dsp-comp-dec-2006 (5)paper solution

8/11/2019 DSP-COMP-DEC-2006 (5)paper solution

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Q 1 Figures show the direct form II realization of IIR filter.

(a) Find the Transfer Function of the filter. [3](b) Find the corresponding difference equation. [2](c) Realize the filter using cascade form using first order modules. [3](d) Realize the filter using parallel form using first order modules. [4](e) Find the impulse response function of the filter. [2]

(f) Show pole zero pattern of the filter. [2](g) State whether the filter is stable. Why or Why not ? [2]

(h) Find the magnitude squared response at w=0 and w=. [2]

Solution : 1(a) ToFind Transfer Function :

22

11

22

110

1)(

++

++=

zaza

zbzbbzH

From realization diagram we get b0 = 1 b1= b2=a1= 5/8 a2 = 1/16

By substituting we get, 21611

85

2811

43

1

1)(

+

++=

zz

zzzH

NS

Solution : 1(b) ToFind Difference Equation :

Now,2161185

2811

43

1

1)(

+

++=

zz

zzzH

21611

85

2811

43

1

1

)(

)(

+

++=

zz

zz

zX

zY

)()()()()()( 2811

432

1611

85 zXzzXzzXzYzzYzzY ++=+

]2[]1[][]2[]1[][81

43

161

85 ++=+ nxnxnxnynyny

]2[]1[][]2[]1[][81

43

161

85 +++= nxnxnxnynyny

NS

BE COMP D S P DEC-2006

By Kiran Talele ([email protected])

5/8 3/4

1/8-1/16

Z

Z

x(n) Y(n)


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B E COMP D S P DEC-2006

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2

Solution : (c) ToRealize the filter in cascade form

( )( )21

81

2

1

4

1

161

852

8

1

4

32

)(++=

+++=

zzzz

zz

zzzH

+

+=

2121

8141

)(z

z

z

zzH

Let X(z) = X1(z) X2(z)

Solution : (d) ToRealize the filter in parallel form

( )( )21

81

21

41

)(

++=

zz

zzzH

( )( )21

81

21

41

)(

++=

zzz

zz

z

zH

21

81

)(

+

+=

z

C

z

B

z

A

z

zH

Where A = == 0

)()(

z

zz

zH

2

B =( )

==

81

81)(

z

zz

zH

5

C =( )

==

21

21)(

z

zz

zH

4

+

+=

21

81

)(Z

zC

z

zBAzH

+

+=

11 5.01

4

125.01

52)(

zzzH

Z1

1 / 8

y[n]x[n]

Z1

0.5 0.50.25


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Solution : (e) ToFind Impulse Response of the system.

Now ,

+

=

21

81

452)(Z

z

z

zzH

By IZT, ( ) ( ) ][4][5][2][21

81 nununnh

nn+=

NS

Solution : (f) To show pole zero plot:

( )( )21

81

21

41

)(

++=

zz

zzzH

( ) ( )( ) ( )5.0125.0

5.025.0)(

++

=zz

zzzH

ZEROS : Z1 = 0.25 Z2= 0.5POLES : P1= 0.125 P2

= 0.5

Solution : (g) Tocheck stability :

All POLES are inside the unit circle.Therefore System is Stable.

Solution : (h) Tofind Magnitude Squared Response :

( )( )21

81

21

41

)(

++=

zz

zzzH

0.5

Z1

Z1

0.125

y[n]x[n]

-5

4

2

Z Z P P


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(i) At w = 0 , z = 1

| H(w)|2=

( ) ( )

( )( )

2

21

81

21

41

11

11

++

=

(ii) At w = , z = 1

| H(w)|2=( ) ( )( )( )

2

21

81

21

41

11

11

++=

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

Q 2 (a) Find z- transform of x[n]= an (coswon) u(n) [8]

Solution :x[n] = ancos (n ) u[n]

][2

][ nuee

ranxjnwjnw

n

+=

[ ]][][2

1][ nueanueanx jnwnjnwn +=

( ) ( )

+= ][][

2

1][ nueanueanx

njwnjw

By ZT,

+

= jwjw

eaz

z

eaz

zzX21][

) )( )( )

+=

jwjw

jwjw

eazeaz

eazzeazzzX

2

1][

+

+=

22

22

2

1][

aezaezaz

ezazezazzX

jwjw

jwjw

++

+=

22

2

)(

)(2

2

1][

aeezaz

eezazzX

jwjw

jwjw

+

=

22

2

cos2

)cos(22

2

1][

azaz

wzazzX

||||;cos2

)cos()(

22

2az

awzaZ

wzaZzX >

+

= NS

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Q 2 (b) The discrete time system is represented by following equation

y(n) = 3/2 y(n-2) + x(n) with initial conditions y(-1) = 0, y(2) = 2 andx(n) = ()nu(n)Determine: i)zero input response. ii)Zero state response. iii) Total responseof the system. [12]

Solution :y(n) = 3/2y(n-2) + x(n)

By ZT, Y(z) = 3/2 { z2 Y(z) + z1 y[1] + y[2] } + X(z)

Y(z) = 1.5 { z2 Y(z) + 0 2 } + X(z)

0.25

18.37


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Y(z) = 1.5 z2 Y(z) 3 + X(z)

Y(z) { 1 1.5 z2 } = 3 + X(z)

22 5.11

)(

5.11

3)(

+

=

z

zX

zzY

5.1

)(

5.1

3)(

2

2

2

2

+

=

z

zXz

z

zzY

)()2244.1()2244.1()2244.1()2244.1(

3)(

22zX

zz

z

zz

zzY

++

+

=

Let )()()( zYzYzY ZSRZIR +=

(1) To find ZIR

)2244.1()2244.1(

3)(

2

+

=zz

zzYZIR

)2244.1()2244.1(

3)(

+=

zz

z

z

zYZIR

2244.12244.1

)(

++

=

z

B

z

A

z

zYZIR

Where A = ==

2244.1

)2244.1()(

z

zz

zYZIR

B = ==

+

2244.1

)2244.1()(

z

zz

zYZIR

++

= 2244.12244.1)( z

zBz

zAzYZIR

By IZT,

( ) ( ) ][2244.1][2244.1)( nuBnuAnY nnZIR += ------(1)

(2) To find ZSR

)()2244.1()2244.1(

)(2

zXzz

zzYZSR

+=

For x(n)= ()nu(n) = (0.25)nu[n]

By ZT,25.0

)(

=z

zzX

By Substrituting,

+=

25.0)2244.1()2244.1()(

2

z

z

zz

zzYZSR

)25.0()2244.1()2244.1(

)( 2

+=

zzz

z

z

zYZSIR

25.02244.12244.1

)(

+++= zC

z

B

z

A

z

zYZSR

Where A = ==

2244.1

)2244.1()(

z

zz

zYZSR

B = ==

+

2244.1

)2244.1()(

z

zz

zYZSR

C ==

=

25.0

)25.0()(

z

zz

zYZSR


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+

+

+

=

25.02244.12244.1)(

z

zC

z

zB

z

zAzYZSR

By IZT,

( ) ( ) ( ) ][25.0][2244.1][2244.1)( nuCnuBnuAnY nnnZSR ++= ------(2)

(3) To find y[n] = ZIR + ZSR

y[n] = yZIR[n] + yZSR[n]

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Q 3 (a) Determine the DFT of a sequence using DIT-FFT Algorithm.

x(n) = (1,2,1,2,0,2,1,2) Using the same result, otherwise find DFT of(0,2,1,2,1,2,1,2) [15]

Solution :

(i) To find X[k]

STAGE-1 Result

(i) A1 =

(ii) B1 =

(iii) C1 =

(iv) D1 =

(v) E1 =

(vi) F1 =

(vi) G1 =

(vii) H1 =

STAGE-2 Result

(i) A2 =

(ii) B2 =

(iii) C2 =

(iv) D2 =

(v) E2 =

(vi) F2 =

(vii) G2 =

(viii)H2 =

X[0]

X[1]

X[2]

X[3]

X[4]

X[5]

X[6]

X[7]

1

1

1

1

1NW

2NW

3NW

x[ 0 ]

x[ 4 ]

x[ 2 ]

x[ 6 ]

x[ 1 ]

x[ 5 ]

x[ 3 ]

x[ 7 ]1

1

1

1 -j

1

11

1

j

A1

B1

C1

D1

A2

B2

C2

D2

H2

G2

F2

E2

E1

F1

G1

H1


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=

=

0

][

k

kP

STAGE-3 Result

(1) X[0] =

(2) X[1] =

(3) X[2] =

(4) X[3] =

(5) X[4] =

(6) X[5] =

(7)

X [6] =

(8) X[7] =

(ii) To find P[k]

Let p[n] = { 0, 2, 1, 2, 1, 2, 1, 2 }

P[n] = x[n-4]By Time shift property of DFT,

( ) ][][ 4 kXWkP kN

=

( ) ][1][ kXkP k=

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

Q 3 (b) Find the DFT of x(n) = (1,2,3,4) [5]

Solution :

][][ 4 kXWkP k

N=

1

3

4

2

4

2

6

2

4

2

6

2j

10

-2+2j

2

2-2jj

x[ 0 ]

x[ 2 ]

x[ 1 ]

x[ 3 ]

X[ 0 ]

X[ 1 ]

X[ 2 ]

X[ 3 ]

1 1

1

1


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Q 4 (a)Design a digital Butterworth filter that satisfies the following constraint usingbilinear transformation. Assume T = 1s

0.9 |H(ejw) | 1 0 w /2|H(ejw)|0.2 3 /4 w [15]

Solution : Refer DSP-MAY-2006 Q 5 (a)-----------------------------------------------------------------------------------------------------------------

Q 4 (b)Determine the energy of the signal given by [5]x(n) = ( 1/4)n n 0

= 2n n < 0

Solution : Refer DSP-MAY-2006 Q 1 (b)-----------------------------------------------------------------------------------------------------------------

Q 5(a) The desired response of a low pass filter is [10]

=

||0)(

43

43

433

w

wforeeH

wjjw

d

Determine H (ejw) for M=7 using Hamming Window.The Hamming Window function is

( )

=

Otherwise

MnforM

n

nw

0

101

2cos46.054.0

][

Solution :

Step 1. Find Hd(w)

Let Hd(w) = | Hd(w) |

Where (i)Magnitude response :

Hd() =

otherwise

wwfor cc

0

1

(ii)Phase Response : (w) = ej

For Linear Phase LPF with symmetric h[n]

= wN2

1 = w

(w) =wje

By substituting,

=

otherwise

wwewH cc

wj

d

0)(

where = 3 and wc=4

3

- -wc 0 wc

1


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Step 2. Find hd[n]

By Inverse DTFT, dwewHnh jnwd )(2

1][

=

dwee jnwwjcw

cw

)(2

1

=

dwe wnj

cw

cw

)(

2

1

=

cw

cw

wnj

jn

e

=

)(2

1 )(

=

j

ee

n

cwnjcwnj

2)(

1 )()(

where = 3 and wc=4

3

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

Step 3. Find h[n]

Linear phase FIR filter with impulse response h[n] is given by,

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

=

4

3)3(

4

3)3sin(

4

3][

n

n

nh

6

2cos46.054.0

n

=

=

075.0

159.0

225.0

750.0

225.0

159.0

075.0

][nh

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

Q 5 (b) The impulse response of a LTI system h(n) = { 1, } [10]Find the response of the system when input x(n) = { 1, 2, 3 }By i)fold, shift, multiply and sum concept.

ii) Tabulation technique.

Solution : (i) To find y[n] by using Time domain formula

x(n) = { 1, 2, 3 } Length L = 3h(n) = { 1, } Length M = 2

c

ccd

w)n(

w)nsin(w]n[h

=


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(1) y(n) = cos[x(n)] (2) y(n) = x(n) cos(won )

Solution :

(1) y(n) = cos[x(n)]

1(i) Static or Dynamic

Output depends on present value. System is Static

1(ii) Linear or Nonlineary1(n) = cos { x1[n] }y2n) = cos { x2[n] }y(n) = cos { x1[n] + x2[n] }

Here y[n] y1[n] + y2[n]

System is NotLinear.

1(iii)Shift Variant or Shift Invariant.

y(n) = cos[x(n)]Delay by input by ky(n) = cos[x(nk)]Here y[n] = y[n-k]System is Shift Invariant

1(iv) Causal or Non Causal.

Output depends on present value of input. System is causal.1(v) Stable or Unstable.

System is BIBOStable.

(2) y(n) = x(n) cos(won )

2(i) Static or Dynamic

Output depends on present value. System is Static2(ii) Linear or Nonlinear

y1(n) = x1(n) cos (won )y2n) = x2(n) cos(won )y(n) = { x1[n] + x2[n] } cos (won)

= x1(n) cos (won ) + x2(n) cos(won )Here y[n] = y1[n] + y2[n]System is Linear.

2(iii)Shift Variant or Shift Invariant.

y(n) = x(n) cos(won )Delay by input by k

y(n) = x(nk) cos(wo[nk])Here y[n] = y[n-k]System is Shift Invariant

2(iv) Causal or Non Causal.Output depends on present value of input. System is causal.

2(v) Stable or Unstable.System is BIBOStable.

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Q 7 Write short note on: [20](i) Fetal ECG monitoring. (ii) DSP processorTMS 320C5X.

(iii) Hilbert transforms.

Solution : 7(ii) DSP processorTMS 320C5X.

1. Bus Structure

Separate program and data buses : Provides a high degree of parallelism.

1) Program Bus (PB) : - Carries instruction code & immediate operands from

programmemory space to the cpu

2) Program address bus (PAB) : - Provides addresses to program memory space for both

reads and writes

3) Data read bus (DB) Interconnects various elements of the CPU to data memory space

4) Data read address bus (DAB) Provides the address to access the data memory space


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2. Central Arithmetic Logic Unit (CALU)

Consists of :-

1)

Hardware multiplier unit performs 16 x 16 multiplication of number representedin 2s complement form.

2) ALU3) Accumulator (ACC) : Stores one of the operand & results of operation4) Accumulator buffer (ACCB) : 32 bit register ACCB is used for temporary

storage of ACC5) Product register (PREG) : The 32-bit PREG holds the result of multiplication

3. Auxiliary Register ALU (ARAU)

Consists of :

1) Eight 16- bit auxiliary registers (ARs) AR0-AR72) 3- bit auxiliary register pointer (ARP)3) Unsigned 16- bit ALU

4) ARAU calculates indirect addresses by using inputs from ARs, 16- bit indexregister (INDX) and auxiliary register compare register (ARCR)

Accessing data does not require CALU for address manipulation; therefore, theCALU is free for other operations in parallel. Thus instructions to be executedfaster compared to conventional processors.

4.

Other Registers

1) INDEX Register (INDEX) (16-bit) is used by ARAU as a step value to modifythe address in the ARs during indirect addressing. It can also map the dimensionof the address block used for bit-reversal addressing.

2) Auxiliary Register Compare Register (ARCR) (16-bit) is used for addressboundary comparison.

3) Block Move Address Register (BMAR) (16-bit) holds an address value to beused with block moves and multiply / accumulate operations. BMAR Provides16-bit address for an indirect-addressed second operand.

4) Block Repeat Register (all 16-bit wide)i) RPTC (Repeat counter register) holds the repeat count in a repeat singleinstruction operation and is loaded by the RPT and RPTZ instructions.ii) BRCR (Block repeat counter register) holds the count value for the blockrepeat feature.

5. Parallel Logic Unit (PLU)Performs Boolean operations or bit manipulations required of high-speedcontrollers. PLU can set, clear, test or toggle bits in a status register control register,

or any data memory location. Performs logic operations directly on data memory.Results of PLU function are written back to the original data memory Memory-Mapped Registers. PLU has 96 registers mapped into page 0 of the data memoryspace.

6. Program Controller

Contains logic circuitry that Decodes the instructions, Manages the CPU pipeline,Stores the status of CPU operations and Decodes the conditional operations.Program controller performs three concurrent memory operations in any givenmachine cycle: Fetch an instruction, Read an operand, Write an operand.


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It consists of 16-bit program counter (PC), 16-bit status registers processor moderegister (PMST) and circular buffer control register (CBCR), Address generationlogic, Instruction register, Interrupt flag register and interrupt mask register

7. On-Chip Memory

Total address range of 224K words x 16 bits divided into :64K-word program memory, 64K-word local data memory, 64K-word I/O ports,32K-word global data memory

It Contains Program Read Only Memory (ROM), Data/Program Dual-AccessRAM (DARAM), Data/Program Single-Access RAM (SARAM), DSPs have amaskable option that protects the contents of non-chip memories.

Solution : 7 (iii) Hilbert transforms. Refer DSP-MAY-2005 Q 7 (b)

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