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Q1. (a) Let h[n] be the unit impulse response of a Low Pass Filter with a cutoff frequency wc, what type of filter has a unit sample response g[n] = (–1)n h[n]. ? [4]

Solution : Consider a Low Pass Filter with h[n]=(0.5)n u[n]

Then 5.0

)(−

=z

zzG POLE = 0.5

g[n] = (–1)n h[n] g[n] = (-1)n (0.5)n u[n] = (–0.5)n u[n]

By ZT, 5.0

)(+

=z

zzG POLE = – 0.5

(i) At w=0, z=1 |H(w)|=0.66 (ii) At w=π, z=–1 |H(w)|=2 That means g[n] is impulse response of High Pass Filter. ANS

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

(b) Assume that a complex multiply takes 1 micro sec and that the amount of time to compute a DFT is determined by the amount of time to perform all the multiplications. [8] (i) How much time does it take to compute a 1024 point DFT directly? (ii) How much time is required if FFT is used. ?

Solution : (i) By DFT Total Complex multiplications = N2 =(1024)2

Time required to compute N2 Multiplications =(1024)2micro sec =1.048576 Sec (ii) By FFT Total Complex multiplications = N/2 log2N=(512) log2(1024) = 5120

Time required to compute 5120 Multiplications =(5120)micro sec =0.005120 Sec --------------------------------------------------------------------------------------------------------

(c) Sequence Xp(n) is periodic repetition of sequence x[n]. What is the relationship between Ck of Discrete Time Fourier Series of xp[n] and DFT X[k] of x[n]. [8]

Solution :

BE EXTC D T S P MAY- 2007 By Kiran Talele ( talelesir@yahoo.com )

0 π

|H(w)|2

0.66

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2

Relationship between DTFS and DFT

The fourier series representation of a periodic sequence xp[n] with fundamental period N is given by,

xp[n] = ∑−

=

1

0

2N

k

Nnkj

k eCπ

, ∞≤≤∞− n

Where the fourier series coefficients are given by,

Ck = ,][1 2N

nkj

p enxN

π−

∑ 1nk0 −≤≤

By DFT, ∑∑−

=

−−

=

==1

0

21

0][][][

N

n

Nnkj

nkN

N

nenxWnxkX

π

By comparing the above equations, If x[n] = xp[n] Then X [k] = N.Ck. --------------------------------------------------------------------------------------------------------

Q2. (a) A digital filter that is implemented on a DSP chip is described by the linear constant coefficient difference equation.

y[n] = ¾ y[n–1] – 1/8 y[n–2] + x[n].

In evaluating the performance of the filter, the unit sample response is measured. The internal storage registers on the chip, however, are not set to zero prior to applying the input. Therefore, the output of the filter contains the effect of the initial conditions, which are y[–1] = –1 and y[–2] = 1. Determine the responses of the filter for all n ≥ 0 and compare it with the zero state response. [8]

Solution : Now, y[n] = ¾ y[n–1] – 1/8 y[n–2] + x[n]

By ZT, [ ] [ ] )()2()1()()1()()( 12811

43 zXyyzzYzyzYzzY +−+−+−−+= −−−

Put y[–1] = –1 and y[–2] = 1. [ ] [ ] )()2()1()()1()()( 12

811

43 zXyyzzYzyzYzzY +−+−+−−+= −−−

[ ] [ ] )(1)(1)()( 12811

43 zXzzYzzYzzY ++−−−= −−−

)()()()( 811

812

81

431

43 zXzzYzzYzzY +−+−−= −−−

)()()1)(( 811

81

432

811

43 zXzzzzY +−+−=+− −−−

)1()(

)1()(

)( 2811

432

811

43

181

87

−−−−

−

+−+

+−+−

=zz

zXzz

zzY

Let Y(z)=Yzi(z)+Yzs(z)

B E EXTC D T S P MAY- 2007

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3

(i) To find ZIR

81

432

812

87

2811

43

181

87

)1()(

)(+−

+−=

+−+−

= −−

−

zzzz

zzz

zYzi

))(()(

21

41

81

87

−−+−

=zz

zzzYzi

By PFE, 21

41

)(−

+−

=z

Bz

AzzYzi

Where

(i) [ ]83

41

41 )()(

===

−= z

zzzY

Azi

(ii) [ ]45

21

21 )()(

−===

−= z

zzzY

Bzi

⎥⎦

⎤⎢⎣

⎡−

−⎥⎦

⎤⎢⎣

⎡−

=21

41 4

583)(

zz

zzzYzi

By IZT, ][21

45][

41

83][ nununy

nn

zi ⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛=

(ii) To find ZSR

)()1(

)()(81

432

2

2811

43

zXzz

zzz

zXzYzs ⎟⎟⎠

⎞⎜⎜⎝

⎛+−

=+−

= −−

For unit sample response x[n] = δ[n] ∴ X(z)= 1

( ) ( )21

41

2

)(−−

==zz

zzYzs

))(()(

21

41 −−

=zz

zzzYzs

By PFE, 21

41

)(−

+−

=z

Bz

AzzYzs

Where

(i) [ ]1)()(

41

41

−===

−= z

zzzY

Azs

(ii) [ ]2)()(

21

21

===

−= z

zzzY

Bzs

( ) ⎥⎦

⎤⎢⎣

⎡−

+⎥⎦

⎤⎢⎣

⎡−

−=21

41

21)(z

zz

zzYzs

By IZT, ][212][

41][ nununy

nn

zs ⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛−=

B E EXTC D T S P MAY- 2007

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4

(iii) To find y[n] = zir + zsr

][212][

41][

21

45][

41

83][ nununununy

nnnn

⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛=

][21

43][

41

85][ nununy

nn

⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛−

= ANS

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

(b) For following linear shift invariant systems, draw pole zero diagram and identify the filter type based on pass band. [12]

(i) 11

)( 1

*1

<−−

= −

−

awherezaaZzH (ii) 44

4

11)( −

−

+−

=za

zAzH

Solution : Q2. (b) (i)

11

)( 1

*1

<−−

= −

−

awherezaaZzH

5.01)(*

=−−

= aLetazzazH

5.05.01

5.05.01)(

*

−−

=−−

=z

zz

zzH

[1] POLE-ZERO diagram

5.05.01)(

−−

==z

zzH

ZEROS: Z1 = 2 POLES : P1= 0.5

[2] Magnitude Spectrum

5.05.01)(

−−

==z

zzH

1. At w=0, z=1 |H(w)|=1 Filter is ALL Pass Filter

2. At w=π, z=–1 |H(w)|=1

0.5 2Real

Imaginary

0 π

|H(w)|1

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5

Real

Imaginary

Solution : Q2. (b) (ii)

44

4

11)( −

−

+−

=za

zAzH

44

4 1)(az

zAzH+−

= Let a =0.5

[1] POLE-ZERO diagram

44

4

)5.0(1)(

+−

=z

zAzH

[2] Magnitude Spectrum

44

4

)5.0(1)(

+−

=z

zAzH Let A =1 Put z = ejw

[ ] [ ][ ] [ ])4sin(0625.0)4cos(

)4sin1)4cos(0625.01)( 4

4

wjwwjw

eewH wj

wj

+++−

==+

−=

Filter is

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

ZEROS POLES

2

24

4

4

101

kjk

kj

ez

ezzz

π

π

=

=

=

=−

k=0 , 10 =z k=1, 2

1

πjez = k=2, πjez =2

k=3, 23

3

πjez =

( )( )( )( ) ( )4

12

5.0

5.0

5.0

05.0

244

44

44

+

=

=

−=

=+

kjk

kjj

ez

eez

z

z

π

ππ

k=0 , 45.00

πjez = k=1, 4

3

5.01

πjez = k=2, 4

5

5.02

πjez = k=3, 4

7

5.03

πjez =

w |H(w)| φ 0 0 0

0.2π 2.002 0.35 0.4π 1.151 -1.000 0.6π 1.151 1.000 0.8π 2.002 -0.353 π 0 0

0 π

|H(w)|1

B E EXTC D T S P MAY- 2007

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6

Q3. (a) Consider the following specifications for a Low Pass Filter.

0.99 ≤ H(e j w) ≤ 1.01 for 0 ≤ w ≤ 0.3 π

H(e j w) ≤ 0.01 for 0.35 π ≤ w ≤ π

Design a Linear Phase FIR filter to meet these specifications using the window design method. [10]

Solution :

(i) Ap = 20log(1.01) – 20log(0.99) == 0.1737 dB

(ii) As = 20log(1.01) – 20log(0.01) == 40.086 dB

(iii) wp = 0.3π (iv) ws = 0.35π (v) Fs = 1 Hz (vi) LPF

(1) Select window function For Hamming window function As= 53 dB > 40.086 dB(required value)

∴ Select Hamming window function. (2) Calculate N

12 ffCN−

≥

(i) For hamming window function C=3.14

(ii) f2 – f1 = fs - fp wp = 0.3π = 2π fp ∴ fp = 0.15 ws = 0.35π = 2π fp ∴ fs = 0.175

15.0175.0

14.3−

≥N

6.125≥N

Let N = 127 (odd)

1.010.99

0.01

0 0.5π 0.75π 0 0.3π 0.35π π

20log(1.01)20log(0.99) 20log(0.01)

0 0.5π 0.75π

B E EXTC D T S P MAY- 2007

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7

(3) Calculate wc

radww

w spc π325.0

2==

+=

-------------------------------------------------------- Step 1. Find Hd(w) Let Hd(w) = | Hd(w) | φ(w)

Where (i) Magnitude response :

Hd(ω) = ⎩⎨⎧ ≤≤−

otherwisewwfor cc ω

01

(ii) Phase Response : φ(w) = ej φ

For Linear Phase LPF with symmetric h[n]

φ = ( ) wN2

1−− = – α w

∴φ(w) = wje α−

By substituting, ⎪⎩

⎪⎨⎧ ≤≤−

=−

otherwisewwewH cc

wjd

ωα

0)(

where α = 63 and wc = 0.325π Step 2. Find hd[n]

By Inverse DTFT, dwewHnh jnwd )(

21][ ∫

−

=π

ππ

dweenh jnwwjw

wd

c

c

)(21][ α

π−

−∫=

dwenh wnjw

wd

c

c

)(

21][ α

π−

−∫=

c

c

w

w

wnj

d jnenh

−

−

⎥⎦

⎤⎢⎣

⎡−

=)(2

1][)(

απ

α

- π -wc 0 wc π

1

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8

⎥⎦

⎤⎢⎣

⎡ −−

=−−−

jee

nnh

cc wnjwnj

d 2)(1][

)()( αα

απ

where α =63 and wc = 0.325π

------------------------------------------------------------------------------------------ Step 3. Find h[n] Linear phase FIR filter with impulse response h[n] is given by,

h[n] = hd[n] w[n]

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−−

=ππ

325.0)83(325.0)63sin(325.0][

nnnh ⎥

⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−

1262cos46.054.0 nπ for 0 ≤ n ≤ 126

---------------------------------------------------------------------------------------------------- (b) Use the Bilinear Transformation to design a Discrete Time Chebyshev High

Pass filter with an equiripple passband with [10]

0 ≤ H(e j w) ≤ 0.1 for 0 ≤ w ≤ 0.1 π

0.9 ≤ H(e j w) ≤ 1.0 for 0.3 π ≤ w ≤ π

Solution :

(i) Ap = 20 log(1.0) – 20 log(0.9) == 0.9151 dB

(ii) As = 20 log(1.0) – 20 log(0.1) == 20 dB

(iii) wp = 0.3π (iv) ws = 0.1π (v) Fs = 1 Hz (vi) HPF Step-1 : Design Analog Chebyshev High Pass filter

(1) Calculate s,p ΩΩ

⎟⎠⎞

⎜⎝⎛=Ω

2wptan

T2p ⎟

⎠⎞

⎜⎝⎛=

23.0tan

10001

2 π = sec05.1019 rad

⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛=Ω

21.0tan

10001

22

tan2 πwsT

s = sec77.316 rad

c

ccd w)n(

w)nsin(w]n[hα−α−

π=

0 0.5π 0.75π 0 0.1π 0.3π π

1.00.9

0.1

0

20 log(1.0)20 log(0.9)

20 log(0.1)

0 0.1π 0.3π π

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9

(2) Calculate Filter order N.

⎥⎦⎤

⎢⎣⎡

⎥⎦

⎤⎢⎣

⎡−

−

=

⎥⎦

⎤⎢⎣

⎡ΩΩ

⎥⎦

⎤⎢⎣

⎡−−

=

−

−

−

−

77.31605.1019cosh

110110cosh

cosh

110110cosh

1

21

09151.0

21

1

21

10/

101

N

ps

NAp

As

N = Let N =

(3) Calculate Normalized HPF (4) Calculate De-normalized HPF Step-2 : Design Digital Chebyshev High Pass filter Ooooooooooooooooooo --------------------------------------------------------------------------------------------------------

Q4. (a) Consider the sequence x[n] = δ[n] + 2 δ[n-2] + δ[n-3]

(i) Find four point DFT of x[n]. [4] (ii) If y[n] is four point circular convolution of x[n] with itself, find y[n] and

the four point DFT Y[k]. [4] (iii) With h[n] = δ[n] + δ[n-1] + 2 δ[n-3] Find four point circular convolution of

x[n] with h[n]. [4] Solution : Q4. (a) (i) To find four point DFT of x[n]

Given x[n] = δ[n] + 2 δ[n-2] + δ[n-3] =

↑1 , 0, 2, 1

By DFT, X[k] = ∑−

=

1

0

N

n x[n] WN

nk

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡=

9630

6420

3210

0000][

NNNN

NNNN

NNNN

NNNN

wwwwwwwwwwwwwwwwKX

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

]3[]2[]1[]0[

xxxx

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10

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−−

−−=

jj

jjkX

111111

111111

][

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

1201

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡ =

−−

+−=

0

1214

][

k

j

jkX

ANS

Solution : Q4. (a) (ii) To find Y[k] y[n] = x[n] ⊗ x[n] By Convolution property of DFT, Y[k] = X[k] X[k]

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

+−=

j

jkY

1214

][

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

+−

j

j

1214 ==

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡−

=

j

jk

242

016 ANS

To find y[n]:

By Inverse DFT, ∑−

=

−=1

0][1][

N

k

nkNWkY

Nny N

jN eWwhere

π21 −=

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡=

−−−

−−−

−−−

9630

6420

3210

00001][

NNNN

NNNN

NNNN

NNNN

wwwwwwwwwwwwwwww

Nny

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

]3[]2[]1[]0[

YYYY

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−−−−

=

jj

jjny

111111

111111

41

][⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡−

=

j

jk

242

016

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡ =

=

254

05

][

n

ny ANS

--------------------------------------------------------------------------------------------------------------- Solution : Q4. (a) (iii) To find circular convolution of x[n] with h[n].

Given x[n] = δ[n] + 2 δ[n-2] + δ[n-3] = ↑1 , 0, 2, 1 Length L = 4

h[n] = δ[n] + δ[n-1] + 2 δ[n-3] = ↑1 , 1, 0, 2 Length M = 4

y[n] = x[n] ⊗ h[n]

∑∞

=

−=0

))][((][][m

mnhmxny

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡ ==

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡=

545

02

1201

1102211002111021][ nny

ANS

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

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11

(b) Explain the POLE ZERO locations for Type-I, Type-II, Type-III, and Type-IV

Linear Phase FIR filter. [8] Solution :

Case-1 : When h[n] is Symmetric with N EVEN [ Type-2 Linear Phase Filter ]

⎟⎠⎞

⎜⎝⎛= −−

ZHzzH N 1)( )1(

⎟⎠⎞

⎜⎝⎛= −

ZHzzH ODD 1)(

(1) At Z = –1 w = π (2) At Z = 1 w = 0

( )1)1()1( −−=− − HH ODD

( )1)1()1( −−=− HH

( )1)1( −−=− HH

So, H(-1) = 0

0)( 1 =−=zzH That means there exists definite ZERO at z = –1

( )1)1()1( HH ODD−= ( )1)1()1( HH = ( )1)1( HH = No definite ZERO exists at z = 1

Case-2 : When h[n] is Symmetric with N ODD [ Type-1 Linear Phase Filter ]

⎟⎠⎞

⎜⎝⎛= −−

ZHzzH N 1)( )1(

⎟⎠⎞

⎜⎝⎛= −

ZHzzH EVEN 1)(

(1) At Z = –1 w = π (2) At Z = 1 w = 0

( )1)1()1( −−=− − HH EVEN

( )1)1()1( −=− HH

( )1)1( HH = No definite ZERO exists at z = –1

( )1)1()1( −−=− − HH EVEN ( )1)1()1( HH = ( )1)1( HH = No definite ZERO exists at z = 1

Case-3 : When h[n] is Anti-Symmetric with N EVEN [ Type-4 Linear Phase Filter ]

⎟⎠⎞

⎜⎝⎛−= −−

ZHzzH N 1)( )1( ⎟

⎠⎞

⎜⎝⎛−= −

ZHzzH ODD 1)(

(1) At Z = –1 w = π (2) At Z = 1 w = 0

( )1)1()1( −−−=− − HH ODD

( )1)1()1( −−−=− HH

( )1)1( −=− HH No definite ZERO exists at z = –1

( )1)1()1( HH ODD−−=

( )1)1( HH −=

So, H(1) = 0 0)( 1 ==zzH

That means there exists definite ZERO at z = 1

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12

Case-4 : When h[n] is Anti-Symmetric with N ODD [ Type-3 Linear Phase Filter ]

⎟⎠⎞

⎜⎝⎛−= −−

ZHzzH N 1)( )1( ⎟

⎠⎞

⎜⎝⎛−= −

ZHzzH EVEN 1)(

(1) At Z = –1 w = π (2) At Z = 1 w = 0

( )1)1()1( −−−=− − HH EVEN

( )1)1( −−=− HH

So, H(-1) = 0

0)( 1 =−=zzH That means there exists definite ZERO at z = –1

( )1)1()1( HH EVEN−−= ( )1)1( HH −=

So, H(1) = 0 0)( 1 ==zzH That means there exists definite ZERO at z = 1

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

Q5. (a) The unit sample response of An FIR filter is , [10]

⎩⎨⎧ ≤≤

=Otherwise

nnh

n

060

][α

(i) Draw the direct form implementation of this system. (ii) Determine system function and use this to draw a flow graph that is cascade

of FIR system with IIR system. (iii) For both of this implementations, determine the number of multiplications

and additions required to compute each output value and the number of storage registers that are required.

Solution : Q5. (a)(i) Direct form implementation

⎩⎨⎧ ≤≤

=Otherwise

nnh

n

060

][α

By ZT,

nn

n

n

n

zzH

znhzH

−

=

−∞

−∞=

∑

∑

=

=

α6

0)(

][)(

( )nn

zzH 16

0)( −

=∑= α

( )1

71

11)( −

−

−−

=z

zzHαα

1

77

11)( −

−

−−

=zzzH

αα

z–1

y[n]

α

x[n]

z–1

z–1

z–1

z–1

z–1

z–1 -α7

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13

Solution : Q5. (a)(ii) Cascade form implementation

1

77

11)( −

−

−−

=zzzH

αα

[ ] ⎥⎦

⎤⎢⎣

⎡−

−= −−

177

111)(

zzzH

αα

Let )()()( 21 zHzHzH = Solution : Q5. (a)(iii)

Sr No Parameter Direct Form Cascade form 1 Number of

Mltiplications 2 2

2 Number of Additions 2 2 3 Number of Delay Blocks 7 7

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(b) Draw a lattice filter implementation for the all POLE filter. [10]

321 6.04.02.011)( −−− ++−

=zzz

zH

And determine the number of multiplications, additions and delays required to implement the filter. Compare this structure to a direct form realization of H(z) in terms of multiplies and adds and delays.

Solution :

)(1

6.04.02.011)( 321 zAzzz

zHN

==++−

= −−− where N= 3

All pole system

For N = 3, A3(z) = 1 – 0.3z–1 + 0.4 z–2 + 0.6 z–3

z–1

y[n]

α

z–1

z–1

z–1

z–1

z–1

z–1

z–1 -α7

x[n]

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14

(i) To find k3

A3(z) = 1 – 0.3z–1 + 0.4 z–2 + 0.6 z–3 ∴ =3k 0.6

B3(z) = 0.6 + 0.4 z–1 – 0.3 z–2 + z–3

(ii) To find k2

23

3332 1

)()()(k

zBkzAzA−−

=

( ) ( )2

321321

2 )6.0(13.04.06.0)6.0(6.04.03.01)(

−+−+−++−

=−−−−−− zzzzzzzA

212 625.00937.01)( −− +−= zzzA ∴ =2k 0.625

B2(z) = 0.625 – 0.0937 z–1 + z–2

(ii) To find k!,

22

2221 1

)()()(k

zBkzAzA−−

=

( ) ( )2

2121

1 )625.0(10937.0625.0)625.0(625.00937.01)(

−+−−+−

=−−−− zzzzzA

11 0576.01)( −−= zzA ∴ =1k –0.0576

Lattice Realization Diagram of All POLE IIR filter :

Sr No Parameter Direct Form Lattice form

1 Number of Mltiplications

3 6

2 Number of Additions 2 6 3 Number of Delay Blocks 3 3

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g1[n]

fo[n] f1[n] f3[n]

x[n] ++

k1 –k1 k2 –k2

Z–1 Z–1

[ ]

y[n]

+ + Z–1

–k3 k3

f2[n]

g2[n] g3[n] +

+

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15

Q6. (a) Determine 8 point DFT for a continuous time signal x[t]= sin(2πFt) with FS = 50

Hz using DIF algorithm. [10] Solution :

x[t]= sin(2πFt) Let F=1 Hz

By sampling, ⎟⎠⎞

⎜⎝⎛ π

=⎟⎠⎞

⎜⎝⎛ π

=⎟⎠⎞

⎜⎝⎛ π

=25nsin

50n2sin

Fsn2sin]n[x

⎟⎠⎞

⎜⎝⎛ π

==25nsin]n[x

7705.0,6845.0,5877.0,4817.0,3681.0,2486.0,1253.0,0]n[x↑

==

∑−

==

1

0][][

N

n

nkNWnxkX N

jN eWiiNiwhere

π21)(8)(

−==

X[k] = X[0]+ X[1] WNk + X[2] WN

2k + X[3] WN3k + X[4] WN

4k + X[5]WN5k + X[6] WN

6k + X[7] WN7k

X[k] = 0.1253 WNk + 0.2486 WN

2k + 0.3681WN3k + 0.4817WN

4k + 0.5877 WN5k + 0.6845 WN

6k + 0.7705 WN7k

================== k=0, X[0] = 3.2664 k=1, X[1] =0.1253 WN

1 + 0.2486 WN2 + 0.3681WN

3 + 0.4817WN4 + 0.5877 WN

5 + 0.6845 WN6 + 0.7705 WN

7

X[1] =

k=2, X[2] =0.1253 WN2 + 0.2486 WN

4 + 0.3681WN6 + 0.4817WN

8 + 0.5877 WN10 + 0.6845 WN

12 + 0.7705 WN14

X[2] =

k=3, X[3] =0.1253 WN3 + 0.2486 WN

6 + 0.3681WN9 + 0.4817WN

12 + 0.5877 WN15 + 0.6845 WN

18 + 0.7705 WN21

X[3] =

k=4, X[4] =0.1253 WN4 + 0.2486 WN

8 + 0.3681WN12 + 0.4817WN

16 +0.5877 WN20 + 0.6845 WN

24+0.7705 WN28

X[4] = k=5, X[5] = k=6, X[6] = k=7, X[7] =

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(b) With the help of block diagram, explain architecture of TMS320C5X series of processor. [10] Soln : Refer DTSP-Dec-2004 Q7 (b)(i)

-------------------------------------------------------------------------------------------------------- Q7. (a) Explain the Goertzel Algorithm. [6] Soln : Refer DTSP-May-2005 Q7 (b)

(b) Write short note on applications of DCT. [6]

(c) Explain briefly any one method of long data filtering. [8] Soln : Refer DTSP-May-2005 Q7 (c)

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