODE’s)
Let’s compute the transfer function: H(s) =
Y (s)
U(s)
— take Laplace transform of both sides (zero I.C.’s):
90. Example
Consider the system
ÿ + 3ẏ + 2y = u, y(0) = ẏ(0) = 0
(need two I.C.’s for 2nd-order ODE’s)
Let’s compute the transfer function: H(s) =
Y (s)
U(s)
— take Laplace transform of both sides (zero I.C.’s):
s2
Y (s) + 3sY (s) + 2Y (s) = U(s)
91. Example
Consider the system
ÿ + 3ẏ + 2y = u, y(0) = ẏ(0) = 0
(need two I.C.’s for 2nd-order ODE’s)
Let’s compute the transfer function: H(s) =
Y (s)
U(s)
— take Laplace transform of both sides (zero I.C.’s):
s2
Y (s) + 3sY (s) + 2Y (s) = U(s) H(s) =
Y (s)
U(s)
=
1
s2 + 3s + 2
93. Example (continued)
ÿ + 3ẏ + 2y = u, y(0) = α, ẏ(0) = β
Compute the step response, i.e., response to u(t) = 1(t)
94. Example (continued)
ÿ + 3ẏ + 2y = u, y(0) = α, ẏ(0) = β
Compute the step response, i.e., response to u(t) = 1(t)
Caution!! Y (s) = H(s)U(s) no longer holds if α 6= 0 or β 6= 0
95. Example (continued)
ÿ + 3ẏ + 2y = u, y(0) = α, ẏ(0) = β
Compute the step response, i.e., response to u(t) = 1(t)
Caution!! Y (s) = H(s)U(s) no longer holds if α 6= 0 or β 6= 0
Again, take Laplace transforms of both sides, mind the I.C.’s:
96. Example (continued)
ÿ + 3ẏ + 2y = u, y(0) = α, ẏ(0) = β
Compute the step response, i.e., response to u(t) = 1(t)
Caution!! Y (s) = H(s)U(s) no longer holds if α 6= 0 or β 6= 0
Again, take Laplace transforms of both sides, mind the I.C.’s:
s2
Y (s) − sy(0) − ẏ(0) + 3sY (s) − 3y(0) + 2Y (s) = U(s)
97. Example (continued)
ÿ + 3ẏ + 2y = u, y(0) = α, ẏ(0) = β
Compute the step response, i.e., response to u(t) = 1(t)
Caution!! Y (s) = H(s)U(s) no longer holds if α 6= 0 or β 6= 0
Again, take Laplace transforms of both sides, mind the I.C.’s:
s2
Y (s) − sα − β + 3sY (s) − 3α + 2Y (s) = U(s)
98. Example (continued)
ÿ + 3ẏ + 2y = u, y(0) = α, ẏ(0) = β
Compute the step response, i.e., response to u(t) = 1(t)
Caution!! Y (s) = H(s)U(s) no longer holds if α 6= 0 or β 6= 0
Again, take Laplace transforms of both sides, mind the I.C.’s:
s2
Y (s) − sα − β + 3sY (s) − 3α + 2Y (s) = U(s)
U(s) = L {1(t)} = 1/s, which gives
s2
Y (s) − sα − β + 3sY (s) − 3α + 2Y (s) =
1
s
99. Example (continued)
ÿ + 3ẏ + 2y = u, y(0) = α, ẏ(0) = β
Compute the step response, i.e., response to u(t) = 1(t)
Caution!! Y (s) = H(s)U(s) no longer holds if α 6= 0 or β 6= 0
Again, take Laplace transforms of both sides, mind the I.C.’s:
s2
Y (s) − sα − β + 3sY (s) − 3α + 2Y (s) = U(s)
U(s) = L {1(t)} = 1/s, which gives
s2
Y (s) − sα − β + 3sY (s) − 3α + 2Y (s) =
1
s
Y (s) =
αs + (3α + β) + 1
s
s2 + 3s + 2
100. Example (continued)
ÿ + 3ẏ + 2y = u, y(0) = α, ẏ(0) = β
Compute the step response, i.e., response to u(t) = 1(t)
Caution!! Y (s) = H(s)U(s) no longer holds if α 6= 0 or β 6= 0
Again, take Laplace transforms of both sides, mind the I.C.’s:
s2
Y (s) − sα − β + 3sY (s) − 3α + 2Y (s) = U(s)
U(s) = L {1(t)} = 1/s, which gives
s2
Y (s) − sα − β + 3sY (s) − 3α + 2Y (s) =
1
s
Y (s) =
αs + (3α + β) + 1
s
s2 + 3s + 2
=
αs2 + (3α + β)s + 1
s(s + 1)(s + 2)
101. Example (continued)
ÿ + 3ẏ + 2y = u, y(0) = α, ẏ(0) = β
Compute the step response, i.e., response to u(t) = 1(t)
Caution!! Y (s) = H(s)U(s) no longer holds if α 6= 0 or β 6= 0
Again, take Laplace transforms of both sides, mind the I.C.’s:
s2
Y (s) − sα − β + 3sY (s) − 3α + 2Y (s) = U(s)
U(s) = L {1(t)} = 1/s, which gives
s2
Y (s) − sα − β + 3sY (s) − 3α + 2Y (s) =
1
s
Y (s) =
αs + (3α + β) + 1
s
s2 + 3s + 2
=
αs2 + (3α + β)s + 1
s(s + 1)(s + 2)
Note: if α = β = 0, then Y (s) =
1
s(s + 1)(s + 2)
= H(s)U(s)
103. Example (continued)
Compute the step response of
ÿ + 3ẏ + 2y = u, y(0) = α, ẏ(0) = β
Y (s) =
αs2 + (3α + β)s + 1
s(s + 1)(s + 2)
y(t) = L −1
{Y (s)}
104. Example (continued)
Compute the step response of
ÿ + 3ẏ + 2y = u, y(0) = α, ẏ(0) = β
Y (s) =
αs2 + (3α + β)s + 1
s(s + 1)(s + 2)
y(t) = L −1
{Y (s)}
Use the method of partial fractions:
αs2 + (3α + β)s + 1
s(s + 1)(s + 2)
=
a
s
+
b
s + 1
+
c
s + 2
105. Example (continued)
Compute the step response of
ÿ + 3ẏ + 2y = u, y(0) = α, ẏ(0) = β
Y (s) =
αs2 + (3α + β)s + 1
s(s + 1)(s + 2)
y(t) = L −1
{Y (s)}
Use the method of partial fractions:
αs2 + (3α + β)s + 1
s(s + 1)(s + 2)
=
a
s
+
b
s + 1
+
c
s + 2
— this gives a = 1/2, b = 2α + β − 1, c = −α − β + 1/2
106. Example (continued)
Compute the step response of
ÿ + 3ẏ + 2y = u, y(0) = α, ẏ(0) = β
Y (s) =
αs2 + (3α + β)s + 1
s(s + 1)(s + 2)
y(t) = L −1
{Y (s)}
Use the method of partial fractions:
αs2 + (3α + β)s + 1
s(s + 1)(s + 2)
=
a
s
+
b
s + 1
+
c
s + 2
— this gives a = 1/2, b = 2α + β − 1, c = −α − β + 1/2
Y (s) =
1
2s
+ (2α + β − 1)
1
s + 1
+
−α − β + 1/2
s + 2
107. Example (continued)
Compute the step response of
ÿ + 3ẏ + 2y = u, y(0) = α, ẏ(0) = β
Y (s) =
αs2 + (3α + β)s + 1
s(s + 1)(s + 2)
y(t) = L −1
{Y (s)}
Use the method of partial fractions:
αs2 + (3α + β)s + 1
s(s + 1)(s + 2)
=
a
s
+
b
s + 1
+
c
s + 2
— this gives a = 1/2, b = 2α + β − 1, c = −α − β + 1/2
Y (s) =
1
2s
+ (2α + β − 1)
1
s + 1
+
−α − β + 1/2
s + 2
y(t) = L −1
{Y (s)} =
1
2
1(t) + (2α + β − 1)e−t
+ (1/2 − α − β)e−2t
108. Example (continued)
The step response of
ÿ + 3ẏ + 2y = u, y(0) = α, ẏ(0) = β
is given by
y(t) =
1
2
1(t) + (2α + β − 1)e−t
+ (1/2 − α − β)e−2t
109. Example (continued)
The step response of
ÿ + 3ẏ + 2y = u, y(0) = α, ẏ(0) = β
is given by
y(t) =
1
2
1(t) + (2α + β − 1)e−t
+ (1/2 − α − β)e−2t
What are the transient and the steady-state terms?
110. Example (continued)
The step response of
ÿ + 3ẏ + 2y = u, y(0) = α, ẏ(0) = β
is given by
y(t) =
1
2
1(t) + (2α + β − 1)e−t
+ (1/2 − α − β)e−2t
What are the transient and the steady-state terms?
I The transient terms are e−t
, e−2t
(decay to zero at exponential
rates −1 and −2)
111. Example (continued)
The step response of
ÿ + 3ẏ + 2y = u, y(0) = α, ẏ(0) = β
is given by
y(t) =
1
2
1(t) + (2α + β − 1)e−t
+ (1/2 − α − β)e−2t
What are the transient and the steady-state terms?
I The transient terms are e−t
, e−2t
(decay to zero at exponential
rates −1 and −2)
Note the poles of H(s) =
1
(s + 1)(s + 2)
at s = −1 and s = −2
112. Example (continued)
The step response of
ÿ + 3ẏ + 2y = u, y(0) = α, ẏ(0) = β
is given by
y(t) =
1
2
1(t) + (2α + β − 1)e−t
+ (1/2 − α − β)e−2t
What are the transient and the steady-state terms?
I The transient terms are e−t
, e−2t
(decay to zero at exponential
rates −1 and −2)
Note the poles of H(s) =
1
(s + 1)(s + 2)
at s = −1 and s = −2
— these are stable poles (both lie in LHP)
113. Example (continued)
The step response of
ÿ + 3ẏ + 2y = u, y(0) = α, ẏ(0) = β
is given by
y(t) =
1
2
1(t) + (2α + β − 1)e−t
+ (1/2 − α − β)e−2t
What are the transient and the steady-state terms?
I The transient terms are e−t
, e−2t
(decay to zero at exponential
rates −1 and −2)
Note the poles of H(s) =
1
(s + 1)(s + 2)
at s = −1 and s = −2
— these are stable poles (both lie in LHP)
I the steady-state part is 1
2 1(t)
114. Example (continued)
The step response of
ÿ + 3ẏ + 2y = u, y(0) = α, ẏ(0) = β
is given by
y(t) =
1
2
1(t) + (2α + β − 1)e−t
+ (1/2 − α − β)e−2t
What are the transient and the steady-state terms?
I The transient terms are e−t
, e−2t
(decay to zero at exponential
rates −1 and −2)
Note the poles of H(s) =
1
(s + 1)(s + 2)
at s = −1 and s = −2
— these are stable poles (both lie in LHP)
I the steady-state part is 1
2 1(t) — converges to steady-state value
of 1/2
115. DC Gain
u y
h
Definition: the steady-state value of the step response is called
the DC gain of the system.
116. DC Gain
u y
h
Definition: the steady-state value of the step response is called
the DC gain of the system.
DC gain = y(∞) = lim
t→∞
y(t) for u(t) = 1(t)
117. DC Gain
u y
h
Definition: the steady-state value of the step response is called
the DC gain of the system.
DC gain = y(∞) = lim
t→∞
y(t) for u(t) = 1(t)
In our example above, the step response is
y(t) =
1
2
1(t) + (2α + β − 1)e−t
+ (1/2 − α − β)e−2t
118. DC Gain
u y
h
Definition: the steady-state value of the step response is called
the DC gain of the system.
DC gain = y(∞) = lim
t→∞
y(t) for u(t) = 1(t)
In our example above, the step response is
y(t) =
1
2
1(t) + (2α + β − 1)e−t
+ (1/2 − α − β)e−2t
therefore, DC gain = y(∞) = 1/2
122. Steady-State Value
u y
h
u(t) = 1(t) U(s) =
1
s
=⇒ Y (s) =
H(s)
s
— can we compute y(∞) from Y (s)?
Let’s look at some examples:
123. Steady-State Value
u y
h
u(t) = 1(t) U(s) =
1
s
=⇒ Y (s) =
H(s)
s
— can we compute y(∞) from Y (s)?
Let’s look at some examples:
I Y (s) =
1
s + a
, a 0 (pole at s = −a 0)
124. Steady-State Value
u y
h
u(t) = 1(t) U(s) =
1
s
=⇒ Y (s) =
H(s)
s
— can we compute y(∞) from Y (s)?
Let’s look at some examples:
I Y (s) =
1
s + a
, a 0 (pole at s = −a 0)
y(t) = e−at
=⇒ y(∞) = 0
125. Steady-State Value
u y
h
u(t) = 1(t) U(s) =
1
s
=⇒ Y (s) =
H(s)
s
— can we compute y(∞) from Y (s)?
Let’s look at some examples:
I Y (s) =
1
s + a
, a 0 (pole at s = −a 0)
y(t) = e−at
=⇒ y(∞) = 0
I Y (s) =
1
s + a
, a 0 (pole at s = −a 0)
126. Steady-State Value
u y
h
u(t) = 1(t) U(s) =
1
s
=⇒ Y (s) =
H(s)
s
— can we compute y(∞) from Y (s)?
Let’s look at some examples:
I Y (s) =
1
s + a
, a 0 (pole at s = −a 0)
y(t) = e−at
=⇒ y(∞) = 0
I Y (s) =
1
s + a
, a 0 (pole at s = −a 0)
y(t) = e−at
=⇒ y(∞) = ∞
127. Steady-State Value
u y
h
u(t) = 1(t) U(s) =
1
s
=⇒ Y (s) =
H(s)
s
— can we compute y(∞) from Y (s)?
Let’s look at some examples:
I Y (s) =
1
s + a
, a 0 (pole at s = −a 0)
y(t) = e−at
=⇒ y(∞) = 0
I Y (s) =
1
s + a
, a 0 (pole at s = −a 0)
y(t) = e−at
=⇒ y(∞) = ∞
I Y (s) =
1
s2 + ω2
, ω ∈ R (poles at s = ±jω, purely imaginary)
128. Steady-State Value
u y
h
u(t) = 1(t) U(s) =
1
s
=⇒ Y (s) =
H(s)
s
— can we compute y(∞) from Y (s)?
Let’s look at some examples:
I Y (s) =
1
s + a
, a 0 (pole at s = −a 0)
y(t) = e−at
=⇒ y(∞) = 0
I Y (s) =
1
s + a
, a 0 (pole at s = −a 0)
y(t) = e−at
=⇒ y(∞) = ∞
I Y (s) =
1
s2 + ω2
, ω ∈ R (poles at s = ±jω, purely imaginary)
y(t) = sin(ωt) =⇒ y(∞) does not exist
129. Steady-State Value
u y
h
u(t) = 1(t) U(s) =
1
s
=⇒ Y (s) =
H(s)
s
— can we compute y(∞) from Y (s)?
Let’s look at some examples:
I Y (s) =
1
s + a
, a 0 (pole at s = −a 0)
y(t) = e−at
=⇒ y(∞) = 0
I Y (s) =
1
s + a
, a 0 (pole at s = −a 0)
y(t) = e−at
=⇒ y(∞) = ∞
I Y (s) =
1
s2 + ω2
, ω ∈ R (poles at s = ±jω, purely imaginary)
y(t) = sin(ωt) =⇒ y(∞) does not exist
I Y (s) =
c
s
(pole at the origin, s = 0)
130. Steady-State Value
u y
h
u(t) = 1(t) U(s) =
1
s
=⇒ Y (s) =
H(s)
s
— can we compute y(∞) from Y (s)?
Let’s look at some examples:
I Y (s) =
1
s + a
, a 0 (pole at s = −a 0)
y(t) = e−at
=⇒ y(∞) = 0
I Y (s) =
1
s + a
, a 0 (pole at s = −a 0)
y(t) = e−at
=⇒ y(∞) = ∞
I Y (s) =
1
s2 + ω2
, ω ∈ R (poles at s = ±jω, purely imaginary)
y(t) = sin(ωt) =⇒ y(∞) does not exist
I Y (s) =
c
s
(pole at the origin, s = 0)
y(t) = c1(t) =⇒ y(∞) = c
131. The Final Value Theorem
We can now deduce the Final Value Theorem (FVT):
If all poles of sY (s) are strictly stable or lie in the open left
half-plane (OLHP), i.e., have Re(s) 0, then
y(∞) = lim
s→0
sY (s).
132. The Final Value Theorem
We can now deduce the Final Value Theorem (FVT):
If all poles of sY (s) are strictly stable or lie in the open left
half-plane (OLHP), i.e., have Re(s) 0, then
y(∞) = lim
s→0
sY (s).
In our examples, multiply Y (s) by s, check poles:
133. The Final Value Theorem
We can now deduce the Final Value Theorem (FVT):
If all poles of sY (s) are strictly stable or lie in the open left
half-plane (OLHP), i.e., have Re(s) 0, then
y(∞) = lim
s→0
sY (s).
In our examples, multiply Y (s) by s, check poles:
I Y (s) =
1
s + a
sY (s) =
s
s + a
134. The Final Value Theorem
We can now deduce the Final Value Theorem (FVT):
If all poles of sY (s) are strictly stable or lie in the open left
half-plane (OLHP), i.e., have Re(s) 0, then
y(∞) = lim
s→0
sY (s).
In our examples, multiply Y (s) by s, check poles:
I Y (s) =
1
s + a
sY (s) =
s
s + a
if a 0, then y(∞) = 0; if a 0, FVT does not give correct
answer
135. The Final Value Theorem
We can now deduce the Final Value Theorem (FVT):
If all poles of sY (s) are strictly stable or lie in the open left
half-plane (OLHP), i.e., have Re(s) 0, then
y(∞) = lim
s→0
sY (s).
In our examples, multiply Y (s) by s, check poles:
I Y (s) =
1
s + a
sY (s) =
s
s + a
if a 0, then y(∞) = 0; if a 0, FVT does not give correct
answer
I Y (s) =
1
s2 + ω2
sY (s) =
s
s2 + ω2
136. The Final Value Theorem
We can now deduce the Final Value Theorem (FVT):
If all poles of sY (s) are strictly stable or lie in the open left
half-plane (OLHP), i.e., have Re(s) 0, then
y(∞) = lim
s→0
sY (s).
In our examples, multiply Y (s) by s, check poles:
I Y (s) =
1
s + a
sY (s) =
s
s + a
if a 0, then y(∞) = 0; if a 0, FVT does not give correct
answer
I Y (s) =
1
s2 + ω2
sY (s) =
s
s2 + ω2
poles are purely imaginary (not in OLHP), FVT does not give
correct answer
137. The Final Value Theorem
We can now deduce the Final Value Theorem (FVT):
If all poles of sY (s) are strictly stable or lie in the open left
half-plane (OLHP), i.e., have Re(s) 0, then
y(∞) = lim
s→0
sY (s).
In our examples, multiply Y (s) by s, check poles:
I Y (s) =
1
s + a
sY (s) =
s
s + a
if a 0, then y(∞) = 0; if a 0, FVT does not give correct
answer
I Y (s) =
1
s2 + ω2
sY (s) =
s
s2 + ω2
poles are purely imaginary (not in OLHP), FVT does not give
correct answer
I Y (s) =
c
s
sY (s) = c
138. The Final Value Theorem
We can now deduce the Final Value Theorem (FVT):
If all poles of sY (s) are strictly stable or lie in the open left
half-plane (OLHP), i.e., have Re(s) 0, then
y(∞) = lim
s→0
sY (s).
In our examples, multiply Y (s) by s, check poles:
I Y (s) =
1
s + a
sY (s) =
s
s + a
if a 0, then y(∞) = 0; if a 0, FVT does not give correct
answer
I Y (s) =
1
s2 + ω2
sY (s) =
s
s2 + ω2
poles are purely imaginary (not in OLHP), FVT does not give
correct answer
I Y (s) =
c
s
sY (s) = c
poles at infinity, so y(∞) = c – FVT gives correct answer
140. Back to DC Gain
u y
h
Step response: Y (s) =
H(s)
s
141. Back to DC Gain
u y
h
Step response: Y (s) =
H(s)
s
— if all poles of sY (s) = H(s) are strictly stable, then
y(∞) = lim
s→0
H(s)
by the FVT.
142. Back to DC Gain
u y
h
Step response: Y (s) =
H(s)
s
— if all poles of sY (s) = H(s) are strictly stable, then
y(∞) = lim
s→0
H(s)
by the FVT.
Example: compute DC gain of the system with transfer function
H(s) =
s2 + 5s + 3
s3 + 4s + 2s + 5
143. Back to DC Gain
u y
h
Step response: Y (s) =
H(s)
s
— if all poles of sY (s) = H(s) are strictly stable, then
y(∞) = lim
s→0
H(s)
by the FVT.
Example: compute DC gain of the system with transfer function
H(s) =
s2 + 5s + 3
s3 + 4s + 2s + 5
All poles of H(s) are strictly stable (we will see this later using
the Routh–Hurwitz criterion), so
y(∞) = H(s)
s=0
=
3
5
.
