$$left( {{{sin left( {{{pi t} over 5}}
ight)} over {{{pi t} over 5}}}{e^{j{pi over 4}}}}
ight)$$
$$left( {{{sin left( {{{pi t} over 5}}
ight)} over {{{pi t} over 5}}}{e^{ - j{pi over 4}}}}
ight)$$
$$sqrt 2 left( {{{sin left( {{{pi t} over 5}}
ight)} over {{{pi t} over 5}}}{e^{j{pi over 4}}}}
ight)$$
$$sqrt 2 left( {{{sin left( {{{pi t} over 5}}
ight)} over {{{pi t} over 5}}}{e^{ - j{pi over 4}}}}
ight)$$
Answer is Wrong!
Answer is Right!
The correct answer is $\boxed{\sqrt{2} \left( \frac{\sin \left( \frac{\pi t}{5} \right)}{\frac{\pi t}{5}} e^{j\frac{\pi}{4}} \right)}$.
The complex envelope of a bandpass signal is given by
$$E(t) = A(t) e^{j\phi(t)}$$
where $A(t)$ is the envelope and $\phi(t)$ is the phase. The envelope is the real part of the complex envelope, and the phase is the imaginary part.
In this case, the
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