25 redrawn above a signal g t xt 1 is symmetric about

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: imilar to the previous condition, X (j� ) will have no imaginary part only if x(t) is an even function, i.e x(t) has only an even component � x(−t) = x(t). By inspection, x2 (t) is the only even signal, and hence this condition is true for x2 (t) and false for x1 (t) and x3 (t). (3) There exists a real � such that ej�� X (j� ) is real Again, the only way for a real signal to have a real Fourier transform is if that signal is purely even. The complex exponential that is multiplied by X (j� ) hints to time-shifting (see O & W, Section 4.3.2, p. 301). So what this condition tests is whether the signal can b e made even by shifting it in time. For an aperiodic signal this condition can be true only if the signal has both even and o dd components. The reason is that for an aperiodic signal to be even or odd, it must have symmetry. Time-shifting such a signal will destroy that symmetry so that the signal can not b e described as even or odd afterward. On the other hand, this is not true for p eriodic signals. For example the cosine and sine functions are even and...
View Full Document

Ask a homework question - tutors are online