Z Transform f2014

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S Y S C 4 4 0 5 D i g i t a l S i g n a l P r o c e s s i n g Z _ T r a n s f o r m _ f 2 0 1 4 . f m

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Z- Transform

Definition of (the bilateral) z-transform

The z-transform of the sequence x(n) is denoted by X(z), and is defined as:

The inverse transform of X(z) is x(n). This is expressed as:

The Region of Convergence

Since the z-transform is an infinite power series, it exists only for those values of z for which this series converges.The region of convergence (ROC) of X(z) is the set of all values of z for which X(z) attains a finite value. Thus any

time a z-transform is cited, the ROC should also be indicated.Example

Determine the z-transforms of the following finite-duration sequences:

a)

b)

c)

d)

e)

f) ,

X z x n z n

n =

=

Z1

X z x n =

x1 n 1 2 5 7 0 1 =

x2 n 1 2 5 7 0 1 =

x3 n 0 0 1 2 5 7 0 1 =

x4 n 2 4 5 7 0 1 =

x5 n n =

x6 n n k = k 0


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g) ,

Solutiona) , ROC: entire z-plane except

b) , ROC: entire z-plane except and

c) , ROC: entire z-plane except

d) , ROC: entire z-plane except and

e) , ROC: entire z-plane

f) , ROC: entire z-plane except

g) , ROC: entire z-plane except

From the above example it can be concluded that the ROC of a finite duration signal is the entire z-plane exceptpossibly for the points and/or

In many cases the sum of the infinite series X(z) can be expressed in a closed form expression.

Example

Determine the z-transform of the real exponential sequence

Solution

x6 n n k+ = k 0

X1 z 1 2z1 5z 2 7z 3 z 5+ + + += z 0=

X2 z z2 2z 5 7z 1 z 3+ + + += z 0= z =

X3 z z2 2z 3 5z 4 7z 5 z 7+ + + += z 0=

X4 z 2z2 4z1 5 7z 1 z 3+ + + += z 0= z =

X5 z 1=

X6 z z k= z 0=

X7 z zk= z =

z 0= z =

x n 12---

nu n =

X z x n z n

n =

12---

n

z n

n 0=

12---z 1

n

n 0=

112---z 1

12---z 1

2

12---z 1

n

+ + + + += = = =


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The right hand side of the above equation is an infinite geometric series, the sum of which converges when is

less than 1.

, ROC:

For this example we notice that 1) the sequence is causal, and 2) the region of convergence is all points in the z-plane

that fall outsideof a circle of radius

Example

Determine the z-transform of the real exponential sequence

Solution

ROC

For this example we notice that 1) the sequence is anti-causal, and 2) the region of convergence is all points in the z-plane that fall insideof a circle of radius

12---z 1

X z 1

112---z 1

-------------------=12---z 1 1 z

12---

r1

2

---=

x n 2nu n 1 =

X z x n z n

n =

2nz n

n =

1

2 1z ll 1=

2 1z 2 1z 2 2 1z n + + + +

2 1z 1 2 1z 2 1z 2 2 1z n + + + + + 21z

1 2 1z-------------------

1

1 2z 1-------------------

= = = = =

= =

2 1z 1 z 2

r 2=


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More on the Region of Convergence:

In the case of an infinite sequence, the infinite series that defines X(z) can be split into two parts as shown above. Thefirst sum accounts for the samples of x(n) for , which converges only for points in the z-plane where issmall enough (< some positive real number r). The second sum accounts for the samples of x(n) for , which

converges only for points in the z-plane where is large enough (> some positive real number r). The region ofconvergence for X(z) is the annular region between the two circles with radii r1and r2provided that .

X z x n z n

n =

x n z n

n =

1

x n z n

n 0=

+ x n zn

n 1=

x n z n

n 0=

+= = =

X z x n zn

n 1=

x n z n

n 0=

+ x n zn

n 1=

x n z n

n 0=

+=

n 1 z0 n

z

r1 r2

X z x n zn

n 1=

x n

zn-------------

n 0=

+

converges for |z| < r2 converges for |z| > r1


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Example

Determine the z-transform of the signal

Solution

r1r2 r1

r2

ROC for

x n z n

n =

1

ROC for

x n z n

n 0=

ROC for

x n z n

n =

1

x n z n

n 0=

+

x n anu n bnu n 1 +=

X z x n z n

n =

bnz n

n =

1

anz n

n 0=

+ b nzn

n 1=

anz n

n 0=

+= = =


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ROC1:

ROC2:

ROC: annular region of the z-plane between a circle of radius |a| and a circle of radius |b| provided that . If thiscondition does not hold, X(z) will not exist.

Z-transform of some Special Sequences

The unit sample sequence

Shifted Unit Sample

b nzn

n 1=

b 1z 1 b 1z b 1z 2 b 1z n + + + + + b 1z

1 b 1z------------------- 1

1 bz 1-------------------= = =

b 1z 1 z b

anz n

n 0=

1 az 1 az 1 2 az 1 n + + + + +1

1 az 1-------------------= =

az 1 1 z a

X z 11 bz 1-------------------

1

1 az 1-------------------+=

b a

X z Z n z0

1= = =

X z Z n m z m

= =


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Right Sided (aka causal) Sequences

RSS are zero for

The general complex exponential sequence where A, a and 0are real constants.

with a ROC

.

The transforms for a number of special sequences can be obtained from the general exponential sequence discussedabove.

n 0

x n Aanej0n

u n =

X z Aanej0n

z n

n 0=

A aej0

z1

n

n 0=

A

1 aej0

z1

-----------------------------= = =

aej0

z1

1 z a

ROC

imag

real

|a|Z-plane


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Unit Step Sequence :

Substitute , to obtain:

with a region of convergence (ROC) described by

Real Exponential Sequence

To find X(z) substitute . The transform of the real exponential is

with a ROC

General Oscillatory Sequence

with a ROC

x n u n =

A a 0 1 1 0 =

X z u n z n

n 0=

1

1 z1

----------------= =

z 1

x n anu n =

A 0 1 0=

X z anz n

n 0=

A az1

n

n 0=

1

1 az1

-------------------= = =

z a

x n Aan 0n cos u n A

2---a

nej0nu n A

2---a

ne

j0n u n += =

X z Z A2---a

nej0n

u n

z A

2---a

ne

j0nu n

+

A 2

1 aej0z 1----------------------------- A 2

1 ae j0 z 1-------------------------------+

A 1 az1 0cos

1 2az 1 0cos a2z 2+----------------------------------------------------------

= =

=

z a


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Table of transform pairs (reproduced from table 3.3, Proakis and Manolakis)

time sequence x(n) z-Transform X(z) ROC

n 1 allz

u n 11 z 1---------------- z 1

anu n 11 az 1------------------- z a

nanu n az 1

1 az 1 2--------------------------- z a

anu n 1 11 az 1------------------- z a

nanu n 1 az 1

1 az1

2

--------------------------- z a

0ncos u n 1 z 1 0cos

1 2z 1 0cos z2+

------------------------------------------------- z 1

0sin n u n z 1 0sin

1 2z 1 0cos z2+

------------------------------------------------- z 1

an 0ncos u n 1 az1

0cos1 2az 1 0cos a

2z 2+---------------------------------------------------------- z a

an 0nsin u n az 1 0sin

1 2az 1 0cos a2z 2+

---------------------------------------------------------- z a

A

1 pz 1-------------------

A

1 pz 1----------------------+2A p

n n p A+ cos u n z a


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Properties of the Z-transform

1)Linearity

The transform is linear in the sense that if , then

where {Ck} are arbitrary constants, and where .

To prove it:

The region of convergence of Y(z) is that part of the z-plane over which all of {Fk(z)} converge; i.e. it is theintersection of all ROCs associated with {Fk(z)}.

Example

Starting with the transform pair , with a ROC , determine the transform of the

sequence

Solution:

y n Ckfkn k

=

Y z Z Ckfkn k

CkZ fkn k

CkFk z k

= = =

Z fkn Fk z =

Y z Ckfkn

k

z n

n =

Ck fkn z n

n =

k

Ck Fk z k

= = =

Aan

e

j0n

u n A

1 aej0

z1

----------------------------- z a

x n an 0nsin u n =

x n an 0nsin u n an e

j0n e j0n

2j

------------------------------- u n 1

2j

-----anej0nu n 1

2j

-----ane j0n u n = = =


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,

2) Time Shifting

The sequence y(n) is a time-shifted version of another sequence f(n) according to the relation: wherem is an integer.

The transform of y(n) is related to that of f(n) according to:

To prove it:

In the above equation change variables according to and therefore

The ROC for Y(z) is the same as the ROC of F(z), with the addition of the constraint if m is positive or if mis negative.

X z Z 12j-----anej0nu n

Z12j-----ane j0n u n

+=

X z

12j-----

1 aej0z 1-----------------------------

12j-----

1 ae j0 z 1-------------------------------

12j----- ej0 e j0 az 1

1 a ej0 e j0+ a2z 2+----------------------------------------------------------------

az 1 0sin

1 2az 1 0cos a2z 2+

----------------------------------------------------------= = = z a

y n f n m =

Y z Z f n m z m F z = =

Y z f n m z n

n =

=

k n m= n k m+=

Y z f k z k m+

k =

z m f k z kk =

z m F z = = =

z 0 z


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Example

A discrete time system is described by the input/output relationship

Determine the z-transform of the output, Y(z), in terms of the transform of the input, X(z).

Solution

3)Differentiation in the z-domain

If y(n) is related to f(n) according to: , then the transform of y(n) is given by:

Proof

The ROC for Y(z) is the same as the ROC for X(z)

y n bkx n k

k 0=

8

8=

Y z Z bkx n k

k 0=

8

8

18--- bkZ x n k

k 0=

8

18--- bkz

kX z

k 0=

8

bkz k

k 0=

8

8----------------------- X z = = = =

y n nx n =

Y z Z nx n zzdd

X z = =

zdd

X z zdd

x n z n

n =

x n zdd

z n

n =

nx n z n 1n =

z 1 nx n z nn =

z 1 F z = = = = =


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4) Time Reversal

Assume that ; i.e. y(n) is a time reversed version of x(n). Also assume that the ROC for X(z) is

The transform of y(n) is related to the transform of x(n) according to:

, ROC:

Proof

ROC:

5) Convolution of two sequences

Let y(n) be: . The transform of y(n) is then given by the product of transforms of both x(n) andh(n). i.e.

The ROC of Y(z) is the part of the z-plane over which both X(z) and H(z) do converge.

y n x n = r1 z r2

Y z X 1z---

=

1r2---- z

1r1----

Y z x n z n

n =

x k zkk =

x k z 1 kk =

X z 1 X1z---

= = = =

r11z--- r2

1r2---- z

1r1----

y n x n h n =

Y z Z x n h n Z x n Z h n X z H z = = =


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Proof

6) Initial Value Theorem

If x(n) is causal (i.e. for ), then

Proof

Since x(n) is causal,

as , for and the right hand side of the equation above is reduced to x(0).

Y z x k h n k k =

z n

n =

x k h n k z n

k =

k =

H z x k z k

k =

= = =

x n h n z kH z

x n 0= n 0

x 0 X z z lim=

X z x n z n

n 0=

x 0 x 1 z 1 x 2 z 2 x n z n + + + + += =

z z n 0 n 1


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Inverse Z-transform

Inverse Transform by Power Series Expansion

The basic idea is that we are given a transform X(z) with a specified region of convergence. We can expand X(z) intothe form:

which converges in the given ROC. By the uniqueness of the transform, , for .

When X(z) is rational (i.e. ratio of two polynomials in z), the power series expansion can be achieved by long division.

Example

Determine the inverse transform of when the ROC is

Solution: Since the ROC is , we know we are looking at a causal sequence. As such, we are looking for a series

expansion of the form

X z cnz n

n =

=

x n cn= n

X z az1

1 az 1 2---------------------------= z a

z a

X z cnz n

n 0=

=


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Therefore

, for

az 1

az 1 2a2z 2 a3z 3+

2a2z 2 a3z 3

2a2z 2 4a3z 3 2a4z 4+

3a3z 3 2a4z 4

3a3z 3 6a4z 4 3a5z 5+

4a4z 4 3a5z 5

4a4z 4 8a5z 5 4a6z 6+

5a5z 5 4a6z 6

az 1 2a2z 2 3a3z 3 4a4z 4 5a5z 5 + + + + +

1 2az 1 a2z 2+

5a5z 5 10a6z 6 5a7z 7+

X z az 1 2a2z 2 3a3z 3 4a4z 4 5a5z 5 nanz n + + + + + + +=

x n nan= 0 n


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ExampleDetermine the inverse transform of when the ROC is

Since the ROC is , we know we are looking at an anti-causal sequence. As such, we are looking for a series

expansion of the form

X z az 1

1 az 1 2---------------------------= z a

z a

X z cnzn

n 0=

=

az 1

a1z 1 2 a 1z1+

2 a 1z1

2 4a 1z1 2a 2z2+

3a 1z1 2a 2z2

3a 1z1 6a 2z2 3a 3z3+

4a 2z2 3a 3z3

4a 2z2 8a 3z3 4a 4z4+

5a 3z3 4a 4z4

a 1z1 2a 2z2 3a 3z3 4a 4z4 5a 5z5 + + + + +

a2z 2 2az 1 1+

5a 3z3 10a 4z4 5a 5z5+


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for

, for

Others:read example 3.4.2, text

The inverse by Partial Fraction Expansion

X(z) is a rational function of the form:

The rational function above is proper if and . Note that an improperrational function (because

can always be written as the sum of a rational function and a polynomial. For example,

In general terms, a rational function X(z) can always be written as a sum of a polynomial and a proper rationalfunction. The inverse transform of X(z) is equal to the sum of:

The inverse transform of the polynomial part which is obtained by inspection.

X z a 1z1 2a 2z2 3a 3z3 4a 4z4 5a 5z5 na nzn + + + + + + +=

x n na n= 0 n

x n nan= 1 n

x n nan u n 1 =

X z b0 b1z

1 b2z1 bMz

M+ + + +

1 a1z 1 22z 1 aNz N+ + + +-------------------------------------------------------------------------------=

aN 0 M N M N

2 z 2

1 z 1---------------- 1 z 1

1

1 z 1----------------+ +=

properimproper

polynomial


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The inverse of the proper rational function. It is for this part that we need to do partial fraction expansion.

How does Partial Fraction Expansion Work

1) To start, assume that we have a proper rational function in of the form:

2) To simplify the steps to follow, re-write X(z) as a function of (as opposed to :

We notice that as given above is a proper rational function.

3) Next we determine the roots of the denominator polynomial, by solving:

Assume that the roots determined above are located at

At this point we need to consider two different scenarios

The case of distinct roots

We assume that all roots are distinct. In this case, , can be written as a sum of N first order fractions

z 1

X z b0 b1z

1 b2z1 bMz

M+ + + +

1 a1z1 22z

1 aNz N+ + + +

-------------------------------------------------------------------------------=

z z 1

X z b0z

N b1zN 1 b2z

N 2 bMzN M+ + + +

zN a1zN 1 a2z

N 2 aN 1 z aN+ + + + +-----------------------------------------------------------------------------------------------------=

X z z

----------- b0z

N 1 b1zN 2 b2z

N 3 bMzN M 1+ + + +

zN a1zN 1 a2z

N 2 aN 1 z aN+ + + + +------------------------------------------------------------------------------------------------------------------=

X z z-----------

zN a1zN 1 a2z

N 2 aN 1 z aN+ + + + + 0=

z p1 p2 p3 pN =

X z z

-----------


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The constants are determined as follows:

After the constants have been determined, write X(z) in the form

Aided with knowledge of the ROC, the inverse transform of each fraction can be easily determined.

Example

Determine the inverse transform of

with a ROC

X z z

----------- A1

z p1-------------

A2

z p2-------------

A3

z p3-------------

AN

z pN--------------+ + + +

b0zN 1 b1z

N 2 b2zN 3 bMz

N M 1+ + + +

z p1 z p2 z p3 z pN ------------------------------------------------------------------------------------------------------------------= =

A1 A2 A3 AN

A1X z

z----------- z p1

z p1=

b0zN 1 b1z

N 2 b2zN 3 bMz

N M 1+ + + +

z p2 z p3 z pN ------------------------------------------------------------------------------------------------------------------

z p1=

= =

A2

X z

z----------- z p

2

z p2=

b0zN 1 b1z

N 2 b2zN 3 bMz

N M 1+ + + +

z p1 z p3 z pN ------------------------------------------------------------------------------------------------------------------

z p2== =

ANX z

z----------- z pN

z pN=

b0zN 1 b1z

N 2 b2zN 3 bMz

N M 1+ + + +

z p1 z p2 z p3 z pN 1 ------------------------------------------------------------------------------------------------------------------

z pN=

= =

A1 A2 A3 AN

X z A1z

z p1-------------

A2z

z p2-------------

ANz

z pN--------------+ + +

A1

1 p1z1

---------------------- A2

1 p2z1

---------------------- AN

1 pNz1

----------------------+ + += =

Y z 1 z2+

1 0.5z 1+ 1 z 1 0.5z 2+ + ------------------------------------------------------------------------= z 0.707


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Solution

is first re-written as a rational function of (as opposed to )

Notice that the roots of the denominator polynomial are located at:

, and . For later use, we need the magnitude and angle of

, which are:

Knowing that is in the second quadrant of the complex plane (because of the negative real part and a positive

imaginary part), its angle is computed as follows:

The magnitude and angle of are given by:

Y z z z 1

Y z z

3z+

z 0.5+ z2 z 0.5+ + -----------------------------------------------------=

Y z z

---------- z2

1+z 0.5+ z 0.5 j0.5+ z 0.5 j0.5+ + -----------------------------------------------------------------------------------------------

A1

z 0.5+ ---------------------

A2

z 0.5 j0.5+ ------------------------------------

A3

z 0.5 j0.5+ + -------------------------------------+ +

= =

p1 0.5= p2 0.5 j0.5+= p3 0.5 j0.5 conjugate of p2= =

p2

p2 0.52 0.52+ 0.5 0.707= = =

p2

p2 1 atan34

------ 2.3562 radians= = =

A1z

21+

z 0.5 j0.5+ z 0.5 j0.5+ + -------------------------------------------------------------------------

z 0.5=

5= =

A2z

21+

z 0.5+ z 0.5 j0.5+ + ----------------------------------------------------------

z 0.5 j0.5+=

2 j+= =

A2


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Therefore

It now remains to look-up the inverse transform for each term:

where , , and

Therefore

Case of repeated roots

If there is a repeated root, p1, say of order two, then:

A2 2 2 12+ 5 2.236= = =

A2 12--- atan 2.6779 radians= =

A3z

21+

z 0.5+ z 0.5 j0.5+ ----------------------------------------------------------

z 0.5 j0.5=

2 j conjugate of A2= = =

Y z 5

1 0.5 z 1--------------------------------

A2

1 p2z1

---------------------- A2

1 p2z1

-------------------------+ +=

Z1 5z

z 0.5+ ---------------------

5 0.5 nu n =

Z 1 A2z

z p2 ------------------

A3z

z p3 ------------------+

Z 1 A2

1 p2z1

---------------------- A2

1 p2z1

-------------------------+

2A2 p2n n p2 A2+ cos u n = =

A2 2.236= A2 2.68 radians= p2 0.707= p2 2.68 radians=

y n 5 0.5 nu n 2 2.236 0.707 n 2.3562n 2.6779+ cos u n +=


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and to find we proceed as follows:

differentiate the equation above w.r.t z

If we substitute we end up with

Y z z

---------- A1

z p1-------------

A2

z p1 2

--------------------- A3

z p3-------------

AN

z pN--------------+ + + +

b0zN 1 b1z

N 2 b2zN 3 bMz

N M 1+ + + +

z p1 2 z p3 z pN

------------------------------------------------------------------------------------------------------------------= =

A2Y z z

---------- z p1 2

z p1=

b0zN 1 b1z

N 2 b2zN 3 bMz

N M 1+ + + +

z p3 z pN ------------------------------------------------------------------------------------------------------------------

z p1=

= =

A3Y z z

---------- z p3 2

z p3=

b0zN 1 b1z

N 2 b2zN 3 bMz

N M 1+ + + +

z p1 2 z p4 z pN

------------------------------------------------------------------------------------------------------------------

z p3=

= =

ANY z z

---------- z pN 2

z pN=

b0zN 1 b1z

N 2 b2zN 3 bMz

N M 1+ + + +

z p1 2 z p3 z pN 1

------------------------------------------------------------------------------------------------------------------

z pN=

= =

A1

z p1 2Y z

z---------- A1 z p1 A2 z p1 2

A3

z p3-------------

AN

z pN--------------+ ++ +=

d

dz----- z p1

2Y z z

----------

A1 z p1 2d

dz-----

A3

z p3-------------

AN

z pN--------------+ +

2 z p1 A3

z p3-------------

AN

z pN--------------+ +

+ +=

z p1=

A1d

dz----- z p1

2Y z z

----------

z p1=

d

dz-----

b0zN 1 b1z

N 2 b2zN 3 bMz

N M 1+ + + +

z p3 z pN ------------------------------------------------------------------------------------------------------------------

z p1=

= =


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Example1

Determine the inverse transform of with a ROC

Solution

The denominator polynomial has a repeated root (multiplicity 2) at z = 1. Therefore, the partial fraction expansiontakes the form:

where are given by:

1. Examples 3.4.7 and 3.4.10 of the text, Proakis and Manolakis

X z

1

1 z 1+ 1 z 1 2---------------------------------------------

= z 1

X z z3

z 1+ z 1 2----------------------------------- X z

z

----------- z2

z 1+ z 1 2-----------------------------------= =

X z z

----------- A1

z 1-----------

A2

z 1 2------------------

A3

z 1+-----------+ +=

A1 A2 A3

A2 z2z 1+ ----------------

z 1=

12---= =

A3z2

z 1 2------------------

z 1=

14---= =

A1d

dz----- z2

z 1+ ----------------

z 1=

2z z 1+ z2z 1+ 2---------------------------------

z 1=

34---= = =

S Y S C 4 4 0 5 D i i t l S i l P i


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X z z

-----------

34---

z 1-----------

12---

z 1 2------------------

14---

z 1+-----------+ +=

X z 34--- 1

1 z 1---------------- 1

2---

z 1

1 z 1 2----------------------- 1

4--- 1

1 z 1+----------------+ +=

x n 34---u n 1

2---nu n 1

4--- 1 nu n + +=

S Y S C 4 4 0 5 D i g i t a l S i g n a l P r o c e s s i n g Z T f f 2 0 1 4 f


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Analysis of LTI Systems

Transfer Functions

The transfer function of a discrete-time LTI system is defined as the ratio , where Y(z) is the transform of the

output sequence y(n), which caused by the input sequence x(n). X(z) is the transform of the input sequence.

Since , it follows that . It therefore follows that, the transfer function is equal to the z-

transform of the impulse response h(n).

The transfer function can also be related directly to the system coefficientsas follows.

Y z X z -----------

Y z H z X z = Y z

X z ----------- H z =

y n aky n k

k 1=

N

bkx n k k 0=

M

+=

Y z akz k

Y z

k 1=

N

bkz kX z

k 0=

M

+=

Y z 1 akz k

k 1=

N

+ X z bkz k

k 0=

M

=

Y z X z ----------- H z

bkz k

k 0=

M

1 akz k

k 1=

N

+-------------------------------- zN M

bkzM k

k 0=

M

zN

akzN k

k 1=

N

+--------------------------------------- zN M

b0zM b1z

M 1 bM+ + +

zN a1zN 1 aN+ + +

---------------------------------------------------------------= = = =

S Y S C 4 4 0 5 D i g i t a l S i g n a l P r o c e s s i n g Z T r a n s f o r m f 2 0 1 4 f m


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The system has N poles. Thepolesof the system are the roots of the denominator polynomial:

The system has M zeros located at the roots of the numerator polynomial:

In addition, there are M-N (trivial) zeros located at the origin due to the factor

For systems with real coefficients (which is all we have in this course), complex poles occur in conjugate pairs. Also,complex zeros occur in conjugate pairs.

Example

Consider a causal LTI system with a transfer function of the form:

The poles of this system are given by:

The system zeros are given by:

zN

a1zN 1

a2zN 2 aN+ + + + 0=

b0zN

b1zN 1

b2zN 2 bMz

N M+ + + + 0=

zN M

H z z

4z

2+

z4

0.25+

---------------------=

z4

0.25+ 0=

z

4

0.25 0.25ej 1 2k+ 1

2-------e

j1 2k+

4---------------

4

= = =

z1 z2 z3 z4 1

2-------e

j4---

1

2-------e

j34

------1

2-------e

j54

------1

2-------e

j74

------

1

2-------e

j4---

1

2-------e

j34

------1

2-------e

j34

------ 1

2-------e

j4---

= =

S Y S C 4 4 0 5 D i g i t a l S i g n a l P r o c e s s i n g Z T r a n s f o r m f 2 0 1 4 f m


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Stability

Stability of Recursive SystemsThe impulse response h(n) is the inverse transform of the system transfer function. It contains exponential terms of the

formpnwherepis a system pole. Therefore if any of the system poles falls outside the unit circle, h(n) will not beabsolutely summable and hence the system can not be stable. We therefore conclude thatfor a causal, recursive IIRsystem to be stable all of its poles must have magnitudes less than one.

z4

z2

+ z2

z2

1+ z2 z j+ z j 0= = =

z1 z2 z3 z4 0 0 j j =

-1 -0.5 0 0.5 1

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Real Part

Imaginar

yPart

2

unit circle

pole

zero

Pole-Zero Diagram

conjugate

pair

S Y S C 4 4 0 5 D i g i t a l S i g n a l P r o c e s s i n g Z T r a n s f o r m f 2 0 1 4 . f m


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g g g _ _

29 / 49

Stability of Non-Recursive FIR Systems

The I/O relationship of a non-recursive system is expressed as

where {bk} are the system coefficients. The unit-sample response of such a system is given by:

The transfer function is given by:

We notice that H(z) has all of its poles (M of them) at the origin ( ). As a result, we conclude that non-recursiveFIR systems are always stable.

Solving Difference Equations using the z-Transform

Unilateral Transform

The unilateral transform of an arbitrary sequence

1

is defined as:

1. Note: by definition, the summation is over the range even if happened to be nonzero for

y n bkx n k

k 0=

M

=

h n bk n k k 0=

M

b0 n b1 n 1 b2 n 2 bM n M + + + += =

H z b0 b1z1 b2z

2 bMz M+ + + +

1

zM------ b0z

M b1zM 1 b2z

M 2 bM+ + + + = =

z 0=

n

0 n n n 0

Fu z f n z n

n 0=

=



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The properties we discussed earlier with respect to the bilateral transform also apply to the unilateral transform, withthe exception of the time-shifting property. Therefore, we revisit this one property.

Time Shifting PropertyLet where m is a positive integer. It can be shown that the unilateral transforms of y(n) and f(n) arerelated by:

In the above expression we may identify two parts:

y n f n m =

Yu z f m z0

f m 1+ z 1 f m 2+ z 2 f 1 z m 1+ z m Fu z + + + + +=

-4 -3 -2 -1 n

f(n)

-1 0 1 2 3n

f(n-3)

f n u n Fu z

f n 3 u n 3 z 3 Fu z

f n 3 f 3 f 2 z 1 f 1 z 2 z 3 Fu z + + +

f n m f m z 0 f m 1+ z 1 f m 2+ z 2 f 1 z m 1+ z m Fu z + + + + +



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1) a power series part which depends on past values of thesequence corresponding to . This component is clearly dependent on the value of the time shift

2) The second part is the transform of the original (un-shifted) sequence, multiplied by . This part is

identical to what we would have had if we were using the bi-lateral transform.

To solve a difference equation

1) Given a difference equation for an Nth-order system of the form:

where the input sequence x(n) is assumed to be right-sided.

2) Take the uni-lateral transform of both sides of the difference equation

3) Rearrange results

m z 0 f m 1+ z 1 f m 2+ z 2 f 1 z m 1++ + + +m n 1 m

z m Fu z z

m

y n aky n k

k 1=

N

bkx n k k 0=

M

+=

Yu z aky k y k 1+ z1 y 1 z k 1+ z k Yu z + + + +

k 1=

N

bkz kXu z

k 0=

M

+=

Yu z 1 a

kz

k

k 1=

N

+ a

ky k y k 1+ z 1 y 1 z k 1++ + +

k 1=

N

b

kz

kX

u z

k 0=

M

+=



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The first term to the right is the transform of the initial condition response. The second term is recognized as theproduct of the system transfer function and the transform of the input. This second term is the transform of the forcedresponse.

4) Find the inverse transform to obtain the total system response in the time domain.

Example

A second order system is described by:

where y(-2), y(-1) = 20.25, -8.5

a) Find the unit-sample response of the system

Yu z

aky k y k 1+ z1

y 1 z k 1++ + +

k 1=

N

1 akz k

k 1=

N

+-------------------------------------------------------------------------------------------------------------------------

bkz k

k 0=

M

1 akz k

k 1=

N

+-------------------------------------Xu z +=

y n 2.5y n 1 y n 2 x n x n 1 +=

H z

bkz k

k 0=

M

1 akz k

k 1=

N

+-------------------------------------

1 z1

1 2.5z 1 z 2+ +--------------------------------------

z2

z

z2 2.5z 1+ +------------------------------= = =

H z z

----------- z 1

z2

2.5z 1+ +------------------------------

z 1z 0.5+ z 2+ --------------------------------------

C1

z 0.5+ ---------------------

C2

z 2+ ----------------+= = =



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b) Determine the initial-condition response of the system

C1z 1z 2+ ----------------

z 0.5=

1= =

C2z 1

z 0.5+ ---------------------

z 2=

2= =

H z zz 0.5+ --------------------- 2z

z 2+ ----------------+=

h n 0.5 n 2 2 n+ u n =

Yic z a1 y 1 a2 y 2 y 1 z

1+ +

1 a1z1

a2z2

+ +------------------------------------------------------------------------------------=

Yic z

2.5 8.5 1 20.25 8.5 1 z 1+ +

1 2.5z 1 z 2+ +----------------------------------------------------------------------------------------------------------------

1 8.5z1

+

1 2.5z1

z2

+ +-------------------------------------- z

28.5z+

z2

2.5z 1+ +------------------------------=

= =

Yic z

z--------------

z 8.5+

z2

2.5z 1+ +------------------------------

z 8.5+z 0.5+ z 2+ --------------------------------------

C1

z 0.5+ ---------------------

C2

z 2+ ----------------+= = =

C1z 8.5+z 2+ ----------------

z 0.5=

163------= =

C2z 8.5+z 0.5+ ---------------------

z 2=

133------= =



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,

Other examplesThe following examples relate to a Linear Time Invariant System with a transfer function of the form:

Question 1:Determine the systems impulse response h(n) for using long division.Solution:Using long division as shown above leads to:

Therefore, h(n), for can be expressed as:

Yic z

163------z

z 0.5+ ---------------------

133------z

z 2+ ----------------+=

yic n 163------ 0.5 n 13

3------ 2 n= n 2

H z z 1+z z2 z 0.5+ ----------------------------------=

n 6

H z Z 2 2Z 332---Z 4

12---Z 5

14---Z 6

12---Z 4 1

8---Z 5+

Z3 Z212---Z+

----------------------------------------+ + + +=

n 6

h n n 2 2 n 3 32--- n 4

12--- n 5

14--- n 6 + + +=



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Question 2Find the difference equation of the system.

Z 1+

Z 112---Z 1+

212---Z 1

2 2Z1

Z2

+

32---Z 1 Z 2

32---Z 1

32---Z 2

34---Z 3+

12---Z 2

34---Z 3

12---Z 2

12---Z 3

14---Z 4+

14---Z 3

14---Z 4

14---Z 3 1

4---Z 4 1

8---Z 5+

12---Z 4

18---Z 5+

Z 2 2Z 332---Z 4

12---Z 5

14---Z 6+ + +

Z3 Z212---Z+



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Solve the difference equation iteratively to find h(n) for .Solution

Therefore, the system difference equation is given by

where

The difference equation of the system is given by:

h(n) is the system output, y(n), subject to the assumptions that , and . Subject to these 2conditions we proceed to solve the system equation iteratively as follows:

n 6

H z z 1+

z z2 z 0.5+ ----------------------------------

z 1+

z3 z212---z+

---------------------------- z 2 z 3+

1 z 112---z 2+

--------------------------------- b2z

2 b3z3+

1 a1z1 a2z

2-----------------------------------------= = = =

y n a1y n 1 a2y n 2 b2x n 2 b3x n 3 + + +=

a1 a2 b2 b3 112--- 1 1

=

y n y n 1 12---

y n 2 x n 2 x n 3 + + +=

x n n = y 2 y 1 0= =

y 0 y 1 12--- y 2 2 3 + + + 0 0 0 0+ + + 0= = =

y 1 y 0 12---

y 1 1 2 + + + 0 0 0 0+ + + 0= = =

y 2 y 1 12---

y 0 0 1 + + + 0 0 1 0+ + + 1= = =



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Therefore, h(n), for can be expressed as:

which is identical to the result obtained using long division followed by inverse-z transform.

Question 3:

Use partial fraction expansion to find h(n)Solution1

The constants are evaluated as follows:

y 3 y 2 12---

y 1 1 0 + + + 1 0 0 1+ + + 2= = =

y 4 y 3 12--- y 2 2 1 + + + 2 12--- 1 0 0+ + + 32---= = =

y 5 y 4 12---

y 3 3 2 + + +32---

12---

2 0 0+ + +12---= = =

y 6 y 5 12---

y 4 4 3 + + + 12---

12---

32---

0 0+ + + 14---= = =

n 6

h n n 2 2 n 3 32--- n 4

12--- n 5

14--- n 6 + + +=

H z z

----------- z 1+

z2 z1

2-------e

j4---

z1

2-------e

j4---

----------------------------------------------------------------- c1

z2-----

c4

z-----

c2

z1

2-------e

j4---

-------------------------------

c3

z1

2-------e

j4---

----------------------------+ + += =

c1 c2 c3 c4

c1z 1+

z1

2-------e

j4---

z1

2-------e

j4---

------------------------------------------------------------

z 0=

1

1

2-------e

j4---

1

2-------e

j4---

-------------------------------------------------- 112------ 2= = = =



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After substituting the values of the equation above is simplified to:

Therefore, the constants needed are

Now, H(z) can be re-written as

c2z 1+

z2 z 12

-------ej

4---

---------------------------------

z1

2-------e

j4---

=

11

2-------e

j4---

+

12---e

j2---

1

2-------e

j4--- 1

2-------e

j4---

--------------------------------------------------------------1.5 j

12---

j2--- j22-------- 4---sin

--------------------------------------1.5 j

12---

j2---- j

-------------------- 3 j1+= = = = =

c3z 1+

z2 z 12

-------e j4---

------------------------------------

z1

2-------e

j4---

=

11

2-------e

j4---

+

12---ej

2---

12

-------ej4--- 1

2-------e j

4---

------------------------------------------------------------1.5 j

12---+

j

2--- j2

2------- 4---sin

---------------------------------1.5 j

12---+

j

2--- j

------------------- 3 j1= = = = =

c4

z-----

z 1+

z2 z1

2

-------ej

4---

z1

2

-------ej

4---

----------------------------------------------------------------- c1

z2-----

c2

z1

2

-------ej

4---

-------------------------------

c3

z1

2

-------ej

4---

----------------------------=

c1 c2 c3

c4

z-----

z 1+ 6z3 6z2 2z 1+ +

z2 z1

2-------e

j

4

---

z

1

2-------e

j

4

---

-----------------------------------------------------------------------

6z3

6z2

3z+

z2 z2 z 0.5+ ------------------------------------------

6z---

z2 z 0.5+ z2 z 0.5+ -------------------------------

6z---= = = =

c1 c2 c3 c4 2 3 j1+ 3 j1 6 =



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The inverse transform is given by:

This result produces the same numerical values obtained using long division, and iterative solutions for the systemsdifference equation.

Solution 2:Let us express H(z) as , and therefore , where

Now let us try to find f(n).

H z 2z--- 6

3 j1+

11

2

-------ej

4---z 1

--------------------------------------- 3 j1

11

2

-------ej

4---z 1

-------------------------------------+ + +=

h n 2 n 1 6 n 3 j1+ 1

2-------

nej

4---n

3 j1 1

2-------

nej

4---n

+ u n + +

6 n 2 n 1 2 Real 3 j1+ 1

2-------

nej

4---n

u n

+ +

6 n 2 n 1 2 10 1

2-------

n n

4------ 1

3---atan+

cos u n + +

= =

=

H z z 1 F z = h n f n 1 =

F z z 1+

z1

2-------e

j4---

z1

2-------e

j4---

------------------------------------------------------------=

F z z

----------- z 1+

z z1

2-------e

j4---

z1

2-------e

j4---

-------------------------------------------------------------- c1

z-----

c2

z1

2-------e

j4---

-------------------------------

c3

z1

2-------e

j4---

----------------------------+ += =



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Finally we can write h(n) as:

c1z 1+

z1

2

-------ej

4---

z1

2

-------ej

4---

------------------------------------------------------------

z 0=

1

1

2

-------ej

4---

1

2

-------ej

4---

-------------------------------------------------- 112------ 2= = = =

c2z 1+

z z1

2-------e

j4---

-------------------------------

z1

2-------e

j4---

=

11

2-------e

j4---

+

1

2-------e

j4---

1

2-------e

j4--- 1

2-------e

j4---

------------------------------------------------------------------1.5 j

12---

1

2-------e

j4---

j

--------------------------------

1 j3+ 1

2-------e

j4--- 10

2---------- e

j 4--- 3atan+

=

= = = =

c310

2----------e

j 4--- 3atan+

=

n 2 n 2 Real10

2----------e

j 4--- 3atan+

1

2------- n

e

j4---n

u n

+

2 n 2 10 1

2-------

n 1+ n

4------

4--- 3atan+ +cos u n +

= =



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Since , the equation above can be rewritten as:

Solution 3

Let us express H(z) as , and therefore , where

Now let us try to find f(n).

h n f n 1 2 n 1 2 101

2-------

n n

4------

4---

4--- 3atan+ + +cos u n 1 += =

3atan

2--- 1

3---atan=

h n 2 n 1 2 10 1

2-------

n n

4------ 1

3---atan+cos u n 1 +=

H z z 2 F z = h n f n 2 =

F z z 1+ z

z1

2-------e

j4---

z1

2-------e

j4---

------------------------------------------------------------=

F z z

----------- z 1+

z1

2-------e

j4---

z1

2-------e

j4---

------------------------------------------------------------ c2

z1

2-------e

j4---

------------------------------- c3

z1

2-------e

j4---

----------------------------+= =

c2 z 1+

z1

2-------e

j4---

----------------------------

z1

2-------e

j4---

=

11

2

-------ej

4---

+

1

2-------e

j4--- 1

2-------e

j4---

-------------------------------------------

1.5 j1

2

---

j2

2-------

4---sin

------------------------ 1 j3+2-------------- 102----------ej 3atan= = = = =



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Finally we can write h(n) as:

Since , the equation above can be rewritten as:

Question4

Use a combination of inverse transforms and time-domain convolution to find h(n)

Solution

H(z) can be rewritten as

c3102

----------e j 3atan=

n 2 Real 102

----------ej 3atan1

2-------

nej

4---n

u n

2 10 1

2-------

n 2+ n4

------ 3atan+cos u n = =

h n f n 2 2 101

2

-------

n n

4

------

2

--- 3atan+ +cos u n 2 = =

3atan

2--- 1

3---atan=

h n 2 10 1

2-------

n n4

------ 13---atan+cos u n 2 =

H z z 1+z z2 z 0.5+ ----------------------------------=

H z z 1+

z z1

2-------e

j4---

z1

2-------e

j4---

-------------------------------------------------------------- z 2 z 3+ 1

11

2-------e

j4---z 1

---------------------------------- 1

11

2-------e

j4---z 1

-------------------------------= =



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h2 n Z1 1

1 12

-------e j4--- z 1

----------------------------------

1

2

------- n e

jn4

------ u n = =

h3 n Z1 1

11

2

-------ej

4---z 1

-------------------------------

1

2-------

n ej

n4

------ u n = =

h4 n h2 n h3 n 1

2-------

m e

jm4

------- 1

2-------

n m e

j n m

4---------------------

m 0=

n

1

2-------

n ej

n4

------ e

jm2

-------

m 0=

n

1

2-------

n ej

n4

------ 1 e

j n 1+

2--------------------

1 ej2---

-------------------------------

1

2-------

n ej

n4

------ ej

n 1+ 4

--------------------

ej

4---

---------------------- ej

n 1+ 4

--------------------e

j n 1+

4--------------------

ej

4---

ej

4---

------------------------------------------------- 1

2-------

n n 1+

4--------------------sin

4---sin

----------------------------- 2 1

2-------

n n 1+ 4

--------------------sin

= = =

= =

= =



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The result above is valid for . In other words, h(n) is given by:

Question 5

An LTI system has an impulse response h(n) of the form:

Is this system stable?Solution:

h n h4 n 2 h3 n 3 + 21

2-------

n 2 n 1 4

--------------------sin 21

2-------

n 3 n 2 4

--------------------sin+

2

1

2-------

n

2 n 1

4--------------------sin 2 2

n 2 4--------------------sin+ 2

1

2-------

n

2

n4------sin 2

n4------cos 2 2

n4------cos

21

2-------

n 2 n4

------sin 3 2 n

4------cos 2

1

2-------

n n4

------sin 3 n

4------cos

2 101

2-------

n 1

10---------- n

4------sin

3

10---------- n

4------cos 2 10 1

2-------

n n4

------sinsin n4

------coscos

2 101

2

------- n n

4------ +

cos 2 10 1

2

------- n n

4------ 1

3---atan+

cos

= = =

= =

= =

= =

=

n 2

h n 2 10 1

2-------

n n4

------ 13---atan+

cos u n 2 =

h n 2 101

2-------

n n4

------ 13---atan+

cos u n 2 =



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Therefore, is finite, and the system is stable.

Example

Consider the discrete-time system shown above.

a) what is the difference equation of this system?

h n

n 0=

2 101

2-------

n n4

------ 13---atan+

cos u n 2

n 0=

2 101

2-------

n n

4------ 1

3---atan+

cos

n 2=

2 101

2-------

n n4

------ 13---atan+

cos

n 2=

2 101

2

-------

n

n 2=

2 10

1

1 12

-------

---------------- 11

2

-------

= =

=

h n

n 0=

z 1

z 1

x n y n

2

3

0.9

w n


b) U l di i i t d t i th fi t i l f th i l f th t


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b) Use long division to determine the first six values of the impulse response of the system.c) Determine an analytical expression for the impulse response of the system.d) Determine the unit step response of the system

e) give a pole-zero diagram for the system. Is it stable?part a)

The transfer function of the system is

Therefore, the difference equation is

part b)...see next page

the first six samples are {1, 2.9, 5.61,5.049,4.5441,4.08969}

H z W z X z ------------

Y z W z ------------ 1

1 0.9z 1------------------------ 1 2z 1 3z 2+ + = =

y n 0.9y n 1 x n 2x n 1 3x n 2 + + +=



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1 2z 1 3z 2+ +

1 0.9z 1

2.9z 1 3z 2+

2.9z 1 2.61z 2

5.61z 2

5.61z 2 5.049z 3

5.049z 3

5.049z 3 4.5441z 4

4.5441z 4

4.5441z 4 4.08969z 5

4.08969z 5

1 2.9z 1

5.61z 2

5.049z 3

4.5441z 4

4.8969z 5

+ + + + +

1 0.9z 1


part c)


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part c)

part d)

the unit step response may be evaluated as follows

notice that the RHS above is made of three terms which are identical except for a time delay. Therefore, the inversetransform, which is required to find y(n) can easily be evaluated as:

part e)

This system has two zeros at and one poles at and . The poles are inside the unit circle in the z-plane, which implies that the system is stable. The pole-zero diagram follows:

h n Z 1 H z Z 1 1

1 0.9z 1

------------------------ 2z 1

1 0.9z 1

------------------------ 3z 2

1 0.9z 1

------------------------+ +

0.9 nu n 2 0.9 n 1 u n 1 3 0.9 n 2 u n 2 + +

n 2.9 n 1 0.92 1.8 3+ + 0.9 n 2 u n 2 + + n 2.9 n 1 5.61 0.9 n 2 u n 2 + +

= = =

=

=

Y z H z X z 11 0.9z 1------------------------ 2z

1

1 0.9z 1------------------------ 3z

2

1 0.9z 1------------------------+ +

11 z 1----------------= =

n Z 1 11 0.9z 1------------------------ 1

1 z 1----------------

Z 19

1 0.9z 1------------------------ 10

1 z 1----------------+

9 0.9 nu n 10u n += = =

y n f n 2f n 1 3f n 2 + + 9 0.9 nu n 10u n

2 9 0.9 n 1

u n 1

10u n 1

+

3 9 0.9 n 2

u n 2

10u n 2

+

+ +

+

= =

1 j 2 z 0.9= z 0=



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-1.5 -1 -0.5 0 0.5 1 1.5

-1.5

-1

-0.5

0

0.5

1

1.5

Real Part

Imaginary

Part

Z Transform f2014

Documents

Transcript of Z Transform f2014