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8/19/2019 L8s http://slidepdf.com/reader/full/l8s 1/38 1 Discrete-time System Analysis via the DTFT and DFT
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1
Discrete-time System Analysisvia the DTFT and DFT
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2
ObjectivesUnderstand the systems
representations
Model linear discrete-time systems usingfrequency
domain approaches
Analyze systems using frequency domain
techniquesDiscrete-time Fourier series for continuous-timeperiodic signal
Discrete-time Fourier transform for continuous-
time aperiodic
signal
Discrete Fourier transform
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3
Recall
linear time-invariant discrete-time systemsConvolution representationTime domain system analysis
Output = input convolve with unit-pulse response
Frequency domain system analysis?
[ ]∑−
=
−=1
0
][][ M
i
in xihn y
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4
Simple input: exponentialinput
h[n] Output?n j je Aen x
ω φ ˆ][ =
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5
Response to ComplexExponentials
[ ] [ ] n j j M
i
i j M
i
in j j e Aeeihe Aeihn y ω φ ω ω φ ˆ1
0
ˆ1
0
)(ˆ][⎟⎟
⎠
⎞
⎜⎜⎝
⎛ == ∑∑
−
=
−
−
=
−
n j j e Aen x ω φ ˆ][ =
( ) ( ) [ ]n x H e Ae H n y n j j ω ω ω φ ˆˆ][ ˆ ==
Frequency response function for the LTI system
∑
−
=
Ω−=Ω1
0 ][)(
M
i
i j
eih H
[ ]∑
−
=−=
1
0][][
M
iin xihn y
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Simple input: exponentialinput
h[n]n j j
e Aen x ω φ ˆ][ = ( ) n j j
e Ae H n y ω φ ω
ˆˆ][ =
∑−
=
Ω−=Ω1
0
][)( M
i
i jeih H
Frequency response function for the LTI system
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System: unit-pulse + frequencyresponse
=Ω)( H
[ ]
∑
−
=
−=1
0
][][ M
i
inihnh δ
Unit pulse response: h[n] = {1, 2, 1}
=][nh
∑
−
=Ω−=Ω
1
0][)(
M
i
i jeih H
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System: unit-pulse + frequencyresponse
Unit impulse response:h[n] = {1, 2, 1}
Frequency response)(2 )(21)( Ω∠Ω−Ω− Ω=++=Ω H j j j e H ee H
=++=Ω Ω−ΩΩ− j j jeee H 2)(
=Ω∠
=Ω
)(
)(
H
H
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Exampleh[n] = {1, 2, 1}
Input
Output ( ) n j je Ae H n y
ω φ ω ˆˆ][ =
3/ˆ4/
2][ 3/4/ˆ
π ω π φ
π π ω φ
==== n j jn j j eee Aen x
=)3/(π H
=][n y
Output = 3*input and phase shift by /3
[ ]Ω+=Ω Ω− cos22)( je H
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Simple input: Periodic input(Discrete-time Fourier Series)
h[n] ?][ =n y
∑−
=
Ω−=Ω1
0
][)( M
i
i jeih H
Frequency response function for the LTI system
( )
( )[ ]∑
∑
=
+−+
=
++=
++=
N
i
n jn ji
N
i
iii
iiii ee A
A
n A A
n x
1
)ˆ(ˆ
0
1
0
2
ˆcos
][
φ ω φ ω
φ ω
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Response to Periodic Inputs
( )( ) ( ) ( )
( ) ( )
( )( ) ( ) ( )
( ) ( ) ( )
⊕
⎪⎪⎪
⎪⎪⎪
⎭
⎪⎪⎪⎪⎪
⎪
⎬
⎫
⎪⎪⎪
⎪⎪⎪
⎩
⎪⎪⎪⎪⎪
⎪
⎨
⎧
−−
=
−−
=
+−+−
++
+−+−
++
M
2ˆ22ˆ2
2ˆ2
2ˆ2
1
ˆ1
1
ˆ1
1ˆ1
1ˆ1
00
ˆ2
ˆ2
ˆ2
ˆ2
2
ˆ2ˆ2
ˆ2
ˆ2
100
2222
2222
1111
1111
ω ω
ω ω
ω ω
ω ω
φ ω φ ω
φ ω φ ω
φ ω φ ω
φ ω φ ω
H e Ae A
H e A
e A
k
H e
A
e
A
H e A
e A
k
H A A
output Freqinput
n jn j
n jn j
n jn j
n jn j
n j j e Aen x ω φ ˆ
][ = ( ) n j j e Ae H n y ω φ ω ˆˆ][ =
( )
[ ]∑=+−+
++=
N
i
n jn ji iiii
ee
A
An x1
)ˆ(ˆ
0 2][ φ ω φ ω
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Response to Periodic Inputs
( ) ( ) ( ) ( )[ ]∑=
+−+ −++= N
i
in j
in ji H e H e
A H An y iiii
1
)ˆ(ˆ0 ˆˆ
2
0][ ω ω φ ω φ ω
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ...ˆ2
ˆ2
ˆ2
ˆ2
0][
2ˆ2
2ˆ2
1ˆ11ˆ10
2222
1111
+−+
+−++=
+−+
+−+
ω ω
ω ω
φ ω φ ω
φ ω φ ω
H e A
H e A
H e A H e A H An y
n jn j
n jn j
( )Ω==Ω−=Ω ∑∑ −
=
Ω−
=
Ω− *1
0
1
0
][)(][)( H eih H eih H M
i
i j M
i
i j
( ) ( ) ( ) ( )[ ]∑=
+−+ ++= N
i
in j
in ji H e H e
A H An y iiii
1
*)ˆ(ˆ0 ˆˆ
2
0][ ω ω φ ω φ ω
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Response to Periodic Inputs
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )[ ]
( ) ( ) ( )[ ]∑
∑
=
=
∠−+−∠+
∠−∠
∠+++=
++=
==
N
i
iiiii
N
i
H jn j H jn jii
H jii
H jii
H n H A H A
eeee H A
H A
n y
e H H e H H
iiiiii
ii
1
0
1
ˆ)ˆ(ˆˆ0
ˆ*ˆ
ˆˆcosˆ0
2
ˆ0
][
ˆˆˆˆ
ω φ ω ω
ω
ω ω ω ω
ω φ ω ω φ ω
ω ω
( ) ( ) ( ) ( )[ ]∑=
+−+ ++= N
i
in j
in ji H e H e
A H An y iiii
1
*)ˆ(ˆ0 ˆˆ
2
0][ ω ω φ ω φ ω
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Simple input: Periodic input
(Discrete-time Fourier Series)
h[n]
∑
−
=
Ω−=Ω1
0
][)( M
i
i jeih H
Frequency response function for the LTI system
( )
( )[ ]∑
∑
=
+−+
=
++=
++=
N
i
n jn ji
N
i
iii
iiii ee A
A
n A A
n x
1
)ˆ(ˆ
0
1
0
2
ˆcos
][
φ ω φ ω
φ ω ( )
( ) ( )[ ]∑=
∠+++
= N
i
iiiii H n H A
H An y
1
0
ˆˆcosˆ
0][
ω φ ω ω
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Example: Periodic Inputh[n] = {1, 2, 1}
Input: x[n] = 1 + 2 cos
(πn/3 -
π/2)Output?
Solution:[ ]
Ω−=∠Ω+=Ω
Ω+=Ω Ω−
)ˆ(cos22)(
cos22)(
w H H
e H j
( )∑=++=
N
i
iii n A An x
10 ˆcos][ φ ω
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Example: Periodic Input
=][n y
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Exercise 1
Solution
[ ] ⎟ ⎠
⎞⎜⎝
⎛ +⎟
⎠
⎞⎜⎝
⎛ −+=
21
20cos3
23cos34
π π π nnn x
[ ] ?=n y
h[n] = {1, 2, 1}
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18
Simple input: Aperiodic
input
(Discrete Fourier Transform)
h[n]
∑
−
=
Ω−=Ω1
0
][)( M
i
i jeih H
Frequency response function for the LTI system
?][ =n y
∑
∑−
=
−
=
−
=
=
1
0
2
1
0
2
][1
][
][][
N
k
n N
k j
N
n
n N
k j
ek X N
n x
en xk X
π
π
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19
Responses to Aperiodic
Inputs
Discrete Fourier Transform:
Output
∑∑ −
=
−
=
−==
1
0
21
0
2
][1
][][][ N
k
n N
k j N
n
n N
k j
ek X N
n xen xk X π π
[ ]
∑ ∑
∑ ∑
∑ ∑∑
−
=
−−
=
−
=
−−
=
−
=
−
=
−−
=
=
=
=−=
1
0
21
0
)(2
1
0
)(21
0
1
0
1
0
)(21
0
][][1][
][][1
][
][1
][][][
N
k
m
N
k j M
m
n
N
k j
N
k
mn N
k j M
m
M
m
N
k
mn N
k j M
m
emhek X N
n y
emhk X N
n y
ek X N
mhmn xmhn y
π π
π
π
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20
Responses to Aperiodic
Inputs
Discrete Fourier Transform:
Output
∑∑ −
=
−
=
−==
1
0
21
0
2
][1
][][][ N
k
n N
k j N
n
n N
k j
ek X N
n xen xk X π π
{ }∑
∑
∑ ∑
−
=
−
=
−
=
−−
=
=
=
=
1
0
)(2
1
0
)(2
1
0
21
0
)(2
][][1
][
][][1
][
][][1
][
N
k
n
N
k j
N
k
n N
k j
N
k
m N
k j M
m
n N
k j
ek H k X N
n y
k H ek X N
n y
emhek X N
n y
π
π
π π
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Responses to Aperiodic
Inputs
For aperiodic
inputs, output can be obtainedby:DFT the input X [k ]
DFT the impulse response H [k ]Called the frequency response
Multiply X [k ] and H [k ]
Inverse DFT { }∑
−
=
=1
0
)(2
][][1
][ N
k
n N
k j
ek H k X N
n yπ
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22
Recall
LTI System
[ ] [ ] [ ] [ ]
[ ] [ ]∑
∑
=
=
−=
⊗=−=
M
k
M
k
k nnhnh
n xnhk n xnhn y
0
0
][
][
δ
LTI h[n] x[n] y[n]
Length of convolution output:
Length of h + Length of x - 1
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23
Example: Time domain
x [n ] = {2, 4, 6, 4, 2}h [n ] = {3, -1, 2, 1}
[ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ]3322110][ −+−+−+= n xhn xhn xhn xhn y
Length of h = 4
Length of x = 5
Length of y = 8
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24
Example: DFT of input
DFT of input: x [n ] = {2, 4, 6, 4, 2}8 point DFT x [n ] = {2, 4, 6, 4, 2, 0, 0, 0}
π
π π π
π π π π π
jk
k j
k j
k j
k j
k j
k j
k j
n
nk
j
eeeek X
eeeeen xk X
−−−−
−−−−
=
−
++++=
++++== ∑
24642][
24642][][
4
3
24
48
23
8
22
8
2
8
27
0
82
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25
Example: DFT of h
DFT of filter: h [n ] = {3, -1, 2, 1}8 point DFT h [n ] = {3, -1, 2, 1, 0, 0, 0, 0}
4
3
24
38
22
8
2
8
27
0
82
23][
23][][
k j
k j
k j
k j
k j
k j
n
nk
j
eeek H
eeeenhk H
π π π
π π π π
−−−
−−−
=
−
++−=
++−== ∑
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Example: DFT of output
DFT of output:y [n ] = {6, 10, 18, 16, 18, 12, 8, 2}
8 point DFT
78
2
2
3
4
5
4
3
24
78
26
8
25
8
24
8
23
8
22
8
2
8
2
7
0
82
2812181618106][
2812181618106][
][][
k j
k j
k j
jk
k j
k j
k j
k j
k j
k j
k j
k j
k j
k j
n
nk
j
eeeeeeek Y
eeeeeeek Y
en xk Y
π π π
π
π π π
π π π π π π π
π
−−−−
−−−
−−−−−−−
=
−
+++++++=
+++++++=
= ∑
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27
78
2
2
3
4
5
4
3
24 2812181618106][
k j
k j
k j
jk
k j
k j
k j
eeeeeeek Y
π π π
π
π π π −−−
−−−−
+++++++=
4
7
2
3
4
5
2
3
4
5
4
3
4
5
4
3
2
4
3
24
4
3
24
4
3
244
3
24
2426
48412
612618
48412
2426
2324642][][
k j
k j
k j
k j
k j
k j
k j
k j
k j
k j
k j
k j
k j
k
j
k
j
k
j
k j
k j
k j
k j
k j
k j
jk
k j
k j
k j
eeee
eeee
eeee
eeee
eee
eeeeeeek H k X
π π π
π
π π
π
π
π
π
π π
π
π π π
π π π
π π π
π
π π π
−−−−
−−−
−
−−
−−
−−−−
−−−
−−−−
−−−
−
−
−
−
−
⎟⎟ ⎠
⎞
⎜⎜⎝
⎛
++−⎟⎟ ⎠
⎞
⎜⎜⎝
⎛
++++=
4
7
2
3
4
5
4
3
24
2812181618106][][
k j
k j
k j
k j
k j
k j
k j
eeeeeeek H k X
π π π
π
π π π −−−
−−−−
+++++++=
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Example][][][ k Y k H k X =[ ] [ ] [ ] [ ]n xnhk n xnhn y
M
k
⊗=−= ∑=0
][
Y [k ]
X [k ]
H [k ]
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Example
][][][ k Y k H k X =[ ] [ ] [ ] [ ]n xnhk n xnhn y
M
k ⊗=−= ∑=0
][
What happens if N is not 8?
5 point DFT
5
14
5
12
5
10
5
8
5
6
5
4
5
2
5
6
5
4
5
2
4
5
23
5
22
5
2
5
2
812181618106][][
23][
24642][
k j
k j
k j
k j
k j
k j
k j
k j
k j
k j
k j
k j
k j
k j
eeeeeeek H k X
eeek H
eeeek X
π π π π π π π
π π π
π π π π
−−−−−−−
−−−
−−−−
+++++++=
++−=
++++=
12
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Example5 point DFT
5
8
5
6
5
4
5
2
5
4
5
2
5
8
5
6
5
4
5
2
1816201818][][
2812181618106][][
k j
k j
k j
k j
k j
k j
k j
k j
k j
k j
eeeek H k X
eeeeeek H k X
π π π π
π π π π π π
−−−−
−−−−−−
++++=
+++++++=
y[n] = {6, 10, 18, 16, 18, 12, 8, 2}
y[n] = {6, 10, 18, 16, 18,
12, 8, 2}
Circular convolution!
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Linear and Circular Convolution
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Convolution
Linear/Circular convolutionRelates three signals of interest
The input signal, the output signal, and the
impulse responseProvides a mathematical framework for DSP
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DFT and Convolution
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Exercise 1
By using DFT, determine the responseof the FIR filter with impulseresponse h [n ] = {2, 1, 3} to the input
sequence x [n ] = {3, 2, 2, 1}. (Considerlinear convolution)
If one uses 4 point DFT, what will bethe output?
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Solution: Exercise 1
Input sequence: length =Impulse response: length =
Output sequence: length =
The size of DFT: at leasth [n ] =
x [n ] =Choose _______-point DFT
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Solution: Exercise 1
=][k Y
=
=
=
][][
][][
k H k X
k H k X
y[n] =
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Solution: Exercise 14-pt DFT
=][k Y
=
=
=
][][
][][
k H k X
k H k X
y[n] =
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References
Fundamentals of Signals &SystemsM.J. Roberts, McGraw Hill, 2008
Chapter 13: Discrete-time FourierTransform Analysis of Signals and
Systems
