Download - DTFT & zT The z Transform
T. Y. Choi, ECE, Ajou University DTFT & zT 47
DTFT & zT zT
| | 1 ( ) ( ) | j
j
X z ez R X e X z
C is a closed counterclockwise
contour in the complex z-plane
lying entirely within the ROC RX
The z Transform
z-Transform
2
( ) [ ]
1[ ] ( )
2
j j n
n
j j n
X e x n e
x n X e e d
1
( ) [ ] ,
1[ ] ( )
2
n
X
n
n
c
X z x n z R
x n X z z dzj
DTFT
jz e jdz je d
T. Y. Choi, ECE, Ajou University DTFT & zT 48
DTFT & zT zT
0
1
1
1
( )
1 ( )
1
1, | | | |
1
n n
n
X z a z
az
a z
z aa z
Example: The z transform of [ ] [ ]nx n a u n
1 1| | | | 1, ( ) 0az
If az az
ROC (Region of Convergence)
jIm
Re| |a
Unit circle of radius 1
zT: Circular disk, ring
LT: Vertical line, strip
If |a|<1, |z|=1 is included in ROC
( ) ( ) |
1
1
j
j
z e
j
X e X z
ae
T. Y. Choi, ECE, Ajou University DTFT & zT 49
DTFT & zT zT
( ) [ ]
[ ]( )
[ ]
jj
n
z re nz re
j n
n
n j n
n
X z x n z
x n re
x n r e
1. Relationship between the DTFT and the z Transform
Proof:
0 2
DTFT [ ] nx n r
( ) ( ) [ ]j
j n
z reX z X re x n r
jIm
Re
jz re
(1) If a circle of radius r is included within the ROC RX in the z
plane, the DTFT of exists along the circle, i.e., [ ] nx n r
T. Y. Choi, ECE, Ajou University DTFT & zT 50
DTFT & zT zT
( [ ] ) [ ] [ ]( )n n j n j n
n nx n r x n r e x n re
Proof:
( ) ( [ ] ) , | |j
n
re zX z x n r z r
0 2
[ ] nx n r
( [ ] ) [ ] ( ), | |j
n n
nre zx n r x n z X z z r
( ), | |X z z r
(2) Reversely, if the DTFT of exists, the zT of
exists along a circle of radius r and that is equivalent to
the substitution of in the DTFT of , i.e.,
[ ] nx n r [ ]x n
jre z
( )X z
( )X z
[ ] nx n r
jIm
Re
jz re
T. Y. Choi, ECE, Ajou University DTFT & zT 51
DTFT & zT zT
If (| | 1) , ( ) ( ) | j
j
X z ez R X e X z
jIm
Re
jz e
| | 1z
1
( ) ( ) | , | | 1j
j
e zX z X e z
(3) If the unit circle (|z|=1) is included within the ROC RX in the
z plane, the DTFT of x[n] exists along the circle, i.e., ( )jX e
Reversely, if the DTFT exists, must exist along
the unit circle and , i.e.,
( )jX e ( )X z
( ) ( ) | j
j
e zX z X e
T. Y. Choi, ECE, Ajou University DTFT & zT 52
DTFT & zT zT
Example: DTFT and zT of [ ] 2 [ ]nx n u n
0 2
10
1( ) 2 , | | 2
1 2
n n
nX z z z
z
(2) zT:
0( ) 2j n j n
nX e e
(1) DTFT: diverges
[ ] 2 [ ]n n nx n r r u n (3) DTFT of
1
( ) ( [ ] )
1, | | 2
1 2
j
n
re zX z x n r
z rz
[ ] nx n r ( ), | |X z z r
If r>2, the DTFT of exists. Therefore,
[ ] nx n r
jIm
Re
2 r1
1
10
1
1 (2 )( [ ] ) (2 )
1 2
1, 2
1 2
jn j n
jn
j
r ex n r r e
r e
rr e
jz e
T. Y. Choi, ECE, Ajou University DTFT & zT 53
DTFT & zT zT
1
10
1
1 ( )( )
1
1, | | 1
1
n n
n
azX z a z
a z
a
a z z
jIm
Re
| | | |z a
z a
( | | 1)a
-1 0 1 2 n
1
-1 0 1 2 n
( | | 1 )a 1
If |a|<1, ROC includes |z|=1 and then we have the DTFT from the zT as
1( ) ( ) |
1j
j
jz eX e X z
ae
Proof:
elsewhere (|a|>1), |z|=1 is not included in ROC and the DTFT diverges.
If a=1, i.e., x[n]=u[n], X(z) = ?
1
1[ ] , | | | |
1
zna u n z aa z
(1) Right sided signal: [ ] [ ]nx n a u n
RX is the open exterior of a circle lying a pole
2. zTs of right, left, and double sided sequences
T. Y. Choi, ECE, Ajou University DTFT & zT 54
DTFT & zT zT
1
1
1
1
( )
, | | 11
n n n n
n n
X z a z a z
a z z
a z a
jIm
Re
| | | |z a
z a
ROC is the open interior of a circle lying a pole
( | | 1)a
-3 -2 -1 0 1 n
a-1
-3 -2 -1 0 1 n
( | | 1 )a
a-1
If |a|>1, ROC includes |z|=1 and then we have the DTFT from the zT as
1( ) ( ) |
1jw
j
jz eX e X z
ae
Proof:
elsewhere (|a|<1), |z|=1 is not included in ROC and the DTFT diverges.
If a=1, i.e., x[n]=u[-n -1]., X(z) = ?
1
1[ 1] , | | | |
1
zna u n z aa z
(2) Left sided signal: [ ] [ 1]nx n a u n
T. Y. Choi, ECE, Ajou University DTFT & zT 55
DTFT & zT
-3 -2 -1 0 1 2 3 n
1
2 n
-3 -2 -1 0 1 2 3 n
1
2n
1
1 1
12 [ 1] , | | 2
1 2
zn u n zz
1
1 1
12 [ ] , | | 2
1 2
zn u n zz
1
12 [ 1] , | | 2
1 2
zn u n zz
1
12 [ ] , | | 2
1 2
zn u n zz
Right/left sided
Right/left sided
± sign, ROC(outer/inner)
T. Y. Choi, ECE, Ajou University DTFT & zT 56
DTFT & zT zT
2 [ ] 3 [ 1]n nu n u n Ex-1.
2 [ ] 3 [ 1]n nu n u n
jIm
Re
| | | | | |a z b
z a
z b
Common ROC is an open ring
between two adjacent poles.
1/2<|z|<3
2<|z|<3
0 n
2n
3n
1
Ex-2.
0 n
1 ROC
Ex-3. 3 3 ( [ ] [ 1])n n u n u n 3<|z|<3 NO common ROC ?
[ ] [ ] [ 1]n ny n a u n b u n (3) Double sided signal:
1 1
1 1( ) , | | | | | |
1 1Y z a z b
az bz
T. Y. Choi, ECE, Ajou University DTFT & zT 57
DTFT & zT zT
3. Confliction between DTFT and zT on the circle lying a pole
It is not obvious whether the zT can be expressed on the circle lying a pole z=1.
Which is the superset? The DTFT or the ZT ?
The DTFT of u[n] is expressed even with an impulse train.
In the same way, its z transform could be expressed.
jz e
jz e
is not allowed since |z|>1.
Substitution with
DTFT 1[ ] ( 2 )
1 jku n k
e
1
1[ ] ( 1) , | | 1
1
zu n z zz
?
( 1)je
See Property of Delta function
1
1[ ] , | | 1
1
zu n zz
T. Y. Choi, ECE, Ajou University DTFT & zT 58
DTFT & zT
( ) diric( , 2 1)N
j j n
n N
X e e N
DTFT: 1 0 1N N n
1
(2 1) 1 ( 1/2) ( 1/2)
1 1/2 1/2
(1 )( ) ,
1 1
N N N N N NNn
n N
z z z z z zX z z z
z z z z
z-T:
The limiting value 2N+1 of X(z) as z approaches 1 is finite unless N is infinite. Thus if N is finite, we regard there is a pole-zero cancellation at z=1.
(2 1)/2 (2 1)/2
/2 /2
sin((2 1) / 2)( ) diric( , 2 1)
sin( / 2)j
j n j n
j jz e
e e NX z N
e e
(0<|z|<∞)
DTFT and zT of [ ] [ ]N
k Nx n n k
lim ( ) diric( , ) 2 ( 2 ) 2 ( 1)j j
odd nNX e n e
lim ( ) lim ( ) 2 ( 1), | | 1j
j
N N e z
X z X e z z
(Around the unit circle)
As N→∞, lim [ ] 1N
x n
(dc sequence 1)
T. Y. Choi, ECE, Ajou University DTFT & zT 59
DTFT & zT zT
1( )
2X f
a j f
1 1( )
2 1
aT jj
aT j
e eX e
e e
4. Summary of X(f), Xs(f), DTFT, and zT
1( ) ( ), 0atx t e u t a
t0
12
[ ] ( )
[ ] [ ]
anT
anT
x n e u n
e u n n
12
0
( ) ( ) ( )anT
s
n
x t t e t nT
n
12
0 t nT1 2 3 4
( 2 )
0
2
2
1( )
2
1 1
2 1
aT j Tf n
s
n
aT j Tf
aT j Tf
X f e
e e
e e
1
1
1 1( ) , | | 1
2 1
aTaT
aT
e zX z z e
e z
1
1[ ] , | | 1
1
zaTn aT
aTe u n z e
e z
2j Tfe z
2 Tf
je z
n
12
0 1 2 3 4
Red dots
+ 1/2
With repetition of X(f)
T. Y. Choi, ECE, Ajou University DTFT & zT 60
DTFT & zT zT
Red
[ ]anTe u n
1
10
1
1 ( )[ ]
1
1, | |
1
aTzaTn aTn n
aTn
aT
aT
e ze u n e z
e z
z ee z
n
12
0 1 2 3 4
( ) ( ), 0atx t e u t a
( )anTe u n
10
1
1
1 1 1( )
2 2 1
1 1, | |
2 1
zaTn aTn n
aTn
aTaT
aT
e u n e ze z
e zz e
e z
Blue
T. Y. Choi, ECE, Ajou University DTFT & zT 61
DTFT & zT zT
x[n - N]
- a
Xs(f)
xs(t)
s=j2f
(1)
Xs(s)
xs(t)
esT = z
(4)
X(ej)
x[n]
z = ei
(2)
X(z)
x[n]
2 fT =
(3)
ej2 fT = z(6)
sT=j
(5)
( ) [ ] ( )s
n
x t x n t nT
( ) [ ] sTn
s
n
X s x n e
2( ) [ ] j nTf
s
n
X f x n e
( ) [ ]j j n
n
X e x n e
( ) [ ]n
n
X z x n z
[ ] ( )x n x nT
( )X fIf exists, all of do too. ( ), ( ), ( ), ( ), ( )j
s sX s X f X s X e X z
T. Y. Choi, ECE, Ajou University DTFT & zT 62
DTFT & zT zT
Basically, the discrete-time system does a weighted sum of shifted input sequences. Thus the time shift is the fundamental algorithm in the discrete-time system analysis.
( )[ ] [ ] ( )o on n nn
o
n n
x n n z x n z X z z Proof:
Verification:
(1) Time shift [ ] ( ),onz
o Xx n n z X z R
[ ] ( ) ojnj
ox n n X e e
[ ] ( )on
ox n n z X z
5. Properties of the z transform Pole-zero cancellation
je z
1
1
1
1[ ] ( 1) , | | 1
1
[ 1] ( 1) , | | 11
u n z zz
zu n z z
z
1[ ] ( 1)
1
j
ju n e
e
[ 1] ( 1) ( 1)1 1
j jj j j
j j
e eu n e e e
e e
T. Y. Choi, ECE, Ajou University DTFT & zT 63
DTFT & zT zT
1
1
[ ] [ ] ( )
( ), | |
n n n
o o
n n
o o X
z x n z x n z z
X z z z R
(2) Modulation 1[ ] ( ), | |zn
o o o Xz x n X z z z R
( )[ ] ( )o oj n j wDTFTe x n X e
1[ ] ( ), | |zn
o o o Xz x n X z z z R
oj
oe z
je z
Proof:
Verification:
( )
( 1) [ ] ( ),
( 1) [ ] ( ) ( )
n
X
n j j
x n X z R
x n X e X e
( 1)n j n n
oz e
T. Y. Choi, ECE, Ajou University DTFT & zT 64
DTFT & zT
1[ ] ( ), | |zn
o o o Xz x n X z z z R
1
1
1
1
1
1 2
1
1 2
1[ ] , | | 1
1
1[ ] , | | | |
1
1[ ] , | | 1
1
1[ ] , | | 1
1
1 cos( )cos( ) [ ] , | | 1
1 2cos( )
sin( )sin( ) [ ] , | | 1
1 2cos( )
o
o
o
o
n
j n
j
j n
j
oo
o
oo
o
u n zz
a u n z aaz
e u n ze z
e u n ze z
zn u n z
z z
zn u n z
z z
T. Y. Choi, ECE, Ajou University DTFT & zT 65
DTFT & zT zT
(3) Convolution sum [ ]* [ ] ( ) ( ),z
X Yx n y n X z Y z R R R
[ ]* [ ] ( ) ( )DTFT j jx n y n X e Y e
[ ]* [ ] ( ) ( ),z
X Yx n y n X z Y z R R R je z
Proof:
Verification:
[ ] [ ]* [ ] [ ] [ ]
( ) [ ] [ ] [ ] [ ]
( ) [ ] ( ) ( )
m
n n
n m m n
m
m
c n x n y n x m y n m
C z x m y n m z x m y n m z
Y z x m z Y z X z
T. Y. Choi, ECE, Ajou University DTFT & zT 66
DTFT & zT zT
( ) [ ]n
n
X z x n z
1' 1( ) [ ] [ ]n n
n n
X z n x n z z n x n z
'( ) [ ]n
n
z X z n x n z
Differentiating the above defining sum with respect to z on both sides,
( )[ ]
jDTFT dX e
n x n jd
'[ ] ( ),z
Xn x n z X z R
( ) ( ) ( )
/
jdX e dX z dX zjz
d dz jz dz
(4) Differentiation: '[ ] ( ),z
Xn x n z X z R
je z
Proof:
Verification:
,jje d dz jzd dz
' 1
1 1 2
1[ ] , | | | |
1 (1 )
n azna u n z z a
az az
T. Y. Choi, ECE, Ajou University DTFT & zT 67
DTFT & zT zT
2
1
( ) [ ] , 1 | | 2n
n
n n
X z x n z r z r
[ ] ( )DTFT jx n X e
1 1[ ] ( ),z
Xx n X z R
1 21 1
2 1
1
[ ] [ ] ( ), 1 | | 2
1 1( ), | |
2 1
n nn n
zn n n n
x n z x n z X z r r
X z zr r
(5) Folding: 1 1[ ] ( ),z
Xx n X z R
je z
1
1
1
[ 1] ( 1) , | | 11
1[ 1] ( 1) , | | 1
1
zu n z z
z
u n z zz
Proof:
Verification:
T. Y. Choi, ECE, Ajou University DTFT & zT 68
DTFT & zT zT
Useful z transform pairs
[ ]N
k N
n k
1
1
1
1 2
1
1 2
1
1 2
(2 1)/2 (2 1)/2
2 1 1/2
1[ ] , | | | |
1
1[ ] , | | 1
1
1 cos( )cos( ) [ ] , | | 1
1 2 cos( )
sin( )sin( ) [ ] , | | 1
1 2 cos( )
[ ] , | | | |(1 )
o
o
n
j n
j
oo
o
oo
o
n
N N
nN
a u n z aaz
e u n ze z
zn u n z
z z
zn u n z
z z
azna u n z a
az
z z
z z
1/2, 0 | |z
or z
T. Y. Choi, ECE, Ajou University DTFT & zT 69
DTFT & zT zT
6. The Inverse z Transform
( ) [ ] , {| | }n
X
n
X z x n z z r R
When |z| = r is contained in the ROC, i.e.,
( ) ( ) [ ] DTFT( [ ] )i
i n j n n
z ren
X z X re x n r e x n r
2
| |
1[ ] ( )
2
1( )
2
n j j n
n n
z r
x n r X re e d
dzX z z r
jz
, (0 2 )
,
j
j
re z
dzjre d dz d
jz
Taking the IDTFT,
Therefore
1
| |
1[ ] ( )
2
n
z r
x n X z z dzj
|z|=r is a closed counterclockwise contour in the complex z-plane lying entirely within the ROC RX
T. Y. Choi, ECE, Ajou University DTFT & zT 70
DTFT & zT zT
7. DLTI System analysis in the z transform domain
( )jX e DLTI h[n]
( ) ( ) ( )j j jY e H e X e
[ ]x n [ ] [ ]* [ ]y n x n h n
( )X z ( ) ( ) ( )Y z H z X z
Causality: [ ] 0 for 0h n n
20
[1] [2]( ) [ ] [0]n
n
h hH z h n z h
z z
Right-sided signal
H(z) does not contain any positive power of z. The system is causal if and only if
Stability: Absolutely summable h[n] leads to the convergence of H(ejω). The system is stable if and only if
the RH is the exterior of a circle including infinity.
the RH contains the unit circle.
|z|>r
T. Y. Choi, ECE, Ajou University DTFT & zT 71
DTFT & zT zT
Causality and Stability depending on possible ROCs
{| | 1} Xz R
jIm
Re1
{| | 1} Xz R
jIm
Re1
jIm
Re1
jIm
Re1
| | 0.7z 0.7 | | 2z
| | 0.7z | | 1.5z
Stable, causal Stable, anticausal
Unstable, anticausal Unstable, causal
Causal: ROC contains
a unit circle including infinity
T. Y. Choi, ECE, Ajou University DTFT & zT 72
DTFT & zT zT
0
1 0
[ ] [ ] [ ], ( 0)N M
k k
k k
y n a y n k b x n k b
Transfer Function Characterization of LTI Difference Systems
1 0
( )(1 ) ( )N M
j jk j jk
k k
k k
Y e a e X e b e
0
1
( )
1
Mjk
kj k
Njk
k
k
b e
H e
a e
For a given difference equation:
Taking the DTFT or the zT,
The transfer function is then given by the DTFT (zT) of the output divided by the DTFT (zT) of the input:
1 0
( )(1 ) ( )N M
k k
k k
k k
Y z a z X z b z
0
1
( )
1
Mk
k
k
Nk
k
k
b z
H z
a z
T. Y. Choi, ECE, Ajou University DTFT & zT 73
DTFT & zT zT
x[n] y[n]
y[n-1]
z-1
-a
+
+
한 표본지연
[ ] [ 1] [ ]y n ay n x n
Design of the first-order system
1
1( ) , | | | |
1H z z a
az
[ ] ( ) [ ]nh n a u n
This is the IIR (infinite impulse response) system.
or, the unit circle is not contained in the ROC and therefore the system is unstable
If |a|>1,
the system may be saturated and so the system is unstable.
T. Y. Choi, ECE, Ajou University DTFT & zT 74
DTFT & zT zT
1 2[ ] [ 1] [ 2] [ ]y n a y n a y n x n
1 2
1 2
1( )
1H z
a z a z
Design of the second-order system
[ ]y n[ ]x n
1z
1z
-a1
-a2
[ 1]y n
[ 2]y n
T. Y. Choi, ECE, Ajou University DTFT & zT 75
DTFT & zT zT
1 2
1 2
1 2
1 2
( )1
M
o M
N
N
b b z b z b zH z
a z a z a z
1
1( ) ( )M
o MY z b b z b z W z
1
1( ) 1 ( )N
NW z a z a z X z
1[ ] [ ] [ 1] [ ]o My n b w n b w n b w n M
1[ ] [ 1] [ ] [ ]Nw n a w n a w n N x n
Canonical form [ ]y n[ ]x n
1z
-a1
[ ]w n
1z
-a2
1z
-aN
-aN-1
b0
b1
b2
bM
(1) y[n] with w[n] (Forward):
(2) x[n] with w[n] (Feedback):
---- (A)
---- (B)
H(z)=(A)/(B)
Numerator polynomial of H(z)
Denominator polynomial of H(z)
T. Y. Choi, ECE, Ajou University DTFT & zT 76
DTFT & zT zT
[ ] 0.9 [ 4] [ ]y n y n x n
41/ 4
4 4
1( ) , | | 0.9
1 0.9 0.9
zH z z
z z
Example:
[ ]y n[ ]x n
4z
1/4 /2 40.9 ( 0,1,2,3) 0.9jk
kz e k z
4
4( )
0.9
jj
j
eH e
e
0.9
Comb filter (빗살여파기)
H(z) has 4 poles located at intervals of π/2 on the circle of radius 0.91/4 such as
T. Y. Choi, ECE, Ajou University DTFT & zT 77
DTFT & zT zT
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 10
2
4
6
8
10
12Comb filter: y[n] - 0.9 y[n-4] = x[n]
in units
4
.974r 1r
[ ] 0.9 [ 4] [ ]y n y n x n
Pole-zero diagram
| ( )|jH e
1/40.9 0.974
T. Y. Choi, ECE, Ajou University DTFT & zT 78
DTFT & zT
Impulse train of period N (N=3)
8. Sampling Techniques
Upsampling
Downsampling
Impulse Train Sampling
… … n
3[ ]x n
… …
[ ]x n
n
… …
3[ ]x n
n
… …
3[ ] [ ]x n i n
n
…
0 3 6 9
… 3 3[ ] [ ]i n i n
n
3[ ] [ ]x n i n
3[ ]x n
3[ ]x n
3 3[ ] [ ]i n i n
T. Y. Choi, ECE, Ajou University DTFT & zT 79
DTFT & zT
(1) Upsampling
[ ] [ ] [ ] ( )j n j Nm jN
Nn mN m
nx n x e x m e X e
N
[ ] [ ] [ ] ( )n Nm N
Nn mN m
nx n x z x m z X z
N
Expansion in Time Insertion of (N-1) zeros (No loss in signal) Compression in Frequency
… … n
[ ]x n
… … n
3[ ]x n
2 0
A )( jeX
max
3
3( ) ( )j jX e X e
2
0
A
max
3
2
3
4
3
,[ / ],
[ ] ( ), ( )0,
jN N
N
x n N n mNx n X e X z
n mN
T. Y. Choi, ECE, Ajou University DTFT & zT 80
DTFT & zT
Example of upsampling: Periodic impulse train of period N
3
3
2 2( ) ( ) ( )
3 3
j j
k
kI e I e
3 3[ ] [ ] [ 3 ]
k
i n i n n k
[ ] [ ]k
i n n k
2
3/43/2
2
3
0
… …
3( )jI e
… n
0 1 2 3 4 5 6
…
… n
0 3 6 9
… 3[ ]i n
[ ]i n
( ) [ ]
diric( , ) 2 ( 2 )
j j n j n
n k n
odd
k
I e n k e e
k
2 2[ ] [ ] ( ) ( )j
N N
k k
ki n n kN I e
N N
Impulse train
3 3
3( ) diric(3 , ) ( )j j k j
odd
k
I e e I e
Or, direct evaluation:
T. Y. Choi, ECE, Ajou University DTFT & zT 81
DTFT & zT
(2) Impulse training sampling
Shift 2k/3, k=0,1,2
Gain 1/3
… … n
[ ]x n
… … n
3[ ] [ ]x n i n
21
( )
0
1[ ] [ ] ( )
k
N
Nj
N
k
x n i n X eN
23
23
( )132 2
2 212 3 32
2( ) 21
3 320
2( )1
3
0
( ) ( ) ( )
( ) ( )
( ) ( )
( )
k
k
j j j
j k
k
j k
k
j
k
Y e X e I e d
X e d
X e d
X e
Let 3[ ] [ ] [ ]y n x n i n
Periodic convolution
For every k=0,1,2, unit area
T. Y. Choi, ECE, Ajou University DTFT & zT 82
DTFT & zT
(3) Downsampling (decimation):
Expansion in Frequency
Upsampling
3
3
[ ] [ ]
( ) ( )j j
y n d n
Y e D e
Compression in Time 21
( )
0
1[ ] [ ] ( )
k
N
Nj
Nk
x n x Nn X eN
… … n
3[ ] [ ] [3 ]d n x n x n
… … n
[ ]x n
… … n
3[ ] [ ] [ ]y n x n i n Impulse train sampling 2
3
2( )
0
1( ) ( )
3
kjj
k
Y e X e
2 2
3 3
2 1( ) ( )
0 0
1 1( ) ( ) ( ) ( )
3
k k
N
Nj j jj
k k
D e Y e X e X eN
T. Y. Choi, ECE, Ajou University DTFT & zT 83
DTFT & zT
Spectrum of downsampled sequence
2 0
A )( jeX
21
0
1( ) ( )
kN
Njj
Nk
X e X eN
20
A/N
max
maxN
max
N
To prevent aliasing,
21( )
0
1[ ] [ ] ( )
k
N
Nj
Nk
x n x Nn X eN
T. Y. Choi, ECE, Ajou University DTFT & zT 84
DTFT & zT
Impulse sampling, upsampling, downsampling
… …
n
… …
… …
3[ ]x n
[ ]x n
… …
3[ ]x n
3[ ] [ ] [ ]y n x n i n
2 3/43/2
/ 3A
0
BW
2
/ 3A
0
3BW
2
)( jeXA
0
BW
21( )
0
1( ) ( )
kN j wj N
k
Y e X eN
2 3/43/2
A
0
/ 3BW( ) ( )j jN
NX e X e
21
0
1( ) ( ) ( )
k
N N
Nj jj
Nk
X e Y e X eN
Bandwidth
T. Y. Choi, ECE, Ajou University DTFT & zT 85
DTFT & zT zT
21
( ) ( )k
jj N
Nk
X e X eN
(4) Reconstruction of decimated signals [ ]
Nx n
( )jH e N N
[ ]N
x n
[ ]x n [ ]y n[ ]uy n
2( )1
( ) ( ) ( )k
jj jN N
u Nk
Y e X e X eN
( ) ( )* ( 2 )2
j
k
NH e N k
N
N
N
( )jH e
( ) ( ) ( ) ( )j j j j
uY e Y e H e X e
Assume that a signal x[n] was downsampled to without aliasing.
That is, , or x[n] was N times over sampled.
Then x[n] can be recovered from by first upsampling by N followed by an ILPF of cutoff frequency .
N max
[ ]N
x n
c N
[ ]N
x n
N max
Perfect reconstruction
T. Y. Choi, ECE, Ajou University DTFT & zT 86
DTFT & zT
2
)( jeX
3BW ( )j
NX e
0
A
2
0
max 4
maxN
/A N
No aliasing.
max4 3
f
( )X f
0
AT
1khzmf
1 8khzs Tf
max 42 mTf
( )j
UY e
2
0
/A N
max 2
N
4
N
2
N
2
( )jH e
0max
N
N
(N=3)
T. Y. Choi, ECE, Ajou University DTFT & zT 87
DTFT & zT zT
[ ] [ ] [ ]y n ay n N x n
1/1( ) , | | | |
1
N
N NH z z a
az
[ ]y n[ ]x n
Nz
a
1 1
1( ) ( ) , | | | |
1H z H z z a
az
(N=1):
1[ ] [ ] [ ]nh n h n a u n
1( ) ( ) ( )N N
NH z H z H z (N>1):
1[ ] [ ] [ ]N N N
h n h n h n
Upsampled h1[n] by N
(insert N-1 zeros between samples)
(1) Unit sample response
→ hN = 1, z, a, z, a2, z, a3, . . .
z= (N-1) zeros: z= 0,0,. . .,0
→ h = 1, a, a2, a3, . . .
9. Feedback comb filter analysis
T. Y. Choi, ECE, Ajou University DTFT & zT 88
DTFT & zT
(2) Unit step response (N=1)
n 0 1 2 3 …
h[n] 1 a a2 a3 …
h[n-1] 1 a a2 …
h[n-2] 1 a …
… …
s[n] 1 1+a 1+a+a2 1+a+a2+a3 …
Running sum
Sum of geometric series
10 0
1 0 0
2
1[ ] [ ] [ ]* [ ] [ ] [ ]
1
[ ] [ 1] [ 2]
(1 ) [ ]
nn n k
k k
n
as n s n h n u n h k a u n
a
h n h n h n
a a a u n
T. Y. Choi, ECE, Ajou University DTFT & zT 89
DTFT & zT
Unit step response (N>1)
Ex.: (N=2)
n 0 1 2 3 4 5 …
h2[n] 1 0 a 0 a2 0 …
h2[n-1] 1 0 a 0 a2 …
h2[n-2] 1 0 a 0 …
h2[n-3] 1 0 a …
h2[n-4] 1 0 …
…
sN[n] 1 1 1+a 1+a 1+a+a2 … …
[ ] [ ]*ones(1, )Ns k s k N : N times repetition of s[k]
0
0
[ ] [ ]* [ ] [ ] [ 1] [ 2]
[ ]
[0] [1] [2] [ ]
N N N N N
n
Nk
N N N N
s n h n u n h n h n h n
h k
s s s s n
T. Y. Choi, ECE, Ajou University DTFT & zT 90
DTFT & zT zT
1 1
1( ) ( ) , | | | |
1H z H z z a
az
(N=1):
(3) Unit step response in zT domain
Sum of geometric series
1
1 0
1[ ] [ ] [ ] [ ]
1
nn k
k
as n s n u n a u n
a
1 1 1
11 1 1
1 2
1 1( ) ( )
1 1
1
1 1
1
1 (1 )
a
S z S zz az
a
z az
a z az
n 0 1 2 3 …
s[n] 1 1+a 1+a+a2 1+a+a2 +a3 …
1[ ]s n[ ]n
1z
1z
1+a
-a
T. Y. Choi, ECE, Ajou University DTFT & zT 91
DTFT & zT zT 1/1
( ) , | | | |1
N
N NH z z a
az
(N>1):
[ ] [ ] [ 1] [ 2] [ 1]N N N N Ns n s n s n s n s n N
1 1
1 2 ( 1)
1
1 1 1 1 1( )
1 1 1 1 1
1( ) 1 ( )
1
N
N N N N
NN N N
zS z
z az z z az
zS z z z z S z
z
Unit step response in zT domain
1. S↑N[n]: Upsampling s[n] by N
2. Sum of all shifted S↑N[n-k] for k=0:N-1, i.e.,
Replacing N-1 zeros occurred by upsampling with S↑N[n].
n 0 1 2 3 …
s[n] 1 1+a 1+a+a2 1+a+a2 +a3 …
s↑2[n] 1 0 1+a 0 1+a+a2 0 …
s↑2[n-1] 1 0 1+a 0 1+a+a2 …
s2[n] 1 1 1+a 1+a 1+a+a2 1+a+a2 …
EX.: (N=2)
T. Y. Choi, ECE, Ajou University DTFT & zT 92
DTFT & zT
0 5 10 15 20 250
0.2
0.4
0.6
0.8
1
Unit sample response h[n]
(1) y[n] - 0.8y[n - 1] = x[n]
0 2 4 6 8 100
2
4
6
5
Unit step response uo[n]
0 5 10 15 20 250
0.2
0.4
0.6
0.8
1
Unit sample response h[n]
(2) y[n] - 0.8y[n - 4] = x[n]
0 5 10 15 20 25 30 35 400
2
4
6
5
Unit step response uo[n]
Unit sample and step responses
(the 1st and 4 th comb filters): class_zt05.m
T. Y. Choi, ECE, Ajou University DTFT & zT 93
DTFT & zT
Frequency responses
(the 1st and 4 th comb filters)
-1 -0.5 0 0.5 10
2
4
6
in units
(1) |H(e j)|: H(z)=1/(1-0.8z-1)
-1 -0.5 0 0.5 10
2
4
6
in units
(2) |H(e j)|: H(z)=1/(1-0.8z-4)
1 1 1
1 1(1) | ( ) | | |: ( )
1 0.8 1 0.8
j
jH e H z
e z
4 4
4 1 4 14 4
1 1(2) | ( ) | | | | ( ) |: ( ) ( )
1 0.8 1 0.8
j j
jH e H e H z H z
e z
T. Y. Choi, ECE, Ajou University DTFT & zT 94
DTFT & zT
= 0
|H(ej
)|
= -/2
0.8<r1
Mesh surface of X(rej
) for 0.8<r1
unit circle |z| = 1
= /2
|H(rej
)|: H(z) = 1/(1 + 0.8z-1
)
T. Y. Choi, ECE, Ajou University DTFT & zT 95
DTFT & zT zT
1 0
[ ] [ ] [ ]N M
k k
k k
y n a y n k b x n k
1 2
1 2
1 2
1 2
( )1
M
o M
N
N
b b z b z b zH z
a z a z a z
10. DLTI system Analysis with MATLAB
y = filter(b, a, x)
1D filtering of x, column by column
r = filter( a, b, y)
1D inverse filtering of y, column by column
x[n] r[n] = x [n] y[n]
(b, a) (a, b)
Cascaded system
1 2
1 2
[ , , , , ]
[1, , , , ]
o M
N
b b b b b
a a a a
Coefficient vectors
T. Y. Choi, ECE, Ajou University DTFT & zT 96
DTFT & zT zT
(1) Pole-zero plot
zplane(b,a)
(2) Frequency response H(ejw) for |w|pi
H = freqz(b,a, pi*[-100:100]/100)
(3) Impulse response h[n] for n=0:49
h = impz(b,a,50)
h = filter(b,a,[1, zeros(1,49)] ) % size of h: 1x50
h = filter(b,a,[1, zeros(1,49)]’ ) % size of h: 50x1
(4) Step response s[n] for n=0:49
s = stepz(b,a,50);
s = filter(b,a,ones(1,50) ) % size of s: 1x50
s = filter(b,a,ones(1,50)’ ) % size of s: 50x1
Plotting the pole-zero, frequency response, h[n], s[n]
T. Y. Choi, ECE, Ajou University DTFT & zT 97
DTFT & zT zT
(1) Image reading
f = imread('flower.jpg’), s = size(f), % (502x753x3): 3D color
(2) Color components
fr = f(:,:,1); size(fr), % Red component (502x753)
fg = f(:,:,2); % Green
fb = f(:,:,3); % Blue
(3) 1D filtering of each color component (Column by column)
yr = filter(b,a,fr)
yg = filter(b,a,fg)
yb = filter(b,a,fb)
(4) Concatenate each component.
y = cat(3,yr,yg,yb)
Image processing
y = filter(b,a,f)
T. Y. Choi, ECE, Ajou University DTFT & zT 98
DTFT & zT zT
save class y % Store y in the file class.mat
load class % Take y from class.mat again
whos % List current variables
imshow(y) % Display y
Image saving and loading