[amp_mcq option1=”$${{\\sin \\left( t \\right)} \\over {\\pi t}}$$” option2=”$${{\\sin \\left( {2t} \\right)} \\over {2\\pi t}}$$” option3=”$${{2\\sin \\left( t \\right)} \\over {\\pi t}}$$” option4=”$${\\left( {{{\\sin \\left( t \\right)} \\over {\\pi t}}} \\right)^2}$$” correct=”option4″]<\/p>\n
The correct answer is $\\boxed{{2\\sin \\left( t \\right)} \\over {\\pi t}}$.<\/p>\n
The convolution of two signals $x(t)$ and $h(t)$ is defined as:<\/p>\n
$$x(t)*h(t) = \\int_{-\\infty}^{\\infty} x(\\tau)h(t-\\tau) d\\tau$$<\/p>\n
In this case, we have $x(t) = {{\\sin \\left( t \\right)} \\over {\\pi t}}$ and $h(t) = {{\\sin \\left( t \\right)} \\over {\\pi t}}$. Substituting these into the convolution formula, we get:<\/p>\n
$$x(t)*h(t) = \\int_{-\\infty}^{\\infty} {{\\sin \\left( \\tau \\right)} \\over {\\pi \\tau}} {{\\sin \\left( t-\\tau \\right)} \\over {\\pi (t-\\tau)}} d\\tau$$<\/p>\n
We can simplify this by multiplying the numerators and denominators by $\\pi t$:<\/p>\n
$$x(t)*h(t) = \\int_{-\\infty}^{\\infty} {{\\sin^2 \\left( \\tau \\right)} \\over {t^2-\\tau^2}} d\\tau$$<\/p>\n
This integral can be evaluated using the following identity:<\/p>\n
$$\\int_{-\\infty}^{\\infty} {{\\sin^2 \\left( \\tau \\right)} \\over {t^2-\\tau^2}} d\\tau = {2 \\over \\pi} \\left[ {1 \\over t} + {1 \\over t+i\\sqrt{t^2-1}} + {1 \\over t-i\\sqrt{t^2-1}} \\right]$$<\/p>\n
Substituting this into the convolution formula, we get:<\/p>\n
$$x(t)*h(t) = {2 \\over \\pi} \\left[ {1 \\over t} + {1 \\over t+i\\sqrt{t^2-1}} + {1 \\over t-i\\sqrt{t^2-1}} \\right]$$<\/p>\n
We can simplify this by multiplying the numerator and denominator by $t$:<\/p>\n
$$x(t)*h(t) = {2 \\over \\pi t} \\left[ {1} + {1 \\over t+i\\sqrt{t^2-1}} + {1 \\over t-i\\sqrt{t^2-1}} \\right]$$<\/p>\n
Finally, we can simplify this by using the following identities:<\/p>\n
$$1 + {1 \\over t+i\\sqrt{t^2-1}} + {1 \\over t-i\\sqrt{t^2-1}} = 2 \\cos \\left( {\\arctan \\left( t \\right) \\over 2} \\right)$$<\/p>\n
$${1 \\over t} = \\lim_{n \\to \\infty} \\left( 1 – {t^2 \\over n^2} \\right)^n$$<\/p>\n
Substituting these into the convolution formula, we get:<\/p>\n
$$x(t)*h(t) = {2 \\over \\pi t} \\lim_{n \\to \\infty} \\left( 1 – {t^2 \\over n^2} \\right)^n \\cos \\left( {\\arctan \\left( t \\right) \\over 2} \\right)$$<\/p>\n
As $n \\to \\infty$, the term $\\left( 1 – {t^2 \\over n^2} \\right)^n$ approaches $1$. Therefore, we get:<\/p>\n
$$x(t)*h(t) = {2 \\over \\pi t} \\cos \\left( {\\arctan \\left( t \\right) \\over 2} \\right)$$<\/p>\n
Finally, we can simplify this by using the following identity:<\/p>\n
$$\\cos \\left( {\\arctan \\left( t \\right) \\over 2} \\right) = {2 \\sin \\left( t \\right) \\over \\sqrt{1+t^2}}$$<\/p>\n
Substituting this into the convolution formula, we get:<\/p>\n
$$x(t)*h(t) = {4 \\sin \\left( t \\right) \\over \\pi t \\sqrt{1+t^2}}$$<\/p>\n
Therefore, the convolution of $x(t) = {{\\<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”$${{\\sin \\left( t \\right)} \\over {\\pi t}}$$” option2=”$${{\\sin \\left( {2t} \\right)} \\over {2\\pi t}}$$” option3=”$${{2\\sin \\left( t \\right)} \\over {\\pi t}}$$” option4=”$${\\left( {{{\\sin \\left( t \\right)} \\over {\\pi t}}} \\right)^2}$$” correct=”option4″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[959],"tags":[],"class_list":["post-56209","post","type-post","status-publish","format-standard","hentry","category-signal-processing","no-featured-image-padding"],"yoast_head":"\n