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*Answer can only contain numeric values

QUESTION: 1

** **A sequence x(n) with the z-transform X(z) = z^{3} + 2z^{2} - 3z + z^{-1} - 4z^{-3} is applied as an input to a linear, time-invariant system with the impulse response h(n) = 28(n - 2) where

The output at n = 2 is__________________________________

Solution:

The output y(n) can **be drived as, **

**y(z) = I-1(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 z-transform 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

QUESTION: 2

then causal sequence f(n) is

Solution:

On equating both sides we get. A 4- C = 3.

4A- 2B + C D = 5,

48 + D = 0

QUESTION: 3

**Q. **The value of X(e^{iπ}) = ?

Solution:

Since

∵

**=**** -1, when n = odd ****= +1, when rt = even**

**So,
**X(e

**[1**** -**** 1**** -**** 1 - 1 - 1 + 1]**

**4- 2**** = 2**

QUESTION: 4

Solution:

∵

**at n 0, we have**

**= 2π X 2 = 4 π**

QUESTION: 5

The Laplace transform of the waveform shown in the figure below is

Then what is the value of 'C'?

Solution:

**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

QUESTION: 6

Fourier transform of the signal shown in figure below is,

Solution:

Also, from the given plot we have

x(t) = u(t 2) + u(t + 1) - u(t - 1) - u(t - 2)

••• (I)

QUESTION: 7

**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

Solution:

For the given input x[n] cos 2n

response, y[n]

Now

QUESTION: 8

**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

Solution:

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

QUESTION: 9

** **Fourier analysis of waveform shown in figure

**Q.** The waveform is

Solution:

**wave form is neither odd nor even.**

QUESTION: 10

** **Fourier analysis of waveform shown in figure

**Q. **In the waveform shown

1.

2.

3.

4.

5.

Solution:

QUESTION: 11

Causal sequence f(n) if F(z) =

, is given by

Solution:

........................(i)

........................(ii)

From 0) and 0) **we **get

A 1, B = 1, C = -1

Also **we **know, f(n) = Inverse z-transform of **F(z)**

**f(n) = [(-1) ^{n} (-3)**

QUESTION: 12

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

Solution:

QUESTION: 13

After some arithmetic operation flag register of 8085 microprocessor becomes BBH. The contents of the accumulator after the operation may be

Solution:

Flag registers in 808511P is as,

Since, sign bit of 138.1-1. 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.

QUESTION: 14

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

Solution:

∴ [H] <— 24

and [L] <— FF

INH H means increment the content of 'H' by 01.

∴ [H] <— 25

and [L] <— FE

⇒ Content of H-L pair = 25FF H

L means increment the content of 'L'by 01.

∴ [H] <— 24

and [L] <— [FF] + [01]

i.e. [L] <— 00

⇒ Content of IH-L pair = 2400 H

QUESTION: 15

If the HLT instruction of a 8085 microprocessor is executed then,

Solution:

HLT → is a 1-byte 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

QUESTION: 16

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

Solution:

**Line Cy Z**

**1**

**2 0 0**

**3 0 0**

**4 0 0**

**0 0**

**6 1 0**

**7 1**

QUESTION: 17

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

Solution:

**∴ 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

QUESTION: 18

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?

Solution:

QUESTION: 19

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?

Solution:

Program is searching the value higher than the content of accumulator.

QUESTION: 20

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 T-states for this above programme is given by

Solution:

MVI C, FFH → 7 T-states

LOOP: DCR C → 5 T-states

JNZ LOOP → 10 T-states

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