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Q2. Consider an LTI system with impulse response $h[n]$: $h[n] = 4^{-n}u[n + 3]$ Calculate the output of the system in the input is x[n] = 2u[-n - 3] - 2u[-n - 7] Good Luck Dr. Evren DE??RMENC? 

          Q2. Consider an LTI system with impulse response $h[n]$:
$h[n] = 4^{-n}u[n + 3]$
Calculate the output of the system in the input is
x[n] = 2u[-n - 3] - 2u[-n - 7]
Good Luck
Dr. Evren DE??RMENC?
Q2. Consider an LTI system with impulse response h[n]: h[n] = 4^-nu[n + 3] Calculate the output of the system in the input is x[n] = 2u[-n - 3] - 2u[-n - 7] Good Luck Dr. Evren DE??RMENC?

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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Q2.Consider an LTI system with impulse response h[n] h[n]=4-nu[n+3] Calculate the output of the system in the input is x[n]=2u[-n-3]-2u[-n-7] Good Luck Dr.Evren DEGiRMENCi
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00:01 To this question the given circuit can be redrawn as here this is x of n this h1 and this is h2 multiplied by h3 and this is y of n, y of n is equal to h1 plus h2 multiplied by h3 multiplied by x of n therefore h of n is equal to h1 of n plus h2 of n multiplied by h3 of n this is the answer of first part the second part here given h1 of n is equal to 1 by 2 whole raised to n u of n h2 of n is equal to 3 u of n minus 5 and h3 of n is equal to 1 by 3 whole raised to n u of n here h2 of n multiplied by h3 of n is up to n equal to 5 its value will be only h3 and n equal to 5 above the value will be 1 by 3 whole raised to n minus 1 therefore h2 of n multiplied by h3 of n is equal to 3 u of n minus 5 multiplied by 1 by 3 whole raised to n u of n therefore h2 of n multiplied by h3 of n is equal to summation of n equal to 0 to 4 1 by 3 whole raised to n plus 3 multiplied by summation of n equal to 5 to infinity 1 by 3 whole raised to n minus 1 multiplied by u of n which is equal to 1 by 3 raised to 0 plus 1 by 3 raised to 1 plus 1 by 3 raised to 2 plus 1 by 3 raised to 3 plus 1 by 3 raised to 4 plus 3 multiplied by summation of n equal to 5 to infinity 1 by 3 whole raised to n minus 1 multiplied by u of n therefore it is equal to 1 plus 1 by 3 plus 1 by 3 square plus 1 by 3 cube plus 1 by 3 raised to 4 plus 3 multiplied by summation over n equal…
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