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Applications of Z-Transform in Signal Processing - II

Yogananda Isukapalli

Frequency response estimation

h[n]x[n] y[n]

The frequency response of a system can be obtained from its z-transform as:Given {h[n]},

We have:

nj

n

j

n

n

enheH

znhzH

qq -¥

-¥=

¥

-¥=

-

å

å

=

=

][)(

].[)(

Implies that :

qq

jezj zHeH

== )()( (1)

Discrete Time System

h[n] Example : A Comb Filter

y[n] = x[n] - x[n-12]

zeros zm = 1ejm(2p/12)m=0, 1, 2, ….1112 zeros and 12 poles

H(ejq) = 1 - e-j12q

= j2 e-j6qsin(6q)

=2sin(6q)ej(p /2 - 6q)

12

1212 11)(

zzzzH -=-= -

………….

0 p/6 p/3 p/2 p q

|H(ejq)|

Plot of H(ejq)

3

Frequency response function properties :

(1) H(ejq) is a periodic function.

1

.].[)(

].[)(

2

2)2(

)2()2(

=

=

=

-

-¥

-¥=

-+

¥

-¥=

+-+

å

å

mj

mj

m

mjj

m

mjj

e

eemheH

emheH

p

pqpq

pqpq

for all integer values of m

)()(

].[)(

:

)2(

)2(

qpq

qpq

jj

m

mjj

eHeH

emheH

Therefore

=

=

+

¥

-¥=

-+ å

(2) M = |H(ejq)| = |H(e-jq)|)()( qq jj eHeHP --==

(3) MdB = 20log10M

Frequency units used in discrete-time systems:

and then evaluating the z- transfer function, H(z), in the following intervals:

• Two frequency units are normally used to describe the frequency response of discrete-time systems: w (rad s-1) and f (Hz)

• The frequency response is found by letting fTjTjj eeez 2pwq ===

• The following figure shows how wT and z change as w varies from 0 to ws.

• As the angle q=wT goes from 0 to 2p the value of z varies from 1 through j and back to 1.

Fig: Units of frequency used in discrete-time systems and their relationship to points on the unit circle

• The following figure also makes it clear that the frequency response of a discrete-time system is cyclic.

Fig: z-plane unit circle showing critical frequency points

• As we go round the circle one or more revolutions the values of z simply repeat.

Example)

Soln)

Geometric evaluation of frequency response:

We have

Õ

Õ

=

=

-

-

=------

=N

ii

N

ii

N

N

pz

zzK

pzpzpzzzzzzzK

zH

1

1

21

21

)(

)(

))...()(())...()((

)(

The frequency response is given by

2/0 sww ££

Õ

Õ

=

=

-

-

==N

ii

Tj

N

ii

Tj

Tjj

pe

zeK

eHeH

1

1

)(

)(

)()(w

w

wq for

(2)

(3)

• For a z-transform with only two zeros and two poles, the frequency response is given by

2211

2211

21

21

))(())((

)()(

ffqq

ww

wwwq

ÐÐÐÐ

=

--

--==

VVUKU

pepezezeK

eHeHTjTj

TjTjTjj

(4)

where U1 and U2distances from the zeros to the point z=ejwT

V1 and V2 distances from the poles to the point z=ejwT

Fig: Geometric evaluation of the frequency response from pole-zero diagram

•

[ ] )()(1 ,)(

2121

21

21

ffqqw

w

+-+=Ð

==

Tj

Tj

eH

KVVUU

eH(5)

• In general, in the geometric method, the frequency response at a given frequency w (at an angle wT) is given by

poles ofnumber ,...1 zeros, ofnumber ,...1 , ==ÐÐ

jiVU

jj

ii

fq

Example: Determine the frequency response at dc, 1/8, 1/4, 3/8 and 1/2 the sampling frequency of the causal discrete-time system with the following z-transform:

7071.01)(

-+

=zzzH

Soln) H(z) has single pole and single zero. So,

)sin(7071.0)cos()sin()cos(1

7071.01

)(TjT

TjTee

VUeH

Tj

TjTj

wwww

fq

w

ww

+-++

=-

+=

ÐÐ

= (6)

At dc, wT=0 and the zero and pole vectors to the point z=0 are ando02Ð o02929.0 Ð

o0828.62929.0/2)( Ð==TjeH w

Fig: Frequency response estimation using geometric method and pole-zero diagram

Thus the frequency response is given by

At 4/8/ ,8/ pwwww === sss FT

oo

o5.676131.2

907071.05.228477.1

)4/sin(7071.0)4/cos()4/sin()4/cos(1

)(

-Ð=Ð

Ð=

+-++

=pp

ppwj

jeH Tj

The responses at the remaining frequencies are summarized below:

Fig: A sketch of the frequency response

Example : The Graphical Design of a Comb filter :

• In medical applications, the 60Hz frequency of the power supply is often “picked up” by the test equipment (EKG recorder)

• Also harmonically related frequencies such as f2 = 2x60 = 120Hz, and f3 = 3x60 = 180 Hz are generated because of non-linear phenomena.

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• It is observed that the magnitude response is symmetrical about half the sampling frequency (Nyquist frequency), and the phase response antisymmetrical about the same frequency.

• The frequency response is periodic with a period of ws.

• The object of a digital filter design is to eliminate or suppress these unwanted frequencies which distort or mask up the signals of interest.

Thus the desired response of the filter would be :

0 60 120 180 f Hz

p/3 2p/3 p q rad

M

Thus we require a nonrecursive (FIR) comb filter:

Note : Digital Frequencies

Sampling frequency 360 Hz15

Thus we have :

q1 = w1T = 2p(60)/360 = p/3

q2 = w2T = 2p(120)/360 = 2p/3

q3 = w3T = 2p(180)/360 = p

Proposed pole - zero design :

Note :

1.) Complex zeros must occur in conjugate pairs.

2.) q = 0 is added to eliminate any DC component in the signal.

0 60 120 180 f Hz

p/3 2p/3 p q rad

Actual Response

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]6[][][

)1()(

1)(

][]6[][

1)())()()()()(1()(

6

6

6

6

35

34

32

3

--=

-=\Þ

-=

-+=

-=

------=

-

nxnxny

zzHzzzH

nxnxny

zzH

ezezezezezzzHjjjjjpp

ppp

Implies :

But the above obtained filter is non-causal !! To make it causal filter we place six poles at z = 0.

Thus the required causal FIR comb filter is:

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Design of a Notch Filter

SignalNoise

2px60 2px1000 2px1250 wwn w1 w2

Analog

SignalNoise

-0.03p 0.03p 0.5p 0.625p wwn w1 w2

Digital Ts=0.25x10-3sq=wTs

-0.03p 0.03p 0.5p 0.625p wwn w1 w2

Proposed Filter:

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.03p

Pole - Zero Locations

Follows :

and

pq

q

p

03.0

1|)(|

0|)(| 03.0

±¹

»

=±

foreH

eH

j

j

-0.03 p -0.03 p ….. p

Designed Filter|H(ejq)|

X19

Sinusoidal Steady state response of discrete -time systems :

h[n]x[n] y[n]

Convolution : å¥

-¥=

-=m

mnxmhny ][][][

Input x[n]: Complex exponential ejnq

Output :

å

å

å

¥

-¥=

-

¥

-¥=

-

¥

-¥=

-

=

=

=

=

m

mjj

njjss

m

mjnjss

m

mnjss

emheH

WhereeeHny

eemhny

emhny

qq

qq

qq

q

].[)(

:)(][

].[][

].[][ )(

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Note: q=wT (rad)

Discrete Time System

h[n]

å¥

-¥=

-=m

mjj emheH qq ].[)(

H(ejq) is the frequency response function.

Example : h[n] = d[n] + d[n-1]

H(ejq) = 1 + e-jq

Now consider the input to the system to be :

)10

cos(101][pn

nx +=

Re-writing the input in terms of complex exponentials:

][][][][

55.1][

321

10100

nxnxnxnx

neeenx

njnjnj

++=

¥££¥-++=

-pp

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Discrete Time System

h[n]

Using the Superposition principle :

We have :

16.01010

16.01010

00

321

97.11)()(

97.11)()(

21)()(:

)()()(][

3

2

1

332211

jjjj

jjjj

jjj

jjjjjjss

eeeHeH

eeeHeH

eeHeH

WhereeceHeceHeceHny

=+==

=+==

=+==

++=

-

--

-

ppq

ppq

q

qqqqqq

We have c1 = 1, c2 = 5, c3 = 5

)16.010

cos(7.192][

)97.1.(52][

)(][

)16.010()16.0

10(

3

1

-+=

úû

ùêë

é++=

=

---

=å

p

pp

qq

nny

eeny

eceHny

ss

njnj

ss

njk

k

kjss

k

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Discrete Time System

h[n]

Generalization:

h[n]x[n] = Acos(nq) yss[n] = AMcos(nq+P)

)(

|)(|

].[)(

q

q

qq

j

j

jP

m

mjj

eHP

eHM

MeemheH

=

=

== å¥

-¥=

-

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Discrete Time System

h[n]

Example :

h[n] = 2d[n] - 3d[n-1] + 4d[n-2]

H(z) = 2 - 3z-1 + 4z-2

H(ejq) = 2 - 3e-jq + 4e-j2q

Example:y[n] + 0.25y[n-4] = x[n] -x[n-2]

4)2(

41

414

)2(4

4

4

22

)25.0()(

25.025.0

25.0)1()(

pp

pp

mj

mj

ez

ez

z

zzzzH

+

+

=

=

-=

+-

=

Filter Poles:

43

2

41

707.0

707.0p

p

j

j

ep

ep

=

=

25.0)1()(

707.0

707.0

4

22

47

4

45

3

+-

=

=

=

q

qqq

p

p

j

jjj

j

j

eeeeH

ep

epPoles are:

Filter is stable :

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Discrete Time System

h[n] Only Stable systems have Sinusoidal Steady State Response.

H(z)X(z) Y(z)

Y(z) = H(z)X(z)

qq jjin

ezC

ezC

pzzC

pzzCzY in--

+-

+-

+-

=*

2

2

1

1)(

Sinusoidal Input

y[n] = C1P1n + C2P2n + Cinejqn + C*ine-jqn

System Poles Input Poles

y[n] = yss[n] = Cinejqn + C*ine-jqn

if |Pi| < 1 (stable)

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Discrete Time System

h[n] Generalization for Stable System

H(z)Acos(nq) yss[n]

yss[n] = A|H(ejq)|cos(nq +

A Typical Signal Processing Example :

Ås[n] Desired Signal

r[n] Signal received at sensor

Digital Filter

g[n] Unwanted Signal due to medium y[n]

• The purpose of designing a digital filter is to minimize the effect of the unwanted signal or to completely eliminate it from y[n].

• Specify for illustration :s[n] = Acos(nq) = 10cos(pn/20)

g[n] = Bcos(10qn + f) = 4cos(pn/2 + p/6 )

r[n] = 10cos(pn/20) + 4cos(pn/2 + p/6 )

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Propose :

y[n] = r[n] + 0.9r[n-2]

Then :

)62

cos(4.0][

1.0)(

)62

cos(4][

)15.020

cos(8.18][

88.19.01)(

)20

cos(10][

9.01)(

9.01)(

1

02

1

15.01020

2

2

pp

pp

p

p

p

pp

qq

+=

=

+=

-=

=+=

=

+=

+=

--

-

-

nny

eeH

nnr

nny

eeeH

nnr

eeH

zzH

ss

jj

ss

jjj

jj

By Superposition :

For an Input of :

For an input of :

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)62

cos(4.0)15.020

cos(8.18][ppp

++-=

\Þnnnyss

Notes:

1.) Desired Signal is amplified by a factor of 1.88 when passed through the filter.

2.) Filter response :

p/10 p/2 3p/4 p q

2.0

1.75

0.25

3.) Sketch the Pole zero locations of the filter.

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