Z-TRANSFORM - Roma Tre Universityhost.uniroma3.it/.../sp4bme/documents/LectureZtransform.pdf ·...
Transcript of Z-TRANSFORM - Roma Tre Universityhost.uniroma3.it/.../sp4bme/documents/LectureZtransform.pdf ·...
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Z-TRANSFORM
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In today’s class…
Z-transform Unilateral Z-transform Bilateral Z-transform Region of Convergence
Inverse Z-transform Power Series method Partial Fraction method
Solution of difference equations
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Need for transformation?
Why do we need to transform our signal from one domain to another?
Information available in one domain is not sufficient for complete analysis Looking at a sine wave in time-domain, we cannot really
know the frequency content
So we have to look into the frequency domain
An alternate domain may express the information more comprehensively A pole-zero map easily tells whether a systems is stable or
not
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Z-transform
Digital counterpart for the Laplace transform used for analog signals
Mathematically defined as,
X (z) x[n] z n
n
This equation is in general a power series, where z is a complex variable.
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Derivation…
The continuous-time Fourier transform of x(t) is given as,
Fxt xt e j 2ft dt
And the discrete-time Fourier transform of x[nT] is given as,
n
D xnT e F xnT j 2fnT
The Z-transform of x[n] is given as the Fourier transform of x[n] multiplied by rn
nT D D n x F r x nT
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D x nT
D x nT F D r x nT n
r n x nT e j 2 fnT
n
Bi-lateral Z-transform
n
D
n
n
D
n D
x nT z n x nT
z re
x nT re x nT
x nT
j 2 fT
j 2 fT
x nT r n e j 2 fnT
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Uni-lateral Z-transform
n
n D xnT z xnT
n0
X z xnT z n
n0
xnT xnT z n
where z-1 would show a delay by one sample time
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Example 1: Find the z-transform of the following finite-length sequence
y nT
- 2 - 1 4 5 6 7
4 3 . 5
3 2 . 5
2 1 . 5
1 0 .5
0 0 1 2 3
0 0 0 2 0 4 3 2 0 0 ynT ynT 2 n 1T 4 n 3T 3 n 4T 2 n 5T
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2 0 4 3 2
2 0 4 3 2 0 0
Y z 0
ynT 0
0 0z n
Y z ynT z n
n0
n0
Y z 0z 0 2z 1 0z 2 4z 3 3z 4 2z 5 0z 6 0z 7
Y z 2z 1 4z 3 3z 4 2z 5
So, Y(z) would exist on the entire z-plane except the point z=0
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Z-transform as Rational Function
Q(z)
Often it is convenient to represent Z-transform X(z) as a rational function
X (z) P(z)
Where P(z) and Q(z) are two polynomials The values of z at which X(z) becomes zero (X(z) = 0) are called zeros The values of z at which X(z) becomes infinite (X(z) = ∞), are called poles
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Significance of Poles & Zeros
Poles Roots of the denominator Q(z)
The point where H(z) becomes infinite
The point where H(ejw) shows a peak value
System may become unstable
Zeros Roots of the numerator P(z)
The point where H(z) becomes zero
The point where H(ejw) shows maximum attenuation
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Convergence issues
A power series may not necessarily converge The infinite sum may not always be finite
The set of values of z for which the z-transform
converges is called Region of Convergence (RoC)
The convergence of X(z) depends only on z and it converges for
n | x [ n ] z | n
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Replacing
n
n jw x [ n ] re
jw re z re j 2 fn
x [ n ] r n e jwn
| x [ n ] z n | n
X ( z )
1
n
x[n]r n n0
X (z) x[n]r n
n
This equation can be segmented into two parts, one for the right-sided (causal) signal and second for the left-sided (non-causal) signal
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n0 n r
x[n] X (z) x[n]rn
n1
For X(z) to exist in a particular region (for certain values of z), both summations must be finite in that region
so that For the first summation, r should be small enough |x[-n]rn| converges when summed to infinite terms
For the second summation, r should be large enough so that |x[n]/rn| converges when summed to infinite terms
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So, there are two circles with radius rL & rR for the sequence x[n]
If it is defined as a left-sided sequence (non-causal), then the second summation becomes zero (by definition), and the radius rL should be small enough to make the first summation converge
If x[n] is defined as a right-sided sequence (causal), then the first summation becomes zero (by definition), and the radius rR
should be large enough to make the second summation converge
rrRR
rr L
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Example 2: Find the z-transform of the following finite-length sequence
0.2
0.4
0.6
0.8
1
x[n] 2 [n 2] [n 1] 2 [n] [n 1] 2 [n 2] The z-transform of this sequence is given as,
it is clear to see that the sequence does not have any poles
-1 -0.5 0.5 1
-1
-0.8
-0.6
-0.4
-0.2
0 4
0 Real Part
Imag
inar
y P
art
(denominator is 1), it has 4 zeros
It can be observed that X(z) becomes undetermined at z = 0 and z = ∞, so the RoC is entire z-plane except at z = 0 and z = ∞
X ( z) 2 z 2 z 2 z 1 2 z 2
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Example 3: Find the z-transform of the following right-sided sequence
x[n] anu[n]
X (z) anu(n)z n
n
an z n
n0
(az 1 )n
n0
For convergence we require X (z)
z 1
n0 1 az z a
1 (az 1 )n X (z)
Now, X(z) will not exist for z=a & RoC is entire z-plane except z=aHowever, since the z-plane is a circle so we have to use the following condition (the sequence is right-sided)
| z || a |
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Example 4: Find the z-transform of the following right-sided sequence
x[n] anu[n 1]
X (z) anu(n 1)z n
n
1
an z n
n
an zn
n1
1 an zn
n0
z
1 n0 1 a z z a
1 (a1 z)n 1 X (z) 1
For convergence we require X (z)
Now, X(z) will not exist for z=a & RoC is entire z-plane except z=a However, since the z-plane is a circle so we have to use the following condition (the sequence is left-sided)
| z || a |
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Concepts
From the two examples we observe that the closed form equations for the z-transform of causal & non- causal signals come out to be same
This creates an ambiguity about the existence of
their z-transform
Therefore, we require complimentary information
apart from the closed form equations, i.e. the RoC
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Properties of RoC
Property 1: The RoC is a ring or disk in the z-plane centred at the origin; i.e., 0 rR z rL
Property 2: The RoC cannot contain any poles
Property 3: If x[n] is a finite-duration sequence i.e.
a sequence that is zero except in a finite interval
N1 n N2 , then the RoC is the entire z-plane except possibly z=0 and z=∞
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Property 4: If x[n] is a right-sided sequence i.e. a sequence that is zero for, , the RoC extends outward from the outermost (i.e. largest magnitude) finite pole in X(z) to z=∞
Property 5: If x[n] is a left-sided sequence i.e. a
n N1
sequence that is zero for, , the RoC extends outward from the outermost (i.e. largest magnitude) finite pole in X(z) to z=0
n N2
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( n ) 1
( n m ) z m All z except 0 (if m>0) or (if m<0)
Z-transform pairs
Sequence z-transform RoC
All z
a n u ( n ) 1
1 az 1 | z | | a |
anu(n 1) 1
1 az 1 | z || a |
nanu(n) (1 az 1 )2
az 1
| z || a |
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1
[sin 0 n]u(n) 0
0
1 [2 cos ]z 1 z 2
[sin ]z 1 | z | 1
[cos 0 n]u(n) 0
1[2 cos ]z 1 z 2
0
1[cos ]z 1 | z | 1
Sequence z-transform RoC
n [r sin 0n]u(n) 0 1 2 2
1[2r cos 0 ]z r z [r sin ]z 1
| z | r
n [r cos 0 n]u(n) 0
0
1[2r cos ]z 1 r 2 z 2
1[r cos ]z | z | r
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Example 5: Find the RoC of x[n] (0.5)n u[n] (0.4)n u[n]
Using the properties of z-transform we get
1 1
1 0.4z 1 1 0.5z 1 X (z)
z z z(z 0.4) z(z 0.5)
z 0.5 z 0.4 (z 0.5)(z 0.4)
It is clear that the RoC is given by | z || 0.4 | and | z || 0.5 |
So we can conclude that the RoC is | z || 0.5 |
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Example 6: Find the RoC of x[n] (0.5)n u[n] (0.9)n u[n 1]
Using the properties of z-transform we get
1 1
1 0.9z 1 1 0.5z 1 X (z)
z z z(z 0.9) z(z 0.5)
z 0.5 z 0.9 (z 0.5)(z 0.9)
The RoC due to the first part is | z || 0.5 | since it is a right-sided sequence
however, the second part is a left-hand sequence, therefore its RoC is
| z || 0.9 |
So we can conclude that the RoC for X(z) is | 0 .5 | | z | | 0 .9 |
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Inverse Z-transform Power Series method
Simple Tedious for large n Not accurate
Partial Fraction method
Complicated More accurate
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IZT: Power Series method
In this method we divide the numerator of a rational Z-transform by its denominator
The basic idea is
“Given a Z-transform X(z) with its corresponding RoC, we can expand X(z) into a power series of the form
n
which converges in the given RoC”
n X (z) cn z
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Since RoC is the exterior of the circle, so we expect a right-sided sequence, so we seek an expansion in the negative powers of z By dividing the numerator of X(z) by its denominator, we obtain the power series
Example 7: Find the Inverse Z-transform of X(z) 1
11.5z10.5z2 X(z) RoC |z|>1
1 1 3 z 1 7 z 2 15 z 3 31 z 4 ...
2 4 8 16 1 3 z 1 1 z 2
2 2
x[n] = [1, 3/2, 7/4, 15/8, 31/16,…. ]
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Since RoC is the interior of the circle, so we expect a left-sided sequence, so we seek an expansion in the positive powers of z By dividing the numerator of X(z) by its denominator, we obtain the power series
Example 8: Find the Inverse Z-transform of X(z) 1
11.5z10.5z2 X(z) RoC |z|<1
1 2 z 2 6 z 3 14 z 4 30 z 5 ... 1 3 z 1 1 z 2
2 2
x[n] = […., 30, 14, 6, 2, 0, 0]
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IZT: Partial Fraction method
Steps to follow Eliminate the negative powers of z for the z-transform
function X(z)
Determine the rational function X(z)/z (assuming it is
proper), and apply the partial fraction expansion to the determined rational function X(z)/z using formulae in table (next slide)
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Partial fraction(s) and formulas for constant(s)
Partial fraction with the first-order real pole:
A z p z z p A (z p) X (z) |
Partial fraction with the first-order complex poles:
Az A* z
z p z p * z z p A (z p) X (z) |
Partial fraction with mth-order real poles:
Ak A1
(z p)k
z p (z p)2
Ak 1 k z z p
A
X (z) | (z p)k 1 d
(k 1)! dzk 1
k 1
, A A*
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An example for Simple Real Poles
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An example for Multiple Real Poles
𝑓 𝑛 =[9(0.3)n8(0.2)n+2n(0.2)n]u(n)
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Pulse Transfer Function
Pulse transfer function H(z) is defined as the ratio of the Z-transform of the input x[n] to the Z-transform of the output y[n]
Y z X z H z
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Derivation
l k
y n bi xn i ai y n i i 0 i 1
Applying Z-transform and moving the terms of ‘y’ to one side
i i i b X zz Y za z Y z
l
i i0
k
i1
l i i
l i i
k i i
b z a z X z
b z a z X z z Y z Y
i i0
k
i i1
i0 i1
Y z1
i
l i bi z
X z Y z
H z k
i1
i0
1 ai z
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l ibi z
Y z H z i0 k
a2 0.02 a1 0.1
b0 2 b1 1 1
Example 8: Find the Pulse Transfer function of the difference equation
yn 0.1yn 1 0.02 yn 2 2xn xn 1
i i X z 1 a z
i1
1
i 2
i i1
i
a z
bi z H z i0
1 0 1
1 0.1z 1 0.02z 2
2 z 1 0.1z 1 0.02z 2
2z 1z H z