Fourier Transform

INTRODUCTION

Fourier series enable us to represent a periodic function as a sum of sinusoids and to obtain the frequency spectrum from the series. We begin by using a Fourier series as a stepping stone in defining the Fourier transform. Then we develop some of the properties of the Fourier transform. Next, we apply the Fourier transform in analyzing circuits. We discuss Parseval's theorem, compare the Laplace and Fourier transforms, and see how the Fourier transform is applied in amplitude modulation and sampling.

DEFINITION OF THE FOURIER TRANSFORM

We saw in the previous chapter that a nonsinusoidal periodic function can be represented by a Fourier series, provided that it satisfies the Dirichlet conditions. What happens if a function is not periodic? Unfortunately, there are many important nonperiodic functions-such as a unit step or an exponential function-that we cannot represent by a Fourier series. As we shall see, the Fourier transform allows a transformation from the time to the frequency domain, even if the function is not periodic. Suppose we want to find the Fourier transform of a nonperiodic function $ p(t) $, shown in Fig. 1(a). We consider a periodic function $ f(t) $ whose shape over one period is the same as $ p(t) $, as shown in Fig. 1(b). If we let the period $ T \rightarrow \infty $, only a single pulse of width $ \tau $ the desired nonperiodic function in Fig. 1(a) remains, because the adjacent pulses have been moved to infinity. Thus, the function $ f(t) $ is no longer periodic. In other words, $ f(t)=p(t) $ as $ T \rightarrow \infty $. It is interesting to consider the spectrum of $ f(t) $ for $ A=10 $ and $ \tau=0.2 $. Figure $ 2 $ shows the effect of increasing $ T $ on the spectrum. First, we notice that the general shape of the spectrum remains the same, and the frequency at which the envelope first becomes zero remains the same. However, the amplitude of the spectrum and the spacing between adjacent components both decrease, while the number of harmonics increases. Thus, over a range of frequencies, the sum of the amplitudes of the harmonics remains almost constant. Since the total "strength" or energy of the components within a band must remain unchanged, the amplitudes of the harmonics must decrease as $ T $ increases. Since $ f=1 / T $, as $ T $ increases, $ f $ or $ \omega $ decreases, so that the discrete spectrum ultimately becomes continuous. To further understand this connection between a nonperiodic function and its periodic counterpart, consider the exponential form of a Fourier series in Eq. (A), namely, where The fundamental frequency is and the spacing between adjacent harmonics is Substituting Eq. (2) into Eq. (1) gives If we let $ T \rightarrow \infty $, the summation becomes integration, the incremental spacing $ \Delta \omega $ becomes the differential separation $ d \omega $, and the discrete harmonic frequency $ n \omega_{0} $ becomes a continuous frequency $ \omega $. Thus, as $ T \rightarrow \infty $, so that Eq. (5) becomes The term in the brackets is known as the Fourier transform of $f (t)$ and is represented by $ F(\omega) $. Thus $$F(\omega)=\mathcal{F}[f(t)]=\int_{-\infty}^{\infty} f(t) e^{-j \omega t} d t \tag{8}$$ where $ \mathcal{F} $ is the Fourier transform operator. It is evident from Eq. (8) that: In general, $ F(\omega) $ is a complex function; its magnitude is called the amplitude spectrum, while its phase is called the phase spectrum. Thus $ F(\omega) $ is the spectrum. Equation (7) can be written in terms of $ F(\omega) $, and we obtain the inverse Fourier transform as The function $ f(t) $ and its transform $ F(\omega) $ form the Fourier transform pairs: $$f(t) \quad \Longleftrightarrow \quad F(\omega)$$ since one can be derived from the other. The Fourier transform $ F(\omega) $ exists when the Fourier integral in Eq. (8) converges. A sufficient but not necessary condition that $ f(t) $ has a Fourier transform is that it be completely integrable in the sense that $$\int_{-\infty}^{\infty}|f(t)| d t<\infty$$ For example, the Fourier transform of the unit ramp function $ t u(t) $ does not exist, because the function does not satisfy the condition above.
To avoid the complex algebra that explicitly appears in the Fourier transform, it is sometimes expedient to temporarily replace $ j \omega $ with $ s $ and then replace $ s $ with $ j \omega $ at the end.