LN5 Frequency Domain Representation Part2 FourierTransform v2
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Transcript of LN5 Frequency Domain Representation Part2 FourierTransform v2
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7/25/2019 LN5 Frequency Domain Representation Part2 FourierTransform v2
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ilolu 1
FREQUENCY DOMAIN REPRESENTATION OF LTI SYSTEMS
DICRETE TIME FOURIER TRANSFORM (DTFT)
EXISTENCE
MEAN SQUARE CONVERGENCE
FOURIER TRANSFORM OF A CONSTANT SEQUENCEFOURIER TRANSFORM OF A COMPLEX EXPONENTIAL SEQUENCE
FOURIER TRANSFORM OF A SINUSOIDAL SEQUENCE
FOURIER TRANSFORM OF UNIT STEP SEQUENCE
SYMMETRY PROPERTIES OF FOURIER TRANSFORM
REAL SEQUENCES
FOURIER TRANSFORM THEOREMS
FOURIER TRANSFORM PAIRS
IDEAL LOWPASS FILTER (IMPULSE RESPONSE)MOVING AVERAGE FILTER
EXAMPLES
LCCDES AND FREQUENCY RESPONSE
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ilolu 2
DICRETE TIME FOURIER TRANSFORM (DTFT)
The Fourier transform of a sequence x n is defined as
j j nn
X e x n e
If the FT exists (summation converges) the sequence can be obtained from its FT
as
1
2
j j n
x n X e e d
Fourier Transform is periodic with 2.
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ilolu 3
LTI SYSTEMS
The frequency response function, ()is the FT of the impulse response
[]
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EXISTENCE
FT of a sequence []exists, i.e., []=
converges to a continuous function of ,
if []is absolutely summable.(sufficient condition)
Proof: Exercise
[]= []cos sin
=
[] cos
=
[] sin
=
Both sums have to converge
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ilolu 5
All stable LTI systems have frequency response functions.
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ilolu 6
Ex: n
x n a u n
0 01
if 1 or 11
nj n j n j j
jn n
X e a e ae ae aae
cos21
1
sincos1
1
1
1
2
2
2
2
aa
ja
aeeX
j
j
cos1
sintan
sincos10
11
1
a
a
jaa
aeeX jj
-4 -3 -2 -1 0 1 2 3 40.5
1
1.5
2
2.5
3
3.5
4
4.5
5
jX e 8.0a
-4 -3 -2 -1 0 1 2 3 4-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
4
4
jeX
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ilolu 7
MEAN SQUARE CONVERGENCE
Some sequences, which are not absolutely summable but square summable
2
n
x n
can still be represented by Fourier Transform, but
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ilolu 8
Ex: Ideal lowpass filter.
1
0
cj
c
H e
Lets find h n !
[] 1
2
12 ( ) sin
Note that,
[] sin is not absolutely summable!
Then, one may question the Fourier transform of [], sin
= ?
c
c
jH e
1
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ilolu 9
Define =
Even if you take M oscillations do not die to zero.
However 2
lim 0j j
MM
H e H e d
. This is called mean square convergence.
The oscillatory behavior aroundc
is called the Gibbs phenomenon.
MATLAB codeclear all; close all;
precision =0.0001
w = [-pi:precision:pi];
ideaL = zeros(1,length(w));
wc = pi/2;
orta = round(length(w)/2);
ideaL((orta-round(wc/precision)):(orta+round(wc/precision)))=1;
M = 10000;
H = 0;
forn = -M:-1
H = H+(sin(wc*n)/(pi*n))*exp(-i*w*n);
end
forn = 1:M
H = H+(sin(wc*n)/(pi*n))*exp(-i*w*n);
end
H = H+(wc/pi);
plot(w,H); hold on;
plot(w,idea,'r')
grid
detail
-4 -3 -2 -1 0 1 2 3 4-0.2
0
0.2
0.4
0.6
0.8
1
1.2
1M 2
c
-4 -3 -2 -1 0 1 2 3 4-0.2
0
0.2
0.4
0.6
0.8
1
1.2
2c
4M
-4 -3 -2 -1 0 1 2 3 4-0.2
0
0.2
0.4
0.6
0.8
1
1.2
10M 2
c
-4 -3 -2 -1 0 1 2 3 4-0.2
0
0.2
0.4
0.6
0.8
1
1.2
100M 2
c
-4 -3 -2 -1 0 1 2 3 4-0.2
0
0.2
0.4
0.6
0.8
1
1.2
1000M 2
c
-1.571 -1.5708-1.5706-1.5704 -1.5702 -1.57 -1.5698-1.5696 -1.5694
0.5
0.6
0.7
0.8
0.9
1
1.1
10000M
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ilolu 10
FOURIER TRANSFORM OF A CONSTANT SEQUENCE
1x n 2 2jr
X e r
not absolutely
summable
or we can write as
202 jeX
keeping in mind that FT is periodic with 2.
-4 -2 2 4
(2)
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ilolu 11
FOURIER TRANSFORM OF A COMPLEX EXPONENTIAL SEQUENCE
0j nx n e 02 2jr
X e r
not absolutely
summable
or we can write as
202 0
j
eX
keeping in mind that FT is periodic with 2
0+4 0 -2 0 0+2 0+4
(2)
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ilolu 12
FOURIER TRANSFORM OF A SINUSOIDAL SEQUENCE
0 001
cos( )2
j n j nx n n e e
r
jrreX
22
00
not absolutely summable
or 2000
jeX since FT is periodic with 2
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ilolu 13
FOURIER TRANSFORM OF UNIT STEP SEQUENCE
x n u n 1
21
j
jr
X e re
not absolutely summable
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ilolu 14
SYMMETRY PROPERTIES OF FOURIER TRANSFORM
Definitions:
Conjugate symmetric (CS) sequence. x n x n
Conjugate antisymmetric (CaS) sequence. x n x n
Using the above definitions, any sequence can be written as
e o
x n x n x n
where
1
2e
x n x n x n
is the CS part
and
12
ox n x n x n
is the CaS part.
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SYMMETRY PROPERTIES
Fundamental relations
Let j
x n X e
be a FT pair. Then, the following hold:
jx n X e since j j n
n
X e x n e
[] (
) since
(
) []
=
Above yields
jx n X e
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The two relations above also yield:
1)
Re2 2
j j
j
e
X e X ex n x nx n X e
( CS part of jX e )
2)
22Im
**
jjj
o
eXeXeX
nxnxnxj
( CaS part of jX e )
3)
Re2 2
j j
j
e
X e X ex n x nx n X e
( real part of jX e )
Therefore FT of an even seq. is real!
4) Im2 2
j jj
o
X e X ex n x nx n j X e
( imag. part of jX e )
Therefore FT of an odd seq. is purely imaginary!
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ilolu 17
Ex:Let []and []be two real sequences with their DTFTs()and(), respectively.Let
[] [] []Then,
() ( )()Note that
()is NOT the real part of
().
However,
() () ()2 Since
() ()() ()() Similarly,
(
) () ()
2
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ilolu 18
Ex: (contd)
[]: [1 1]
[]: [1 1]
[]: [1 1 ]
() 1
(
) 1
() 1 () 1
() ()2 1 () ()2
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ilolu 19
REAL SEQUENCES
Based on the above relations, for real sequences ( x n x n ):
jj
eXeX
, conjugate symmetrywhich implies
Magnitude is even.... jj eXeX
Phase is odd... jj eXeX Real part is even.. jRjR eXeX Imaginary part is odd......................... jIjI eXeX
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ilolu 20
Verification by an example
Ex: nx n a u n 1
11
j
jX e a
ae
a)FT is conjugate symmetric:
1
11
j j
jX e X e a
ae
b)Real part of FT is an even function:
21 cos
Re1 2 cos
j j j
R R
aX e X e X e
a a
c) jR eX is the FT of nxe :
2
1 cos
1 2 cos
j
R
aX e
a a
1
2
n n
ex n a u n a u n
d)Imaginary part of Ft is an odd function.
2sin
Im1 2 cos
j j j
I I
aX e X e X e
a a
n
e
x n
1 2
1
n
x n
1 2
01
10 a
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ilolu 21
e) jI eX is the FT of nxo :
2
sin
1 2 cos
j
I
aX e
a a
1
2
n n
ox n a u n a u n
f) Magnitude of FT is an even function:
2
1
1 2 cos
j jX e X e
a a
g)Phase of FT is an odd function
jjeX
a
aeX
cos1
sintan
1
n
o
x n
1 2
1
4
-4 -3 -2 -1 0 1 2 3 40.5
1
1.5
2
2.5
3
3.5
4
4.5
5
jX e
8.0a
-4 -3 -2 -1 0 1 2 3 4-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
4
4
jeX
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ilolu 22
FOURIER TRANSFORM THEOREMS
1,F
j j jx n X e x n F X e X e F x n
1) F
j jax n by n aX e bY e Linearity
2) 00F
j n jx n n e X e
Time-shift
3)
00
Fjj n
e x n X e
Freq. shift
4) F
jx n X e
Time reversal
5) jF dX e
n x n jd
Differentiation in frequency
domain
6) F
j jx n y n X e Y e
Convolution
7) 1
2
Fj j jy n x n w n Y e X e W e d
Modulation, windowing
Parsevalstheorem (prove as an exercise)
8) 22 1
2
j
n
x n X e d
energy of the signal
2
jX e
is called the energy density spectrum.
9) 1
2
j j
n
x n y n X e Y e d
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Proof of (6):
Let
[] [][ ]= [] []
() ?() [][ ]=
=
[] [ ]
=
=
[] []=
=
()()
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ilolu 24
Proof of (8) using (6):
Let
[] []
[ ]
= []
[]
() ( )() ( )
[0] []
[]
= |[]|
=
[0] 12 ()
12 ()
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ilolu 25
Proof of (9):
[]
[]
=
12 ()
[]=
1
2 (
)
[]
= ()
12 ()()
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ilolu 26
FOURIER TRANSFORM PAIRS
n 1
0n n 0j ne
1 n 2 2jr
X e r
nx n a u n 1a 1
1 jae
x n u n 1
21
jk
ke
nn a u n 1a
2
1
j
j
ae
ae
1 nn a u n 1a
n n
a u n a u n
2
1
1 jae
2 12
n
n n a u n
3
1
1 jae
1 !1
1 ! !
nn k
a u nk n
11 2 1
1 !
nn k n k n a u n
k
1
1k
jae
1
sin 1sin
n
p
p
r n u n
1r
2 2
1
1 2 cos j j
pr e r e
show using
nx n a u n
sincn
n
1,
0,
cj
c
X e
1, 0
0, otherwise
n Mx n
/ 2
sin 1 / 2
sin / 2
j MM
e
0j n
e 02 2
r
r
0cos n
r
j
r
jrere 22
00
n0
sin
r
j
r
jrejrej 22
00
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ilolu 27
Ex: IDEAL LOWPASS FILTER (IMPULSE RESPONSE)
Plots of sin sequence
They are infinitely long sequences in
< <
Plots are arbitrarily in 2 0 < < 2 0
-20 -15 -10 -5 0 5 10 15 20-0.05
0
0.05
0.1
0.15
0.2
-20 -15 -10 -5 0 5 10 15 20-0.1
-0.05
0
0.05
0.1
0.15
0.2
0.25
-20 -15 -10 -5 0 5 10 15 20-0.2
-0.1
0
0.1
0.2
0.3
0.4
0.5
sin cn
n
4c
c
2c
n
n
n
6c
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ilolu 28
Ex: MOVING AVERAGE FILTER
Magnitude
Phase
DTFT functions are plotted in or in 20
-4 -3 -2 -1 0 1 2 3 40
1
2
3
4
5
6
-4 -3 -2 -1 0 1 2 3 40
2
4
6
8
10
12
-4 -3 -2 -1 0 1 2 3 4-3
-2
-1
0
1
2
3
5M
-4 -3 -2 -1 0 1 2 3 4-3
-2
-1
0
1
2
3
10M
/ 2sin 1 / 2
sin / 2
j MM
e
5M
1M
2
1M
10M
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ilolu 29
M = 1
1
12
h n n n 21
1 cos2 2
jj j
H e e e
1
12
y n x n x n
M = 4
1
1 2 3 45
y n x n x n x n x n x n
1
1 2 3 45
h n n n n n n
2 3 4
2
11
5
11 2 cos 2 cos 2
5
j j j j j
j
H e e e e e
e
or
2 3 41
15
j j j j jH e e e e e
5 5 5
2 2 2
5
2 2 2
1 1 1
5 1 5
j j j
j
jj j j
e e ee
ee e e
2
5sin
1 2
5sin
2
je
-4 -3 -2 -1 0 1 2 3 40
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
frequency
|H|
-4 -3 -2 -1 0 1 2 3 40
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
frequency
|H|
2
5
4
5
2
5
4
5
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ilolu 30
Ex: Express 1j
X e , 2 jX e , in terms of jX e , the DTFT of x n .
1 7 7x n x n x n x n x n 7
1
j j j jX e X e e X e
2 4x n x n x n 4
2
j j j j
X e X e e X e
-2 0 2 4 6 8 100
1
2
3
4
1x n
-2 0 2 4 6 8 100
1
2
3
4
2x n
-2 0 2 4 60
0.5
1
1.5
2
2.5
3
3.5
4
x n
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ilolu 31
One can also write as
() ( ) 7() () 7() since []is real()() 7()
()7 7() 7()
2 Re {()+7}()7() ( ) ( 1 4)
()( ) ()2cos2 2 c o s2 ()()
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ilolu 32
Ex: What is the inverse DTFT, [], of () 38?From the table 1 nn a u n
2
1
1 jae
for 1a
2 1
18
n
n u n 22
11
8
je
2
3
12 3
8
n
n u n
3
21
8
2
1
j
j
e
e
Why does the high frequency gain of MA filter decreaseas M increases?
Comment on the above illustrations.
-5 0 5 10 15 20-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
-5 0 5 10 15 20-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
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LCCDEs AND FREQUENCY RESPONSE
[] [] [] () ( )() () ()
[ ]= [ ]
= ()=
()
=
() = ()
=
() () = =
()