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SYSC5602: Digital Signal Processing
Mohamed ElTanany, Professor
Department of Systems & Computer Engineering, Carleton University, Ottawa, Ontario
1 / 23
FREQUENCY RESPONSE
Defined only for stable systems
Consider an LTI system with impulse response h(n), and an input sequence of the
form
. The system output in the steady state is given by:
Therefore, the phase/amplitude relations between the input and output sequences are
described by the function:
Therefore, the frequency response is given by the transfer function of the system,
evaluated along the unit circle
.
If the input x(n) is a linear combination of complex exponentials:
xn
()
e
j
ω
0
n
=
y n
hn
e
j
ω
0
n
⊗
hm
e
j
ω
0
nm
–
m
∞
–
=
∞
∑
e
j
ω
0
n
e
j
ω
0
m
–
m
∞
–
=
∞
∑
==
=
He
j
ω
0
e
j
ω
0
m
–
m
∞
–
=
∞
∑
z
m
–
m
∞
–
=
∞
∑
ze
j
ω
0
=
Hz
j
ω
0
=
=
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View Full Document SYSC5602: Digital Signal Processing
Mohamed ElTanany, Professor
Department of Systems & Computer Engineering, Carleton University, Ottawa, Ontario
2 / 23
Then the output, in the steadystate, is given by:
Properties Of The Frequency Response:
is periodic with period 2
π
; this means the magnitude has even symmetry, while the phase
exhibits oddsymmetry.
To show above is true:
xn
()
C
1
e
j
ω
1
n
C
2
e
j
ω
2
n
……
C
M
e
j
ω
M
n
++
+
=
y n
C
1
He
j
ω
1
e
j
ω
1
n
C
2
j
ω
2
e
j
ω
2
n
………
C
M
j
ω
M
e
j
ω
M
n
+
=
j
ω
j
ω
H
∗
e
j
ω
–
=
j
ω
hm
e
j
ω
m
–
m
∞
–
=
∞
∑
ω
m
cos
m
∞
–
=
∞
∑
jh
m
ω
m
sin
m
∞
–
=
∞
∑
–
==
j
ω
–
e
j
ω
m
m
∞
–
=
∞
∑
ω
m
cos
m
∞
–
=
∞
∑
m
ω
m
sin
m
∞
–
=
∞
∑
+
j
ω
[]
∗
=
j
ω
ω
m
cos
m
∞
–
=
∞
∑
2
ω
m
sin
m
∞
–
=
∞
∑
2
+
j
ω
–
SYSC5602: Digital Signal Processing
Mohamed ElTanany, Professor
Department of Systems & Computer Engineering, Carleton University, Ottawa, Ontario
3 / 23
The frequency response can also be obtained using the FFT (will elaborate later)
Example
Consider the LTI system with transfer function
a) Calculate, and plot the frequency response
, magnitude and phase, as a
function of frequency for the range
.
b) Assume that the input sequence is of the form:
Calculate the output sequence in the steady state.
Solution
The frequency response may be calculated by hand as follows.
He
j
ω
()
∠
hm
ω
m
sin
m
∞
–
=
∞
∑
⎝⎠
⎜⎟
⎛⎞
ω
m
cos
m
∞
–
=
∞
∑
⁄
atan
–
j
ω
–
∠
–
==
Hz
z
4
z
2
+
z
4
0.25
+

=
j
ω
0
ωπ
≤≤
xn
0.532
π
n
4

cos
1.125
π
n
2
cos
0.17
π
n
π
4

+
cos
++
=
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View Full Document SYSC5602: Digital Signal Processing
Mohamed ElTanany, Professor
Department of Systems & Computer Engineering, Carleton University, Ottawa, Ontario
4 / 23
The results are shown graphically for both the magnitude and phase functions. The
figure shows conjugate symmetry (magnitude is even, phase is odd) around
π
/2 This
He
j
ω
()
z
4
z
2
+
z
4
0.25
+

ze
j
ω
=
e
j
4
ω
e
j
2
ω
+
e
j
4
ω
0.25
+

==
j
0
e
j
0
e
j
0
+
e
j
0
0.25
+

2
1.25

1.6
e
j
0
=
j
π
4

⎝⎠
⎜⎟
⎛⎞
e
j
π
e
j
π
2
+
e
j
π
0.25
+
1
–
j
+
1
–0
.
2
5
+
1.8856
e
j
0.7854
–
===
j
π
2
e
j
2
π
e
j
π
+
e
j
2
π
0.25
+

11
–
10
.
2
5
+

0
=
j
π
e
j
4
π
e
j
2
π
+
e
j
4
π
0.25
+
2
1.25
1.6
e
j
0
=
SYSC5602: Digital Signal Processing
Mohamed ElTanany, Professor
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This note was uploaded on 10/24/2009 for the course SCE sysc5602 taught by Professor Eltanany during the Spring '09 term at Carleton CA.
 Spring '09
 Eltanany

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