ECE-2025
Spring-2011
Info: Lab & HW
ƒ Quiz #3
ƒ Resolve grades by 11-Apr (Today)
ƒ Lab
L b #10
#10: C
Cochlear
hl
Filt
Filterbank
b k iis FORMAL L
Lab:
b
Lecture 20
Fourier Transform:
f
Frequency Response of
Continuous-Time Systems
11-Apr-2011
p 0
ƒ Demo your working system when you turn in the report
ƒ Open Lab tomorrow: 12
12-4:30+
4:30+
ƒ HW #9 due this week
ƒ Quiz #4 will be 24-Apr (Friday)
ƒ Covers
C
HW #9 and
d #10 (Ch
(Chaps 9
9, 10 and
d 11)
4/11/2011
Stability & Causality
for LTI Systems
ECE-2025
Spr-2011
jMc
2
Convolution GUI: Sinusoid
ƒ A system is stable if every bounded input produces a bounded
output.
ƒ A continuous-time LTI system is stable if and only if
∞
∫ h(t) dt < ∞
−∞
∞
ƒ A system is causal if and only if y(t0) depends only on x(τ) for τ< t0
ƒ An LTI system is causal if and only if
h(t ) = 0 for t < 0
4/11/2011
ECE-2025
Spr-2011
jMc
Lecture
3
Lecture
4/11/2011
ECE-2025
Spr-2011
jMc
4
READING ASSIGNMENTS
LECTURE OBJECTIVES
ƒ This Lecture:
ƒ Review of convolution
ƒ THE operation for LTI Systems
ƒ Chapter 10, all
ƒ Chapter 11: 11-1 to 11-5
ƒ Complex
C
l exponential
ti l iinputt signals
i
l
ƒ Frequency Response
ƒ Cosine
C i signals
i
l
ƒ Other Reading:
ƒ Real part of complex exponential
ƒ Fourier Transform H(jω)
ƒ Recitation: Chapter 11
ƒ Next Lecture: the rest of Chapter 11
4/11/2011
© 2003, JH McClellan & RW Schafer
ƒ Same as Frequency Response
5
LTI Systems
4/11/2011
6
© 2003, JH McClellan & RW Schafer
Complex Exponential Input
x(t ) = A
Ae jϕ e jω t 6 y (t ) = H ( jω ) Ae
A jϕ e j ω t
∞
y (t ) = ∫ h(τ ) Ae
A jϕ e jω ( t −τ ) dτ
−∞
ƒ Convolution defines an LTI system
⎛∞
⎞
y (t ) = ⎜⎜ ∫ h(τ )e − jωτ dτ ⎟⎟ Ae jϕ e jω t
⎝ −∞
⎠
∞
y (t ) =
∫ x(τ )h(t − τ )dτ = x(t ) ∗ h(t )
−∞
∞
ƒ Response
p
to a complex
p
exponential
p
g
gives
frequency response H(jω)
4/11/2011
© 2003, JH McClellan & RW Schafer
H ( jω ) = ∫ h(τ )e − jωτ dτ
7
4/11/2011
−∞
© 2003, JH McClellan & RW Schafer
Frequency
q
y
Response
8
h(t ) = e − at u (t ) ⇔ H ( jω ) =
When does H(jω) Exist?
ƒ When
Wh is
i
ƒ Suppose that h(t) is:
H ( jω ) < ∞ ?
H ( jω ) =
∞
∫ h(τ )e
− jωτ
dτ ≤
−∞
a =1
∞
∫ h(τ )
−∞
∞
H ( jω ) ≤ ∫ | h(τ ) | dτ < ∞
−∞
ƒ Thus the frequency response exists if the LTI
system is a stable system.
system
a>0
H ( jω ) = ∫ e u (t )e
− at
− jω t
−∞
∞
dt = ∫ e −( a + jω ) t dt
0
∞
9
© 2003, JH McClellan & RW Schafer
e − ( a + jω ) t
e − ( a + jω ) ∞ − 1
1
=
=
H ( jω ) =
− ( a + jω ) 0
− ( a + jω ) a + jω
4/11/2011
© 2003, JH McClellan & RW Schafer
10
Cosine Input x(t ) = A cos(ω1t + ϕ )
Magnitude and Phase Plots
1
H ( jω ) =
1 + jω
h(t ) = e − t u (t )
e − jωτ dτ
∞
4/11/2011
1
a + jω
1
1
=
a + jω
a2 + ω 2
x(t ) = A cos(ω1t + ϕ ) = 12 A
Ae jϕ e jω1 t + 12 A
Ae − jϕ e − jω1 t
y (t ) = H ( jω1 ) 12 Ae jϕ e jω1 t + H (− jω1 ) 12 Ae − jϕ e − jω1 t
y (t ) = 12 | H | e j∠H Ae jϕ e jω1 t + 12 | H | e − j∠H Ae − jϕ e − jω1 t
∠H ( jω ) = − tan −1 (ω )
H ( − jω ) = H ( jω )
y (t ) = A H ( jω1 ) cos(ω1t + ϕ + ∠H ( jω1 ))
∗
4/11/2011
© 2003, JH McClellan & RW Schafer
Si
Since
H ( − jω ) = H ∗ ( jω )
11
4/11/2011
© 2003, JH McClellan & RW Schafer
12
Sinusoid in Gives Sinusoid out
Fourier Transform
∞
Frequency
Response
H ( jω ) = ∫ h(τ )e − jωτ dτ
−∞
ƒ TRANSFORMATION
ƒ The impulse response h(t) could be any signal
ƒ H(jω) is the frequency content
ƒ INVERSE TRANSFORM
ƒ Given , H(j
(jω) recover h(t)
()
4/11/2011
13
© 2003, JH McClellan & RW Schafer
4/11/2011
ƒ For
F non-periodic
i di signals
i
l
x(t ) =
1
2π
jω t
X
(
j
ω
)
e
dω
∫
x(t ) = δ (t − t d )
Fourier Synthesis
∞
X ( jω ) = ∫ δ (τ − t d )e − jωτ dτ = e − jωτ δ
−∞
X ( jω ) =
14
Example
p 1: Delayed
y
Impulse
p
Fourier Transform Defined
∞
© 2003, JH McClellan & RW Schafer
−∞
Fourier Analysis
∞
X ( jω ) = e
− jω t
x
(
t
)
e
dt
∫
− jω t d
−∞
4/11/2011
© 2003, JH McClellan & RW Schafer
15
4/11/2011
© 2003, JH McClellan & RW Schafer
16
x(t ) = Aδ (t ) ⇔ X ( jω ) = A
Example 2: X ( jω ) = 2πδ (ω − ω0 )
1
x(t ) =
2π
∞
∫ 2πδ (ω − ω )e
jω t
0
dω = e jω0 t
−∞
x(t ) = e jω0t ⇔ X ( jω ) = 2πδ (ω − ω0 )
x(t ) = 1 ⇔ X ( jω ) = 2πδ (ω )
x(t ) = cos(ω0t ) ⇔
X ( jω ) = πδ (ω − ω0 ) + πδ (ω + ω0 )
4/11/2011
© 2003, JH McClellan & RW Schafer
17
x(t ) = cos(ω0t ) ⇔
4/11/2011
⎧1
x(t ) = ⎨
⎩0
Example 3:
X ( jω ) = πδ (ω − ω0 ) + πδ (ω + ω0 )
X ( jω ) =
T /2
∫ (1)e
− jω t
−T / 2
e − jω t
X ( jω ) =
− jω
© 2003, JH McClellan & RW Schafer
19
4/11/2011
t <T /2
t >T /2
T /2
dt =
− jω t
e
∫ dt
−T / 2
T /2
−T / 2
X ( jω ) =
4/11/2011
18
© 2003, JH McClellan & RW Schafer
e − j ω T / 2 − e + jω T / 2
=
− jω
sin(ωT / 2)
ω/2
© 2003, JH McClellan & RW Schafer
20
⎧1
x(t ) = ⎨
⎩0
t <T /2
sin(ωT / 2)
⇔ X ( jω ) =
t >T /2
ω/2
Example 4:
∞
1
x(t ) =
2π
∫ X ( jω ) e
−∞
1 e jω t
x(t ) =
2π jt
j
ωb
−ωb
x(t ) =
21
© 2003, JH McClellan & RW Schafer
sin(ωbt )
πt
⇔
⎧1
X ( jω ) = ⎨
⎩0
ω < ωb
ω > ωb
4/11/2011
jω t
1
dω =
2π
ωb
jω t
1
e
∫ dω
−ωb
1 e jω b t − e − jω b t
=
2π
jjt
x(t ) =
4/11/2011
ω < ωb
ω > ωb
⎧1
X ( jω ) = ⎨
⎩0
sin(ωbt )
πt
22
© 2003, JH McClellan & RW Schafer
Table of Fourier Transforms
x(t ) = e − a t u (t ) ⇔ X ( jω ) =
⎧1
x(t ) = ⎨
⎩0
x(t ) =
t <T /2
t >T /2
1
a + jω
⇔ X ( jω ) =
⎧1
sin(ωbt )
⇔ X ( jω ) = ⎨
πt
⎩0
sin((ωT / 2)
ω/2
ω < ωb
ω > ωb
x(t ) = δ (t − t0 ) ⇔ X ( jω ) = e − jω t0
4/11/2011
© 2003, JH McClellan & RW Schafer
23
x(t ) = e jω0t ⇔ X ( jω ) = 2πδ (ω − ω0 )
4/11/2011
© 2003, JH McClellan & RW Schafer
24