xxiv | GATE 2017 Solved Paper CS: Set – 1
Solution: For two vectors u and v, || u || = 2 || v || x3 + x + 1 as the generator polynomial to generate the check
⇒ u and 2v will have the same length. bits. In this network, the message 01011011 is transmitted
as
We know that if two vectors v1 and v2 have the same length, (A) 01011011010 (B) 01011011011
then v1 + v2 bisects the angle between v1 and v2. (C) 01011011101 (D) 01011011100
As || u || = 2|| v ||, u + 2v bisects the angle between the vec-
tors u and 2v. Solution: Given polynomial generator as x3 + x + 1, then
generator will be 1011.
Also, we know that the angle between u and v is same as the Given data = 01011011
angle between u and 2v.
Message transmitted will be:
So, u + 2v bisects the angle between the vectors u and v.
∴ W = u + α v bisects the angle between u and v when α 1011 01011011000 01000011
= 2. 0000
1011
Hence, the correct option is (A). 1011
Question Number: 31 Question Type: MCQ 00000
0000
0001
Let A be n × n real valued square symmetric matrix of rank 0000
n n 0011
2 with ∑∑ A
i =1 j =1
2
ij = 50 . Consider the following statements. 0000
0110
0000
(I) One eigenvalue must be in [−5, 5] 1100
1011
(II) The eigenvalue with the largest magnitude must be 1110
strictly greater than 5 1101
101
Which of the above statements about eigenvalues of A is/are
necessarily CORRECT? The transmitted message is 01011011101.
(A) Both (I) and (II)
(B) (I) only Hence, the correct option is (C).
(C) (II) only
Question Number: 33 Question Type: MCQ
(D) Neither (I) nor (II)
Consider a combination of T and D flip-flops connected as
Solution: Case (i):- shown below. The output of the D flip-flop is connected to
the input of the T flip-flop and the output of the T flip-flop is
Let n > 2.
connected to the input of the D flip-flop.
∴ P(A) = 2 < order of A.
⇒ A is a singular matrix.
⇒ O is an eigen value of A and 06[-5, 5]
Q1
∴ I is correct. T
Q0
D
Filp-
5 0 0 Flop
Filp-
Also for A = 0 0 0 , none of the eigen values have
Flop
0 0 −5
magnitude not greater than 5.
Clock
So, II is not correct.
Case (II):- Let n = 2 Initially, both Q0 and Q1 are set to 1 (before the 1st clock
−5 0 cycle). The outputs
Consider the matrix B =
0 5 (A) Q1 Q0 after the 3rd cycle are 11 and after the 4th
Clearly the eigen values of B lie in [-5, 5]. cycle are 00 respectively
But none of its eigen values have magnitude greater than 5. (B) Q1 Q0 after the 3rd cycle are 11 and after the 4th
cycle are 01 respectively
Hence, the correct option is (B).
(C) Q1 Q0 after the 3rd cycle are 00 and after the 4th
Question Number: 32 Question Type: MCQ cycle are 11 respectively
A computer network uses polynomials over GF(2) for (D) Q1 Q0 after the 3rd cycle are 01 and after the 4th
error checking with 8 bits as information bits and uses cycle are 01 respectively
