22 BME 1062
Digital Assignment – 3
Solve the following difference equation using Z transform:
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Kedar
Ponagin
22
2C yn 2
Yaz
224
22 412
Y
22
22
212 1
2
24127
324227
212
22
32 2
2
21
z
22 32 27
2
yo
2
6 2
2722 1322
2
22 2
2
2
22 4 42
262 1
276 1
Cate
TE
Ez
D
TESTED
A Cz 2 Cz 1
B Cz
1
YE cathcz 1
I
12
I
Bet
1
231 3
Un
A
É
2
2n
t
n t
222
in
Ii
232 c
22 4C 042
2244C C 42
A 22 2 22 27 B2 B
A22 32 At 2A t B2 B
22 Atc
213A B 4C
At c o
2
2A B 4C
c
224127
22 22 2
412
2
22
1
4
2
1
IA 3
y
1
22
2
1
1 t B
TB
0
T
B
0
A
2
2
3A
T
t 1
I
2
Eye It
2
412
22
332
2
12 172
A 2 17212 1 t B
a
2
4127
22412
1
k
2
112
O
t
k
40
Ix
I
17 1
13
D
A
AC 4
2
z 172
AD
D
2
C 2
17 1
2
TO
Ein TLE
I
112
C 2 1
O
t
B
2
E1
9D
I
L At
C
11
C1
I or
fax
k
1k
I
4
z 172
224127 22
412
4127
22
10412
32
I 22 32 107
22
0
0
32
22 3 1 0
32
10
45
2
2 5
2357
5
2
32412
5 3
22 22 52 10
2
21
2 2
t
2
2
2
2 3
A
4
1
4127
Yn
2
32 2
4
I
15J
s
t
5
t
Ey
I
2
2
BC
B
5 2
A
Bez 5
A 2 2
E
z
2
5
412
22 422
412
22 1
412
L
24
yet
A2 113
z
A
I
2
2
1
2
1
t
c
1
E
335,1
oosny
Isin
E
yh
KID
Ecosny
22 1
It
I 341
If
C
It
ELSE
22 12
un
D
I
KEI
II
2 1
AZ t BL
BAC 224 C
22C A C ZC ATB
Bt f
45
met
1
A 22
L
Btc
412
BEEB
peak
221
Sinn
IN
2
EF cong
2
Ey
z
Cosmo
3
2
Csinno
Edinger
sinny
3
90 1
Since
2
no 2
Kntz
t
fkn
3k ntl
221422
3 22
ZY
223
z
22
32 4
22
2
42
12 1712 4
U z
A
413
4127
In
2
2
4422
92
44123 0
322 22 92
322
0
72
322 72
2 4
23
2
32 422
3
1 2 4
2 1
47
412
z
4
2
4
2
t 3
4
3
32 7
ATB
E
7
7
t
t
a
A 2 4 1 B Cz 1
AZ Bz 4A B
4A 13 7
13 1
1
E
2 1
4
t
it less
0
22
2 32
412
hake
It
Et
can
O
confusing