The complex envelope of the bandpass signal \(x\left( t \right) =
A. <span class="math-tex">\(\left( {\frac{{\sin \left( {\frac{{\pi t}}{5}} \right)}}{{\frac{{\pi t}}{5}}}} \right){e^{j\frac{\pi }{4}}}\)</span>
B. <span class="math-tex">\(\left( {\frac{{\sin \left( {\frac{{\pi t}}{5}} \right)}}{{\frac{{\pi t}}{5}}}} \right){e^{ - j\frac{\pi }{4}}}\)</span>
C. <span class="math-tex">\(\sqrt 2 \left( {\frac{{\sin \left( {\frac{{\pi t}}{5}} \right)}}{{\frac{{\pi t}}{5}}}} \right){e^{j\frac{\pi }{4}}}\)</span>
D. <span class="math-tex">\(\sqrt 2 \left( {\frac{{\sin \left( {\frac{{\pi t}}{5}} \right)}}{{\frac{{\pi t}}{5}}}} \right){e^{ - j\frac{\pi }{4}}}\)</span>
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Right Answer is: C
SOLUTION
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.