[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″]<\/p>\n
The correct answer is $\\boxed{\\sqrt{2} \\left( \\frac{\\sin \\left( \\frac{\\pi t}{5} \\right)}{\\frac{\\pi t}{5}} e^{j\\frac{\\pi}{4}} \\right)}$.<\/p>\n
The complex envelope of a bandpass signal is given by<\/p>\n
$$E(t) = A(t) e^{j\\phi(t)}$$<\/p>\n
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.<\/p>\n
In this case, the bandpass signal is given by<\/p>\n
$$x(t) = \\sqrt{2} \\left( \\frac{\\sin \\left( \\frac{\\pi t}{5} \\right)}{\\frac{\\pi t}{5}} \\right) \\sin \\left( \\pi t – \\frac{\\pi}{4} \\right)$$<\/p>\n
The envelope of this signal is given by<\/p>\n
$$A(t) = \\sqrt{2} \\left( \\frac{\\sin \\left( \\frac{\\pi t}{5} \\right)}{\\frac{\\pi t}{5}} \\right)$$<\/p>\n
The phase of this signal is given by<\/p>\n
$$\\phi(t) = \\pi t – \\frac{\\pi}{4}$$<\/p>\n
Therefore, the complex envelope of this signal is given by<\/p>\n
$$E(t) = \\sqrt{2} \\left( \\frac{\\sin \\left( \\frac{\\pi t}{5} \\right)}{\\frac{\\pi t}{5}} \\right) e^{j\\left( \\pi t – \\frac{\\pi}{4} \\right)} = \\sqrt{2} \\left( \\frac{\\sin \\left( \\frac{\\pi t}{5} \\right)}{\\frac{\\pi t}{5}} e^{j\\frac{\\pi}{4}} \\right)$$<\/p>\n","protected":false},"excerpt":{"rendered":"
[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 … <\/p>\n