Question

In binary frequency shift keying (FSK), the given signal waveforms are

u₀(t) = 5 cos(20000πt); 0 ≤ t ≤ T, and

u1(t) = 5 cos(22000πt); 0 ≤ t ≤ T

where T is the bit-duration interval and t is in seconds. Both u₀(t) and u₁(t) are zero outside the interval 0 ≤ t ≤ T. With a matched filter (correlator) based receiver, the smallest positive value of T (in milliseconds) required to have u₀(t) and u₁(t) uncorrelated is

1. 0.25 ms
2. 0.5 ms
3. 0.75 ms
4. 1.0 ms

Answer 1

Correct Answer - Option 2 : 0.5 ms

Concept:

If two signals are uncorrelated then:

\(\mathop \smallint \limits_0^T {u_0}\left( t \right){u_1}\left( t \right) = 0\)

Calculation:

\(\smallint 5\cos \left( {20,000\;\pi t} \right).5\cos \left( {22,000\;\pi t} \right)dt = 0\)

\(\frac{{25}}{2}\smallint \left[ {\cos \left( {42000\;\pi t} \right) + \cos \left( {2000\;\pi t} \right)} \right]dt = 0\)

\(\frac{{25}}{2 }\left[ {\frac{{\sin \left( {42000\;\pi T} \right)}}{{42000\;\pi }} + \frac{{\sin \left( {2000\;\pi T} \right)}}{{2000\;\pi }}} \right] = 0\)

Both terms should be individually zero, i.e.

sin 2000 πT = 0

\(\begin{array}{l} \Rightarrow 2000\;\pi T = \pi \left[ {smallest} \right]\\ T = \frac{1}{{2000}} \end{array}\)

T = 0.5 msec

So, at T = 0.5 msec both terms are zero.