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GATE Electronics and Communications (EC) 2019 Solved Paper

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Question : 65 of 65

Marks: +1, -0

A single bit, equally likely to be 0 and 1 , is to be sent across an additive white Gaussian noise (AWGN) channel with power spectral density

\frac{N_0}{2}.

Binary signaling, with

0 \rightarrow p(t)

1 \rightarrow q(t)

, is used for the transmission, along with an optimal receiver that minimizes the biterror probability.

\phi_{1}(t), \phi_{2}(t)

form an orthonormal signal set.

p(t)=\phi_{1}(t)

q(t)=-\phi_{1}(t)

, we would obtain a certain bit-error probability

P_b

p(t)=\phi_{1}(t)

q(t)=\sqrt{E}

\phi_{2}(t)

, for what value of

E

would we obtain the same bit-error probability

P_b?

3
2
0
1

Solution:

0 is represented by p(t)

And 1 is represented by q(t)

And

\psi_{1}(t) \;\text{and}\; \psi_{2}(t)

are orthogonal signal set

(i)

p(t)=\psi_{1}(t)

and

q(t)=-\psi_{1}(t)

So signal space diagram will be,

\therefore dmin_1 = 2

(ii)

p(t)=\psi_{1}(t)

and

q(t)=\sqrt{E} \psi_{2}(t)

So signal space diagram will be

\therefore dmin_2 = \sqrt{E+1}

Now bit error probability is same in both cases

\therefore dmin_1 = dmin_2

\sqrt{1+E}=2

\therefore E = 3

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