 #1
 80
 0
Hi everyone!
I was trying to solve an old exercise but in a different way. It asks to find the signal x(t) given the graphs of X(f) and arg[X(f)]. Please refer to the attachment.
My stategy this time is as follows:
1) find D(f) by deriving X(f)
2) antitransform D(f) and get d(t)
3) divide by j2πt and finally get x(t)
So the first step is finding D(f). From a similarly worked out problem I gathered it should be
[tex] D(f)= \frac{1}{16} \text{rect}_{4W}(f+4W)  \frac{1}{16} \text{rect}_{4W}(f4W) [/tex]
However I'm not sure whether the following addend should also be included in D(f):
[tex]\frac{1}{4}W \delta(f+2W)+\frac{1}{4}W \delta(f2W) [/tex]
So ultimately it should be
[tex]D(f)= \frac{1}{16} \text{rect}_{4W}(f+4W)  \frac{1}{16} \text{rect}_{4W}(f4W) \frac{1}{4}W \delta(f+2W)+\frac{1}{4}W \delta(f2W)[/tex]
Is it right? Or there should be no dirac?
Thank you for your time.
I was trying to solve an old exercise but in a different way. It asks to find the signal x(t) given the graphs of X(f) and arg[X(f)]. Please refer to the attachment.
My stategy this time is as follows:
1) find D(f) by deriving X(f)
2) antitransform D(f) and get d(t)
3) divide by j2πt and finally get x(t)
So the first step is finding D(f). From a similarly worked out problem I gathered it should be
[tex] D(f)= \frac{1}{16} \text{rect}_{4W}(f+4W)  \frac{1}{16} \text{rect}_{4W}(f4W) [/tex]
However I'm not sure whether the following addend should also be included in D(f):
[tex]\frac{1}{4}W \delta(f+2W)+\frac{1}{4}W \delta(f2W) [/tex]
So ultimately it should be
[tex]D(f)= \frac{1}{16} \text{rect}_{4W}(f+4W)  \frac{1}{16} \text{rect}_{4W}(f4W) \frac{1}{4}W \delta(f+2W)+\frac{1}{4}W \delta(f2W)[/tex]
Is it right? Or there should be no dirac?
Thank you for your time.
Attachments

14.7 KB Views: 365