**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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