Question

Coherent orthogonal binary FSK modulation is used to transmit two equiprobable symbol waveforms s₁(t) = α cos 2πf₁t and s₂(t) = α cos 2πf₂t, where α = 4 mV. Assume an AWGN channel with two-sided noise power spectral density \(\frac{{{N_0}}}{2} = 0.5 \times {10^{ - 12}}\;W/Hz\). Using an optimal receiver and the relation \(Q\left( v \right) = \frac{1}{{\sqrt {2\pi } }}\mathop \smallint \limits_v^\infty {e^{ - {u^2}/2}}du\), the bit error probability for a data rate of 500 kbps is
1. Q(2)
2. \(Q\left( {2\sqrt 2 } \right)\)
3. Q(4)
4. \(Q\left( {4\sqrt 2 } \right)\)

Answer 1

Correct Answer - Option 3 : Q(4)

Concept:

FSK Modulation:

In FSK: transmission of 1 is represented as:

s₁(t) = A_c cos 2π f_Ht

Transmission of 0 is represented as:

s₂(t) = A_c cos 2π f_Lt

and the Bit error probability \(= Q\left[ {\sqrt {\frac{{{E_d}}}{{2{N_0}}}} } \right]\)

Where E_d is the energy of s₁(t) – s₂(t)

\({E_d} = \mathop \smallint \limits_0^{{T_b}} \{ {s_1}\left( t \right) - {s_2}\left( t \right)\}^2\;dt\;\) 

\({E_d} = \mathop \smallint \limits_0^{{T_b}} s_1^2\left( t \right)dt + \mathop \smallint \limits_0^{{T_b}} s_2^2\left( t \right)dt - 2\mathop \smallint \limits_0^{{T_b}} {s_1}\left( t \right) - {s_2}\left( t \right)dt\) 

Since s₁(t) & s₂(t) are orthogonal, we cn write:

\(\therefore \;\mathop \smallint \limits_0^{{T_b}} {s_1}\left( t \right) - {s_2}\left( t \right)dt = 0\) 

\({E_d} = \mathop \smallint \limits_0^{{T_b}} s_1^2\left( t \right)dt + \mathop \smallint \limits_0^{{T_b}} s_2^2\left( t \right)dt\) 

\({E_d} = \frac{{A_c^2{T_b}}}{2} + \frac{{A_c^2{T_b}}}{2} = A_c^3{T_b}\) 

\(BER = Q\left( {\sqrt {\frac{{A_c^2{T_b}}}{{2{N_0}}}} } \right)\) 

Analysis:

Given:

A_c = α = 4 mV

\(\frac{{{N_0}}}{2} = 0.5 \times {10^{ - 12}}\;w/Hz\) 

N₀ = 10^-12 w/Hz

\({T_b} = \frac{1}{{{R_b}}} = \frac{1}{{800 \times {{10}^3}}}\) 

T_b = 0.2 × 10^-5

T_b = 2 × 10^-6 sec.

\(BER = Q\left( {\sqrt {\frac{{A_c^2{T_b}}}{{2{N_0}}}} } \right)\) 

\(BER = Q\left( {\sqrt {\frac{{{{\left( {4 \times {{10}^{ - 3}}} \right)}^2} \times 2 \times {{10}^{ - 6}}}}{{2 \times {{10}^{ - 12}}}}} } \right)\) 

\(BER = Q\left( {\sqrt {\frac{{16 \times {{10}^{ - 6}} \times 2 \times {{10}^{ - 6}}}}{{2 \times {{10}^{ - 12}}}}} } \right)\) 

\(BER = Q\left( {\sqrt {16} } \right)\) 

BER = Q(4)