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Chapter 1: Introduction to z-Transform
1.1 z-Transform1.2 Properties
1.2.1 Linearity1.2.2 Shifting1.2.3 Time Reversal1.2.4 Multiplication by Exponential Sequence1.2.5 Differentiation in the z-domain1.2.6 Discrete Convolution
1.3 Inverse z-transform1.3.1 Relationship between the z-transform and the
Laplace Transform1.4 Frequency Response Estimation1.5 Pole-Zero Description of Discrete-Time Systems1.6 A Second-Order Resonant System1.7 Problem Sheet B1Prof
esso
r E. A
mbikair
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UNSW, A
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Chapter 1: Introduction to z-Transform [1,3]
s
aa f
fT πωθ 2==
Analogue Domain(Continuous-time Domain)
x(t)
X(s)
LaplaceTransform
s=jω
X(ω)
ω-analoguefrequency
jωS-PlaneStable
Region
Discrete-time Domain
x[n]
X(z)
z-transform
z=ejθ
X(ejθ)
-π≤θ≤π
t = nT
Im(z) z-Plane
θ = πθ = -π Re(z)
stable region
|z|=1(unit circle)
θσ
θ
-digital
frequency
θ -digital frequency
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The primary role of the Laplace transformin engineering is transient and stability analysis of causal LTI system described by differential equations.
The primary roles of the z-transform are the study of system characteristics and derivation of computational structures for implementing discrete-time systems on computers. The-transform is also used solve difference equations.
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1.1 z-TransformThe z-transform of a discrete-time signal
In causal systems, x[n] may be zero when n < 0
[ ]∑∞
−∞=
−=n
nznxzX )( z = rejθ, θ = digital frequencywhere z is a complex variable
two-sided transform
[ ]∑∞
=
−=0
)(n
nznxzX one-sided transform
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Clearly, the z-transform is a power series with an infinite number of terms and so may not converge for
all values of z.
The region where the z-transform converges is known as the region of convergence (ROC) and in this region the values of X(z) are finite
The step sequence: [ ]⎩⎨⎧
<∞≤≤
=00
01n
nnx
...11)( 21
0+++=⋅= −−
∞
=
−∑ zzzzXn
n
This is a geometric series with a common ratio of z-1.The series converges if |z-1| < 1 or equivalently if |z| > 1Prof
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In this case, the z-transform is valid everywhere outside a circle of unit radius whose centre is at the origin (see below)
111...1)( 1
21
−=
−=+++= −
−−
zz
zzzzX
Im(z)
region of convergenceRe(z)
|z|=1 is a circle of unit radius referred to as the ‘unit circle’
|z|=1
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221
1...221)(2 121 =
−=+++=⇒= −
−−zXzLet
( ) ( ) ( ) KK +++=+++=⇒= −− 42115.0 1211
21zXzLet
⎭⎬⎫
⎩⎨⎧
−=
1)(
zzzX
|z| > 1 then X(z) converges and |z| < 1 then it diverges
So the Region Of Convergence (ROC) is seen to be bounded
by the circle |z| = 1, the radius of the pole of
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The delta sequence δ[n]:
[ ] [ ] 1}{0
== ∑∞
=
−
n
nznnZ δδ
11
1)(0
<−
=−
==∑∞
=
−
zafor
azz
zazazX
n
nn
1)(
−=
zzzX
The geometric sequence: x[n] = an
or equivalently, |z| > |a|.
when a = 1, x[n] = 1 for n ≥ 0 ie. x[n] = u[n]
ROC: |z|>1Profes
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The complex exponential sequence: x[n] = ejnθ
{ }
1cos2)sincos(
1)()(
))(()(
1
1
2
2
0
+−+−
=
++−−
=−−
−=
−−
×−
=−
=−
==
−
−
−
−
−
−∞
=
−∑
θθθ
θθ
θ
θθ
θ
θ
θ
θθθθθ
zzjzz
zeezezz
ezezezz
ezez
ezz
ezz
ze
zeez
jj
j
jj
j
j
j
jjjn
nnjnj
1cos2sin
1cos2)cos(}sin{cos
1cos2sin
1cos2)cos(}{
22
22
+−+
+−−
=+
+−+
+−−
=
θθ
θθθθ
θθ
θθθ
zzzj
zzzznjnZ
zzzj
zzzzeZ nj
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Exploiting the linearity property
1cos2sin
1cos2)cos(}sin{cos
1cos2sin
1cos2)cos(}{
22
22
+−+
+−−
=+
+−+
+−−
=
θθ
θθθθ
θθ
θθθ
zzzj
zzzznjnZ
zzzj
zzzzeZ nj
1cos2sin}{sin
1cos2)cos(}{cos
2
2
+−=
+−−
=∴
θθθ
θθθ
zzznZ
zzzznZ
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1.2 Properties
1.2.1 Linearity [ ] [ ] )()( zbYzaXnbynax Z +⎯→←+
1.2.2 Shifting Property (Delay Theorem):
[ ] )(zXzknx kZ −⎯→←−
A very important property of the z-transform is the delay theorem. [ ]{ } ( )
[ ]{ } ( )zXznxZzXznxZ
2
1
21
−
−
=−
=−
z-1
Unit delay
x[n]
X(z)
x[n-1]
z-1X(z) Tx[n] x[n-1]
One sample delay
≡
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1.2.3 Time reversal:
[ ] ( )11 −⎟⎠⎞
⎜⎝⎛⎯→←− zXor
zXnx z
Example: Find the z-transform of x[n] = -u[n].
[ ] [ ]zz
zznuz
zznu−
=−
⎯→←−−
⎯→← −
−
11
1;
1 1
1
ROC: |z| < 1
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1.2.4 Multiplication by exponential sequence:
[ ] )( 1zaXnxa zn −⎯→←
[ ] )( zeXznxe jnj θθ −⎯→←
As a special case if x[n] is multiplied by ejnθ
1.2.5 Differentiation in the z-domain:
[ ] )(zXdzdzznnx −⎯→←
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1.2.6 Discrete Convolution:If y[n] = x[n] * h[n] then Y(z) = X(z) . H(z)
[ ] [ ] [ ] [ ] [ ] [ ]
[ ] [ ] [ ] [ ]
[ ] [ ] [ ] [ ]
[ ] [ ] [ ] [ ]⎭⎬⎫
⎩⎨⎧
=
−=⎭⎬⎫
⎩⎨⎧
−−=
⎭⎬⎫
⎩⎨⎧
−−⋅=
⎭⎬⎫
⎩⎨⎧
−−⋅=
∑∑
∑∑
∑ ∑
∑
∞
−=
−−∞
−∞=
∞
=
−∞
−∞=
∞
=
−∞
−∞=
∞
−∞=
km
km
k
n
n
k
n
n
k
k
zmhmukxku
knmLetzknhknukxku
zknhknukxku
knhknukxkuZnhnxZ
0
0
}*{
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[ ] [ ] [ ] [ ]
[ ] [ ]
)()()()(
0 0
zHzXzHzX
zmhzkx
zmhmuzkxku
k m
mk
km
m
k
k
⋅=
⎭⎬⎫
⎩⎨⎧
=
⎭⎬⎫
⎩⎨⎧
=
∑ ∑
∑∑∞
=
∞
=
−−
∞
−=
−∞
−∞=
−
y[n] = x[n] * h[n]Y(z) = X(z) ⋅ H(z)
h[n]H(z)
x[n]
X(z)
y[n]
Y(z)
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Example :
Concept of the transfer function
]2[]1[]2[]1[][][ 21210 −+−+−+−+= nybnybnxanxanxany
Take the z-transform of both sides:
( ) ( ) ( ) ( ) ( ) ( )( )[ ] ( )[ ]( ) ( )
( ) 22
11
22
110
22
110
22
11
22
11
22
110
1
1
−−
−−
−−−−
−−−−
−−++
==
++=−−
++++=
zbzbzazaa
zXzYzH
zazaazXzbzbzY
zzYbzzYbzzXazzXazXazY
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Example :Find the difference-equation of the following transfer function
First rewrite H(z) as a ratio of polynomials in z-1
2325)( 2 ++
+=
zzzzH
( ) ( ) ( ) ( ) ( ) 2121
21
21
2523231
25)()()(
−−−−
−−
−−
+=++
+++
==
zzXzzXzzYzzYzYzz
zzzHzXzY
Take inverse z-transform
[ ] [ ] [ ] [ ] [ ]22132215 −−−−−+−=⇒ nynynxnxny
[ ] [ ] [ ] [ ] [ ]22152213 −+−=−+−+ nxnxnynyny
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Example : If x[n] = u[n] – u[n-10], find X(z)?
1
109
0 11)1()( −
−
=
−
−−
== ∑ zzzzX
n
n
Example:
[ ]
k
k
k
kn
nn
nn
z
zz
znuzY
∑
∑∑
∑
∞
=
∞
=
−
−∞=
∞
−∞=
−
⎟⎠⎞
⎜⎝⎛−=
⎟⎠⎞
⎜⎝⎛−=⎟
⎠⎞
⎜⎝⎛−=
−−−=
0
1
1
1
1)(
α
αα
α
If y[n] = -αnu[-n-1], find Y(z)?
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-1
u[n+1]
n -1
u[-(n+1)]
n
1<αz
||||
||||1
11)( 1
αα
αα
<−
=
<−
−= −
zz
z
zz
zY
The sum converges provided ie. |z| < |α|
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Depict the ROC and pole and zero locations in the z-plane
Im(z)
Re(z)α
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1.2.6 Discrete ConvolutionCompute the convolution y[n] of the digital signals given byx1[n] = [1, -2, 1];x2[n] = 1 for 0 ≤ n ≤ 5, x2[n] = 0 elsewherey[n] = x1[n] * x2[n] ⇒ Y(z) = X1(z) ⋅X2(z)X1(z) = 1 –2z-1 + z-2
X2(z) = 1 + z-1 + z-2 + z-3 + z-4 +z-5
Y(z) = X1(z) ⋅ X2(z)= 1-z-1 – z-6 + z-7
Inverse z-transform
y[n] = [1, -1, 0,0,0,0,-1,1]Profes
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Example :
Determine the system function H(z) of the system shown below:
+
T
x[n]y[n]
aay[n-1]
y[n] = x[n] + ay[n-1]
+
z-1
X(z) Y(z)
aaz-1Y(z)
Y(z) = X(z) + az-1Y(z)
111
)()()( −−==
azzXzYzH Prof
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Basic z-Transforms
|z|>rrnsin(nθ) u[n]
|z| > rrncos(nθ) u[n]
|z| > 1sin(nθ) u[n]
|z|>1cos(nθ) u[n]
|z| > |α|nαn u[n]
|z| > |α|αn u[n]
|z| > 1u[n]
all z1δ[n]
ROCTransformSignal
111
−− z
111
−− zα
21
1
)1( −
−
− zzαα
21
1
cos21cos1
−−
−
+−−
zzz
θθ
21
1
cos21sin
−−
−
+− zzz
θθ
221
1
cos21cos1
−−
−
+−−
zrrzrzθ
θ
221
1
cos21sin1
−−
−
+−−
zrrzrzθ
θProfes
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z-Transform Properties
nx[n]
X(z)⋅Y(z)x[n]*y[n]
x[-n]
anx[n]
z-kX(z)X[n-k]
aX(z) + bY(z)ax[n] + by[n]
X(z)X[n]
Transformsignal
⎟⎠⎞
⎜⎝⎛
azX
⎟⎠⎞
⎜⎝⎛
zX 1
)(zXdzdz−Prof
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Note:
If N →∞ , |z-1| < 1
1)1(21
11...1 −
−−−−−
−−
=++++=zzzzzS
NN
N
111
−∞ −=
zS
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1.3 Inverse z-Transform
The inverse z-transform-Partial fraction
The inverse z-transform allows us to recover the discrete-time sequence.
x[n] = Z-1[X(z)]
where X(z) is the z-transform of x[n].
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Example :Find x[n] for the following:
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
+
⎟⎠⎞
⎜⎝⎛
−−
⎟⎠⎞
⎜⎝⎛
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
+
⎟⎠⎞
⎜⎝⎛−
+−
⎟⎠⎞
⎜⎝⎛
=
⎥⎦⎤
⎢⎣⎡
++
−=
+−=
−−=
−−= −−
−
5.054
75.054
5.054
75.054
5.075.0)5.0)(75.0(375.025.0
375.025.01)(
2
21
1
z
z
z
z
zzz
zB
zAz
zzz
zzz
zzzzX
[ ] 0,)5.0()75.0(54
5.054
75.054][ 11
>−−=
⎥⎦⎤
⎢⎣⎡
+−⎥⎦
⎤⎢⎣⎡
−= −−
n
zzz
zzznx
nnProfes
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Power series Method
Residue Method - evaluating the contour integral
1.3 Inverse z-Transform
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1.3.1 Relationship between the z-transform and Laplace transform.
If we let z = esT, then z = e(σ +jω)T (T is the sampling period)
z = eσ T⋅ ejωT = eσ T⋅ ejθ -π ≤ θ ≤ π
Thus |z| = eσ T and
θπω ===∠sf
fTz 2 (θ = digital frequency)
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As ω varies from -∞ to ∞ the s-plane is mapped to the z-plane as shown in Figure 2.10.
The entire jω axis in the s-plane is mapped onto the unit circle. The left-hand s-plane is mapped to the inside of the unit circle and the right-hand s-plane maps to the outside of the unit circle.
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jω
s-plane
σ
z-plane
θ = 0
If θ = πUnit circle|z|=1
θ = πθ = -π
(Half thesampling frequency)
Figure 2.10. Mapping of the s-plane to the z-plane.
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In terms of frequency response, the jω axis is the most important in the s-plane. In this
case σ = 0 and the frequency points in the s-plane are related to points on the z-plane unit circle by z = eσT.ejωT = 1 ejθ
θjez =
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1.4 Frequency Response Estimation
There are many instances when it is necessary to evaluate the frequency response of discrete-time systems. The frequency response of a system can be readily obtained from its z-transforms.
For example, if we set z = ejθ, that is evaluate the z-transform around the unit circle, we obtain the Fourier Transform of the system.
πθπθ θ <<==
- , |)()( jezzHH
Fourier transform of the discrete-time systemProfes
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Example :
Find H(θ). {H(θ)-the frequency response}
Fourier transform of the discrete-time system.
6.0say 01 , 1
1)( 1 =<<−
= − aaaz
zH
( ) πθπθ θ ≤≤−==
|)( jezzHH
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( ) ( ) ( ) θθθ θθ sincos1
11
1|jaaae
zHH jez j+−
=−
== −=
( ) ( ) 222 cos211
sincos1
1)(aaaa
H+−
=+−
=θθθ
θ
θ = ωT; θ = 2π f/fs:; θ = π ⇒ f = fs/2fs= sampling frequency
|H(θ)|
-π πθ
-fs/2 fs/20 f (analogue frequency)
digital frequency
a−11
a+11
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1.5 Pole-Zero Description of Discrete-Time Systems
The zeros of a z-transform H(z) are the values of zfor which H(z)=0. The poles of a z-transform are
the values of z for which H(z)=∞ . If H(z) is a rational function , then
))......()(()).......()((
......1...
)()()(
21
210
11
110
L
M
LL
MM
pzpzpzzzzzzza
zbzbzazaa
zXzYzH
−−−−−−
=
++++++
== −−
−−
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The complex quantities (or may be real) z1, z2, z3 ….
are called zeros of H(z) and the complex quantities
(or may be real) p1, p2, p3 … are called the poles of
H(z). We thus see that H(z) is completely
determined , except for the constant a0, by the
values of poles and zeros
The information contained in the z-transform can be conveniently displayed as a poles-zero diagram (see figure in the next slide)
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0.75
|z|=1 0.5
-0.5
Im(z)
Re(z)-1
In the diagram, ‘X’ marks the position of a pole and ‘O’denotes the position of a zero.
The poles are located at z = 0.5 ± 0.5j and z = 0.75, a single zero is at z = -1.Prof
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An important feature of the pole-zero diagram is the unit circle |z|=1. The pole-zero diagram provides an insight into the properties of a given discrete-time system.
From the locations of the poles d zeros we can infer the frequency response of the system as well as its degree of stability.
For a stable system, all the poles must lie inside the unit circle. Zeros may lie inside, on, or outside the unit circle.Prof
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Example :Determine the transfer function H(z) of a discrete-time system with the pole-zero diagram shown below:
0.5|z| =1 0.5
-0.5
Im(z)
Re(z)-1
21
2
5.01)1(
)5.05.0()5.05.0()1()1()(
−−
−
−−+
=
+−−−+−
=
zzzK
jzjzjzjzKzH
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Example :
Determine the pole-zero plot:
azzzH−
=)(
|z|=a
Im(z)
Re(z)
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Example :
Determine the pole-zero plot :
121
1
)().........)(()(
)( −−
−−−=
−−
= MM
MM
zMzzzzz
azzazzH
The zero z = a cancels the pole at z = a. Thus, H(z)has M-1 zeros and M-1 poles as shown in the diagram below for M = 8.
|z|=a
Im(z)
Re(z)
z= a
8 polesProfes
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Consider a system, H(z) with two complex conjugate poles in the z-plane :
|z|=1
Im(z)
Re(z)θθ
r p1
p2
θ
θ
j
j
rep
rep−=
=
2
1 Poles
)(01 zeroz =
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A typical transfer function might be:
( ) ( )
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−−
−=
⎥⎦⎤
⎢⎣⎡
−+
−=
−−=
−
−−
θθ
θθθθ
θθjj
jjjj
rezjr
rezjrz
rezB
rezAz
rezrezzzH
sin21
sin21
)(
⎥⎦⎤
⎢⎣⎡
−−
−= −−− 11 1
11
1)sin2(
1)(zrezrerj
zH jj θθθProfes
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( ) ( )[ ][ ]njnj
n
njnj
eerjr
rererj
nh
θθ
θθ
θ
θ
−
−
−=
−=
sin2
sin21)(
θ = frequency of oscillation
[ ]θθ
nrnh n sinsin
1)( 1−=
This is the impulse response of the 2nd order system with complex poles
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We note that the impulse response will decay away to zero provided r is less than one. [ r < 1]
Recall that r is also the distance from the origin in
the z-plane to the poles p1 or p2, so that system will be stable if the poles in the z-plane lie inside the unit circle.
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Example :Poles inside unit circle
θ-θ
Exponential decay sinewave (r<1)Stable system
rn
[ ]θθ
nrnh n sinsin
1)( 1−=
Poles on theunit circle
θ
r = 1
-1
1
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Poles outside unit circleθ
-θ
Exponential increasing (r>1)
rn
[ ]θθ
nrnh n sinsin
1)( 1−=
one real Pole inside the unit circle rn, θ = 0
n
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Note :
A system that is both stable and casual must have all its poles inside that unit circle within the z-plane.
We cannot have a pole outside the unit circle, since the inverse transform of a pole located outside the circle will contribute either a right sided increasing exponential term, which is not stable, or a left-sided decaying exponential term that is not causal.
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1.6 A second order Resonant System (Complex Poles)
z -1
z -1
+y[n]x[n]
-b1
-b2
θ0r
002
001
sincos
sincos0
0
θθ
θθθ
θ
jrrrep
jrrrepj
j
−==
+==−
(A) 1
1)(21
2
2
22
11 bzbz
zzbzb
zH++
=++
= −−Profes
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All pole system has poles only (without counting the zeros at the origin)
(B) cos2)(
)(
))(())(()(
20
2
2
22
2
2
21
2
21
12
2
00
00
rzrzz
rzeerzzzH
rezrezz
pzpzz
bzbzzzH
jj
jj
+−=
++−=
−−=
−−=
++=
−
−−
θθθ
θθ
01 cos2 θrb −= 22 rb =
2
10 2 b
bCos −=∴ θ
sff0
02πθ =
Comparing (A) and (B), we obtain
θ0 = resonant frequencyProfes
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We can derive H(θ) and the magnitude from
22
111
1)( −− ++=
zbzbzH
0.99-1.41
0.9-1.34
0.7-1.16
0.5-0.94
b2 = r2b1
⎥⎥⎦
⎤
⎢⎢⎣
⎡ −=
−==
−
2
110
01
22
2cos
)cos(2
bb
rbrb
θ
θ
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40πθ =4
π− ππ−
IV
I
θ
dB
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-3 -2 -1 0 1 2 3-20
-10
0
10
20
30
40
50dB
theta
Magnitude response
I II IIIIV
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Example :sketch the magnitude response for the system having the transfer function
The system has a zero at z = -1 & poles at
)9.01()9.01(
1)(1414
1
−−−
−
−−
+=
zeze
zzHjj ππ
49.0πj
ez−
=
∴ Magnitude response will be zero at θ = π and large at θ0= ± π /4 because the poles are close to the unit circle.
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-π π-π/4 π/4
|H(θ)|
θ
Magnitude Response π/40.9
θ=0θ = -π
θ = π
θ
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Example :Sketch the approximate magnitude response from the pole-zero map given below:
|z|= 1
+ 0.8
- 0.8Re(z)
Im(z)dB
-π -π/2 0 π/2 π
-fs/2fs/2
θ
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Example :
Sketch an approximate magnitude response from the pole-zero map given below:
+ 0.5
- 0.5 Re(z)1
|z|=1
Im(z)
-π -π/2 0 π/2 π θ
|H(θ)| in dB
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Summary of Part B Chapter 1At the end of this chapter, it is expected that you should know:
The properties of z transforms and their application.
Discrete convolution in the time and z domains.
How to find the inverse z transform, given a transfer function.
The difference between a z transform and a Laplace transform and when to use each.
Estimation of frequency response from a transfer function
Hand-calculate magnitude and phase responses for simple transfer functions and plot them.Prof
esso
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The pole-zero description of a discrete time system.
Given a pole-zero diagram, transfer function or difference equation, how to find any of the other representations and discuss the system’s stability with reference to the pole-zero diagram.
How to derive the resonance frequency equation for a second-order resonance system (a complex pole pair).
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Examples
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A digital filter structure is shown below. Determine the Transfer function H(z).
Z-1
Z-1
X(z) a0
a0
+
-b1
b2
Y(z)
+
+
;1
)()()(
22
11
100
−−
−
−++
==zbzb
zaazHzXzY
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Sketch an approximate magnitude response from the pole-zero map given below:
Im(z)
Re(z)1
-0.5
-1
0.5 x
x
-π -π/2 0 π/2 π θ
|H(θ)|
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Question 5(b)Determine the magnitude response of the following filter and show that it has an all-pass
characteristic.
[4 marks]1)1(
)( 1
1
<−−
= −
−
azazazH
filter Pass -All
⇒
=−−+−−+
==
−−
=→−−
=+=
−
−
−
−−
11
1)()().(
1)(
1)()1(
21)(
2
22*
*2
θθ
θθ
θ
θ
θ
θ
θθθ
θθ
jj
jj
j
j
j
j
aeaeaaeaeaHHH
aeeaH
aeeaHzzH
)(θH
θ-π π
1
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A first-order digital filter is described by the system function :
11
1111)( −−
−++−
=zazb
bazH
Draw a canonic realisation of the transfer function H(z).
Z-1
X(z)
b
+
a
+ Y(z)
1-a/1-b
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Sketch roughly the magnitude response corresponding to the pole-zero pattern given below:
Question 3
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Sketch an approximate magnitude response from the pole-zero map given below:
Im(z)
Re(z)1
-0.5
-1
0.5 x
x0.5
θ
Mag
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The difference equation of a digtal filter is given by y(n) = x(n)-x(n-8)- y(n-2 )
Find the transfer function for the above filter.
28
28
11).1(
)()(
)()()()(
−−
−−
+−=
−−=
zz
zXzY
zzYzzXzXzY
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A first-order digital filter has a transfer function given by
Determine the impulse response of the above digital filter H(z) and show that the filter is stable if a < 1.
11
11)( −−
−+=
zazkzH
⎥⎦
⎤⎢⎣
⎡−
+−
= −
−
− 1
1
1 111)(
azz
azkzH
[ ])1()()( 1 −+= − nuanuaknh nn
if a< 1, then h(n) decays to zero.Therefore, stable system.
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