Solution: Recognizing the integrand as [ sinc (x)] 2, which is related to the Fourier transform of the gate function W (t), we think of using Parseval’s formula. 1 Choose a = 2 in (5.91): 1 F W ( 2 t) = 2 sinc (ω). Thus, by Parseval’s formula: $ 

∞ ∞ 1 2 [2 sinc (ω)] 2dω = W 1 t dt. (5.98) 2π Z Z 2 −∞ −∞  $  On noting that

1 1 if |t| < 1 W t =  2  $   0 if |t| > 1  we obtain  ∞ 1 1 2 W 2 t dt = (1) dt = 2 . Z Z 1 −∞ $  − Thus (5.98) leads to ∞ [sinc (ω)] 2dω = π. Z−∞

Exercise 5.15: Use the Fourier transform 2 F(e−| t|) = 1 + ω2 to show that ∞ 1 π 2 2 dx = . Z−∞ (1 + x ) 2

5.4.5 Relation between the continuous and the discrete spectrum A τ-periodic signal f(t) has a discrete Fourier spectrum determined by the complex Fourier coefficients {cn}n∈Z, with discrete frequencies nω 0, where ω0 = 2 π/τ is the (see Figure 5.11). A non-periodic finite-energy signal f(t) has a continuous Fourier spectrum determined by the complex Fourier transform F (ω), where the frequency ω assumes all real values (see figures 5.21 and 5.23). In this section we show that there is a simple relation between the discrete spectrum and the . For fixed τ > 0 consider a signal f(t) that is non-zero only on a subinterval of the interval τ τ − 2 ≤ t ≤ 2 (see Figure 5.25 a)). The Fourier transform of f(t), as given by equation (5.74), becomes τ 2 F (ω) = f(t)e−iωt dt. (5.99) τ Z− 2

168 Let g(t) be the τ-periodic function such that τ τ g(t) = f(t), if − ≤ t ≤ , (5.100) 2 2 (see Figure 5.25 b)). The Fourier coefficients of g(t) are given by

τ 2 1 −inω 0t cn = g(t)e dt, (5.101) τ τ Z− 2 with ω0 = 2 π/τ being the fundamental frequency (see (5.70)). We can substitute (5.100) in (5.101) and if we then compare (5.101) with (5.99), we obtain

τc n = F (nω 0).

This equation gives the desired relation between the discrete spectrum and the continuous spectrum. Note that the continuous spectrum forms an “envelope” for the discrete spectrum (compare b) and d) in Figure 5.25). As τ → ∞, the discrete spectrum (d) becomes finer and finer (the spacing → 0) and approaches the continuous spectrum b).

Figure 5.25: A non-periodic signal a) and its continuous spectrum b), and the associated periodic signal c) and its discrete spectrum d), scaled by the period τ.

169 5.4.6 Things are simpler in the frequency domain We give three brief illustrations of the claim made in the title of this section.

• Signal separation Consider a signal of finite duration in the time domain consisting of the superposition of two high frequency components of widely differing frequencies:

f(t) = ( A1 cos ω1t + A2 cos ω2t)W (at ), with ω2 ≫ ω1. The graph of this function is complicated and it is not helpful to draw it. On the other hand, the Fourier transform of this signal is relatively simple. Using linearity and the shift formula we obtain

1 ω − ω1 ω + ω1 F (ω) = F(f(t)) = A1 sinc + sinc 2a   2a   2a 

1 ω − ω2 ω + ω2 + A2 sinc + sinc 2a   2a   2a 

(exercise, similar to Example 5.15). The graph of F (ω) is shown in Figure 5.26.

Figure 5.26: Signal separation in the frequency domain.

We say the signals have been separated in the frequency domain .

• Differentiation Differentiation in the time domain becomes multiplication by iω (i.e., a purely algebraic operation) in the frequency domain.

Differentiation formula

If F(f(t)) = F (ω) (5.102) then F(f ′(t)) = iωF (ω),

170 provided that f is a suitably regular function on R.

Derivation: We apply the definition (5.74) and then integrate by parts:

∞ F(f ′(t)) = f ′(t)e−iωt dt Z−∞ ∞ ∞ = f(t)e−iωt +iω f(t)e−iωt dt −∞ Z−∞

= 0 + iωF ( ω). (by the definition (5.74)) .

Note that the first term is zero since we require that lim f(t) = 0 (see equation (5.73)). t→±∞

1 t2 Exercise 5.16: Use (5.102) and a result from Table 5.2 to find F(te − 2 ).

• Response of a linear time-invariant system Many signal processing devices can be modelled by using a so-called linear time-invariant (LTI) system, illustrated schematically below.

input output system L f (t ) f in (t) out

“Linear” means that the output (aka the response) is determined from the input by the action of a linear L: fout (t) = Lf in (t). (5.103) “Time-invariant” means that if the input is translated in time, then the output is translated in time by the same amount. One can show that if the input to an LTI system is a sinusoid eiωt then the output is a sinusoid of the same frequency , but the amplitude and phase may change, depending on the frequency : L(eiωt ) = α(ω)eiωt , (5.104) where α(ω) is a complex function of frequency ω, called the system function. The key point is that the system function α(ω) determines the Fourier transform Fout (ω) of the output fout (t) in a simple algebraic way in terms of the Fourier transform of the Fin (ω) of the input fin (t): Fout (ω) = α(ω)Fin (ω). (5.105)

Derivation: Assuming that fin (t) has a Fourier integral representation

∞ 1 iωt fin (t) = Fin (ω)e dω, 2π Z−∞

171