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Engineering and Comp Sci Homework Help
Sketching the Spectrum of the Signal x(t)
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[QUOTE="rude man, post: 4530874, member: 350494"] You will, many times I'm sure. Reason: a + jb = √(a[SUP]2[/SUP] + b[SUP]2[/SUP]) exp[ tan[SUP]-1[/SUP](b/a)] in other words, this is how you change from cartesian to polar & back. The second line does not follow from the first. Following what I wrote above, 1/(1 + jπk) = 1/√[1 + (πk)[SUP]2[/SUP]] exp[j tan[SUP]-1[/SUP](-πk)] so 1/(1 + jπk) exp(j4πkt) = 1/√[1 + (πk)[SUP]2[/SUP]] exp[j tan[SUP]-1[/SUP](-πk)] exp(j4πkt) = 1/√[1 + (πk)[SUP]2[/SUP]] exp j[4πkt + tan[SUP]-1[/SUP](-πk)]. When you do any arc tan function, remember arc tan(-b/a) is not the same angle as arc tan(b/-a) so don't leave the expression as exp[-j arc tan(πk)] as you did in your post #1. It should be exp[j arc tan(-πk)]. As I said, arc tan(-πk) is the phase angle, and arc tan(-πk/1) and arc tan(πk/-1) are 180 degrees apart! [/QUOTE]
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Sketching the Spectrum of the Signal x(t)
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