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Indian Institute of Technology Mumbai (IIT-M) 2009 Other Bachelor

Wednesday, 23 January 2013 02:20Web
38. Refer to the NAND and NOR latches shown in the figure. The inputs ( ) one two P ,P for
both the latches are 1st made (0, 1) and then, after a few seconds, made (1, 1).
The corresponding stable outputs (Q1, Q2) are
(A) NAND: 1st (0, 1) then (0, 1) NOR: 1st (1, 0) then (0, 0)
(B) NAND: 1st (1, 0) then (1, 0) NOR: 1st (1, 0) then (1, 0)
(C) NAND: 1st (1, 0) then (1, 0) NOR: 1st (1, 0) then (0, 0)
(D) NAND: 1st (1, 0) then (1, 1) NOR: 1st (0, 1) then (0, 1)
39. elaborate the counting stages (Q1, Q2) for the counter shown in the figure below?
(A) 11,10,00,11,10,.... (B) 01,10,11,00,01,....
(C) 00,11,01,10,00,.... (D) 01,10,00,01,10,....
2 P
1 P
2 Q
1 Q
2 P
1 P
2 Q
1 Q
Clock JK Flip Flop JK Flip Flop
1 J
1 K
1 Q
1 Q
2 J
2 K
2 Q
2 Q
2 Q one Q
EC½ GATE Paper 2009 www.gateforum.com
© All rights reserved by GATE Forum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the
written permission. explain this ques. paper at www.gateforum.com. Page 10 of 15
40. A system with transfer function H(z) has impulse response h(.) described as
h(2)=1, h(3)=-1 and h(k)=0 otherwise. Consider the subsequent statements
( )
( )
S1:H z is a low pass filter
S2 :H z is a FIR filter
Which of the subsequent is correct?
(A) Only S2 is actual
(B) Both S1 and S2 are false
(C) Both S1 and S2 are true, and S2 is a cause for S1
(D) Both S1 and S2 are true, but S2 is not a cause for S1
41. Consider a system whose input x and output y are related by the formula
y (t) x (t )h(2 ) d
¥

=  - t t t
Where h(t) is shown in the graph
Which of the subsequent 4 properties are possessed by the system?
BIBO: Bounded input provide a bounded output
Causal: The system is causal
LP: The system is low pass
LTI: The system is linear and time invariant
(A) Causal, LP
(B) BIBO, LTI
(C) BIBO, Causal, LTI
(D) LP, LTI
42. The four point Discrete Fourier Transform (DFT) of a discrete time sequence
{1, 0, 2, 3} is
(A)  0, -2 + 2j,2, -2 - 2j
(B)  2,2 + 2j,6,2 - 2j

(C)  6,1 - 3j,2,1 + 3j



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