A sequence x(n) with the ztransform X(z) = z^{3} + 2z^{2}  3z + z^{1}  4z^{3} is applied as an input to a linear, timeinvariant system with the impulse response h(n) = 28(n  2) where
The output at n = 2 is__________________________________
The output y(n) can be drived as,
y(z) = I1(z) X(z)
1.1(z) = 2z^{2}
Y(z) = 2z^{2} (z^{3} + 2z^{2}  3z + z^{1}  4z^{3}) Y(z) 2(z + 2  3z^{1} +
Taking inverse ztransform of Y(z) y(n) 2[6(n + 1) + 28(n)  36(n  1) +
(n  3) 46(p  511
At n = 2, y(2) = 0
then causal sequence f(n) is
On equating both sides we get. A 4 C = 3.
4A 2B + C D = 5,
48 + D = 0
Q. The value of X(e^{iπ}) = ?
Since
∵
= 1, when n = odd = +1, when rt = even
So,
X(e^{iπ}) = (2 +0+ 2 + 0 +
[1  1  1  1  1 + 1]
4 2 = 2
∵
at n 0, we have
= 2π X 2 = 4 π
The Laplace transform of the waveform shown in the figure below is
Then what is the value of 'C'?
From the waveform shown we have, f(t) t u(t)  (t  1) u(t  1) 
(t  4) u(t  4) + 1 .5(t  6) u(t  6)
 1,5(t  8) LAO  8)
∴ F(s) = LT of f(t)
Comparing this with the given Laplace transform and we get,
C = +1.5
Fourier transform of the signal shown in figure below is,
Also, from the given plot we have
x(t) = u(t 2) + u(t + 1)  u(t  1)  u(t  2)
••• (I)
Answer Questions (22 and 23):
A DTS described by H(z) = when
excited by x[n] = cos(0.2 n + 10°) with discrete frequency Ω = 0.2.
Q. Find H(e^{jΩ}) at Ω = 0.2 and is equals to
For the given input x[n] cos 2n
response, y[n]
Now
Answer Questions (22 and 23):
A DTS described by H(z) = when
excited by x[n] = cos(0.2 n + 10°) with discrete frequency Ω = 0.2.
Q. The steady state response y[n] is given by
Now the steady state response
y[n] = 0.25 cos[0.2 n 10° + 14,1°) [n] = 0.25 cos[0.2 n +24.1°1
Fourier analysis of waveform shown in figure
Q. The waveform is
wave form is neither odd nor even.
Fourier analysis of waveform shown in figure
Q. In the waveform shown
1.
2.
3.
4.
5.
Causal sequence f(n) if F(z) =
, is given by
........................(i)
........................(ii)
From 0) and 0) we get
A 1, B = 1, C = 1
Also we know, f(n) = Inverse ztransform of F(z)
f(n) = [(1)^{n} (3)^{n}  (2)^{n} ] u (n)
A signal x_{1}(t) is having Fourier transform
X_{i}(co). Another signal x_{2}(t) having Fourier transform X_{2}(w) is related to X_{i}(co) by X_{2}(w) = [1 +sg n(w)]X_{i} (co)
Value of x_{2}(t) in terms of x_{1}(t) will be
After some arithmetic operation flag register of 8085 microprocessor becomes BBH. The contents of the accumulator after the operation may be
Flag registers in 808511P is as,
Since, sign bit of 138.11. is 1 so, number is negative from the given choice only (c) or (d) gives the sign bit T.
Here, parity bit = 0 means total number of 1 's present is odd
So, only options (d) contains odd nu ^{}r^{.} her of 1's
Here five l's are present.
The contents of register HL pair is 24FF H. Then the contents of register pair if, 1NR H and INR L operations are performed is
∴ [H] <— 24
and [L] <— FF
INH H means increment the content of 'H' by 01.
∴ [H] <— 25
and [L] <— FE
⇒ Content of HL pair = 25FF H
L means increment the content of 'L'by 01.
∴ [H] <— 24
and [L] <— [FF] + [01]
i.e. [L] <— 00
⇒ Content of IHL pair = 2400 H
If the HLT instruction of a 8085 microprocessor is executed then,
HLT → is a 1byte instruction.
→ processor stops executing and enters wait state
→ no register contents are affected.
→ the address bus and data bus are placed in high impedance state
Line : MVI A, B5H
MVI B, OEH
XRI 69H
ADD
ANI 9BH
CPI 9FH
STA 3010 H
HLT
After execution of STA 3010H the status of Cy and Z flag will be
Line Cy Z
1
2 0 0
3 0 0
4 0 0
0 0
6 1 0
7 1
Three memory chips are of size 1 kB, 2 kB and 4 kB. Their address bus is 10 bits. The data bus sizes of chips are respectively
∴ Memory size
= Add, bus size x Data bus size.
=(2)^{add bus line} x.(Data bus size)
So, for 1 kB Chip.
Data bus size =
For 2 kB chip,
Data bus size =
= 16 bits
For 4 kB chip, Elata bus size
= 32 bits
An 8085 assembly language program is given below:
MV1C, 03H
LXI H, 2000 H
MOV A, M
DCR
LOOP 1 : INX H
MOV B, M CMP B
JNC LOOP 2 MOV A, B
LOOP 2 : DCR C
JNZ LOOP 1 STA 2100 H HLT
Contents of memory location 2000: 18 H
2001 : 10H
2002 : 2BH
Q At the end of the program what will be the condition of carry and zero flags?
An 8085 assembly language program is given below:
MV1C, 03H
LXI H, 2000 H
MOV A, M
DCR
LOOP 1 : INX H
MOV B, M CMP B
JNC LOOP 2 MOV A, B
LOOP 2 : DCR C
JNZ LOOP 1 STA 2100 H HLT
Contents of memory location 2000: 18 H
2001 : 10H
2002 : 2BH
Q What does the program do?
Program is searching the value higher than the content of accumulator.
Consider the given instructions of program for the delay register.
MVI C, FFH
LOOP: DCR C
JNZ LOOP
Assume clock frequency of the system is 2 MHz.
Q. The total Tstates for this above programme is given by
MVI C, FFH → 7 Tstates
LOOP: DCR C → 5 Tstates
JNZ LOOP → 10 Tstates
Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 




