Q.1. If X and Y are random variables such that E[2X + Y] = 0 and E[X + 2Y] = 33, then E[X] + E[Y] = _________.
Given X,Y are Random variables
As per The given data
E[2x + y] = 0
E[x + 2y] = 33
⇒ 2E(X) + E(Y) = 0.........(1)
⇒ E(X) + 2 E(Y) = 33......(2)
Solving (1) and (2);
2E(x) + E(y) = 0
2E(x) + 4E(y) = 66 [equ2 mul by 2 and + to -]
------------------------------
-3E(y) = -66
E(y)=22
From (1); 2E(x) + E(y)= 0
2E(x) = -E(y)
2E(x)= -22
E(x)= -11
By solving (1) and (2)
we got E(Y) = 22
E(X) = −11
∴E(X) + E(Y) = 22 − 11 = 11
Q.2. Consider identical four, 3-faced dice. When the dice are rolled, the faces of the dice appear with probabilities given below. Which distribution has the maximum entropy?
(a) (1/2, 1/4, 1/4)
(b) (1/3, 1/3, 1/3)
(c) (1/6, 2/3, 1/6)
(d) (1/4, 1/6, 7/12)
Correct Answer is Option (B)
The Correct Answer Among All the Options is B
For maximum entropy, all symbols must have equal probability.
Hmax = log2m , m=no. of symbols.
Q.3. Consider a binary code with parity check matrix H given below.
Which of the following is a valid codeward?
(a) [1 0 0 1 0 1]
(b) [1 1 1 1 0 1]
(c) [0 1 0 0 1 0]
(d) [1 1 0 1 1 0]
Correct Answer is Option (D)
The Correct Answer Among All the Options is D
The matrix H is called parity check matrix of any code word only if CHT=0
Q.4. In a digital communication system, the overall pulse shape p(t) at the receiver before the sampler has the Fourier transform P(f). If the symbols are transmitted at the rate of 2000 symbols per second, for which of the following cases is inter symbol interference zero?
(a)
(b)
(c)
(d)
Correct Answer is Option (B)
For Inter symbol interference(ISI) free pulse, If P(t) is having spectrum P(f)
Then = constant
RS= 2.4 kHz
This condition is met by pulse given in option B.
Q.5. Consider a binary channel code in which each codeword has a fixed length of 5 bits. The Hamming distance between any pair of distinct codewords in this code is at least 2. The maximum number of codewords such a code can contain is ___________.
Fixed Length of code n=5
Minimum Hamming distance dmin
Without any constraint, 25 = 32 codewords can be formed.
Following table contains all the possible code words having minimum Hamming distance dmin = 2
Hence, total 16 such code words are possible by a 5 bit code with dmin = 2
Q.6. A binary communication system makes use of the symbols "zero" and "one". There are channel errors. Consider the following events:
x0: a "zero" is transmitted
x1: a "one" is transmitted
y0: a "zero" is received
y1: a "one" is received
The following probabilities are given: The information in bits that you obtain when you learn which symbol has been received (while you know that a "zero" has been transmitted) is
Given Binary communication channel Information content in receiving y it in given that xo is transmitted is
Q.7. The minimum distance of a (n,k) = (7,4) linear block code is upper bounded by:
(a) 1
(b) 2
(c) 3
(d) 4
Correct Answer is Option (C)
linear block code,(7,4)=(n,k)
It means 4 bit data is encoded in 7 bit codeword.it will correct any single bit error.
dmin = 2t + 1
t=no of error correction , here t=1.
Therefore, dmin = 2(1) + 1
dmin = 3
Q.8. A source produces three symbols A, B and C with probabilities andThe source entropy is
(a)
(b) 1 bit/symbol
(c)
(d)
Correct Answer is Option (D)
Q.9. A discrete memoryless source has an alphabet {a1,a2,a3,a4} with corresponding probabilities The minimum required average code world length in bits to represent 2 4 8 8 this source for error-free reconstruction is _______.
Minimum required average code world length in bits for error free reconstruction Lmin = H(Entropy)
Q.10. A voice-grade AWGN (additive white Gaussian noise) telephone channel has a bandwidth of 4.0 kHz and two-sided noise power spectral density Watt per Hz. If information at the rate of 52 kbps is to be transmitted over this channel with arbitrarily small bit error rate, then the minimum bit-energy Eb(in mJ/bit) necessary is ____
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