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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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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
-----------------------------------------------------------------------------------------------------------------
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]
-----------------------------------------------------------------------------------------------------------------
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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