[amp_mcq option1=”$$\left( {{{\sin \left( {{{\pi t} \over 5}} \right)} \over {{{\pi t} \over 5}}}{e^{j{\pi \over 4}}}} \right)$$” option2=”$$\left( {{{\sin \left( {{{\pi t} \over 5}} \right)} \over {{{\pi t} \over 5}}}{e^{ – j{\pi \over 4}}}} \right)$$” option3=”$$\sqrt 2 \left( {{{\sin \left( {{{\pi t} \over 5}} \right)} \over {{{\pi t} \over 5}}}{e^{j{\pi \over 4}}}} \right)$$” option4=”$$\sqrt 2 \left( {{{\sin \left( {{{\pi t} \over 5}} \right)} \over {{{\pi t} \over 5}}}{e^{ – j{\pi \over 4}}}} \right)$$” correct=”option3″]
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 bandpass signal is given by
$$x(t) = \sqrt{2} \left( \frac{\sin \left( \frac{\pi t}{5} \right)}{\frac{\pi t}{5}} \right) \sin \left( \pi t – \frac{\pi}{4} \right)$$
The envelope of this signal is given by
$$A(t) = \sqrt{2} \left( \frac{\sin \left( \frac{\pi t}{5} \right)}{\frac{\pi t}{5}} \right)$$
The phase of this signal is given by
$$\phi(t) = \pi t – \frac{\pi}{4}$$
Therefore, the complex envelope of this signal is given by
$$E(t) = \sqrt{2} \left( \frac{\sin \left( \frac{\pi t}{5} \right)}{\frac{\pi t}{5}} \right) e^{j\left( \pi t – \frac{\pi}{4} \right)} = \sqrt{2} \left( \frac{\sin \left( \frac{\pi t}{5} \right)}{\frac{\pi t}{5}} e^{j\frac{\pi}{4}} \right)$$