lecture #07
DESCRIPTION
Lecture #07. Z-Transform. H. = The impulse response of system H. eigenfunction. eigenvalue. MIT signals & systems. Discrete-time Fourier transform. Where z is complex. if. DTFT is a special case of Z transform. Same as FT is a special case of Laplace transform. - PowerPoint PPT PresentationTRANSCRIPT
meiling chen signals & systems 1
Lecture #07
Z-Transform
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nnk
k
kn
k
k
zzHzzkhzkh
knxkhnxnhnxHny
)(}][{][
][][][][]}[{][
][
Hnznx ][ ][ny
][nh = The impulse response of system H
eigenvalue eigenfunction
meiling chen signals & systems 3MIT signals & systems
meiling chen signals & systems 4
k
kzkhkhZzH ][]}[{)(
Where z is complex
jezif
n
njj
njj
enxeX
deeXnx
][)(
)(2
1][
Discrete-time Fourier transform
k
jkj ekheH )()(
DTFT is a special case of Z transform
Same as FT is a special case of Laplace transform
jez jrez
js js
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k
kzkhzH ][)(
Let be a complex number jrez
nzzHny )(][
))}(sin())({cos()(
)()(
)(][
)()(
)(
)(
)(
znjznrreH
erereH
zezHny
ezHzH
nj
jnnzjj
nzj
zj
n
jnn
n
njj ernhrenhreH )][()]([)(
The DTFT of a signal
dereHrnh jnjn
)(
2
1][
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dzzzHj
nh
derreHnh
n
jnnj
1)(2
1][
)(2
1][
dzzj
ddjredzrez jj 11
The z-transform of an arbitrary signal x[n]
n
nznxzX ][)(
and the inverse z-transform
dzzzXj
nx n 1)(2
1][
)(][ zXnx Notation
j
j
st
st
dsesXj
tx
dtetxsX
)(
2
1)(
)()(
Laplace/inverse laplace transfrom
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Region of convergence (ROC)
n
nznxzX ][)(
Critical question : Does the summation converge to a finite value
In general that depends on the value of z
jrez Since
}][,{
n
nj rnxwhichatrezROC
11][
rrezznx j
n
n
Unique circle
meiling chen signals & systems 8MIT signals & systems
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Example : Z-transform ][]cos[][ 0 nunrnx n
1
0
1
0
0
0
000
00
1
1
)()(}){(][
]}[][{2
1][
]}[][{2
1})()]{([
2
1
][]cos[][
zre
zrezrereZzV
zVzVzX
nvnvrerenu
nunrnx
j
n
nj
n
nnjnj
njnj
n
n
nznxzX ][)(
2210
10
11
)cos(21
)cos(1
)1
1
1
1(2
1][
00
zrzr
zr
zrezrezX jj
1z
r
R.O.C
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Linearity
Right shift in time
Left shift in time
Time Multiplication
Frequency Scaling
Modulation
Properties of Z transform
][][]}[][{ 2121 zFzFkfkfZ
][][1
]}[][{ 200 0zFzF
zkkukkfZ k
]1[]1[]0[][]}[][{ 01
0000 kzffzfzzFzkukkfZ kkk
][]}[{ zFdz
dzkkfZ
][]}[{a
zFkfaZ k
)]()([2
}sin][{
)]()([2
1}cos][{
zeFzeFj
kTkfZ
zeFzeFkTkfZ
TjTj
TjTj
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Convolution
)()(]}[][{ 2121 zFzFkfkfZ
)()()(][
][][
][][]}[][{
210
21
0 021
0 02121
zFzFzFzmf
zmkfmf
zmkfmfkfkfZ
m
m
m k
k
k m
k
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Initial value theorem
)}({lim]0[ zFfz
321 ]3[]2[]1[]0[][)( zfzfzffznfzFn
n
z
0 0 0
)]}0()([{lim]1[
)}({lim]0[
fzFzf
zFf
z
z
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Final value theorem
)}(1
{lim][][lim1
zFz
zfkf
zk
][
])1[]2[(])0[]1[(]0[
)]1[][(
)]1[][(lim]}1[][{lim
)(1
)(1
)(]}1[][{
0
011
f
fffff
kfkf
zkfkfkfkfZ
zFz
zzF
zzFkfkfZ
k
k
kzz
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Some common Z transforms
rzrzr
zrkukr
r
zrzr
zraz
zz
z
kukr
kua
ku
zk
ROCzXkx
zk
z
za
z
k
k
1221
0
10
0
1
1
2210
10
0
)cos(21
)sin(][)sin(
1
1
)cos(21
)cos(1
1
][)cos(
][
][
1][
][][
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Example : Inverse Z-transform
)1)(5.0)(5.0(
)1()(
zzz
zzzF
az
zkuaZ k
]}[{
kkkkf
z
z
z
z
z
zzF
zzz
zzz
z
z
zF
)1(3
8)5.0(
3
1)5.0(3][
138
5.031
5.0
3)(
138
5.031
5.0
3
)1)(5.0)(5.0(
)1()(
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Example : the Z transform can be used to solve difference equations
0]0[,1]1[,0][]1[6]2[8 xxkxkxkx
0][][6]1[8][8 2 zXzzXzxzXz
Taking the Z transform
41
21
2
44][
14
16
12
8
168
8][
z
z
z
zzX
zzzzz
zX
})4
1()
2
1{(4][ kkkx