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Sampling and ReconstructionThe impulse response of an continuous-time ideal low pass filter
is the inverse continuous Fourier transform of its frequency response
Let Hlp(jw) be the frequency response of the ideal low-pass filter with the cut-off frequency being ws/2.
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Remember that when we sample a continuous band-limited signal satisfying the sampling theorem, then the signal can be reconstructed by ideal low-pass filtering.
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The sampled continuous-time signal can be represented by an impulse train:
Note that when we input the impulse function δ(t) into the ideal low-pass filter, its output is its impulse response
When we input xs(t) to an ideal low-pass filter, the output shall be
It shows that how the continuous-time signal can be reconstructed by interpolating the discrete-time signal x[n].
( ) ( ) ( ) [ ] ( )∑∑∞
−∞=
∞
−∞=
−=−=nn
cs nTtnxnTtnTxtx δδ
( ) ( ) ( ) ( )TtTtthth lpr ///sin ππ=≡
( ) [ ] ( )[ ]( )∑
∞
−∞= −−
=n
r TnTtTnTtnxtx
//sin
ππ
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Properties:
hr(0) = 1; hr(nT) = 0 for n=±1, ±2, …;
It follows that xr(mT) = xc(mT), for all integer m.
The form sin(t)/t is referred to as a sinc function. So, the interpolant of ideal low-pass function is a sinc function.
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Illustration of reconstruction
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Ideal discrete-to-continuous (D/C) converter
It defines an ideal system for reconstructing a bandlimitedsignal from a sequence of samples.
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Ideal low pass filter is non-causal
Since its impulse response is not zero when n<0.
Ideal low pass filter can not be realized
It can not be implemented by using any difference equations.
Hence, in practice, we need to design filters that can be implemented by difference equations, to approximate ideal filters.
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• Discrete-time Fourier Transform
• z-transform: polynomial representation of a sequence
Z-Transform
( ) [ ]∑∞
−∞=
−=n
jwnjw enxeX
( ) [ ]∑∞
−∞=
−=n
nznxzX
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• Z-transform operator:
• The z-transform operator is seen to transform the sequence x[n] into the function X{z}, where z is a continuous complex variable.– From time domain (or space domain, n-domain) to the
z-domain
Z-Transform (continue)
{ } ⋅Z
[ ]{ } [ ] ( )zXzkxnxZk
k == ∑∞
−∞=
−
[ ] [ ] [ ] [ ]{ } ( )ZXnxZknkxnxz
k=↔−= ∑
∞
−∞=
δ
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• Two sided or bilateral z-transform
• Unilateral z-transform
Bilateral vs. Unilateral
( ) [ ]∑∞
−∞=
−=n
nznxzX
( ) [ ]∑∞
=
−=0n
nznxzX
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Example of z-transform
( ) 54321 24642 −−−−− +++++= zzzzzzX
01246420x[n]
N>5543210n≤−1n
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• If we replace the complex variable z in the z-transform by ejw, then the z-transform reduces to the Fourier transform.
• The Fourier transform is simply the z-transform when evaluating X(z) only in a unit circle in the z-plane.
• From another point of view, we can express the complex variable z in the polar form as z = rejw. With z expressed in this form,
Relationship to the Fourier Transform
( ) [ ]( ) [ ]( )∑∑∞
−∞=
−−∞
−∞=
−==
n
jwnn
n
njwjw ernxrenxreX
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• In this sense, the z-transform can be interpreted as the Fourier transform of the product of the original sequence x[n] and the exponential sequence r−n.– For r=1, the z-transform reduces to the Fourier transform.
Relationship to the Fourier Transform (continue)
The unit circle in the complex z plane
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Relationship to the Fourier Transform (continue)
– Beginning at z = 1 (i.e., w = 0) through z = j (i.e., w = π/2) to z = −1 (i.e., w = π), we obtain the Fourier transform from 0≤ w ≤ π.
– Continuing around the unit circle in the z-plane corresponds to examining the Fourier transform from w = π to w = 2π.
– Fourier transform is usually displayed on a linear frequency axis. Interpreting the Fourier transform as the z-transform on the unit circle in the z-plane corresponds conceptually to wrapping the linear frequency axis around the unit circle.
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– The sum of the series may not be converge for all z.• Region of convergence (ROC)
– Since the z-transform can be interpreted as the Fourier transform of the product of the original sequence x[n] and the exponential sequence r−n, it is possible for the z-transform to converge even if the Fourier transform does not.
– Eg., x[n] = u[n] is absolutely summable if r>1. This means that the z-transform for the unit step exists with ROC |z|>1.
Convergence Region of Z-transform
( ) [ ] [ ]∑∑∞
−∞=
−∞
−∞=
− ≤=n
n
n
n znxznxzX
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• In fact, convergence of the power series X(z)depends only on |z|.
• If some value of z, say z = z1, is in the ROC, then all values of z on the circle defined by |z|=| z1| will also be in the ROC.
• Thus the ROC will consist of a ring in the z-plane.
ROC of Z-transform
[ ] ∞<∑∞
−∞=
−
n
nznx ||
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ROC of Z-transform – Ring Shape
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Analytic Function and ROC
• The z-transform is a Laurent series of z.– A number of theorems from the complex-variable theory
can be employed to study the z-transform.– A Laurent series, and therefore the z-transform,
represents an analytic function at every point inside the region of convergence.
– Hence, the z-transform and all its derivatives exist and must be continuous functions of z with the ROC.
– This implies that if the ROC includes the unit circle, the Fourier transform and all its derivatives with respect to wmust be continuous function of w.
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• First, we analyze the Z-transform of a causal FIR system
– The impulse response is
• Take the z-transform on both sides
Z-transform and Linear Systems
[ ] [ ]∑=
−=M
mm mnxbny
0
[ ] [ ]∑=
−=M
mm mnbnh
0δ
0bM…b3b2b1b00h[n]N> MM…3210n<0n
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Z-transform of Causal FIR System (continue)
• Thus, the z-transform of the output of a FIR system is the product of the z-transform of the input signal and the z-transform of the impulse response.
( ) [ ]{ } [ ] [ ]
[ ] [ ] ( )
[ ]{ } [ ]{ } [ ] ( ) ( )zXzHzXnhZnxZzb
zmnxzbzmnxb
zmnxbmnxbZnyZzY
M
m
mm
M
m
mn
n
mm
M
m
n
nm
n
M
m
nm
M
mm
≡==
⎟⎟⎠
⎞⎜⎜⎝
⎛−=−=
−=⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
−==
∑
∑ ∑∑ ∑
∑ ∑∑
=
−
=
−−∞
−∞=
−
=
−∞
−∞=
∞
−∞= =
−
=
0
00
00
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Z-transform of Causal FIR System (continue)
• H(z) is called the system function (or transfer function) of a (FIR) LTI system.
( ) ∑=
−=M
m
mm zbzH
0
x[n] y[n]h[n]
X(z) Y(z)H(z)
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Multiplication Rule of Cascading System
X(z) Y(z)H1(z)
Y(z) V(z)H2(z)
X(z) Y(z)H1(z)
V(z)H2(z)≡
X(z) Y(z)H1(z)H2(z)≡
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Example
• Consider the FIR system y[n] = 6x[n] − 5x[n−1] + x[n−2]
• The z-transform system function is
( )
( )( ) 211
21
21
31
623
56
z
zzzz
zzzH
⎟⎠⎞
⎜⎝⎛ −⎟⎠⎞
⎜⎝⎛ −
=−−=
+−=
−−
−−
![Page 24: Sampling and Reconstruction - 國立臺灣大學 · 2006. 11. 2. · Sampling and Reconstruction The impulse response of an continuous-time ideal low pass filter is the inverse continuous](https://reader034.vdocument.in/reader034/viewer/2022051902/5ff15f6dea4f632e1468ea1b/html5/thumbnails/24.jpg)
Delay of one Sample
• Consider the FIR system y[n] = x[n−1], i.e., the one-sample-delay system.
• The z-transform system function is
( ) 1−= zzH
z−1
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Delay of k Samples
( ) kzzH −=
• Similarly, the FIR system y[n] = x[n−k], i.e., the k-sample-delay system, is the z-transform of the impulse response δ[n − k].
z−k
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System Diagram of A Causal FIR System
• The signal-flow graph of a causal FIR system can be re-represented by z-transforms.
x[n]
TD
TD
TD
x[n-2]
x[n-1]
x[n-M]
+
+
+
+
b1
b0
b2
bM
y[n]x[n]
z−1
z−1
z−1
x[n-2]
x[n-1]
x[n-M]
+
+
+
+
b1
b0
b2
bM
y[n]
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Z-transform of General Difference Equation (IIR system)
• Remember that the general form of a linear constant-coefficient difference equation is
• When a0 is normalized to a0 = 1, the system diagram can be shown as below
[ ] [ ]∑∑==
−=−M
mm
N
kk mnxbknya
00for all n
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Review of Linear Constant-coefficient Difference Equation
x[n]
TD
TD
TD
x[n-2]
x[n-1]
x[n-M]
+
+
+
+
b1
b0
b2
bM
+
+
+
+
y[n]
TD
TD
TD
− a1
− a2
− aN y[n-N]
y[n-2]
y[n-1]
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Z-transform of Linear Constant-coefficient Difference Equation
• The signal-flow graph of difference equations represented by z-transforms.
X(z)
z−1
z−1
z−1
+
+
+
+
b1
b0
b2
bM
+
+
+
+
Y(z)
z−1
z−1
z−1
− a1
− a2
− aN
![Page 30: Sampling and Reconstruction - 國立臺灣大學 · 2006. 11. 2. · Sampling and Reconstruction The impulse response of an continuous-time ideal low pass filter is the inverse continuous](https://reader034.vdocument.in/reader034/viewer/2022051902/5ff15f6dea4f632e1468ea1b/html5/thumbnails/30.jpg)
• From the signal-flow graph,
• Thus,
• We have
Z-transform of Difference Equation (continue)
( ) ( ) ( )∑∑=
−−
=
−=N
k
kk
mM
mm zzYazzXbzY
10
( ) ( ) mM
mm
N
k
kk zzXbzzYa −
==
− ∑∑ =00
( )( )
∑
∑
=
−
−
== N
k
kk
mM
mm
za
zb
zXzY
0
0
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• Let
– H(z) is called the system function of the LTI system defined by the linear constant-coefficient difference equation.
– The multiplication rule still holds: Y(z) = H(z)X(z), i.e., Z{y[n]} = H(z)Z{x[n]}.
– The system function of a difference equation (or a generally IIR system) is a rational form X(z) = P(z)/Q(z).
– Since LTI systems are often realized by difference equations, the rational form is the most common and useful for z-transforms.
Z-transform of Difference Equation (continue)
[ ] ∑∑=
−−
=
=N
k
kk
mM
mm zazbzH
00/
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• When ak = 0 for k = 1 … N, the difference equation degenerates to a FIR system we have investigated before.
• It can still be represented by a rational form of the variable z as
Z-transform of Difference Equation (continue)
( ) mM
mmzbzH −
=∑=
0
( ) M
mMM
mm
z
zbzH
−
=∑
= 0
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• When the input x[n] = δ[n], the z-transform of the impulse response satisfies the following equation:
Z{h[n]} = H(z)Z{δ[n]}.• Since the z-transform of the unit impulse δ[n] is
equal to one, we have Z{h[n]} = H(z)
• That is, the system function H(z) is the z-transform of the impulse response h[n].
System Function and Impulse Response
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• Convolution
• Take the z-transform on both sides:
Z-transform vs. Convolution
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Time domain convolution implies Z-domain multiplication
Z-transform vs. Convolution (con’t)
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• Generally, for a linear system,y[n] = T{x[n]}
– it can be shown thatY{z} = H(z)X(z).
where H(z), the system function, is the z-transform of the impulse response of this system T{⋅}.
– Also, cascading of systems becomes multiplication of system function under z-transforms.
System Function and Impulse Response (continue)
X(z) Y(z) (= H(z)X(z))H(z)/H(ejw)
X(ejw) Y(ejw) (= H(ejw)X(ejw))
Z-transform
Fourier transform
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• Pole:– The pole of a z-transform X(z) are the values of z for
which X(z)= ∞.• Zero:
– The zero of a z-transform X(z) are the values of z for which X(z)=0.
• When X(z) = P(z)/Q(z) is a rational form, and both P(z) and Q(z) are polynomials of z, the poles of are the roots of Q(z), and the zeros are the roots of P(z), respectively.
Poles and Zeros
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• Zeros of a system function– The system function of the FIR system y[n] = 6x[n] −
5x[n−1] + x[n−2] has been shown as
• The zeros of this system are 1/3 and 1/2, and the pole is 0.
• Since 0 and 0 are double roots of Q(z), the pole is a second-order pole.
Examples
( ) ( )( )zQzP
z
zzzH =
⎟⎠⎞
⎜⎝⎛ −⎟⎠⎞
⎜⎝⎛ −
= 221
31
6
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Example: Finite-length Sequence (FIR System)
( )⎪⎩
⎪⎨⎧ −≤≤=
otherwiseNnanx
n
010Given
Then ( ) ( ) ( )
azaz
z
azazazzazX
NN
N
N
n
NnN
n
nn
−−
=
−
−===
−
−
=−
−−
−
=
− ∑∑
1
1
01
11
1
0
1
11
There are the N roots of zN = aN, zk = aej(2πk/N). The root of k = 0cancels the pole at z=a. Thus there are N−1 zeros, zk = aej(2πk/N), k = 1 …N, and a (N−1)th order pole at zero.
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Pole-zero Plot
![Page 41: Sampling and Reconstruction - 國立臺灣大學 · 2006. 11. 2. · Sampling and Reconstruction The impulse response of an continuous-time ideal low pass filter is the inverse continuous](https://reader034.vdocument.in/reader034/viewer/2022051902/5ff15f6dea4f632e1468ea1b/html5/thumbnails/41.jpg)
• Right-sided sequence: – A discrete-time signal is right-sided if it is nonzero
only for n≥0.• Consider the signal x[n] = anu[n].
• For convergent X(z), we need
– Thus, the ROC is the range of values of z for which |az−1| < 1 or, equivalently, |z| > a.
Example: Right-sided Exponential Sequence
( ) [ ] ( )∑∑∞
=
−∞
−∞=
− ==0
1
n
n
n
nn azznuazX
( ) ∞<∑∞
=
−
0
1
n
naz
![Page 42: Sampling and Reconstruction - 國立臺灣大學 · 2006. 11. 2. · Sampling and Reconstruction The impulse response of an continuous-time ideal low pass filter is the inverse continuous](https://reader034.vdocument.in/reader034/viewer/2022051902/5ff15f6dea4f632e1468ea1b/html5/thumbnails/42.jpg)
• By sum of power series,
• There is one zero, at z=0, and one pole, at z=a.
Example: Right-sided Exponential Sequence (continue)
( ) ( ) azaz
zaz
azzXn
n>
−=
−==
−
∞
=
−∑ 1
11
0
1 ,
: zeros× : poles
Gray region: ROC
![Page 43: Sampling and Reconstruction - 國立臺灣大學 · 2006. 11. 2. · Sampling and Reconstruction The impulse response of an continuous-time ideal low pass filter is the inverse continuous](https://reader034.vdocument.in/reader034/viewer/2022051902/5ff15f6dea4f632e1468ea1b/html5/thumbnails/43.jpg)
• Left-sided sequence: – A discrete-time signal is left-sided if it is nonzero only
for n ≤ −1.• Consider the signal x[n] = −anu[−n−1].
– If |az−1| < 1 or, equivalently, |z| < a, the sum converges.
Example: Left-sided Exponential Sequence
( ) [ ]
( )∑∑
∑∑∞
=
−∞
=
−
−
−∞=
−∞
−∞=
−
−==
−=−−−=
0
1
1
1
1
1
n
n
n
nn
n
nn
n
nn
zaza
zaznuazX
![Page 44: Sampling and Reconstruction - 國立臺灣大學 · 2006. 11. 2. · Sampling and Reconstruction The impulse response of an continuous-time ideal low pass filter is the inverse continuous](https://reader034.vdocument.in/reader034/viewer/2022051902/5ff15f6dea4f632e1468ea1b/html5/thumbnails/44.jpg)
• By sum of power series,
• There is one zero, at z=0, and one pole, at z=a.
Example: Left-sided Exponential Sequence (continue)
( ) azaz
zzaza
zazX <
−=
−
−=
−−=
−
−
−
1111 1
1
1 ,
The pole-zero plot and the algebraic expression of the system function are the same as those in the previous example, but the ROC is different.
![Page 45: Sampling and Reconstruction - 國立臺灣大學 · 2006. 11. 2. · Sampling and Reconstruction The impulse response of an continuous-time ideal low pass filter is the inverse continuous](https://reader034.vdocument.in/reader034/viewer/2022051902/5ff15f6dea4f632e1468ea1b/html5/thumbnails/45.jpg)
Hence, to uniquely identify a sequence from its z-transform, we have to specify additionally the ROC of the z-transform.
( ) ( ) ( )nununxnn
⎟⎠
⎞⎜⎝
⎛−+⎟⎠
⎞⎜⎝
⎛=31
21
( ) ( ) ( )
⎟⎠⎞
⎜⎝⎛ +⎟⎠⎞
⎜⎝⎛ −
⎟⎠⎞
⎜⎝⎛ −
=+
+−
=
⎟⎠
⎞⎜⎝
⎛−+⎟⎠
⎞⎜⎝
⎛=
⎟⎠
⎞⎜⎝
⎛−+⎟⎠
⎞⎜⎝
⎛=
−−
∞
=
−∞
=
−
∞
−∞=
−∞
−∞=
−
∑∑
∑∑
31
21
1212
311
1
211
1
31
21
31
21
11
00
zz
zz
zz
zz
znuznuzX
n
nn
n
nn
n
nn
n
nn
Then
Another example: given
![Page 46: Sampling and Reconstruction - 國立臺灣大學 · 2006. 11. 2. · Sampling and Reconstruction The impulse response of an continuous-time ideal low pass filter is the inverse continuous](https://reader034.vdocument.in/reader034/viewer/2022051902/5ff15f6dea4f632e1468ea1b/html5/thumbnails/46.jpg)
( )21
211
1 21
1>
−⎟⎠
⎞⎜⎝
⎛−
↔ zz
nuzn
,
( )31
311
1 31
1>
+⎟⎠
⎞⎜⎝
⎛−−
↔ zz
nuzn
,
Thus
( ) ( )21
311
1
211
1 31
21
11>
++
−⎟⎠
⎞⎜⎝
⎛−+⎟⎠
⎞⎜⎝
⎛−−
↔ zzz
nunuznn
,
![Page 47: Sampling and Reconstruction - 國立臺灣大學 · 2006. 11. 2. · Sampling and Reconstruction The impulse response of an continuous-time ideal low pass filter is the inverse continuous](https://reader034.vdocument.in/reader034/viewer/2022051902/5ff15f6dea4f632e1468ea1b/html5/thumbnails/47.jpg)
![Page 48: Sampling and Reconstruction - 國立臺灣大學 · 2006. 11. 2. · Sampling and Reconstruction The impulse response of an continuous-time ideal low pass filter is the inverse continuous](https://reader034.vdocument.in/reader034/viewer/2022051902/5ff15f6dea4f632e1468ea1b/html5/thumbnails/48.jpg)
Consider another two-sided exponential sequence
( ) ( ) ( )121
31
−−⎟⎠
⎞⎜⎝
⎛−⎟⎠
⎞⎜⎝
⎛−= nununxnn
Given
Since ( )31
311
1 31
1>
+⎟⎠
⎞⎜⎝
⎛−−
↔ zz
nuzn
,
and by the left-sided sequence example
( )21
211
1 121
1<
−−−⎟
⎠
⎞⎜⎝
⎛−−
↔ zz
nuzn
,
![Page 49: Sampling and Reconstruction - 國立臺灣大學 · 2006. 11. 2. · Sampling and Reconstruction The impulse response of an continuous-time ideal low pass filter is the inverse continuous](https://reader034.vdocument.in/reader034/viewer/2022051902/5ff15f6dea4f632e1468ea1b/html5/thumbnails/49.jpg)
( )⎟⎠⎞
⎜⎝⎛ −⎟⎠⎞
⎜⎝⎛ +
⎟⎠⎞
⎜⎝⎛ −
=−
++
=−−
21
31
1212
211
1
311
111 zz
zz
zzzX
Again, the poles and zeros are the same as the previous example, but the ROC is not.
![Page 50: Sampling and Reconstruction - 國立臺灣大學 · 2006. 11. 2. · Sampling and Reconstruction The impulse response of an continuous-time ideal low pass filter is the inverse continuous](https://reader034.vdocument.in/reader034/viewer/2022051902/5ff15f6dea4f632e1468ea1b/html5/thumbnails/50.jpg)
Some Common Z-transform Pairs
[ ]
[ ]
[ ]
[ ]
[ ]
[ ]
[ ] ( ) . :ROC 1
. :ROC 1
11
. :ROC 1
1. 0) (if or 0) (if 0except all :ROC
1. :ROC 1
11
1. :ROC 1
1. all :ROC 1
21
1
1
1
1
1
azaz
aznuna
azaz
nua
azaz
nua
mmzzmn
zz
nu
zz
nu
zn
n
n
n
m
>−
↔
<−
↔−−−
>−
↔
<∞>↔−
<−
↔−−−
>−
↔
↔
−
−
−
−
−
−
−
δ
δ
![Page 51: Sampling and Reconstruction - 國立臺灣大學 · 2006. 11. 2. · Sampling and Reconstruction The impulse response of an continuous-time ideal low pass filter is the inverse continuous](https://reader034.vdocument.in/reader034/viewer/2022051902/5ff15f6dea4f632e1468ea1b/html5/thumbnails/51.jpg)
Some Common Z-transform Pairs (continue)
[ ]( )
[ ] [ ] [ ][ ]
[ ] [ ] [ ][ ]
[ ] [ ] [ ][ ]
[ ] [ ] [ ][ ]
.0 :ROC 1
10
10
. :ROC 21
. :ROC 211
1. :ROC 21
1. :ROC 211
. :ROC 1
1
1
2210
10
0
2210
10
0
210
10
0
210
10
0
21
1
>−
−↔
⎪⎩
⎪⎨⎧ −≤≤
>+−
↔
>+−
−↔
>+−
↔
>+−
−↔
<−
↔−−−
−
−
−−
−
−−
−
−−
−
−−
−
−
−
zaz
zaotherwise
Nna
rzzrzwr
zwrnunwr
rzzrzwr
zwrnunwr
zzzw
zwnunw
zzzw
zwnunw
azaz
aznuna
NNn
n
n
n
cossin
sin
coscos
cos
cossin
sin
coscos
cos
![Page 52: Sampling and Reconstruction - 國立臺灣大學 · 2006. 11. 2. · Sampling and Reconstruction The impulse response of an continuous-time ideal low pass filter is the inverse continuous](https://reader034.vdocument.in/reader034/viewer/2022051902/5ff15f6dea4f632e1468ea1b/html5/thumbnails/52.jpg)
Properties of the ROC
• The ROC is a ring or disk in the z-plane centered at the origin; i.e., 0 ≤ rR < |z| ≤ rL ≤ ∞.
• The Fourier transform of x[n] converges absolutely iff the ROC includes the unit circle.
• The ROC cannot contain any poles• If x[n] is a finite-length (finite-duration) sequence,
then the ROC is the entire z-plane except possible z = 0 or z = ∞.
• If x[n] is a right-sided sequence, the ROC extends outward from the outermost (i.e., largest magnitude) finite pole in X(z) to (and possibly include) z=∞.
![Page 53: Sampling and Reconstruction - 國立臺灣大學 · 2006. 11. 2. · Sampling and Reconstruction The impulse response of an continuous-time ideal low pass filter is the inverse continuous](https://reader034.vdocument.in/reader034/viewer/2022051902/5ff15f6dea4f632e1468ea1b/html5/thumbnails/53.jpg)
Properties of the ROC (continue)
• If x[n] is a left-sided sequence, the ROC extends inward from the innermost (i.e., smallest magnitude) nonzero pole in X(z) to (and possibly include) z = 0.
• A two-sided sequence x[n] is an infinite-duration sequence that is neither right nor left sided. The ROC will consist of a ring in the z-plane, bounded on the interior and exterior by a pole, but not containing any poles.
• The ROC must be a connected region.
![Page 54: Sampling and Reconstruction - 國立臺灣大學 · 2006. 11. 2. · Sampling and Reconstruction The impulse response of an continuous-time ideal low pass filter is the inverse continuous](https://reader034.vdocument.in/reader034/viewer/2022051902/5ff15f6dea4f632e1468ea1b/html5/thumbnails/54.jpg)
Example
A system with three poles
![Page 55: Sampling and Reconstruction - 國立臺灣大學 · 2006. 11. 2. · Sampling and Reconstruction The impulse response of an continuous-time ideal low pass filter is the inverse continuous](https://reader034.vdocument.in/reader034/viewer/2022051902/5ff15f6dea4f632e1468ea1b/html5/thumbnails/55.jpg)
Different possibilities of the ROC. (b) ROC to a right-sided sequence. (c) ROC to a left-handed sequence.
![Page 56: Sampling and Reconstruction - 國立臺灣大學 · 2006. 11. 2. · Sampling and Reconstruction The impulse response of an continuous-time ideal low pass filter is the inverse continuous](https://reader034.vdocument.in/reader034/viewer/2022051902/5ff15f6dea4f632e1468ea1b/html5/thumbnails/56.jpg)
Different possibilities of the ROC. (b) ROC to a two-sided sequence. (c) ROC to another two-sided sequence.
![Page 57: Sampling and Reconstruction - 國立臺灣大學 · 2006. 11. 2. · Sampling and Reconstruction The impulse response of an continuous-time ideal low pass filter is the inverse continuous](https://reader034.vdocument.in/reader034/viewer/2022051902/5ff15f6dea4f632e1468ea1b/html5/thumbnails/57.jpg)
ROC vs. Linear System
• Consider the system function H(z) of a linear system:– If the system is stable, the impulse response h(n) is
absolutely summable and therefore has a Fourier transform, then the ROC must include the unit circle,.
– If the system is causal, then the impulse response h(n) is right-sided, and thus the ROC extends outward from the outermost (i.e., largest magnitude) finite pole in H(z) to (and possibly include) z=∞.
![Page 58: Sampling and Reconstruction - 國立臺灣大學 · 2006. 11. 2. · Sampling and Reconstruction The impulse response of an continuous-time ideal low pass filter is the inverse continuous](https://reader034.vdocument.in/reader034/viewer/2022051902/5ff15f6dea4f632e1468ea1b/html5/thumbnails/58.jpg)
Inverse Z-transform
• Given X(z), find the sequence x[n] that has X(z) as its z-transform.
• We need to specify both algebraic expression and ROC to make the inverse Z-transform unique.
• Techniques for finding the inverse z-transform:– Investigation method:
• By inspect certain transform pairs.• Eg. If we need to find the inverse z-transform of
From the transform pair we see that x[n] = 0.5nu[n].
( ) 15011
−−=
zzX
.
![Page 59: Sampling and Reconstruction - 國立臺灣大學 · 2006. 11. 2. · Sampling and Reconstruction The impulse response of an continuous-time ideal low pass filter is the inverse continuous](https://reader034.vdocument.in/reader034/viewer/2022051902/5ff15f6dea4f632e1468ea1b/html5/thumbnails/59.jpg)
Inverse Z-transform by Partial Fraction Expansion
• If X(z) is the rational form with
• An equivalent expression is
( )∑
∑
=
−
−
== N
k
kk
mM
mm
za
zbzX
0
0
( )∑
∑
∑
∑
=
−
−
=
=
−−
−
=
−
== N
k
kNk
M
mMM
mm
N
N
k
kNk
N
mMM
mm
M
zaz
zbz
zaz
zbzzX
0
0
0
0
![Page 60: Sampling and Reconstruction - 國立臺灣大學 · 2006. 11. 2. · Sampling and Reconstruction The impulse response of an continuous-time ideal low pass filter is the inverse continuous](https://reader034.vdocument.in/reader034/viewer/2022051902/5ff15f6dea4f632e1468ea1b/html5/thumbnails/60.jpg)
Inverse Z-transform by Partial Fraction Expansion (continue)
• There will be M zeros and N poles at nonzero locations in the z-plane.
• Note that X(z) could be expressed in the form
where ck’s and dk’s are the nonzero zeros and poles, respectively.
( )( )
( )∏
∏
=
−
=
−
−
−
= N
mk
M
mm
zd
zc
ab
zX
1
1
1
1
0
0
1
1
![Page 61: Sampling and Reconstruction - 國立臺灣大學 · 2006. 11. 2. · Sampling and Reconstruction The impulse response of an continuous-time ideal low pass filter is the inverse continuous](https://reader034.vdocument.in/reader034/viewer/2022051902/5ff15f6dea4f632e1468ea1b/html5/thumbnails/61.jpg)
Inverse Z-transform by Partial Fraction Expansion (continue)
• Then X(z) can be expressed as
Obviously, the common denominators of the fractions in the above two equations are the same. Multiplying both sides of the above equation by 1−dkz−1 and evaluating for z = dk shows that
( ) ∑=
−−=
N
k k
k
zdA
zX1
11
( ) ( )kdzkk zXzdA =
−−= 11
![Page 62: Sampling and Reconstruction - 國立臺灣大學 · 2006. 11. 2. · Sampling and Reconstruction The impulse response of an continuous-time ideal low pass filter is the inverse continuous](https://reader034.vdocument.in/reader034/viewer/2022051902/5ff15f6dea4f632e1468ea1b/html5/thumbnails/62.jpg)
Example
• Find the inverse z-transform of
X(z) can be decomposed as
Then
( )( )( ) ( )( ) 2
1 211411
111 >
−−=
−−z
zzzX
//
( )( ) ( )( )( ) ( ) 2211
1411
211
2
411
1
=−=
−=−=
=−
=−
/
/
/
/
z
z
zXzA
zXzA
( )( )( ) ( )( )1
21
1
211411 −− −+
−=
zA
zAzX
//
![Page 63: Sampling and Reconstruction - 國立臺灣大學 · 2006. 11. 2. · Sampling and Reconstruction The impulse response of an continuous-time ideal low pass filter is the inverse continuous](https://reader034.vdocument.in/reader034/viewer/2022051902/5ff15f6dea4f632e1468ea1b/html5/thumbnails/63.jpg)
Example (continue)
• Thus
From the ROC, we have a right-hand sequence. So
( )( )( ) ( )( )11 211
24111
−− −+
−
−=
zzzX
//
[ ] [ ] [ ]nununxnn
⎟⎠
⎞⎜⎝
⎛−⎟⎠
⎞⎜⎝
⎛=41
212
![Page 64: Sampling and Reconstruction - 國立臺灣大學 · 2006. 11. 2. · Sampling and Reconstruction The impulse response of an continuous-time ideal low pass filter is the inverse continuous](https://reader034.vdocument.in/reader034/viewer/2022051902/5ff15f6dea4f632e1468ea1b/html5/thumbnails/64.jpg)
• Find the inverse z-transform of
Since both the numerator and denominator are of degree 2, a constant term exists.
B0 can be determined by the fraction of the coefficients of z−2, B0 = 1/(1/2) = 2.
Another Example
( ) ( )( )( )( ) 1
12111
11
21>
−−
+=
−−
−
zzz
zzX/
( )( )( ) ( )1
21
10 1211 −− −
+−
+=z
Az
ABzX/
![Page 65: Sampling and Reconstruction - 國立臺灣大學 · 2006. 11. 2. · Sampling and Reconstruction The impulse response of an continuous-time ideal low pass filter is the inverse continuous](https://reader034.vdocument.in/reader034/viewer/2022051902/5ff15f6dea4f632e1468ea1b/html5/thumbnails/65.jpg)
From the ROC, the solution is right-handed. So
Another Example (continue)
( )( )( ) ( )
( )( )( ) ( )( )
( )( )( )( ) 811211
512
92111211
512
12112
11
11
1
2
211
11
1
1
12
11
=−−−
+−+=
=−−−
+−+=
−+
−+=
=−
−−
−
=−
−−
−
−−
z
z
zzz
zA
zzz
zA
zA
zAzX
/
//
/
/
( )( )( ) ( )
[ ] [ ] ( ) [ ] [ ]nununnx
zzzX
n 82192
18
21192 11
+−=
−+
−−=
−−
/
/
δ
![Page 66: Sampling and Reconstruction - 國立臺灣大學 · 2006. 11. 2. · Sampling and Reconstruction The impulse response of an continuous-time ideal low pass filter is the inverse continuous](https://reader034.vdocument.in/reader034/viewer/2022051902/5ff15f6dea4f632e1468ea1b/html5/thumbnails/66.jpg)
• We can determine any particular value of the sequence by finding the coefficient of the appropriate power of z−1 .
Power Series Expansion
( ) [ ]
[ ] [ ] [ ] [ ] [ ] ...... ++++−+−+=
=
−−
∞
−∞=
−∑212 21012 zxzxxzxzx
znxzXn
n
![Page 67: Sampling and Reconstruction - 國立臺灣大學 · 2006. 11. 2. · Sampling and Reconstruction The impulse response of an continuous-time ideal low pass filter is the inverse continuous](https://reader034.vdocument.in/reader034/viewer/2022051902/5ff15f6dea4f632e1468ea1b/html5/thumbnails/67.jpg)
• Find the inverse z-transform of
By directly expand X(z), we have
Thus,
Example: Finite-length Sequence
( ) ( )( )( )1112 11501 −−− −+−= zzzzzX .
( ) 12 50150 −+−−= zzzzX ..
[ ] [ ] [ ] [ ] [ ]1501502 −+−+−+= nnnnnx δδδδ ..
![Page 68: Sampling and Reconstruction - 國立臺灣大學 · 2006. 11. 2. · Sampling and Reconstruction The impulse response of an continuous-time ideal low pass filter is the inverse continuous](https://reader034.vdocument.in/reader034/viewer/2022051902/5ff15f6dea4f632e1468ea1b/html5/thumbnails/68.jpg)
• Find the inverse z-transform of
Using the power series expansion for log(1+x) with |x|<1, we obtain
Thus
Example
( ) ( ) azazzX >+= − 1 1log
( ) ( )∑∞
=
−+−=
1
11
n
nnn
nzazX
[ ] ( )⎪⎩
⎪⎨⎧
≤≥−=
+
0011 1
nnnanx
nn /
![Page 69: Sampling and Reconstruction - 國立臺灣大學 · 2006. 11. 2. · Sampling and Reconstruction The impulse response of an continuous-time ideal low pass filter is the inverse continuous](https://reader034.vdocument.in/reader034/viewer/2022051902/5ff15f6dea4f632e1468ea1b/html5/thumbnails/69.jpg)
• Suppose
• Linearity
Z-transform Properties
[ ] ( )
[ ] ( )
[ ] ( )2
1
ROC
ROC
ROC
22
11
x
z
x
zx
z
RzXnx
RzXnx
RzXnx
=↔
=↔
=↔
[ ] [ ] ( ) ( )21
ROC2121 xx
zRR z bXz aX nb x nax ∩=+↔+
![Page 70: Sampling and Reconstruction - 國立臺灣大學 · 2006. 11. 2. · Sampling and Reconstruction The impulse response of an continuous-time ideal low pass filter is the inverse continuous](https://reader034.vdocument.in/reader034/viewer/2022051902/5ff15f6dea4f632e1468ea1b/html5/thumbnails/70.jpg)
• Time shifting
• Multiplication by an exponential sequence
Z-transform Properties (continue)
[ ] ( ) xn
zR zX z nnx ROC0
0 =↔− − (except for the possible addition or deletion of z=0or z=∞.)
[ ] ( ) x
zn Rz zz X nxz 000 ROC =↔ /
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• Differentiation of X(z)
• Conjugation of a complex sequence
Z-transform Properties (continue)
[ ] ( )x
zR
dzzdXz nnx ROC =−↔
[ ] ( ) x
zR z X nx ROC =↔ ***
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• Time reversal
If the sequence is real, the result becomes
• Convolution
Z-transform Properties (continue)
[ ] ( )x
z
R z X nx 1 ROC1 =↔− /
[ ] ( )x
z
R z X nx 1 ROC1 =↔− */**
[ ] [ ] ( ) ( )21
contains ROC2121 xx
zRR zXz X nxnx ∩↔∗
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• Initial-value theorem: If x[n] is zero for n<0 (i.e., if x[n] is causal), then
• Inver z-transform formula:
Z-transform Properties (continue)
[ ] ( )zX xz ∞→
= lim0
Find the inverse z-transform by Integration in the complex domain
dzzzXj
zXZ n 11 )(21)}({ −− ∫=π
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1. A finite-length sequence is non-zero only at a finite number of positions. If m and n are the first and last non-zero positions, respectively, then we call n−m+1 the length of that sequence. What maximum length can the result of the convolution of two sequences of length k and l have?
2. An LTI system is described by the following difference equation: y[n] = 0.3y[n-1] + y[n-2] − 0.2y[n-3] + x[n]. Find the system function and frequency response of this system.
3. Find the inverse z-transform of the following system:
Homework #2
( )31||
)31)(1(
)1(<
++
−= z
zz
zzzY