chapter # 3 : the z-transform and its application to the...

EELE 4310: Digital Signal Processing (DSP)

Chapter # 3 : The z-transform and Its Application to the

Analysis of LTI Systems(Part One)

Spring, 2012/2013

EELE 4310: Digital Signal Processing (DSP) - Ch.3 Dr. Musbah Shaat 1 / 23

Outline

1 The z-Transform

2 Properties of the z-Transform


The z-TransformThe Direct z-Transform ... 1

Direct Transform Relation

The z-transform of a discrete-time signal x(n) is defined as the powerseries

X (z) ≡ Zx(n) ≡∑∞

n=−∞ x(n)z−n

where z is a complex variable. (The notation x(n)z↔X (z) is also used).

The z-transform is an infinite power series ⇒ it exists only for thosevalues of z for which this series converges.

The region of convergence (ROC) of X (z) is the set of all values of z forwhich X (z) attains a finite value.⇒ Any time we cite a z-transform, we should also indicate its ROC.



Ex. 3.1.1: Determine the z-transform of the following signals(a) x1(n) = {2, 4, 5

↑, 7, 0, 1} , (b) x2(n) = δ(n)

(c) x3(n) = δ(n − k), k > 0, (d) x4(n) = δ(n + k), k > 0

(a) X1(z) = 2z2 + 4z + 5 + 7z−1 + z−3

Roc: entire z-plane except z = 0 and z =∞.

(b) X2(z) = 1 [i.e., δ(n)z↔ 1]

Roc: entire z-plane.

(c) X3(z) = z−k [i.e., δ(n − k)z↔ z−k , k > 0]

Roc: entire z-plane except z = 0.

(d) X4(z) = zk [i.e., δ(n + k)z↔ zk , k > 0]

Roc: entire z-plane except z =∞.



Ex. 3.1.2: Determine the z-transform of the following signalx(n) = 1

2

nu(n)

The signal consists of an infinite number of nonzero values

x(n) = {1, 12 ,(12

)2,(12

)3, · · · ,

(12

)n, · · · }

The z-transform of x(n) is the infinite power series

X (z) = 1 + 12z−1,(12

)2z−2,

(12

)nz−n + · · ·

=∑∞

n=0

(12

n)z−n =

∑∞n=0

(12z−1)n

This is an infinite geometric series. recall that:1 + A + A2 + A3 + · · · = 1

1−A if |A| < 1

Consequently, for∣∣12z−1∣∣ < 1, or equivalently |z | > 1

2 , X (z) converges to

X (z) =1

1− 12z−1 , ROC: |z | > 1

2



Ex. 3.1.3: Determine the z-transform of the following signalx(n) = αnu(n)

X (z) =∑∞

n=0 αnz−n =

∑∞n=0

(αz−1

)⇒ this is a geometric series converges when

∣∣αz−1∣∣ < 1, or equivalently,when |z | > |α|.

x(n) = αnu(n)z↔X (z) =

1

1− αz−1,ROC: |z | > |α|



Ex. 3.1.4: Determine the z-transform of the following signalx(n) = −αnu(−n − 1)

X (z) =∑−1

n=−∞ (−α)n z−n = −∑∞

l=1

(α−1z

)lUsing the formula,A + A2 + A3 + · · · = A(1 + A + A2 + · · · ) = A

1−Aprovided that |A| < 1.⇒ when

∣∣α−1z∣∣ < 1, or equivalently, when |z | < |α|.

x(n) = −αnu(−n − 1)z↔X (z) =

α−1z

1− α−1z=

1

1− αz−1,ROC: |z | < |α|



The uniqueness of the z-transform

A discrete-time signal x(n) is uniquely determined by its z-transformX (z) and the ROC of X (z). Ex.Z{αnu(n)} = Z{−αnu(−n − 1)} = 1

1−αz−1 ⇒ different in ROC only.

The ROC of a causal signal is the exterior of a circle of some radius r2,while the ROC of an anticausal signal is the interior of a circle of someradius r1

z-transform is used to refer both the closed-form expression and thecorresponding ROC



Ex. 3.1.5: Determine the z-transform of the following signalx(n) = αnu(n) + bnu(−n − 1)

X (z) =∑∞

n=0 αnz−n +

∑−1n=−∞ bnz−n =

∑∞n=0

(αz−1

)n+∑∞

l=0

(b−1z

)lThe first power series converges if

∣∣αz−1∣∣ < 1 or |z | > |α|.The second power series converges if

∣∣b−1z∣∣ < 1 or |z | > |b|

Case 1 |b| < |α|Two ROC do not overlap ⇒ X (z)does not exist.

Case 2 |b| > |α|The ROC is the ring|α| < |z | < |b|


The z-TransformFamilies of Signals with Their ROCs ... 1


The z-TransformFamilies of Signals with Their ROCs ... 2

Right-sided Signal

Is a signal that has infinite duration on the right side but not on the left.ROC: r0 < |z | <∞

Left-sided Signal

Is a signal that has infinite duration on the left side but not on the right.ROC: 0 < |z | < r1

Finite duration two-sided Signal

Is a signal that has finite duration on both the left and right sides.ROC: 0 < |z | <∞

One-sided or unilateral z-transform

X+(z) =∞∑n=0

x(n)z−n

(Identical to the two-sided transform when x(n) is causal)EELE 4310: Digital Signal Processing (DSP) - Ch.3 Dr. Musbah Shaat 11 / 23

The z-TransformThe inverse z-Transform

The signal x(n) can be obtained form X (z) using the inversion formula

x(n) =1

2πj

∫CX (z)zn−1dz


Properties of the z-TransformLinearity and Time shifting ... 1

When we combine several z-transforms, the ROC of the overall transformis, at least, the intersection of the ROC of the individual transform.

Linearity Property

If x1(n)z↔X1(z) and x2(n)

z↔X2(z)thenx(n) = a1x1(n) + a2x2(n)

z↔X (z) = a1X1(z) + a2X2(z)

Time shifting Property

If x(n)z↔X (z)

thenx(n − k)

z↔ z−kX (z)The ROC of z−kX (z) is the same as the ROC of X (z) except for z = 0if k > 0 and z =∞ if k < 0



Ex. 3.2.1: Determine the z-transform and the ROC of the signalx(n) = [3(2n)− 4(3n)]u(n)

From linearity, X (z) = 3X1(z)− 4X2(z)

where X1(z) = Z{x1(n)} = Z{3(2n)u(n)}

and X2(z) = Z{x2(n)} = Z{4(3n)u(n)}

Recall that x(n) = αnu(n)z↔X (z) = 1

1−αz−1 ,ROC: |z | > |α|

By setting α = 2 and α = 3, we obtain

x1(n) = 2nu(n)z↔X (z) = 1

1−2z−1 ,ROC: |z | > 2

x2(n) = 3nu(n)z↔X (z) = 1

1−3z−1 ,ROC: |z | > 3

Thus, the overall transform

X (z) = 31−2z−1 − 4

1−3z−1 ,ROC: |z | > 3



Ex. 3.2.2: Determine the z-transform and the ROC of the signalx(n) = (cosω0n)u(n)

By using Euler identity, x(n) can be expressed as

x(n) = (cosω0n)u(n) = 12

[e jω0nu(n) + e−jω0nu(n)

]Thus, X (z) = 1

2

[Z{e jω0u(n)}+ Z{e−jω0u(n)}

]If we set α = e±jω0n(|α| = |e±jω0n| = |1|), we obtain

e±jω0nu(n)z↔X (z) = 1

1−e±jω0z−1 ,ROC: |z | > 1

Thus, X (z) = 12

[1

1−e jω0nz−1 + 11−e−jω0z−1

],ROC: |z | > 1

After some manipulations

(cosω0n)u(n)z↔ 1−z−1 cosω0

1−2z−1 cosω0+z−2 ,ROC: |z | > 1



Ex. 3.2.4: Determine the z-transform and the ROC of the signalx(n) = u(n)− u(n − N)

X (z) = Z{u(n)} − Z{u(n − N)}

=(1− z−N

)Z{u(n)}

=(1− z−N

)1

1−z−1 ,ROC: |z | > 1

Time Scaling Property

If x(n)z↔X (z),ROC: r1 < |z | < r2

then

anx(n)z↔X (a−1z),ROC: |a|r1 < |z | < |a|r2

for any constant a, real or complex.


Properties of the z-TransformScaling Property

Ex. 3.2.5: Determine the z-transform and the ROC of the signalx(n) = an(cosω0n)u(n)

Given that

(cosω0n)u(n)z↔ 1−z−1 cosω0

1−2z−1 cosω0+z−2 ,ROC: |z | > 1

we easily obtain

an(cosω0n)u(n)z↔ 1−az−1 cosω0

1−2az−1 cosω0+a2z−2 ,ROC: |z | > |a|


Properties of the z-TransformTime Reversal Property

Time Reversal Property

If x(n)z↔X (z),ROC: r1 < |z | < r2

then

x(−n)z↔X (z−1),ROC: 1

r1< |z | < 1

r2

Ex. 3.2.6: Determine the z-transform and the ROC of the signalx(n) = u(−n)

Given that

u(n)z↔ 1

1−z−1 ,ROC: |z | > 1

we easily obtain

u(−n)z↔ 1

1−z ,ROC: |z | < 1


Properties of the z-TransformDifferentiation in the z-Domain ... 1

Differentiation in the z-Domain

If x(n)z↔X (z)

then

nx(n)z↔−z dX (z)

dz

where both transforms have the same ROC.

Ex. 3.2.7: Determine the z-transform and the ROC of the signalx(n) = nanu(n)

Given that

anu(n)z↔ 1

1−az−1 ,ROC: |z | > |a|

we easily obtain

nanu(n)z↔−z dX (z)

dz = az−1

(1−az−1)2,ROC: |z | > |a|


Properties of the z-TransformDifferentiation in the z-Domain ... 2

Ex. 3.2.8: Determine the z-transform and the ROC of the signalX (z) = log(1 + az−1),ROC: |z | > |a|

Taking the first derivative of X (z), we obtain

dX (z)dz = −az−2

(1+az−1)

Thus

−z dX (z)dz = az−1

[1

(1−(−a)z−1)

], |z | > |a|

The term in the brackets is the inverse z-transform of (−a)nu(n) and themultiplication of z−1 implies a time delay by one sample (time shiftingproperty), hence, from the differentiation property

nx(n) = a(−a)n−1u(n − 1)

⇒ x(n) = (1−)n+1 an

n u(n − 1)


Properties of the z-TransformConvolution of Two Sequences

Convolution of Two Sequences


z↔X2(z)

then

x(n) = x1(n) ∗ x2(n)z↔X (z) = X1(z)X2(z)

where the ROC is, at least, the intersection of that of X1(z) and X2(z).

Ex. 3.2.9: Compute the convolution x(n) of the signalsx1(n) = {1,−2, 1} and x2(n) = u(n)− u(n − 6)

X1(z) = 1− 2z−1 + z−2

X2(z) = 1 + z−1 + z−2 + z−3 + z−4 + z−5

then, X (z) = X1(z)X2(z) = 1− z−1 − z−6 + z−7

Hence, x(n) = {1↑,−1, 0, 0, 0, 0,−1, 1}


Properties of the z-TransformCorrelation of Two Sequences and Initial Value Theorem

Correlation of Two Sequences


z↔X2(z)

then

rx1x2(l) =∑∞

n=−∞ x1(n)x2(n − l)z↔Rx1x2(z) = X1(z)X2(z−1)

where the ROC is, at least, the intersection of that of X1(z) and X2(z−1).

Initial Value Theorem

If x(n) is causal [i.e., x(n) = 0 for n < 0],

then

x(0) = limz→∞ X (z).

The properties are summarized in Table 3.2 page 164.


Properties of the z-TransformSome common z-Transform Pairs


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