3 3 2 2 dt 0 31 1 xt2 2t2 4 t2 1 3 2 1 0 1 2 t 3 note

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Unformatted text preview: � −� −� � � � � 2ω 1 j�t = j �X (j� )e d� j 2ω −� t=0 � � 2ω d = x(t)� , from the Differentiation in Time property � j dt t=0 By checking the condition that x√i (0) = 0, we can see that it is true for x1 (t) and x2 (t) and false for x3 (t). (6) X (j� ) is p eriodic A p ossible way to check this condition is using Parseval’s Relation for Aperiodic Signals: �� � � 1 2 |x(t)| dt = |X (j� )|2 d� (O & W, Section 4.3.7, p. 312) 2ω −� −� The value computed by each side in the equation above is the energy of the signal. In general, a real finite signal would have finite energy if the signal were time-limited, i.e. there exists a real � > 0, such that x(t) = 0, for |t| > �. In contrast, a real finite signal that is p eriodic has finite power and infinite energy. Similarly, a signal with a p eriodic Fourier transform has an infinite energy. This means, by Parseval’s theorem, that a signal that has finite energy will not have a periodic Fourier transform. For a related discussion, see Section 4.1.2, O & W, p.289. Using this approach, we can see that x2 (t) and x3 (t) both have finite energy, i.e. �� |x(t)|2 d < ∗ , and hence do not have a periodic Fourier transform. −� An impulse is not finite, so we cannot apply the previous test on x1 (t). However, notice that x(t) has a close resemblance to the Fourier transform of a sine wave (see Table 4.3, O & W, p. 329). By duality, this ind...
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This document was uploaded on 02/09/2014.

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