Digital Signal Processing
Lecture 10 – Z Transform
Alp Ertürk
alp.erturk@kocaeli.edu.tr
Z Transform
• Fourier Transform does not converge for all sequences
• Z transform encompasses a broader class of signals
• In analytical problems, Z transform often has a more
convenient notation than Fourier transform
• Z transform for discrete-time signals is the counter-part of
the Laplace transform for continuous-time signals
Z Transform
• Fourier transform:
𝑋 𝑒 𝑗𝜔 = 𝑥[𝑛]𝑒 −𝑗𝜔𝑛
𝑛=−∞
• Z transform:
𝑋 𝑧 = 𝑥[𝑛]𝑧 −𝑛
𝑛=−∞
Z Transform
• Z transform can be considered to be related to Fourier
transform by 𝑧 = 𝑒 𝑗𝜔
Im
𝑧 = 𝑒 𝑗𝜔
r=1
Ω 0
Re
2
Z Transform
• Considering the relationship 𝑧 = 𝑒 𝑗𝜔 corresponds to
restricting z to have unity magnitude, i.e. 𝑧 = 1
• In more general terms, 𝑧 = 𝑟𝑒 𝑗𝜔 , which equates to:
∞
−𝑛
𝑋 𝑟𝑒 𝑗𝜔 = 𝑥[𝑛] 𝑟𝑒 𝑗𝜔
𝑛=−∞
= 𝑥[𝑛]𝑟 −𝑛 𝑒 −𝑗𝜔𝑛
𝑛=−∞
• In other words, the Fourier transform of 𝑥[𝑛]𝑟 −𝑛
Z Transform
• Similar to Fourier transform, Z transform does not converge
for all values of z
• The set of values of z for which z transform converges is
called the region of convergence (ROC)
• For convergence, we have to have absolute summability:
𝑥[𝑛]𝑟 −𝑛 < ∞
𝑛=−∞
Z Transform
• If the ROC for z transform includes the unit circle, it means
that the Fourier transform of the sequence converges
• For 𝑢[𝑛], the Fourier transform does not converge as 𝑢[𝑛] is
not absolutely summable
• However, 𝑟 −𝑛 𝑢 𝑛 is absolutely summable for 𝑟 > 1
• This means that z transform for the unit step signal exists,
with a ROC of 𝑧 > 1, which does not include the unit circle
Z Transform
• Typical ROCs:
(a) R.O.C.: z a (b) R.O.C.: z a
(c) R.O.C.: a z b
Z Transform
• Example: Z transform of unit impulse signal
𝑥 𝑛 =𝛿 𝑛
∞ ∞
𝑋 𝑧 = 𝑥[𝑛]𝑧 −𝑛 = 𝛿 𝑛 𝑧 −𝑛
𝑛=−∞ 𝑛=−∞
= 𝛿 0 𝑧 −0 = 1
• ROC is all z values
Z Transform
• Example: Z transform of shifted unit impulse signal
𝑥 𝑛 = 𝛿 𝑛 − 𝑛0
∞ ∞
𝑋 𝑧 = 𝑥[𝑛]𝑧 −𝑛 = 𝛿 𝑛 − 𝑛0 𝑧 −𝑛
𝑛=−∞ 𝑛=−∞
= 𝛿 0 𝑧 −𝑛0 = 𝑧 −𝑛0
• ROC is all z values except 𝑧 = 0 if 𝑛0 > 0, all z values except
∞ if 𝑛0 < 0
Z Transform
• Example: Z transform of unit step signal 𝑥 𝑛 = 𝑢 𝑛
∞ ∞ 𝑁
𝑋 𝑧 = 𝑥[𝑛]𝑧 −𝑛 = 𝑢 𝑛 𝑧 −𝑛 = lim 𝑧 −𝑛
𝑁→∞
𝑛=−∞ 𝑛=−∞ 𝑛=0
If 𝑧 −1 < 1 ≡ 𝑧 > 1
0−1 1
= −1 = ,
𝑧 − 1 1 − 𝑧 −1 1
(a) R.O.C.: z a (
Z Transform
• Example: Z transform of 𝑥 𝑛 = 𝑎𝑛 𝑢 𝑛
∞ ∞ ∞
𝑋 𝑧 = 𝑥[𝑛]𝑧 −𝑛 = 𝑎𝑛 𝑢 𝑛 𝑧 −𝑛 = 𝑎𝑧 −1 𝑛
𝑛=−∞ 𝑛=−∞ 𝑛=0
If 𝑎𝑧 −1 < 1 ≡ 𝑧 > 𝑎 :
0−1 1 𝑧
= −1
= −1
=
𝑎𝑧 − 1 1 − 𝑎𝑧 𝑧−𝑎
(a) R.O.C.: z a
Z Transform
• Example: Z transform of 𝑥 𝑛 = −𝑎𝑛 𝑢 −𝑛 − 1
∞ ∞
𝑋 𝑧 = 𝑥[𝑛]𝑧 −𝑛 = −𝑎𝑛 𝑢 −𝑛 − 1 𝑧 −𝑛
𝑛=−∞ 𝑛=−∞
−1 ∞ ∞
= − 𝑎𝑛 𝑧 −𝑛 = − 𝑎−1 𝑧 𝑛 = 1 − 𝑎−1 𝑧 𝑛
𝑛=−∞ 𝑛=1 𝑛=0
If 𝑎−1 𝑧 < 1 ≡ 𝑧 < 𝑎 :
1 −𝑎−1 𝑧 1 𝑧
𝑋 𝑧 =1− −1
= −1
= −1
=
1 − 𝑎 𝑧 1 − 𝑎 𝑧 1 − 𝑎𝑧 𝑧−𝑎
Z Transform
• Example (continued):
• Note that the Z transform of both 𝑎𝑛 𝑢 𝑛 and −𝑎𝑛 𝑢 −𝑛 − 1
1
are , bu the ROCs are 𝑧 > 𝑎 and 𝑧 < 𝑎 ,
1−𝑎𝑧 −1
respectively !!
Z Transform
1 𝑛 1 𝑛
• Example: Z transform of 𝑥 𝑛 = 𝑢𝑛 + − 𝑢𝑛
2 3
∞ ∞ 𝑛 𝑛
1 1
𝑋 𝑧 = 𝑥[𝑛]𝑧 −𝑛 = 𝑢𝑛 + − 𝑢𝑛 𝑧 −𝑛
2 3
𝑛=−∞ 𝑛=−∞
∞ 𝑛 ∞ 𝑛
1 −𝑛
1
= 𝑧 + − 𝑧 −𝑛
2 3
𝑛=0 𝑛=0
∞ 𝑛 ∞ 𝑛
1 −1 1 −1
= 𝑧 + − 𝑧
2 3
𝑛=0 𝑛=0
Z Transform
• Example (continued):
∞ 𝑛 ∞ 𝑛
1 −1 1 −1
𝑋 𝑧 = 𝑧 + − 𝑧
2 3
𝑛=0 𝑛=0
1 1
= +
1 −1 1 −1
1− 𝑧 1+ 𝑧
2 3
1 −1 1
2− 𝑧 2𝑧 𝑧 −
= 6 = 12
1 −1 1 −1 1 1
1− 𝑧 1+ 𝑧 𝑧− 𝑧+
2 3 2 3
Z Transform
• Example (continued):
1 −1 1
• For convergence, 𝑧 < 1 and − 𝑧 −1 < 1
2 3
1 1
• Or, equivalently, 𝑧 > and 𝑧 >
2 3
1
• The resulting ROC is the region of overlap, 𝑧 >
2
Z Transform
• Example (continued):
Z Transform
1 𝑛 1 𝑛
• Example: Z transform of 𝑥 𝑛 = − 𝑢 −𝑛 − 1 + − 𝑢 𝑛
2 3
𝑛
1 𝑧 1 1
− 𝑢𝑛 → , 𝑧 >
3 1 −1 3
1+ 𝑧
3
𝑛
1 𝑧 1 1
− 𝑢 −𝑛 − 1 → , 𝑧 <
2 1 2
1 − 𝑧 −1
2
1
1 1 2𝑧 𝑧 −
𝑋 𝑧 = + = 12
1 1 1 1
1 − 𝑧 −1 1 + 𝑧 −1 𝑧− 𝑧+
2 3 2 3
Z Transform
• Example (continued):
1 1
• The resulting ROC is the region of overlap, < 𝑧 <
3 2
Z Transform
𝑎𝑛 , 0 ≤ 𝑛 ≤ 𝑁 − 1
• Example: Z transform of 𝑥 𝑛 = ቊ
0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
𝑁−1 𝑁−1
𝑋 𝑧 = 𝑎𝑛 𝑧 −𝑛 = 𝑎𝑧 −1 𝑛
𝑛=0 𝑛=0
𝑎𝑧 −1 𝑁 − 1 1 𝑧 𝑁 − 𝑎𝑁
= −1
= 𝑁−1
𝑎𝑧 −1 𝑧 𝑧−𝑎
• The sum will be finite as long as 𝑎𝑧 −1 is finite, which requires
𝑎 < ∞ and 𝑧 ≠ 0. Assuming 𝑎 is finite, ROC is the entire z-
plane, with the exception of 𝑧 = 0
Z Transform
• Example (continued):
• If 𝑁 = 16 and a is real and between zero and 1, the roots are:
𝑧 = 𝑎𝑒 𝑗(2𝜋𝑘/𝑁) , 𝑘 = 0,1, … , 𝑁 − 1
Z Transform: Properties of ROC
1. The ROC is a ring or disk centered at the origin, 0 ≤ 𝑟𝑅 ≤ 𝑧 ≤ 𝑟𝐿 ≤ ∞
2. DTFT exists if and only if the ROC includes the unit circle
3. The ROC cannot contain any poles
4. The ROC for finite-length sequences is the entire z-plane, except possibly
𝑧 = 0 or 𝑧 = ∞
5. The ROC for a right-handed sequence, i.e. 𝑥 𝑛 = 0 𝑓𝑜𝑟 𝑛 < 𝑁1 < ∞,
extends outward from the outermost pole, possibly including 𝑧 = ∞
6. The ROC for a left-handed sequence , i.e. 𝑥 𝑛 = 0 𝑓𝑜𝑟 𝑛 > 𝑁1 > −∞,
extends inward from the innermost pole, possibly including 𝑧 = 0
7. The ROC of a two-sided sequence is a ring bounded on the interior and
exterior by a pole, and not containing any poles (in accordance with
property 3)
8. The ROC must be a connected region
Z Transform: Properties of ROC
• If a system is stable, it means h[n] is absolutely summable, or that the system’s
Fourier transform exists. For the Fourier transform to exist, the ROC must
include the unit circle. Therefore,
• If a system is stable, ROC includes the unit circle
• If a system is causal, it means h[n] is right sided. In this case, ROC extends
outward from the outermost pole
Z Transform: Properties of ROC
1
• If the system with the pole-zero graph given above is stable, ROC: < 𝑧 < 2
2
• If the system with the pole-zero graph given below is causal, ROC: 𝑧 > 2
Inverse Z Transform
• If X(z) is expressed as a ratio of polynomials:
σ𝑀 𝑏
𝑘=0 𝑘 𝑧 −𝑘 𝑧 −𝑀 σ𝑀 𝑏 𝑧 𝑀−𝑘
𝑘=0 𝑘 𝑧 𝑁 σ𝑀 𝑏 𝑧 𝑀−𝑘
𝑘=0 𝑘
𝑋 𝑧 = 𝑁 = =
σ𝑘=0 𝑎𝑘 𝑧 −𝑘 𝑧 −𝑁 σ𝑁 𝑎
𝑘=0 𝑘 𝑧 𝑁−𝑘 𝑧 𝑀 σ𝑁 𝑎 𝑧 𝑁−𝑘
𝑘=0 𝑘
• There will be M zeros, and N poles at nonzero locations in the z-plane
• In addition, there will be 𝑀 − 𝑁 poles at 𝑧 = 0 if 𝑀 > 𝑁, or 𝑁 − 𝑀 zeros at
𝑧 = 0 if 𝑀 < 𝑁
• Z transform always has the same number of poles and zeros in the finite z-
plane !
Inverse Z Transform
• X(z) can also be expressed in the form of
𝑏0 ς𝑀 −1
𝑘=1(1 − 𝑐𝑘 𝑧 )
𝑋 𝑧 =
𝑎0 ς𝑁 −1
𝑘=1(1 − 𝑑𝑘 𝑧 )
• 𝑐𝑘 ’s are the nonzero zeros of X(z), and 𝑑𝑘 ’s are the nonzero poles of X(z)
• If 𝑀 < 𝑁, and the poles are all first order, X(z) can be expressed as:
𝑁
𝐴𝑘
𝑋 𝑧 = , 𝐴𝑘 = 1 − 𝑑𝑘 𝑧 −1 𝑋(𝑧)|𝑧=𝑑𝑘
1 − 𝑑𝑘 𝑧 −1
𝑘=1
Inverse Z Transform
• Example: Consider a sequence x[n] with the z-transform:
1 1
𝑋 𝑧 = , 𝑧 >
1 1 2
1 − 𝑧 −1 1 − 𝑧 −1
4 2
Inverse Z Transform
• Example (continued):
1 𝐴1 𝐴2
𝑋 𝑧 = = +
1 1 1 −1 1
1 − 𝑧 −1 1 − 𝑧 −1 1− 𝑧 1 − 𝑧 −1
4 2 4 2
1 −1 1 1
𝐴1 = 1 − 𝑧 𝑋 𝑧 |𝑧=1/4 = | = = −1
4 1 −1 𝑧=1/4 1
1− 𝑧 1− 4
2 2
1 −1 1 1
𝐴2 = 1 − 𝑧 𝑋 𝑧 |𝑧=1/2 = | = =2
2 1 −1 𝑧=1/2 1
1− 𝑧 1− 2
4 4
Inverse Z Transform
• Example (continued):
1 −1 2
𝑋 𝑧 = = +
1 1 1 −1 1
1 − 𝑧 −1 1 − 𝑧 −1 1− 𝑧 1 − 𝑧 −1
4 2 4 2
𝑛 𝑛
1 1
𝑥 𝑛 =2 𝑢𝑛 − 𝑢𝑛
2 4
Inverse Z Transform
• Example: Find the sequence x[n] with the z-transform:
1
𝑋 𝑧 =
1 − 3𝑧 −1 + 2𝑧 −2
1 1
𝑋 𝑧 = =
1 − 3𝑧 −1 + 2𝑧 −2 1 − 𝑧 −1 1 − 2𝑧 −1
−1 2
= −1
+
1−𝑧 1 − 2𝑧 −1
Inverse Z Transform
• Example (continued):
• Note that the ROC is not provided in the question. Thefore, we
have to compute for all possible ROCs. The system has the poles
at 1 and 2.
• i) ROC: 𝑧 < 1
−1 2
𝑋 𝑧 = −1
+
1−𝑧 1 − 2𝑧 −1
𝑥 𝑛 = 𝑢 −𝑛 − 1 − 2 2 𝑛 𝑢[−𝑛 − 1]
Inverse Z Transform
• Example (continued):
• ii) ROC: 1 < 𝑧 < 2
−1 2
𝑋 𝑧 = −1
+
1−𝑧 1 − 2𝑧 −1
𝑥 𝑛 = −𝑢 𝑛 − 2 2 𝑛 𝑢[−𝑛 − 1]
• iii) ROC: 𝑧 > 2
−1 2
𝑋 𝑧 = −1
+
1−𝑧 1 − 2𝑧 −1
𝑥 𝑛 = −𝑢 𝑛 + 2 2 𝑛 𝑢[𝑛]
Inverse Z Transform
• We stated that if 𝑀 < 𝑁, and the poles are all first order, X(z) can be expressed
as:
𝑁
𝐴𝑘
𝑋 𝑧 = , 𝐴𝑘 = 1 − 𝑑𝑘 𝑧 −1 𝑋(𝑧)|𝑧=𝑑𝑘
1 − 𝑑𝑘 𝑧 −1
𝑘=1
• If 𝑀 > 𝑁, and the poles are all first order, X(z) should be expressed as:
𝑀−𝑁 𝑁
𝐴𝑘
𝑋 𝑧 = 𝐵𝑟 𝑧 −𝑟 +
1 − 𝑑𝑘 𝑧 −1
𝑟=0 𝑘=1
• 𝐵𝑟 ’s can be obtained by long division, whereas 𝐴𝑘 ’s are obtained as before
Inverse Z Transform
• Example: Consider a sequence x[n] with the z-transform:
1 + 2𝑧 −1 + 𝑧 −2
𝑋 𝑧 = , 𝑧 >1
3 −1 1 −2
1− 𝑧 + 𝑧
2 2
1 + 2𝑧 −1 + 𝑧 −2 1 + 𝑧 −1 2
𝑋 𝑧 = =
3 −1 1 −2 1
1− 𝑧 + 𝑧 1 − 𝑧 −1 1 − 𝑧 −1
2 2 2
Inverse Z Transform
• Example (continued)
1 + 𝑧 −1 2
𝑋 𝑧 =
1
1 − 𝑧 −1 1 − 𝑧 −1
2
• M=N=2 and all the poles are first order. Therefore,
𝐴1 𝐴2
𝑋 𝑧 = 𝐵0 + +
1 −1 1 − 𝑧 −1
1− 𝑧
2
𝐵0 can be obtained by long division: 1+2𝑧 −1 +𝑧 −2
ൗ1−3𝑧 −1+1𝑧 −2 ≈ 2
2 2
Inverse Z Transform
• Example (continued)
𝐴1 𝐴2
𝑋 𝑧 = 𝐵0 + +
1 −1 1 − 𝑧 −1
1− 𝑧
2
−1 + 5𝑧 −1
=2+
1
1 − 𝑧 −1 1 − 𝑧 −1
2
1 −1 1 −1 −1 + 5𝑧 −1
𝐴1 = 1 − 𝑧 𝑋 𝑧 |𝑧=1/2 = 2 1 − 𝑧 + −1
|𝑧=1/2 = −9
2 2 1−𝑧
−1 + 5𝑧 −1
𝐴2 = 1 − 𝑧 −1 𝑋 𝑧 |𝑧=1 = 2 1 − 𝑧 −1 + |𝑧=1 = 8
1
1 − 𝑧 −1
2
Inverse Z Transform
• Example (continued)
9 8
𝑋 𝑧 =2− +
1 −1 1 − 𝑧 −1
1− 𝑧
2
1 𝑛
𝑥 𝑛 = 2𝛿 𝑛 − 9 2
𝑢 𝑛 + 8𝑢[𝑛]
Inverse Z Transform
• If 𝑀 > 𝑁, and X(z) has multiple order poles:
𝑀−𝑁 𝑁 𝑠
𝐴𝑘 𝐶𝑚
𝑋 𝑧 = 𝐵𝑟 𝑧 −𝑟 + +
1 − 𝑑𝑘 𝑧 −1 1 − 𝑑𝑖 𝑧 −1 𝑚
𝑟=0 𝑘=1,𝑘≠𝑖 𝑚=1
• In this case, 𝐵𝑟 ’s and 𝐴𝑘 ’s are obtained as before, whereas 𝐶𝑚 coefficents are
obtained as:
1 𝑑 𝑠−𝑚
𝐶𝑚 = 𝑠−𝑚
𝑠 − 𝑚 ! −𝑑𝑖 𝑑𝑤 𝑠−𝑚
Inverse Z Transform
• If the z transform of a sequence is given as a power series expansion,
∞
𝑋 𝑧 = 𝑥[𝑛]𝑧 −1 = ⋯ + 𝑥 −1 𝑧1 + 𝑥 0 + 𝑥 1 𝑧 −1 + 𝑥 2 𝑧 −2 + ⋯
𝑛=−∞
• We can determine any particular value of the sequence by finding the
coefficient of the appropriate power of 𝑧 −1
Inverse Z Transform
• Example: Consider a sequence x[n] with the z-transform given by
1 −1
𝑋 𝑧 = 𝑧2 1− 𝑧 1 + 𝑧 −1 1 − 𝑧 −1
2
X(z) is rational, but its only poles are at z = 0. Therefore, we cannot use partial
fraction expansion. However, by multiplication, we obtain:
1 1
𝑋 𝑧 = 𝑧 2 − 𝑧 − 1 + 𝑧 −1
2 2
Using the coefficients of the z terms, we obtain:
Inverse Z Transform
• Example (continued):
1 1 −1
𝑋 𝑧 = 𝑧2 − 𝑧−1+ 𝑧
2 2
1, 𝑛 = −2
1
− , 𝑛 = −1
2
𝑥 𝑛 = −1 ,
1 𝑛=0
, 𝑛=1
2 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
0,
1 1
𝑥 𝑛 =𝛿 𝑛+2 − 𝛿 𝑛+1 −𝛿 𝑛 + 𝛿 𝑛−1
2 2
Properties of Z Transform
• 1) Linearity
𝑧
𝑥1 𝑛 ՞ 𝑋1 (𝑧)
𝑧
𝑥2 𝑛 ՞ 𝑋2 (𝑧)
𝑧
𝑎𝑥1 𝑛 + 𝑏𝑥2 𝑛 ՞ 𝑎𝑋1 𝑧 + b𝑋2 𝑧
𝑅𝑂𝐶 𝑐𝑜𝑛𝑡𝑎𝑖𝑛𝑠 𝑅𝑥1 ∩ 𝑅𝑥2
(𝑅𝑂𝐶 𝑖𝑠 𝑒𝑥𝑎𝑐𝑡𝑙𝑦 𝑅𝑥1 ∩ 𝑅𝑥2 𝑖𝑓 𝑡ℎ𝑒𝑟𝑒 𝑖𝑠 𝑛𝑜 𝑝𝑜𝑙𝑒 − 𝑧𝑒𝑟𝑜 𝑐𝑎𝑛𝑐𝑒𝑙𝑙𝑎𝑡𝑖𝑜𝑛)
Properties of Z Transform
• 2) Time Shifting
𝑧
𝑥1 𝑛 ՞ 𝑋1 (𝑧)
𝑧
𝑥1 𝑛 − 𝑛0 ՞ 𝑧 −𝑛0 𝑋1 (𝑧)
𝑅𝑂𝐶 = 𝑅𝑥
(𝑒𝑥𝑐𝑒𝑝𝑡 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑎𝑑𝑑𝑖𝑡𝑖𝑜𝑛 𝑜𝑟 𝑑𝑒𝑙𝑒𝑡𝑖𝑜𝑛 𝑜𝑓 𝑧 = 0 𝑜𝑟 𝑧 = ∞)
Inverse Z Transform
• Example: Consider a sequence x[n] with the z-transform given by
1 1
𝑋 𝑧 = , 𝑧 >
1 4
𝑧−
4
𝑧 −1 1
= , 𝑧 >
1 −1 4
1− 𝑧
4
4 1
= −4 + , 𝑧 >
1 −1 4
1− 𝑧
4
𝑛
1
𝑥 𝑛 = −4𝛿 𝑛 + 4 𝑢[𝑛]
4
Inverse Z Transform
• Example (continued):
• Alternatively,
1 1
𝑋 𝑧 = 𝑧 −1 , 𝑧 >
1 −1 4
1− 𝑧
4
• By the time-shifting property,
𝑛−1
1
𝑥𝑛 = 𝑢[𝑛 − 1]
4
• which is actually equal to the signal found on the previous slide
Properties of Z Transform
• 3) Multiplication by an Exponential Sequence
𝑧
𝑥1 𝑛 ՞ 𝑋1 (𝑧)
𝑧
𝑧0𝑛 𝑥1 𝑛 ՞ 𝑋1 (𝑧/𝑧0 )
𝑅𝑂𝐶 = 𝑧0 𝑅𝑥
• ROC and all pole-zero locations are scaled by 𝑧0
• If 𝑧0 is a real number: ROC shrinks or expands
• If 𝑧0 is a complex number with unit magnitude, ROC rotates (same ROC, but
pole-zero locations rotate)
Inverse Z Transform
• Example: Starting with the z-transform pair
𝑧 1
𝑢𝑛 ՞ , 𝑧 >1
1 − 𝑧 −1
• Let’s determine the z-transform of x n = 𝑟 𝑛 𝑐𝑜𝑠 𝜔0 𝑛 𝑢 𝑛
1 𝑛 1 𝑛
xn = 𝑟 𝑛 𝑐𝑜𝑠 𝜔0 𝑛 𝑢 𝑛 = 𝑟𝑒 𝑗𝜔0 𝑢 𝑛 + 𝑟𝑒 −𝑗𝜔0 𝑢𝑛
2 2
1/2 1/2
𝑋 𝑧 = + , 𝑧 >𝑟
1 − 𝑟𝑒 𝑗𝜔0 𝑧 −1 1 − 𝑟𝑒 −𝑗𝜔0 𝑧 −1
1 − 𝑟 cos 𝜔0 𝑧 −1
= , 𝑧 >𝑟
1 − 2𝑟 cos 𝜔0 𝑧 −1 + 𝑟 2 𝑧 −2
Properties of Z Transform
• 4) Differentiation of X(z)
𝑧
𝑥1 𝑛 ՞ 𝑋1 (𝑧)
𝑧 𝑑𝑋(𝑧)
𝑛𝑥1 𝑛 ՞ − 𝑧
𝑑𝑧
𝑅𝑂𝐶 = 𝑅𝑥
• Proof:
∞ ∞
−𝑛
𝑑𝑋 𝑧
𝑋 𝑧 = 𝑥𝑛𝑧 ⇒ −𝑧 = −𝑧 −𝑛 𝑥 𝑛 𝑧 −𝑛−1 +
𝑑𝑧
𝑛=−∞ ∞ 𝑛=−∞
= 𝑛𝑥[𝑛]𝑧 −𝑛
𝑛=−∞
Inverse Z Transform
• Example: Find the z-transform of 𝑥 𝑛 = 𝑛𝑎𝑛 𝑢[𝑛]
𝑑 1
𝑋 𝑧 = −𝑧 , 𝑧 > 𝑎
𝑑𝑧 1 − 𝑎𝑧 −1
𝑎𝑧 −1
= , 𝑧 > 𝑎
1 − 𝑎𝑧 −1 2
Properties of Z Transform
• 5) Conjugation
𝑧
𝑥1 𝑛 ՞ 𝑋1 (𝑧)
𝑧
𝑥1∗ 𝑛 ՞ 𝑋1∗ (𝑧 ∗ )
𝑅𝑂𝐶 = 𝑅𝑥
Properties of Z Transform
• 6) Time Reversal
𝑧
𝑥1 𝑛 ՞ 𝑋1 (𝑧)
𝑧
𝑥1∗ −𝑛 ՞ 𝑋1∗ (1/𝑧 ∗ )
𝑅𝑂𝐶 = 1/𝑅𝑥
• ROC is inverted, such that if for 𝑅𝑥 𝑟𝑅 < 𝑧 < 𝑟𝐿 , then the ROC becomes
1/𝑟𝐿 < 𝑧 < 1/𝑟𝑅
Inverse Z Transform
• Example: Find the z-transform of 𝑥 𝑛 = 𝑎−𝑛 𝑢[−𝑛]
∗
1
𝑋 𝑧 =
1 − 𝑎 1/𝑧 ∗ −1
1 −𝑎−1 𝑧 −1
= = , 𝑧 < 𝑎−1
1 − 𝑎𝑧 1 − 𝑎−1 𝑧 −1
Alternatively,
∞ ∞ 0 ∞
𝑋 𝑧 = 𝑥[𝑛]𝑧 −𝑛 = 𝑎−𝑛 𝑢[−𝑛] 𝑧 −𝑛 = 𝑎𝑧 −𝑛 = 𝑎𝑧 𝑛
𝑛=−∞ 𝑛=−∞ 𝑛=−∞ 𝑛=0
1
= , 𝑎𝑧 < 1 ≡ 𝑧 < 𝑎−1
1 − 𝑎𝑧
Properties of Z Transform
• 7) Convolution
𝑧
𝑥1 𝑛 ՞ 𝑋1 𝑧
𝑧
𝑥2 𝑛 ՞ 𝑋2 (𝑧)
𝑧
𝑥1 𝑛 ∗ 𝑥2 𝑛 ՞ 𝑋1 𝑧 𝑋2 𝑧
𝑅𝑂𝐶 𝑐𝑜𝑛𝑡𝑎𝑖𝑛𝑠 𝑅𝑥1 ∩ 𝑅𝑥2
(𝑅𝑂𝐶 𝑖𝑠 𝑒𝑥𝑎𝑐𝑡𝑙𝑦 𝑅𝑥1 ∩ 𝑅𝑥2 𝑖𝑓 𝑡ℎ𝑒𝑟𝑒 𝑖𝑠 𝑛𝑜 𝑝𝑜𝑙𝑒 − 𝑧𝑒𝑟𝑜 𝑐𝑎𝑛𝑐𝑒𝑙𝑙𝑎𝑡𝑖𝑜𝑛)
Properties of Z Transform
• Example: 𝑥1 𝑛 = 𝑎𝑛 𝑢[𝑛], 𝑥2 𝑛 = 𝑢[𝑛], determine the z-transform of
𝑥1 𝑛 ∗ 𝑥2 𝑛
1
𝑋1 𝑧 = −1
, 𝑧 > 𝑎
1 − 𝑎𝑧
1
𝑋2 𝑧 = , 𝑧 >1
1 − 𝑧 −1
1 𝑧2
𝑋1 𝑧 𝑋2 𝑧 = = , 𝑧 >1
1 − 𝑎𝑧 −1 1 − 𝑧 −1 𝑧−𝑎 𝑧−1
Properties of Z Transform
• 8) Initial Value Theorem
• If x[n] is zero for n<0, i.e. if x[n] is causal, then:
𝑥 𝑛 = lim 𝑋(𝑧)
𝑧→∞
∞ ∞
lim 𝑋(𝑧) = lim 𝑥[𝑛]𝑧 −𝑛 = lim 𝑥[𝑛]𝑧 −𝑛
𝑧→∞ 𝑧→∞ 𝑧→∞
𝑛=−∞ 𝑛=0
= lim 𝑥 0 + 𝑥 1 𝑧 −1 + 𝑥 2 𝑧 −2 + ⋯ = 𝑥[0]
𝑧→∞
