Electronics and Communication Engineering (ECE) Exam  >  Electronics and Communication Engineering (ECE) Tests  >  GATE ECE (Electronics) Mock Test Series 2025  >  Test: Signals And Systems & Microprocessors - 2 - Electronics and Communication Engineering (ECE) MCQ

Test: Signals And Systems & Microprocessors - 2 - Electronics and Communication Engineering (ECE) MCQ


Test Description

20 Questions MCQ Test GATE ECE (Electronics) Mock Test Series 2025 - Test: Signals And Systems & Microprocessors - 2

Test: Signals And Systems & Microprocessors - 2 for Electronics and Communication Engineering (ECE) 2024 is part of GATE ECE (Electronics) Mock Test Series 2025 preparation. The Test: Signals And Systems & Microprocessors - 2 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The Test: Signals And Systems & Microprocessors - 2 MCQs are made for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Signals And Systems & Microprocessors - 2 below.
Solutions of Test: Signals And Systems & Microprocessors - 2 questions in English are available as part of our GATE ECE (Electronics) Mock Test Series 2025 for Electronics and Communication Engineering (ECE) & Test: Signals And Systems & Microprocessors - 2 solutions in Hindi for GATE ECE (Electronics) Mock Test Series 2025 course. Download more important topics, notes, lectures and mock test series for Electronics and Communication Engineering (ECE) Exam by signing up for free. Attempt Test: Signals And Systems & Microprocessors - 2 | 20 questions in 60 minutes | Mock test for Electronics and Communication Engineering (ECE) preparation | Free important questions MCQ to study GATE ECE (Electronics) Mock Test Series 2025 for Electronics and Communication Engineering (ECE) Exam | Download free PDF with solutions
1 Crore+ students have signed up on EduRev. Have you? Download the App
*Answer can only contain numeric values
Test: Signals And Systems & Microprocessors - 2 - Question 1

 A sequence x(n) with the z-transform X(z) = z3 + 2z2 - 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__________________________________


Detailed Solution for Test: Signals And Systems & Microprocessors - 2 - Question 1

The output y(n) can be drived as,
y(z) = I-1(z) X(z)

1.--1(z) = 2z-2

Y(z) = 2z-2 (z3 + 2z2 - 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

Test: Signals And Systems & Microprocessors - 2 - Question 2

then causal sequence f(n) is

Detailed Solution for Test: Signals And Systems & Microprocessors - 2 - Question 2




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

4A- 2B + C D = 5,

48 + D = 0




Test: Signals And Systems & Microprocessors - 2 - Question 3

Q. The value of X(e) = ?

Detailed Solution for Test: Signals And Systems & Microprocessors - 2 - Question 3

Since 

∵ 

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

So,
 
X(e) = (2 +0+  2 + 0 +

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

4- 2 = 2

 

Test: Signals And Systems & Microprocessors - 2 - Question 4

Detailed Solution for Test: Signals And Systems & Microprocessors - 2 - Question 4


at n 0, we have

= 2π X 2 = 4 π

Test: Signals And Systems & Microprocessors - 2 - Question 5

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

Then what is the value of 'C'?

Detailed Solution for Test: Signals And Systems & Microprocessors - 2 - Question 5

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

Test: Signals And Systems & Microprocessors - 2 - Question 6

Fourier transform of the signal shown in figure below is,

Detailed Solution for Test: Signals And Systems & Microprocessors - 2 - Question 6

Also, from the given plot we have

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

••• (I)



 

Test: Signals And Systems & Microprocessors - 2 - 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) at Ω = 0.2 and is equals to

Detailed Solution for Test: Signals And Systems & Microprocessors - 2 - Question 7

For the given input x[n] cos 2n

response, y[n] 

Now

Test: Signals And Systems & Microprocessors - 2 - 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

Detailed Solution for Test: Signals And Systems & Microprocessors - 2 - Question 8

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

Test: Signals And Systems & Microprocessors - 2 - Question 9

 Fourier analysis of waveform shown in figure

Q. The waveform is

Detailed Solution for Test: Signals And Systems & Microprocessors - 2 - Question 9

wave form is neither odd nor even.

Test: Signals And Systems & Microprocessors - 2 - Question 10

 Fourier analysis of waveform shown in figure

Q.  In the waveform shown

1.

2.
3.
4.
5.

Test: Signals And Systems & Microprocessors - 2 - Question 11

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

, is given by

Detailed Solution for Test: Signals And Systems & Microprocessors - 2 - Question 11



........................(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)n - (2)n ] u (n)

Test: Signals And Systems & Microprocessors - 2 - Question 12

A signal x1(t) is having Fourier transform

Xi(co). Another signal x2(t) having Fourier transform X2(w) is related to Xi(co) by X2(w) = [1 +sg n(w)]Xi (co)

Value of x2(t) in terms of x1(t) will be

Detailed Solution for Test: Signals And Systems & Microprocessors - 2 - Question 12

Test: Signals And Systems & Microprocessors - 2 - Question 13

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

Detailed Solution for Test: Signals And Systems & Microprocessors - 2 - Question 13

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.

Test: Signals And Systems & Microprocessors - 2 - 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

Detailed Solution for Test: Signals And Systems & Microprocessors - 2 - Question 14

∴  [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

Test: Signals And Systems & Microprocessors - 2 - Question 15

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

Detailed Solution for Test: Signals And Systems & Microprocessors - 2 - Question 15

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

Test: Signals And Systems & Microprocessors - 2 - 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

Detailed Solution for Test: Signals And Systems & Microprocessors - 2 - Question 16

 

Line    Cy        Z

1

2           0         0

3           0         0

4           0         0

0          0

6           1         0

7         1

Test: Signals And Systems & Microprocessors - 2 - 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

Detailed Solution for Test: Signals And Systems & Microprocessors - 2 - Question 17

∴ 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

Test: Signals And Systems & Microprocessors - 2 - 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?

Test: Signals And Systems & Microprocessors - 2 - 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

What does the program do?

Detailed Solution for Test: Signals And Systems & Microprocessors - 2 - Question 19

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

Test: Signals And Systems & Microprocessors - 2 - 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

Detailed Solution for Test: Signals And Systems & Microprocessors - 2 - Question 20

MVI C, FFH  →  7 T-states

LOOP: DCR C  → 5 T-states

JNZ LOOP → 10 T-states

24 docs|263 tests
Information about Test: Signals And Systems & Microprocessors - 2 Page
In this test you can find the Exam questions for Test: Signals And Systems & Microprocessors - 2 solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Signals And Systems & Microprocessors - 2, EduRev gives you an ample number of Online tests for practice

Up next

Download as PDF

Up next

Download the FREE EduRev App
Track your progress, build streaks, highlight & save important lessons and more!