16.362 Signal and System I The unit step response of an LTI system ] [ n ] [ n h ] [ n y ] [ n u ] [ n h ] [ n s k k n h k n y ] [ ] [ ] [ ] [ ] [ ] [ ] [ n h k n k h n y k n k k k h k n u k h n s ] [ ] [ ] [ ] [
The unit step response of an LTI system
Jan 06, 2016
The unit step response of an LTI system. Linear constant-coefficient difference equations. +. delay. When n 1,. Causality. Linear constant-coefficient difference equations. +. delay. Determine A by initial condition:. When n = 0 ,. A = 1. - PowerPoint PPT Presentation
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16.362 Signal and System I • The unit step response of an LTI system
][n ][nh ][ny
][nu ][nh ][ns
k
knhkny ][][][
][
][][][
nh
knkhnyk
n
k
k
kh
knukhns
][
][][][
16.362 Signal and System I
• Linear constant-coefficient difference equations
][]1[2
1][ nxnyny
][]1[2
1][ nnhnh
]1[2
1][ nhnh
][]1[2
1][ nnhnh
When n 1, 2
1
]1[
][
nhnh
n
Anh
2
1][
][2
1][ nuAnh
n
Causality
][n
][nh
][nh
2
1
+
delay
16.362 Signal and System I
• Linear constant-coefficient difference equations
][n
][nh
][nh
2
1
][]1[2
1][ nxnyny +
][]1[2
1][ nnhnh delay
][]1[2
1][ nnhnh
][2
1][ nuAnh
n
Determine A by initial condition:
When n = 0, 1]0[]0[ h
]0[2
1]0[
0
uAh
A = 1
16.362 Signal and System I
• Linear constant-coefficient difference equations
]1[ n ][]1[2
1][ nxnyny
][]1[2
1][ nnhnh
][2
1][ nunh
n
?][ ny
Two ways:
(1) Repeat the procedure
(2) ][][][ nhnxny
]1[2
1
]1[
][]1[][
1
nu
nh
nhnny
n
][nh
][nh
2
1
+
delay
16.362 Signal and System I
• Linear constant-coefficient difference equations
)(t
When t>0,dt
dyty
2
1)( tAety 2)(
Determine A by initial condition:
)()( 2 tuAeth t
Causality
)(2
1
2
1)( tx
dt
dyty
)(2
1
2
1)( t
dt
dyty
)(2
1
2
1)( t
dt
dyty
)(th
)(ty
2
1
+
dt
d
2
1
16.362 Signal and System I • Linear constant-coefficient difference equations
Determine A by initial condition:
)()( 2 tuAeth t
)(2
1)()
2
1()()2(
2
1)( 222 ttAetuAetuAe ttt
A = 1 )()( 2 tueth t
)(t)(
2
1
2
1)( tx
dt
dyty
)(2
1
2
1)( t
dt
dyty
)(th
)(ty
2
1
+
dt
d
2
1
16.362 Signal and System I • Linear constant-coefficient difference equations
)()( 3 tuKetx t
][5
][
)]()[(
)()(
)()()(
23
52
)(23
)(23
tt
t
o
t
t
o
t
t
eeK
deKe
deKe
dtueuKe
dthx
thtxty
)(th
)(ty
2
1
)(2
1
2
1)( tx
dt
dyty +
dt
d
2
1
)()( 2 tueth t
)(][5
)( 23 tueeK
ty tt
16.362 Signal and System I
• Fourier series representation of continuous-time periodical signal)()( Ttxtx for all tPeriodic signal
tjk
kkeatx 0)(
tjke 0
k is an integer
form a complete and orthogonal bases
Complete: no other basis is needed.
Fourier series
Orthogonal:
),(),(2
11
0
2
0
)(
00 0
)(
0
0
)(
0
0
000
mkTmk
detde
dtedtee
mkjT tmkj
T tmkjT tjmtjk
),(1
0
00 mkdteeT
T tjmtjk
Orthogonal:
mk
mkmk
0
1),(Kronecker Delta
16.362 Signal and System I
• Fourier series representation of continuous-time periodical signal)()( Ttxtx for all tPeriodic signal
tjk
kkeatx 0)(
k is an integer
m
kk
T tmkj
kk
T tjmtjk
kk
T tjmtjk
kk
T tjm
a
mka
dteT
a
dteeT
a
dteeaT
dtetxT
)(
1
1
1)(
1
0
)(
0
00
0
00
000
T tjm
m dtetxT
a0
0)(1
16.362 Signal and System I
• Fourier series representation of continuous-time periodical signal)()( Ttxtx for all tPeriodic signal
tjk
kkeatx 0)(
k is an integer T tjm
m dtetxT
a0
0)(1
e.g.
k
Tk
Tk
Tk
jk
ee
T
dteT
dteT
dtetxT
a
TjkTjk
T
TT
tjkT tjk
T tjkk
10
0
10
0
0
0
sinsin2
1
11
)(1
1010
1
01
0
0
T
20
1T
2
T
2
T1TT
T
T
1T
16.362 Signal and System I • Fourier series representation of continuous-time periodical signal
k
Tkak
10sin
1T
2
T
2
T1TT
T
T
1T
T tjk
k dtetxT
a0
0)(1
T
Tdt
Ta
T
T
10
21 1
1
tjk
kkeatx 0)(
0 00
16.362 Signal and System I • The response of system to complex exponentials
)(tx )(ty
Band limited channel
Bandwidth 0100
-2 -1.5 -1 -0.5 0 0.5 1 1.5 20
0.2
0.4
0.6
0.8
1
1.2
1.4
Bandwidth 010
16.362 Signal and System I • Fourier series representation of discrete-time periodical signal
][][ Nnxnx
N
20
for all tPeriodic signal
1
0
0][N
k
njkkeanx
1
0
0][1 N
n
njkk enx
Na
kNk aa
16.362 Signal and System I • Example #1
)sin(][ 0nnx N
20
1
0
0][1 N
n
njkk enx
Na
nNjn
Nj
njnj
ej
ej
j
ee
nnx
22
0
2
1
2
1
2
)sin(][00
ja
2
11 j
a2
11
Nk
njkkeanx 0][
16.362 Signal and System I • Properties of discrete-time Fourier series
(1) Linearity
ka
kb
][][][ nBynAxnz kk BbAa
1
0
0][1 N
n
njkk enx
Na
Nk
njkkeanx 0][
Nk
njkkeanx 0][
Nk
njkkebny 0][
16.362 Signal and System I
Nk
njknjkk
Nk
nnjkk
eea
eannx
000
00 )(0 ][
(2) Time shifting
0njkkea
(3) Time reversal
kNa
Nk
njkk
Nk
njkk
ea
eanx
0
0 )(][
16.362 Signal and System I
)(1
0
0][ njkN
kkeanx
(4) Time scaling
ka
1
0
1
0
1
0
1
0
)(
1
0
1
0
0
0
00][][
N
l
N
k
njklkl
N
l
N
m
nmljml
njmN
mm
njlN
ll
eba
eba
ebeanynx
(5) multiplication
1
0
N
llklba
0
16.362 Signal and System I
Nk
njkkeanx 0][
(6) Conjugation and conjugate symmetry
ka
ka
Real signal
][][ nxnx
kk aa
Even
kk aa
kk aa
][][ nxnx
Real & Even kk aa
njkN
kkeanx 0
1
0
*][*
16.362 Signal and System I
Nk Nmmk
Nk Nm
N
n
nmkjmk
Nm
njmm
N
n Nk
njkk
N
n
N
n
mkNaa
eaa
eaea
nxnxnx
),(
][*][][
*
1
0
)(*
*1
0
1
0
1
0
2
0
00
(7) Parseval’s relation
Nk
k
N
n
anxN
21
0
2][
1
16.362 Signal and System I
k
njkk
j
k
njkk
k
njkk
eae
eaeanxnx
00
00
1
]1[][ )1(
(8) Time difference
kj ae 01
kjk
njk
k
k
n
m
mjkk
n
m k
mjkk
n
m
e
ea
ea
eanx
0
0
0
0
1
][
(9) Running sum
01 jkk
e
a
0k
00 a
16.362 Signal and System I
][1 nx
Example
ka
N = 4
][2 nx
[1, 2, 2, 1]
kb [1, 1, 1, 1]
][][][ 11 nxnxny ?kc
1
0
N
llklk bac
132231000 babababac
233201101 babababac
330211202 babababac
031221303 babababac
16.362 Signal and System I • Fourier series and LTI system
)(tx )(th )(ty
Periodic signalSystem response doesn’t have to be periodic.
Output periodic?
)(
)(
)()( )(
sHe
dehe
dehty
st
sst
ts
stetx )( )(th
dehsH s)()(
16.362 Signal and System I
)()( jHety tjtjetx )( )(th
dehjH j)()(
k
tjkk jkHeaty )()( 0
0 tjk
kkeatx 0)(
)(th
dehjkH jk 0)()( 0
16.362 Signal and System I
)(th)(tx )(ty
Filtering
• Frequency-shaping filters
• Frequency-selective filters
(1) Frequency-shaping filters
dt
tdxty
)()(
)(th
kk ajkb 0
00 )( jkjkH
16.362 Signal and System I
)(th)(tx )(ty
(1) Frequency-shaping filters
dt
tdxty
)()(
)(th
kk ajkb 0
jjH )(
16.362 Signal and System I
)(th)(tx )(ty
(2) Frequency-selective filters
Low-pass
high-pass
band-pass
16.362 Signal and System I
][nynjenx ][ ][nh
Discrete-time
k
kjnj
k
knj
k
ekhe
ekh
knxkhny
][
][
][][][
)(
k
kjekhjH ][)(
][ny
Nk
njkkeanx 0][
][nh
Nk
njkk ejkHany 0)(][ 0
ka
][ny
][nx
][nh )( 0jkH
)( 0jkHak
16.362 Signal and System I
][nx ][nh
Example: averaging
]1[][2
1][ nxnxny
Nk
jknjk
k
Nk
njkk
njkk
eea
eaea
nxnxny
2
1
2
1
]1[][2
1][
0
0
00 )1(
2
1 0jk
kk
eab
)2/cos(
2
1)(
02/
0
0
0
ke
ejkH
jk
jk
16.362 Signal and System I • Continuous-time Fourier transform
)(txAperiodic signal
tjk
kkeatx 0)(
k is an integer
T tjk
k dtetxT
a0
0)(1
)()( Ttxtx Periodic signal
dejXtx tj)(
2
1)(
dtetxjX tj )()(
)'(
)'()(
2
1)(
)(2
1)(
)'(
''
jX
djX
dtedjX
dtedejXdtetx
tj
tjtjtj
kajkX 2)( 0
16.362 Signal and System I • Continuous-time Fourier transform
)(txAperiodic signal
tjk
kkeatx 0)(
k is an integer
T tjk
k dtetxT
a0
0)(1
)()( Ttxtx Periodic signal
2
)( 0jkXak
dejXtx tj)(
2
1)(
dtetxjX tj )()(
16.362 Signal and System I • Examples
)()( tuetx at
ja
dtee
dtetxjX
tjat
tj
1
)()(
0
jajX
1)(
2/1
22
1)(
ajX
a
1tan
dejXtx tj)(
2
1)(
dtetxjX tj )()(
16.362 Signal and System I • Properties of continuous-time Fourier transform
)(tx
(1) Linearity
)(ty
)( jX
)( jY
)()()( tBytAxtz )()()( jBYjAXjZ
dejXtx tj)(
2
1)(
dtetxjX tj )()(
16.362 Signal and System I • Properties of continuous-time Fourier transform
dejXtx tj)(
2
1)(
dtetxjX tj )()(
)( 0ttx
(2) Time shifting
0)( tjejX
(3) Time reversal
')'(
)(
)()('
')(
))((
dtetx
dtetx
dtetxjX
tj
tj
tj
)( tx
)( jX
16.362 Signal and System I • Properties of continuous-time Fourier transform
dejXtx tj)(
2
1)(
dtetxjX tj )()(
)( tx
(4) Time scaling
tdetx
dtetxjX
tj
tj
)(1
)()('
jX
1
16.362 Signal and System I • Properties of continuous-time Fourier transform
dejXtx tj)(
2
1)(
dtetxjX tj )()(
(5) Conjugation and conjugate summary
)(tx
)(
)(
)()('
)(
jX
dtetx
dtetxjX
tj
tj
)(tx Real
)()( txtx )()( jXjX
)()( jXjX
16.362 Signal and System I Example
)()( tuetx t
)()( jXjX
)()( jXjX
jjX
1)(
)( tx even
Even and real)()( jXjX
16.362 Signal and System I Differential
)(tx
dejXtx tj)(
2
1)(
)( jX
dt
tdxtg
)()(
dejXj
dt
tdx tj)(2
1)(
)( jG
)()( jXjjG
16.362 Signal and System I Integral)(tx
j
edjX
jj
ejdjX
dedjddjX
dededjX
dedjX
ddejXdx
tj
tj
t j
t jj
t j
t jt
)()(2
1
1)()(
2
1
)sin()cos()(2
1
)(2
1
)(2
1
)(2
1)(
0
00
0
0
)( jX
tdxtg )()( )( jG
)0()()(
)( Xj
jXjG
16.362 Signal and System I
Example
)(tx )( jX
tdxtg )()( )0()(
)()( X
j
jXjG
dt
tdxtf
)()( )()( jXjjF
)(ty)(tx )(th
)()( tuetx bt 0b
)()( tueth at 0a
)()()( thtxty
jbjX
1)(
jajH
1)(
jbjaab
jbjajY
111
11)(
)(1
)(1
)( tueab
tueab
ty btat
ab
16.362 Signal and System I Example
)(ty)(tx )(th
)()( tuetx at
)()( tueth at
0a
)()()( thtxty
jajX
1)(
jajH
1)(
jad
dj
jajY
1
1)(
2
)(
)(2
)(2
)(2
)(2
)(
tute
djXet
djXjtej
jdXej
dejXd
djty
at
tj
tj
tj
tj
16.362 Signal and System I Example
dejXtx tj)(
2
1)(
dtetxjX tj )()(
21
1)(
jX?)( tx
)1(
1
)1(
1
2
1
)1)(1(
11
1)(
2
jj
jj
jX)(
2
1)(
2
1)( tuetuetx tt
tetx 2
1)(
deetx tjt
21
1
2
1
2
1)(
16.362 Signal and System I Example
dejXtx tj)(
2
1)(
dtetxjX tj )()(
?)( jX21
1)(
ttx
deetx tjt
21
1
2
1
2
1)(
dte
te tj
21
1
2
1
2
1
dte
te tj
21
1
2
1
2
1
16.362 Signal and System I Parseval’s relation
djXjX
djXdjX
djXdjXdte
dtdejXdejX
dttxtxdttx
tj
tjtj
)'()(2
1
)'(')'()(2
1
')'(2
1)(
2
1
')'()(2
1
)()()(
)'(
'
2
djXdttx22)(
2
1)(
16.362 Signal and System I Parseval’s relation for continuous-time Fourier series
k
k
Tadttx
T2
0
2)(
1
djXdttx22)(
2
1)(
Parseval’s relation for continuous-time Fourier transfer
16.362 Signal and System I Example
0.5 1.0-0.5-1.0
2/
?)(2
dttx
)( jX
djXdttx22)(
2
1)(
0|)( xtxdt
dD
16.362 Signal and System I Example
0.5 1.0-0.5-1.0
2/
?)(2
dttx
)( jX
0|)( xtxdt
dD
dejXtx tj)(2
1)(
dejXjtxdt
d tj)()(2
1)(
0
)()(2
1
|)()(2
1|)( 00
djXj
dejXjtxdt
dt
tjt
16.362 Signal and System I Example, P. 4.14
)( jX)(tx
(1) real
(2) )()( 2 tuAetg t)()1()( jXjjG
(3) 2)(2
djX
?)( tx
Solution:
)()( 2 tuAetg t
j
AjG
2
)(
)()1()( jXjjG
)2)(1(
)1(
)()(
jj
A
j
jGjX
16.362 Signal and System I Example, P. 4.14
)( jX)(tx
(1) real
(2) )()( 2 tuAetg t)()1()( jXjjG
(3) 2)(2
djX
?)( tx
Solution:
)2)(1()(
jj
AjX
2)(
2
djX
dA
dA
djX
22
2
22
22
4
1
1
1
3
41)(
16.362 Signal and System I Example, P. 4.14
Solution:
2
2
2
222
tantan1
1
1
1
d
dd
tan
2
2
1
tantan1
1
2
1
2
21
1
2
1
4
1
2
2
2
22
22
d
d
dd
tan
2
16.362 Signal and System I Example, P. 4.14
)( jX)(tx
(3) 2)(2
djX
Solution:
)2)(1()(
jj
AjX
2)(
2
djX
6
4
1
1
1
3)(
2
22
22
A
dA
djX
122 A jeA 12
16.362 Signal and System I Example, P. 4.14
)( jX)(tx
(3) 2)(2
djX
Solution:
)2)(1()(
jj
AjX
jeA 12
)2(
1
)1(
112)(
jjejX j )()(12)( 2 tuetueetx ttj
(1) real
)()(12)( 2 tuetuetx tt
16.362 Signal and System I • Multiplication
dejXtx tj)(
2
1)(
dtetxjX tj )()(
111
212211
)(2211
2211
2211
))(()(2
1
)(2)(2
1)(
2
1
)(2
1)(
2
1
)(2
1)(
2
1
)(2
1)(
2
1
)()()(
21
21
21
djYjX
djYdjX
dtedjYdjX
dteeedjYdjX
dejYdejXdte
dtetytxjZ
tj
tjtjtj
tjtjtj
tj)()()( tytxtz
)()(2
1)(
jYjXjZ
16.362 Signal and System I
Example #1
dejXtx tj)(
2
1)(
dtetxjX tj )()(
)()()( tytxtz )()()( jYjXjZ
ttp 0cos)( )()( 00
)(ts )( jS
ttsty 0cos)()( )( jY
))((2
1))((
2
1
))'(()'()'(2
1
))'(()'(2
1
)()(2
1)(
00
00
jSjS
djS
djSjP
jPjSjY
16.362 Signal and System I
Example #2
dejXtx tj)(
2
1)(
dtetxjX tj )()(
)()()( tytxtz )()()( jYjXjZ
ttp 0cos)( )()( 00
)(ts )( jS
ttsty 0cos)()(
)2(()(2)2((4
1
)()(2
1)(
00
jSjSjS
jPjYjG)()()( tptytg
))((2
1))((
2
1
)()(2
1)(
00
jSjS
jPjSjY
16.362 Signal and System I • Frequency-selective filtering with variable center frequency
x)(tx Low pass filter x
tj ce tj ce
)(ty )(tw)(tf
c
0
1
0
c cc
0 c
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