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

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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 &amp; 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...
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