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Chapter 2 Discrete System Analysis – Discrete Signals. Continuous-time analog signal. Sampling of Continuous-time Signals. sampler. Output of sampler. T. How to treat the sampling process mathematically ?. - PowerPoint PPT Presentation
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Robotics Research Labo-ratory
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Chapter 2
Discrete System Analysis – Discrete Signals
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Sampling of Continuous-time Signals
How to treat the sampling process mathematically ?
Output of sampler
( )f t *( )pf t
Continuous-time analog signal
T
For convenience, uniform-rate sampler(1/T) with finite sampling duration (p) is assumed.
T
p1
time
( )f t*( )pf t
where p(t) is a carrier signal (unit pulse train)
sampler
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This procedure is called a pulse amplitude modulation (PAM)carrier signal p(t)
( )f t*( ) ( ) ( )pf t f t p t
PAM
The unit pulse train is written as
kp t u t kT u t kT p p T
( ) ( ) ( ) where
By Fourier series
0
2( ) , ( )
1and ( )
s
s
jnω tn sn
T jnω tn
πp t C e ω sampling frequencyT
C p t e dtT
0
1 1 ss
jnω pp jnω tn
s
eC e dtT jnω p
sjnω p
sn
s
nω ppC eT nω p
2sin( / 2)
/ 2
magnitude phase
or
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0
sin( / 2),
/ 2s
ns
nω pp pC CT nω p T
* ( ) ( ) ( ) ( ) sjnω tp nn
f t f t p t C f t e
* *
*
( ) { ( )}
( )
( )
( )
s
s
p p
jωtp
jnω t jωtnn
jnω t jωtnn
F jω f t
f t e dt
C f t e e dt
C f t e e dt
F
s 2s 3s 4s0-s-2s-3s-4s2T
2p
ω
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Let { ( )} ( ) where { ( )} ( )sjnω tse f t F jω jnω f t F jω F F
* *( ) { ( )} ( )p p n snF jω f t C F jω jnω
F
c s/20-c-s /2
1
|F(j)|
c : Cutoff frequency
* ( ) ( ) sjnω t jωtp nn
F jω C f t e e dt
Spectrum of Input Signal
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0-s2πp
* ( )pF jω12 s cω ω
s0-s
* ( )pF jω12 s cω ω
ω
frequency folding
s ω
Spectrum of Output Signal
2πp
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Theorem : Shannon’s Sampling Theorem
To recover a signal from its sampling, you must sample at least twice the highest frequency in the signal.
2 2since and c sc s
π πω ωT T
or 22
sc c s
ωω ω ω
c s c si e T T T T 1. ., 2 or 2/ 2 : Nyquist frequency (or folding frequency)N sω ω
Remarks:i) A practical difficulty is that real signals do not have Fourier transforms that vanish outside a given frequency band. To avoid the frequency folding (aliasing) problem, it is necessary to filter the analog signal before sampling.
Note: Claude Shannon (1917 - 2001)
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ii) Many controlled systems have low-pass filter characteristics.iii) Sampling rates > 10 ~ 30 times of the BW of the system.iv) For the train of unit impulses ( )δ t kt
1( ) where sjnω tn nn n
δ t kt C e CT
1* ( ) ( )snF jω F jω jnω
T
s0-s
1T
ω-2s
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Remarks :
i) Impulse response of ideal low-pass filter (non-causality)
Ideal low-pass filter is not realizable in a physical system. How to realize it in a physical system ? ZOH or FOH
( )X s( )δ t *( )X s Ideal filter
G(j)( )Y s
samplingc0-c
1
|X(j)|
c0-c
1/T
|X*(j)|
c0-c
|Y(j)|
reconstruction
1/T
T
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0 11 7 18 8Z Z s Zex f H f H f H )
0 1 1 ( )sf f f alias of f
ii) (Aliasing) It is not possible to reconstruct exactly a continuous-time signal in a practical control system once it is sampled.
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If the continuous-time signal involves a frequency compo-nent equal to n times the sampling frequency (where n is an integer), then that component may not appear in the sampled signal.
sω( )f t
iii) (Hidden oscillation)
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2 2) ( ) sin sin3 if 3 / sec 3s s
s
π πex f t t t ω rad Tω
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Signal Reconstruction
How to reconstruct (approximate) the origi-nal signal from the sampled signal?
- ZOH (zero-order hold) - FOH (first-order hold)
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ZOH (Zero-order Hold)2
21
( )( ) ( ) ( )( ) ( )!
where ( )
f kTf t f kT f kT t kT t kT
kT t k T
11
f t f kT kT t k Tf t f kT f kT t kT kT t k T
in ZOH, ( ) ( ) , ( )in FOH, ( ) ( ) ( )( ) , ( )
Ttimek k+1k-1
zero-order hold reconstruction
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/2
( / )
( ) ( ) ( )
1 1( )
1 sin( / 2)( )
sin( / )2 /
1
s
ho
Tsho
jωT jωT
ho
jπ ω ωs
s
Ts
s
g t u t u t T
G s es s
e ωT eG jωjω
e
ωπω ωπ e
ω πω
s
ω
phase lag
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0π
2π
3π
ω
22
2
( ) sin
sin2
ωTjωT
ho
ωT
G jω eωT
ω2ωs ωs2ωs 3ωs0
Ideal low-pass filter2 ( )s
π Tω0.637T
( )hoG jω
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Remarks:
i) The ZOH behaves essentially as a low-pass filter.
ii) The accuracy of the ZOH as an extrapolator depends greatly on the sampling frequency, .
iii) In general, the filtering property of the ZOH is used almost exclusively in practice.
ωs
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( ) ( ) '( )( )( ) (( 1) ) = ( ) ( ), ( 1)
f t f kT f kT t kTf kT f k Tf kT t kT kT t k T
T
k-1 k k+1
T
FOH (First-order Hold)
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When k =0, (0) ( )( ) (0)
impulse (0) 1, ( ) 01 ( ) 1 ( ), 0
hf
f f Tf t f tT
f f T
f t t g t t TT
( ) (0)1, ( ) ( ) ( )
(0) 1, ( ) 01 ( ) 1 ( ), 2
hf
f T fk f t f T t TT
f f T
f t t g t T t TT
2, ( ) 0 ( ) 0, (2 ) 0
k f kTf T f T
( ) 0 ( ), 2
hff t g t t T
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210-1 T 2T 3T time
2
2
( ) 1 ( ) 2 1 ( )
2 1 ( 2 )
1( ) 1
hf
Tshf
t t Tg t u t u t TT T
t T u t TT
TsG s eTs
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2 sω 3 sω ω
2
s
πω
2 sω ω
π
2πfirst
zero
Large lag(delay) in high frequency makes a system unstable
( )hfg jωfirst
zero
3 sω
sω
sω0
0
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Remark:
At low frequencies, the phase lag produced by the ZOH exceeds that of FOH, but as the frequencies become higher, the opposite is true
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Z-transform
( ) ( ) ( )
( ) 1
f t δ t a dt f a
δ τ dτ
* *
*
( ) ( ) 1
( ) ( )
( ) ( )
( ) ( )
sτ
sτ
sτ
skT
δ t δ τ e dτ
r t r t e dτ
r τ δ τ kT e dτ
R s r kT e
L
L
( )r t * ( ) ( ) ( )r t r t δ t kT
T
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1
Since is an irrational function,
define or ln
skT
sTTz e s z
e
T
T
s z
s z
k
R z z - transform of r tr t or r kT
r t
R s
R z r kT z
1
1
*ln
*ln
( ) ( )( ) ( )
[Laplace transform of ( )]
[ ( )]
( ) ( )
Z Z
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iii) If 0, the sequence {rk} is said to be z-transformable.
ii) The series in z-1 has a radius of convergence such that the series converges absolutely when | z-1 |<
1
1.
lim
1 lim
k
kk
k kk
i e zρ
rratio test ρ
r
root test ρr
i) Because R(z) is a power series in z-1 , the theory of power series may be applied to determine the convergence of
the z-transform.
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Z-transform of Elementary Functionsi) Unit pulse function
0
1 , 0( )
0 , 0
( ) ( ) 1kk
ke k
k
E z e k z z
Remark: unit impulse = 1[1]Z
0
1 2
11
1 , 0( )
0 , 0
( ) ( )
11 , 1
1
, 11
k kk k
ke k
k
E z e k z z
z z
zz
z zz
ii) Unit step function
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0 0
1 2 3
1 2 3 4
1 2 3
1 2 3 1
1
1 2 2
, 0( )
0 , 0
( )
( 2 3 )
( ) ( 2 3 4 )
( ) (1 2 3 4 )
( )( 1) (1 ), 1
( )(1 ) ( 1)
k kk k
kT ke k
k
E z kTz T kz
T z z z
E z T z z z zzE z T z z z
E z z T z z z z
z zE z T Tz z
iii) Ramp function
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iv) Polynomial function
10 0
11
, 0( )
0 , 0
( ) ( )
1 , 11
,
k
k k kk k
r ke k
k
E z r z rz
rzrz
z z rz r
v) Exponential function
1 2 2
11
, 0( )
0 , 0
( ) 11 , 1
1
,
rkT
rT rT
rrT
rrT
e ke kk
E z e z e z
e ze zz z e
z e
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0
0
1 1
1
1 2
2
sin , 0( )
0 , 0
( ) sin
2
1 1 1 2 1 1
sin 1 2 cos
sin 2 cos 1
kk
j kT j kTk
k
j T j T
kT ke k
k
E z kTz
e e zj
j e z e zz T
z T zz T
z z T
vi) Sinusoidal function
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Remark: Refer Table 2-1 in pp.29-30 (Ogata) Also, refer Appendix B.2 Table in pp. 702-703(Franklin)
0
0
1 1
1
1 2 2
2 2
cos , 0( )
0 , 0
( ) cos
2
1 1 1 2 1 1
(1 cos ) 1 2 cos
( cos ) 2 cos
k
k kk
j kT j kTk k
k
j T j T
r kT ke k
k
E z r kTz
e er z
e rz e rzz r T
z T r zz z r T
z z T r
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Correspondence with Continuous Sig-nals
s-plane z-planez=esT
( ) ( 2 )sT σ jω T σT jωT σT j ωT πkz e e e e e e
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①
②③
④ ⑤
j
0
j
0
-2 1
ReZ
ImZ
①② ③
④⑤ 0 1-1
ReZ
ImZ
0 1
1Te2Te
2sω
-j
2sω
j
2sω
j
2sω
-j
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j
0 ReZ
ImZ
0
fixed
j
0 ReZ
ImZ
1
2sj
2sj
2sj
2j1j
1j2j
TjTee 2TjTee 1
TjTee 1
2s
43 s
4s
s 0
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Important Properties and Theorems of the z-transform
1. Linearity
1 2
1 2
1 2
1 2
1 2 1 2
( ) ( )
( ) ( )
( ) ( )
( ) { ( )}
Assume ( ) ( )
( ) ( ) ( ) ( )
k
k k
i i
αf k βf k
αf k βf k z
α f k z β f k z
α f k β f kF z f k
αf k βf k αF z βF z
Z
Z{ } ZZ
Z
2. Time Shifting
( )
( ) ( ) ,
( )
( )
k
j n
n
f k n f k n z k n j
f j z
z F z
Z
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k
n k n
n
n k nk
n nn k n k kk k k
nn kk
f k n f k n z
z f k n z
z F z
f k k
f k n z f k n z
z f k n z f k z f k z
z F z f k z
( )
( )0
1 1( )0 0 0
1
0
( ) ( )
( )
( )
If ( ) 0 for 0
( ) ( )
( ) ( ) ( )
( ) ( )
Z
Z
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1 2 1 2( ) ( ) ( ) ( )f k f k l F z F z Z
4. Scaling
( ) ( )kr f k F rz Z5. Initial Value Theorem
0 1 2( ) ( ) (0) (1) (2)
(0) lim ( )
kk
z
F z f k z f z f z f z
f F z
2 2
)( 1) ( ) (0)
( 2) ( ) (0) (1)
exx k zX z zx
x k z X z z x zx
Z Z
3. Convolution
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6. Final Value Theorem
k z z
F z z z F z
f k z F z z F z
1
1 1
If ( ) converges for 1 and all poles of ( 1) ( ) are inside
the unit circle, then
lim ( ) lim(1 ) ( ) lim( 1) ( )
1 2
1 2
11
11
1
1
1
)
( ) (0) (1) (2) ( )
( 1) (0) (1) ( 1)
( )
lim ( ) ( ) ( )
lim ( ) lim lim ( )
k n knk n kn
k nn
k kn nn nz
k nnk k z
pf
f n z f f z f z f k z
f n z f z f z f k z
z f n z
f n z z f n z f k
f k f n z z
1
1 1
1 1
( )
lim ( ) ( ) lim(1 ) ( )
k nn
z z
f n z
F z z F z z F z
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1 1
11
11 1
)
( ) 1 01 1 ( ) z
1 1 ( ) 1
1 ( ) lim(1 ) ( ) lim 1 11
at
aTaT
aTz z
exf t e a
F z ez e z
f
zf z F ze z
Remark: Refer Table 2-2 in p. 38.(Ogata) Also, refer Appendix B.1 Table in p.701(Franklin)
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TContinuous-time
domain
S-plane Z-plane
Discrete-time domain
L Z- 1Z
( )r t *( )r t
*( )R s
sTz e
1( ln )s zT
*( )R z
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1 1
1
1 1
1) ( )( 1)
( ) ( ) 1 , 0
1 1 ( ) 1
(1 ) (1 )
(1 ) (1 ) (1 )(1 ) ( 1)( )
t
tT
T T
T T
ex X ss s
x t X s e t
X z ez e z
e z e zz e z z z e
- 1L
Z
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T
T
Ts
T
s
s
Ts s
exe zX s X z
s s z z e
X sX z X s
z zs ss s s sz e z e
zz e
0 1
)
1 (1 ) ( ) ( )( 1) ( 1)( )
( ) ( ) residue of at pole of ( ) ( )
1 1 lim lim ( 1)( 1) ( 1)( ) ( )
( )
T
T
T
z zz z e
e zz z e
( 1) ( )
(1 ) ( 1)( )
Remark: Refer the derivation of ( ) in pp. 83-85 (Ogata's)
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Inverse z-transform
i) Power Series Method (Direct Division)
ii) Computational Method : - MATLAB Approach
- Difference Equation Approach
iii) Partial Fraction Expansion Method
iv) Inversion Integral Method
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Example 1) Power Series Method
1 2
1
1 2
1 2
( ) (0) (1) (2)
1( )2 1 1.5 0.5
1 2.5 3.252
(0) , (1) (2.5), (2) (3.25)2 2 2
U z u u z u z
T zU zz z
T z z
T T Tu u u
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Example 2) Computational Method
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Example 3) Partial Fraction Expansion Method1
1 2
1
1 1
1 1
1( )2 1 1.5 0.5
12 (1 )(1 0.5 )
1 1 0.532 , 2
1 3( ) 42 2 2
k
k
T zU zz z
T zz z
A Bz z
TA T B
Tu k A B
Remark:
1
1( ) ( )1
kx kT a X zaz
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Example 4) Inverse Integral Method :
1
( ) { ( )} ( )
1( ) ( )2
( ) { ( )} ( )
1( ) ( ) ( )2
st
c j st
c j
k
k
c
F s f t f t e dt
f t F s e dsj
F z f k f k z
f k F z F z z dzj
- 1
L
Z
Z
where c is a circle with its center at the origin of the z plane such that all poles of F(z)zk-1 are inside it.
1
1 2
-1( ) ( ) ( )
ik
m
i
m
f k residue of F z at pole z z of z F z
K K K
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Case 1) simple pole
1( ) ( ) |kz aresidue z a F z z
Case 2) m multiple poles
11
1
2
2
11
2
12
21
1 2
1 ( ) ( ) |( 1)!
)
( )( 1) ( )
( )( 1) ( )
( ) ( 1) ( )
mm k
z am
aT
kk
aT
k
iaTi
dresidue z a F z zm dz
exzF z
z z ezF z z
z z e
zf k residue of at pole z zz z e
K K
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1
1
2
( 1)
2
2
12
21
1
lim ( )( 1) ( )
(1 )
1
1 lim ( 1)(2 1)! ( 1) ( )
lim
aT
aT
kaT
aTz e
a k T
aT
k
aTz
z
K residue of simple pole at z e
zz ez z e
ee
K residue of multiple poles at z
d zzdz z z e
d
1
1
21
2
( 1) ( ) lim( )
1 1 1 (1 )
k
aT
k aT k
aTz
aT aT
zdz z e
k z z e zz e
ke e
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1 2
( 1)
2
( )
1 1 , 0,1,2,1 (1 )
a k T
aT aT
f k K K
k e ke e