Inverse Proportionality of Time and Frequency

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.

The impulse signal x 1 ( t) = δ(t), although not a regular signal, has finite support (its support is only at t = 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 −jt dt =e −j0

δ( t)dt =1 −∞ <  < ∞

displaying infinite support. (The Fourier transform could have also been obtained from the Laplace transform L[δ(t)] = 1 for all values of s. For s = j, we have that F[δ(t)] = 1.) This result means that since δ(t) changes so fast in such a short time, its Fourier transform has all possible frequency components.

Consider then the opposite case: A signal that is constant for all times, that does not change, or

a dc signal x 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 because x 2 ( t) is not absolutely integrable, but we can verify that it is given by X 2 () = 2πAδ() (we will formally show this using the duality property). In fact, the inverse Fourier transform is

1 jt

X 2 jt () e d 2π Aδ()e d

Notice the complementary nature of x 1 ( t) and x 2 ( t): x 1 ( t) = δ(t) has a one-point support, while x 2 ( t) = A has infinite support. Their corresponding Fourier transforms X 1 () = 1 and X 2 () = 2π Aδ() have infinite and one-point support in the frequency domain, respectively.

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) =

e sτ/2

−e

−sτ/2

with the whole s-plane as its region of convergence, so that

X 3 () = X(s)| s =j ( e jτ/2 −e −jτ/2 )

=A

j sin(τ/2)

C H A P T E R 5: Frequency Analysis: The Fourier Transform

or a sinc function where Aτ corresponds to the area under x 3 ( t). The Fourier trans- form X 3 () is an even function of . At  = 0 using L’H ˆopital’s rule we find that

X 3 ( 0) = Aτ . Finally, the Fourier transform of the pulse becomes zero when  = 2kπ/τ , k = ±1, ±2, . . .. If we let A = 1/τ (so that the area of the pulse is unity), and let τ → 0, the pulse x 3 ( t) becomes

a delta function δ(t) in the limit and the sinc function expands (for τ → 0, X 3 () is not zero for any finite value) to become unity. On the other hand, if we let τ → ∞, the pulse becomes

a constant signal A extending from −∞ to ∞, and the Fourier transform gets closer and closer to δ() (the sinc function becomes zero at values very close to zero and the amplitude at  =0 becomes larger and larger, although the area under the curve remains constant). As shown above,

X 3 () = 2πAδ() is the transform of x 3 ( t) = A, −∞ < t < ∞.

To illustrate the transition in the Fourier transform as the time support increases, we used the following MATLAB script to compute the Fourier transform of pulses of the same amplitude A =1 but different time supports 1 and 4. The script below shows the case when the support is 1, but it can be easily changed to get the support of 4. The symbolic MATLAB function fourier computes the Fourier transform. The results are shown in Figure 5.1.

%%%%%%%%%%%%%%%%%%%%% % Time-frequency relation %%%%%%%%%%%%%%%%%%%%% syms t w x = heaviside(t + 0.5) − heaviside(t − 0.5); subplot(211) ezplot(x, [ − 3]);axis([−3 3 − 0.1 1.1]);grid

X = fourier(x) % Fourier transform subplot(212) ezplot(X, [ −50 50]); axis([−50 50 −1 5]);grid

In summary, the support of X() is inversely proportional to the support of x(t). If x(t) has a Fourier transform X() and α 6= 0 is a real number, then x(αt) is an

Contracted (α > 1),

Contracted and reflected (α < −1),

Expanded (0 < α < 1),

Expanded and reflected ( −1 < α < 0), or

Simply reflected (α = −1) signal, and we have the pair

x(αt)

X (5.6)

First let us mention that the symbol ⇔ means that to a signal x(t) in the time domain (on the left) there corresponds a Fourier transform X() in the frequency domain (on the right). This is not an equality—far from it!