Z Transform

Ch3.4: Z-Transform Information source and input transducer

Source Coding

• Ch3.4: z-Transform

Channel Coding

Modulator

Questions to be answered: – z-Transform: A general version of DTFT – Region of Convergence: Where the transform exists.

Channel

– Inverse z-Transform: Back to timedomain – LTI System in Transform Domain: A general version of the frequency response function Information sink and output transducer

Elec3100 Chapter 3.4

Source Decoding

Channel Decoding

Demodulator (Matched Filter) 1

Ch3.4: z-Transform • Definition of z-Transform • Region of Convergence • Inverse z-Transform • Z-Transform Properties • LTI System in Transform Domain

Elec3100 Chapter 3.4

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Generalizing DTFT •

The DTFT provides a frequency-domain representation of discretetime signals and LTI discrete-time systems.

•

However, because of the convergence condition, in many cases, the DTFT of a sequence may not exist.

•

As a result, it is not possible to make use of such frequency-domain characterization in these cases.

•

Question: Can we generalize DTFT?

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z-Transform •

The z-transform X(z) of a sequence x[n] is defined as

where z is a continuous complex variable. Generally, we can where r = express the complex variable z in polar form as |z| > 0 is the magnitude and is the angle of z. •

z-transform gives ∑

•

In particular, when r = 1, then becomes ∑ |

Elec3100 Chapter 3.4

and the above expression

4

z-Transform • The relationship between and can also be illustrated in the z-plane (a real-imaginary plane for complex number).

The z-transform evaluated on the unit circle (as ω varies from 0 to 2π) corresponds to the DTFT of x[n].

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z-Transform • Main advantages of using z-transform over DTFT: – z-transform can encompass a broader class of signals than DTFT – Recall that a sufficient condition for convergence of the DTFT is: |

|

∞

– Example: The DTFT of 0.5 1 does not exist since it is unbounded in the negative direction: -5 -4 -3 -2 -1 0 -8 -16

-4 -2

n

-32 Elec3100 Chapter 3.4

6

z-Transform •

On the other hand, the z-transform exists if ∞ We can choose region of convergence (ROC) for z such that the z-transform converges.

•

Example: z-transform of

•

Proof:

∑ ∑ ∑

0.5

0.5 0.5 0.5

1 exists if 0.5

1

1

. .

Elec3100 Chapter 3.4

.

7

z-Transform • Other advantages of using z-transform: – More convenient notation than DTFT when dealing with analytical problems

– Convenient block diagram representation for implementation of practical systems, Eg., a unit delay system is expressed as: x[n]

z-1

y[ n] = x[n − 1]

– Useful for determining and solving difference equations for discrete-time systems

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Application: Filter Design •

The convolution sum description of an LTI discrete-time system with ∑ . an impulse response is given by

•

The output to an input ∑ ∑ ∑

•

For any given input sequence , the output will be indicating that is an eigen-function for any LTI system.

•

How can we utilize this property?

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is given by

,

9

Use of z-Transform: Filter Design • Can you figure out what kind of filter has the following ztransform and magnitude spectrum?

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Ch3.4: z-Transform • Definition of z-Transform • Region of Convergence • Inverse z-Transform • Z-Transform Properties • LTI System in Transform Domain

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Region of Convergence (ROC) •

Since z can be any point on the z-plane, generally there exists some z which makes X(z) not converge |

•

|

→∞

The set of z values for which X(z) converges ∞ is called the region of convergence (ROC).

•

The ROC must be specified along with X(z) in order for the ztransform to be completely defined.

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Poles and Zeros in z-Transform •

In many applications, X(z) can be expressed as a rational function: M

M

P( z ) X ( z) = = Q( z )

∑b z

k

∑a z

k

k

k =0 N

k

b ( z − c1 )(z − c2 )L (z − cM ) b0 = 0 = a0 ( z − d1 )( z − d 2 )L (z − d N ) a0

k =0

∏ (z − c ) k

k =1 N

∏ (z − d )

a0 ≠ 0, b0 ≠ 0

k

k =1

•

P(z) and Q(z) are numerator and denominator polynomials in z

•

The values of z for which X(z) = 0, i.e. roots of P(z), are called zeros of X(z)

•

The values of z for which X(z) = ∞, i.e. roots of Q(z), are called poles of X(z)

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Properties of the ROC •

The ROC is ring or disk in the z-plane centered at the origin, i.e. 0 | | ∞

•

The DTFT of converges absolutely iff the ROC of the ztransform includes the unit circle.

•

The ROC cannot contain any poles.

•

Example: The z-transform is given by .

,

of the sequence

0.6

0.6

Here, the ROC is just outside the circle going through the point z 0.6.

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ROC of Finite-Length Sequence • If is a finite-duration sequence, i.e. a sequence that is ∞, then zero except over a finite interval ∞ the ROC is the entire z-plane, except z = 0. • Example: Consider a finite-length sequence defined for −M ≤ n ≤ N, where M and N are non-negative integers and | ∞ | • Its z-transform is given by

• Note: has M poles at ∞ and N poles at 0. • Thus, the ROC is the entire z-plane except possibly at and/or at ∞ Elec3100 Chapter 3.4

N1

N2

n

0

15

ROC of Right-Sided Sequence •

A right-sided sequence is a sequence with nonzero sample values for . A right-sided sequence with nonzero sample values for 0 is called a causal sequence.

•

Consider a causal sequence

•

converges exterior to a circle , including the point ∞. On the other hand, a right-sided sequence with nonzero sample values only for with M nonnegative has a ztransform with M poles at ∞. is exterior to a circle , excluding the point The ROC of ∞.

•

•

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, whose z-transform is given by

16

Example 5.1 Right-Sided Sequence • Determine the z-transform of

•

converges if ∑ 1⇒

|

|

∞. This requires

| |, and gives

Note: If |a| < 1, the ROC contains the unit circle and hence the DTFT of this sequence x[n] exists. Shaded area = ROC

•

Remark: u[n] has z-transform, but not DTFT

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x denotes as the pole of X(z) o denotes as the zero of X(z) 17

ROC of Left-Sided Sequence •

•

A left-sided sequence is a sequence with nonzero sample values for . A left-sided sequence with nonzero sample values for 0 is called an anticausal sequence. Consider an anticausal sequence , whose z-transform is given by

•

It can be shown that converges interior to a circle , including the point 0.

•

On the other hand, a left-sided sequence with nonzero sample values only for with N nonnegative has a z-transform with N poles at 0. is interior to a circle , excluding the point The ROC of 0.

•

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Example 5.2 Left-Sided Sequence •

Determine the z-transform of another signal

1

1 •

converges if ∑ 1⇒

|

|

∞. This requires

| |, and gives

Note: If |a| < 1, the ROC does not contain the unit circle and the DTFT of this sequence x[n] does not exist. x denotes as the pole of X(z) o denotes as the zero of X(z)

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Observation from Example 5.1 and 5.2 •

Different signals can give same z-transform, although the ROCs will differ

x[n] = a n u[n]

x[n] = −a nu[−n − 1]

1 X ( z) = −1 if z > a 1 − az

X ( z) =

( ) exists if a < 1 ( ) does not exist if a > 1

X e jω X e jω Elec3100 Chapter 3.4

1 if z < a −1 1 − az

( ) exists if a > 1 ( ) does not exist if

X e jω X e jω

a

if

1 2

n

1 ⎛ −1 ⎞ Z x2 [n] = ⎜ ⎟ u[n] ←⎯→ X 2 ( z) = 1 + 13 z −1 ⎝ 3 ⎠

X ( z) =

1

1

+

1 − 12 z −1 1 + 13 z −1

(

)

1 2 z z − 12 = z − 12 z + 13

(

Elec3100 Chapter 3.4

)(

)

if

if

z >

1 3

1⎞ ⎛ 1⎞ 1 ⎛ ⎜ z > ⎟ I⎜ z > ⎟ ⇒ z > 3⎠ ⎝ 2⎠ 2 ⎝ The overall ROC is the region of overlap of the ROC for X1(z) and the ROC for X2(z); unless there is pole-zero cancellation. 23

Table 3.4.1: Common z-transform pairs

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Ch3.4: z-Transform • Definition of z-Transform • Region of Convergence • Inverse z-Transform • Z-Transform Properties • LTI System in Transform Domain

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The Inverse z-Transform •

There are 4 commonly used techniques to evaluate the inverse ztransform. They are – 1. Inspection Method – 2. Partial Fraction Expansion – 3. Power Series Expansion – 4. Cauchy Integral Theorem • This is the formal inverse z-transform expression based on Contour integral (an integration techniques together with right-hand rule) onto z-plane. • We will not cover this method in this course.

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Inverse z-transform: Inspection Method •

Inspection Method: Familiar with, or recognizing “by inspection”, certain transform pairs.

•

Example: – Find the inverse z-transform of

– This gives

X ( z) =

1

z X ( z) = z − ( 12 )

z>

1 2

1 z> 2

1 − (12 )z −1

– By making use of the transform pair in Table 3.4.1, n

⎛1⎞ x[n] = ⎜ ⎟ u[n] ⎝ 2⎠ Elec3100 Chapter 3.4

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Partial Fraction Expansion Method •

A rational

•

If

can be expressed as ∑ ∑

, then

can be re-expressed as

where the degree of

is less than N. The rational function

is

called a proper fraction.

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Inverse Transform: Partial-Fraction Expansion •

Simple Poles: In most cases, the rational z-transform of interest is a proper fraction with simple poles. Let the poles of be at ,1 . A partial-fraction expansion of is then of the form 1

•

The constants

•

Each term of the sum in partial-fraction expansion has an ROC given by and, thus has an inverse transform of the form . Therefore, the inverse transform of G(z) is given by ∑ .

•

Reminder: There are cases with multiple poles.

Elec3100 Chapter 3.4

are called the residues and given by 1 |

29

Inverse Transform: Partial-Fraction Expansion •

Example 5.5: Determine whose z-transform is given by 2 1 2 1 0.2 1 0.6 0.2 0.6

•

Solution: A partial-fraction expansion of 1

•

0.2

1

is of the form 0.6

Given

we have

Elec3100 Chapter 3.4

. .

. .

. Thus,

30

Inverse Transform by Long Division •

The z-transform G(z) of a sequence can be expanded in a power series in . In the series expansion, the coefficient multiplying the term is then the n-th sample of .

•

For a rational z-transform expressed as a ratio of polynomials in the power series expansion can be obtained by long division.

•

Example – Consider 1

•

2 0.12

Long division of the numerator by the denominator yields 1

•

1 0.4

,

1.6

0.52

0.4

1 1.6 0.52 0.4 As a result: result with that of Example 5.5.

Elec3100 Chapter 3.4

0.2224

⋯

0.2224 … . Compare this 31

Ch3.4: z-Transform • Definition of z-Transform • Region of Convergence • Inverse z-Transform • Z-Transform Properties • LTI System in Transform Domain

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z-Transform Properties •

Linearity Z Z If x1[n]←⎯→ X 1 ( z ), ROC = Rx1 and x2 [n ]←⎯→ X 2 ( z ), ROC = Rx Z then ax1 [n] + bx2 [n]←⎯→ aX 1 ( z ) + bX 2 ( z ), ROC contains Rx I Rx 1

• •

Example: Consider the two-sided sequence 1. and Two sequences , | | and transforms

•

With linearity property, we arrives at

• •

1 ROC is given by the overlap regions of | |? Question: What happens if

Elec3100 Chapter 3.4

1

2

2

1 have z, | | 1 1 | | and

| |.

33

z-Transform Properties •

Time Shifting Z x[n − n0 ]←⎯→ z − n0 X ( z ),

ROC = Rx

ROC is unchanged, except for the possible addition or deletion of −n the points z = 0 or z = ∞. It is because the factor z 0 can alter the number of poles at z = 0 or z =∞. •

Multiplication by an Exponential Sequence Z z 0n x[n]←⎯→ X ( z z0 ),

ROC = z 0 Rx

where ROC | | denotes that the ROC is scaled by | |; ie. if is the set of values of z such that rR < |z| < rL, then the new ROC is the set of values of z such that |z0|rR < |z| < |z0| rL

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z-Transform Properties •

Differentiation

dX ( z ) , nx[n]←⎯→ − z dz Z

ROC = Rx

ROC is unchanged except for the possible addition or deletion of z = 0. •

Conjugation of a Complex Sequence Z x ∗ [n ]←⎯→ X ∗ ( z ∗ ),

•

Z Time Reversal x[− n ]←⎯→ X (1 z ),

ROC = Rx ROC = 1 Rx

where ROC 1/ denotes that the ROC is inverted; i.e. if is the set of values of z such that rR < |z| < rL, then the ROC is the set of values of z such that 1/rL < |z| < 1/rR

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Example 5.6: Time-Reversal • Find the z-transform of x[ n] = a − nu[− n] • Solution:

1 , a u[ n] ←⎯→ z>a −1 1 − az 1 − a −1 z −1 Z −n = a u[ − n] ←⎯→ , −1 −1 1 − az 1 − a z n

Elec3100 Chapter 3.4

Z

z < a −1

36

z-Transform Properties •

Convolution of Sequences Z Z If x1 [n]←⎯→ X 1 ( z ), ROC = Rx1 and x2 [n ]←⎯→ X 2 ( z ), Z then x1 [n]∗ x2 [n]←⎯→ X 1 ( z ) X 2 ( z ),

ROC = Rx2

ROC contains Rx1 I Rx2

The resulting ROC contains at least the intersection of Rx1 and Rx2 – If there is no pole-zero cancellation in the linear combination, the ROC is exactly equal to Rx1 I Rx2 . – If there is pole-zero cancellation in the linear combination, the ROC may be larger. •

Initial-Value Theorem If x[n] is zero for n < 0 (i.e. if x[n] is causal), then x[0] = lim X ( z ) z →∞

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Example 5.7: Convolution •

Find the convolution sum of 1.

and

, where

•

Solution: – Method 1: direct convolution Z −1 – Method 2: Y ( z ) = X 1 ( z ) X 2 ( z ) ⎯⎯→ y[ n] = x1 [n ]∗ x2 [n ] 1 1 X 1 ( z) = , z > a and X ( z ) = , z >1 2 −1 −1 1 − az 1− z This gives Y ( z) =

1

1 −1

−1

z >1

,

1 − az 1 − z 1 ⎛ 1 a ⎞ = − ⎜⎜ ⎟⎟ − 1 − 1 1− a ⎝1− z 1 − az ⎠

y[n] = Elec3100 Chapter 3.4

(

)

1 u[n] − a n+1u[n] 1− a

38

Table 3.4.2: z-transform properties

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Ch3.4: z-Transform • Definition of z-Transform • Region of Convergence • Inverse z-Transform • Z-Transform Properties • LTI System in Transform Domain

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LTI Systems in the Transform Domain •

An LTI discrete-time system is completely characterized in the timedomain by its impulse response sequence .

•

Each LTI discrete-time representation.

•

Such transform-domain representations provide additional insight into the behavior of such systems. It is easier to design and implement these systems in the transform-domain for certain applications.

•

We consider now the use of the DTFT and the z-transform in developing the transform domain representation of an LTI system.

Elec3100 Chapter 3.4

system

has

a

transform-domain

41

The Transfer Function •

Transfer Function: A generalization of the frequency response function.

•

The convolution sum description of an LTI discrete-time system with ∑ . an impulse response is given by

•

Taking the z-transform of both sides, we get ∑ ∑ ∑ ∑ ∑ ∑ ∑

•

∑ ∑

/ is the z-transform of Here, transfer function or the system function.

Elec3100 Chapter 3.4

and is called the 42

FIR vs. IIR Filter •

A finite-duration impulse response (FIR) filter has a impulse response ⎧bn 0 ≤ n ≤ M − 1

h ( n) = ⎨ ⎩0

•

else

The difference equation representation is

y[ n] = b0 x[ n] + b1x[ n − 1] + L + bM −1 x[ n − M + 1] which is a linear convolution of finite support. • •

IIR filters are characterized by infinite-duration impulse response. The difference equation representation of an IIR filter is expressed as

y[n] =

Elec3100 Chapter 3.4

M −1

N −1

m=0

m =1

∑ bm x[n − m] − ∑ am y[n − m] 43

LTI Systems in the Transform Domain • Consider an LTI discrete-time system characterized by linear constant coefficient difference equations of the form: ∑ ∑ . • Applying the z-transform to both sides of the difference equation, we arrive at ∑ ∑ where and denote the z-transforms of and with associated ROCs, respectively. • A more convenient form of the z-domain representation of the difference equation is given by Elec3100 Chapter 3.4

44

LTI Systems in the Transform Domain •

Thus, ∑ ∑ ∑ ∑ ∏ ∏

• , ,…, are finite zeros, and , , … , are finite poles. • If , there are additional zeros at 0. • If , there are additional poles at 0. • For a causal digital filter, the impulse response is a causal sequence. The ROC of the causal transfer function is exterior to a circle going through the pole furthest from the origin with max | | Elec3100 Chapter 3.4

45

Example 5.8: Moving Average Filter •

•

Consider the M-point moving-average FIR filter with an impulse ,0 1 response, 0, otherwise Its transfer function is given by 1

1

1 1

• • •

1

The transfer function has M zeros on the unit circle at / ,0 1. There is an 1)th order pole at 0 and a single pole at 1. The ROC is the entire z-plane except 0

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Example 5.8: Moving Average Filter

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Frequency Response from Transfer Function •

If the ROC of the transfer function H(z) includes the unit circle, then the frequency response of the LTI digital filter can be obtained | simply as follows:

•

For a stable rational transfer function in the form ∏ ∏ the frequency response is given by ∏ ∏

•

It is convenient to visualize the contributions of the zero factor and the pole factor from the factored form of the frequency response.

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Frequency Response from Transfer Function •

•

The magnitude response function is given by: ∏ | | | | || | ∏ | ∏ | | | | ∏ | |

| |

The phase response is of the form arg arg

•

arg

How can we use this?

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Geometric Interpretation •

The factored form of frequency response ∏ ∏ is convenient to develop a geometric interpretation of the frequency response computation from the pole-zero plot as ω varies from 0 to 2π on the unit circle.

•

A typical factor in the factored form of the frequency response is where is a zero or pole. given by

•

It can be observed from the right figure that represents a vector starting at z

and ending on the unit circle .

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Geometric Interpretation •

As indicated by |

|

|

∏

|∏

| |

| |

, the magnitude

response at a specific frequency is given by the product of the magnitudes of all zero vectors divided by the product of the magnitudes of all pole vectors. • • •

Question: When can we obtain the smallest magnitude? A zero (pole) vector has the smallest magnitude when To attenuate signal components in a specified frequency range, we need to place zeros very close to or on the unit circle in this range. To emphasize signal components in a specified frequency range, we need to place poles very close to or on the unit circle in this range.

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Can we build an ideal bandpass filter?

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Can we build an ideal bandpass filter? •

The z-transform of an FIR impulse response can be expressed as a simple polynomial P(z) of degree L - 1 where L is the number of nonzero taps of the filter.

•

The fundamental theorem of algebra states that such a polynomial has at most L - 1 roots. Thus, the frequency response of an FIR filter can never be identically zero over a frequency interval since, if it were, its z-transform would have an infinite number of roots.

•

The argument can be easily extended to rational transfer functions, confirming the impossibility of a realizable filter whose characteristic is piecewise perfectly flat.

•

Then, how can we design a bandpass filter? (To be continued…)

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Stability Condition •

A causal LTI digital filter is BIBO stable if and only if its impulse response h[n] is absolutely summable, i.e., S ∑ | | ∞.

•

Thus, an FIR digital filter with bounded impulse response is always stable.

•

Question: What is the condition in terms of pole locations of ?

•

The ROC of the for the impulse response sequence is defined by values of where ∑ | | ∞. Thus, if the digital filter is stable, the ROC includes the unit circle.

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Stability Condition in Terms of Pole Position •

Consider the causal IIR digital filter with a rational transfer function H(z) given by

•

• •

∑ ∑

. Its impulse response is right-sided

sequence. The ROC of the is exterior to a circle going through the pole furthest from z 0. But stability requires ROC includes the unit cycle. Conclusion: All poles of a causal stable transfer function H(z) must be strictly inside the unit circle. The stability region (shown shaded) in the z-plane is shown below

Elec3100 Chapter 3.4

z-Transform

55

Example 5.9: Stability Condition IIR Filter

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