GATE Electronics and Communications (EC) 2019 Solved Paper

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

{N}_{0} {/} 2 .

Binary signaling, with

0 → {p}( {t})

and

1 → {q}( {t})

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

ϕ_{1}( {t}), ϕ_{2}( {t})

form an orthonormal signal set.
If we choose

{p}( {t})=ϕ_{1}( {t})

and

{q}( {t})=-ϕ_{1}( {t})

, we would obtain a certain bit-error probability

{P}_{ {b}}

If we keep

p(t)=ϕ_{1}(t)

, but take

q(t)=√{E}

ϕ_{2}( {t})

, for what value of

{E}

would we obtain the same bit-error probability

{P}_{ {b}} ?

Solution:

0 is represented by p(t)
And 1 is represented by q(t)
And

ψ_{1}( {t}) \;\text " and "\; ψ_{2}( {t})

are orthogonal signal set
(i)

p(t)=ψ_{1}(t)

and

q(t)=-ψ_{1}(t)

So signal space diagram will be,

∴ {dmin}_{1}=2

(ii)

{p}( {t})=ψ_{1}( {t})

and

{q}( {t})=√{ {E}} ψ_{2}( {t})

So signal space diagram will be

∴ {dmin}_{2}=√{ {E}+1}

Now bit error probability is same in both cases

∴ {dmin}_{1}= {dmin}_{2}

√{1+E}=2

∴ {E}=3

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