5.5 Linearity, Inverse Proportionality, and Duality 305
way. This is so even though the time signals add correctly tox(t). The Fourier transform of the sine
signal will be found using the periodicity ofx(t)or the duality property. n
5.5.2 Inverse Proportionality of Time and Frequency
It is very important to realize that frequency is inversely proportional to time, and that as such,
time and frequency signal characterizations are complementary. Consider the following examples to
illustrate this.
n The impulse signalx 1 (t)=δ(t), although not a regular signal, has finite support (its support is
only att=0 as the signal is zero everywhere else). It is also absolutely integrable, so it has a
Fourier transform
X 1 ()=F[δ(t)]=
∫∞
−∞
δ(t)e−jtdt=e−j^0
∫∞
−∞
δ(t)dt= 1 −∞< <∞
displaying infinite support. (The Fourier transform could have also been obtained from the
Laplace transformL[δ(t)]=1 for all values ofs. Fors=j, we have thatF[δ(t)]=1.) This result
means that sinceδ(t)changes so fast in such a short time, its Fourier transform has all possible
frequency components.
n Consider then the opposite case: A signal that is constant for all times, that does not change, or
a dc signalx 2 (t)=A,−∞<t<∞. We know that the frequency of=0 is assigned to it since
the signal does not vary at all. The Fourier transform cannot be found by means of the integral
becausex 2 (t)is not absolutely integrable, but we can verify that it is given byX 2 ()= 2 πAδ()
(we will formally show this using the duality property). In fact, the inverse Fourier transform is
1
2 π
∫∞
−∞
X 2 ()ejtd=
1
2 π
∫∞
−∞
2 πAδ()ejtd=A
Notice the complementary nature ofx 1 (t)andx 2 (t):x 1 (t)=δ(t)has a one-point support, while
x 2 (t)=Ahas infinite support. Their corresponding Fourier transformsX 1 ()=1 andX 2 ()=
2 πAδ()have infinite and one-point support in the frequency domain, respectively.
n To appreciate the transition from the dc signal to the impulse signal, consider a pulse signal
x 3 (t)=A[u(t+τ/ 2 )−u(t−τ/ 2 )]. This signal has finite energy, and its Fourier transform can be
found using its Laplace transform. We have
X 3 (s)=
A
s
[
esτ/^2 −e−sτ/^2
]
with the whole s-plane as its region of convergence, so that
X 3 ()=X(s)|s=j
=A
(ejτ/^2 −e−jτ/^2 )
j
=Aτ
sin(τ/ 2 )
τ/ 2
(5.5)
