Consider the non-periodic signal x(t) shown below where t1 =1 and t2 = 3. x(t) 1 -t2 t1 t2 -1 The Fourier transform of the signal is given by: x (w) = 2 sin (dw) – sin(30)] a) O b) X(w) = w? [2 sin...

Consider the non-periodic signal x(t) shown below where t1 =1 and<br>t2 = 3.<br>x(t)<br>1<br>-t2<br>t1<br>t2<br>-1<br>The Fourier transform of the signal is given by:<br>x (w) = 2 sin (dw) – sin(30)]<br>a)<br>O b) X(w) = w? [2 sin (3w) – 3 sin(w)]<br>Oc)<br>x (w) = sin (3u) – 3 sin (w)]<br>- 3 sin ( w<br>d)<br>X(w) = 3 sin (3w) – sin (w)|<br>sin ( w<br>0ª x(x) = 3 [2sin(324) – 3sin (2) ]<br>n(2)]<br>2 sin ( 3w<br>- 3 sin ( 2w<br>

Extracted text: Consider the non-periodic signal x(t) shown below where t1 =1 and t2 = 3. x(t) 1 -t2 t1 t2 -1 The Fourier transform of the signal is given by: x (w) = 2 sin (dw) – sin(30)] a) O b) X(w) = w? [2 sin (3w) – 3 sin(w)] Oc) x (w) = sin (3u) – 3 sin (w)] - 3 sin ( w d) X(w) = 3 sin (3w) – sin (w)| sin ( w 0ª x(x) = 3 [2sin(324) – 3sin (2) ] n(2)] 2 sin ( 3w - 3 sin ( 2w

Jun 11, 2022

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Consider the non-periodic signal x(t) shown below where t1 =1 and t2 = 3. x(t) 1 -t2 t1 t2 -1 The Fourier transform of the signal is given by: x (w) = 2 sin (dw) – sin(30)] a) O b) X(w) = w? [2 sin...

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