Correct Answer - Option 3 :
\(\sqrt 2 \left( {\frac{{\sin \left( {\frac{{\pi t}}{5}} \right)}}{{\frac{{\pi t}}{5}}}} \right){e^{j\frac{\pi }{4}}}\)
Concept: The complex envelope of the bandpass signal is nothing but its low pass equivalent. Relation of Bandpass signal x(t) with its equivalent low pass signal is given by \( \;x\left( t \right) = Re\left[ {\tilde x\left( t \right){e^{j2\pi {f_c}t}}} \right]\)
Calculation: The given bandpass signal is:
\(x\left( t \right) = - \sqrt 2 \left[ {\frac{{\sin \frac{{\pi t}}{5}}}{{\frac{{\pi t}}{5}}}} \right]\sin \left( {\pi t - \frac{\pi }{4}} \right)\) …1)
\(\& \;x\left( t \right) = Re\left[ {\tilde x\left( t \right){e^{j2\pi {f_c}t}}} \right]\) …2)
Where fc = center frequency
And x̃ (t) is the complex envelope of x(t)
So, Let x̃(t) = a + jb. We are to find a & b.
From equation …1)
\(x\left( t \right) = \frac{{\sin \left( {\frac{{\pi t}}{5}} \right)}}{{\frac{{\pi t}}{5}}}\left[ {\cos \pi t - \sin \pi t} \right]\) …3) ({On applying sin (A - B) formula}
From Equation …2)
x(t) = Re[(a + jb)(cos πt + j sin πt)] = a cos πt – b sin πt …4)
On equating 3) & 4) we get
\(a = b = \frac{{\sin \left( {\frac{{\pi t}}{5}} \right)}}{{\frac{{\pi t}}{5}}},\;So\;\tilde x\left( t \right) = a + jb = \frac{{\sin \left( {\frac{{\pi t}}{5}} \right)}}{{\frac{{\pi t}}{5}}}\left( {i + j} \right)\)
\($Hence\;\tilde x\left( t \right) = \frac{{\sin \left( {\frac{{\pi t}}{5}} \right)}}{{\left( {\frac{{\pi t}}{5}} \right)}}\left( {\sqrt 2 } \right){e^{\frac{{j\pi }}{4}}}\)
Hence Option 3 is Correct.
