Fourier Transform



Fourier series enable us to represent a periodic function as a sum of sinusoids and to obtain the frequency spectrum from the series.
The Fourier transform allows us to extend the concept of a frequency spectrum to nonperiodic functions.
The transform assumes that a nonperiodic function is a periodic function with an infinite period.
Thus, the Fourier transform is an integral representation of a nonperiodic function that is analogous to a Fourier series representation of a periodic function.
The Fourier transform is an integral transform like the Laplace transform.
It transforms a function in the time domain into the frequency domain.
The Fourier transform is very useful in communications systems and digital signal processing, in situations where the Laplace transform does not apply.
While the Laplace transform can only handle circuits with inputs for $ t>0 $ with initial conditions, the Fourier transform can handle circuits with inputs for $ t<0 $ as well as those for $ t>0 $.
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.


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.
Dirichlet conditions
1. f (t) is single-valued everywhere.
2. f (t) has a finite number of finite discontinuities in any one period.
3. f (t) has a finite number of maxima and minima in any one period.
4. The integral $\int _{t_0}^{t_0+T} |f (t)| dt < ∞$ for any $t_0$.
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).
Fig. 1: (a) A nonperiodic function, (b) increasing T to infinity makes f (t) become the nonperiodic function in (a).
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 $.
Fig. 2: Effect of increasing T on the spectrum of the periodic pulse trains in Fig. 1(b).
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,
Equations from the other pages
$$\bbox[10px,border:1px solid grey]{f(t)=\sum_{n=-\infty}^{\infty} c_{n} e^{j n \omega_{0} t} }\tag{A}$$
consider the exponential form of a Fourier series in Eq. (A), namely,
$$f(t)=\sum_{n=-\infty}^{\infty} c_{n} e^{j n \omega_{0} t} \tag{1}$$
$$c_{n}=\frac{1}{T} \int_{-T / 2}^{T / 2} f(t) e^{-j n \omega_{0} t} d t \tag{2} $$
The fundamental frequency is
$$\omega_{0}=\frac{2 \pi}{T} \tag{3}$$
and the spacing between adjacent harmonics is
$$\Delta \omega=(n+1) \omega_{0}-n \omega_{0}=\omega_{0}=\frac{2 \pi}{T} \tag{4}$$
Substituting Eq. (2) into Eq. (1) gives
$$\begin{aligned}f(t) &=\sum_{n=-\infty}^{\infty}\left[\frac{1}{T} \int_{-T / 2}^{T / 2} f(t) e^{-j n \omega \omega_{0} t} d t\right] e^{j n \omega \omega_{0} t} \\&=\sum_{n=-\infty}^{\infty}\left[\frac{\Delta \omega}{2 \pi} \int_{-T / 2}^{T / 2} f(t) e^{-j n \omega \omega_{0} t} d t\right] e^{j n \omega_{0} t} \\&=\frac{1}{2 \pi} \sum_{n=-\infty}^{\infty}\left[\int_{-T / 2}^{T / 2} f(t) e^{-j n \omega_{0} t} d t\right] \Delta \omega e^{j n \omega_{0} t}\end{aligned} \tag{5}$$
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 $,
$$\begin{array}{l}\sum_{n=-\infty}^{\infty} \Longrightarrow \int_{-\infty}^{\infty} \\\Delta \omega \quad \Longrightarrow \quad d \omega \\n \omega_{0} \quad \Longrightarrow \quad \omega \\\end{array} \tag{6}$$
so that Eq. (5) becomes
$$f(t)=\frac{1}{2 \pi} \int_{-\infty}^{\infty}\left[\int_{-\infty}^{\infty} f(t) e^{-j \omega t} d t\right] e^{j \omega t} d \omega \tag{7}$$
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:
The Fourier transform is an integral transformation of f (t) from the time domain to the frequency domain.
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
$$f(t)=\mathcal{F}^{-1}[F(\omega)]=\frac{1}{2 \pi} \int_{-\infty}^{\infty} F(\omega) e^{j \omega t} d \omega \tag{9}$$
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.
Example 1: Find the Fourier transform of the following functions:
(a) $ \delta\left(t-t_{0}\right) $,
(b) $ e^{j \omega \omega t} $,
(c) $ \cos \omega_{0} t $
(a) For the impulse function,
$$F(\omega)=\mathcal{F}\left[\delta\left(t-t_{0}\right)\right]=\int_{-\infty}^{\infty} \delta\left(t-t_{0}\right) e^{-j \omega t} d t=e^{-j \omega t_{0}} \tag{1.1}$$
where the sifting property of the impulse function has been applied. For the special case $ t_{0}=0 $, we obtain $$\mathcal{F}[\delta(t)]=1 \tag{1.2}$$ This shows that the magnitude of the spectrum of the impulse function is constant; that is, all frequencies are equally represented in the impulse function.
(b) We can find the Fourier transform of $ e^{j \omega_{0} t} $ in two ways. If we let $$F(\omega)=\delta\left(\omega-\omega_{0}\right)$$ then we can find $ f(t) $ using Eq. (9), writing
$$f(t)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} \delta\left(\omega-\omega_{0}\right) e^{j \omega t} d \omega$$
Using the sifting property of the impulse function gives $$f(t)=\frac{1}{2 \pi} e^{j \omega_{0} t}$$ Since $ F(\omega) $ and $ f(t) $ constitute a Fourier transform pair, so too must $ 2 \pi \delta\left(\omega-\omega_{0}\right) $ and $ e^{j \omega_{0} t} $,
$$\mathcal{F}\left[e^{j \omega_{0} t}\right]=2 \pi \delta\left(\omega-\omega_{0}\right) \tag{1.3}$$
Alternatively, from Eq. (2), $$\delta(t)=\mathcal{F}^{-1}[1]$$ Using the inverse Fourier transform formula in Eq. (9),
$$\delta(t)=\mathcal{F}^{-1}[1]=\frac{1}{2 \pi} \int_{-\infty}^{\infty} 1 e^{j \omega t} d \omega$$
$$\int_{-\infty}^{\infty} e^{j \omega t} d \omega=2 \pi \delta(t) \tag{1.4}$$
Interchanging variables $ t $ and $ \omega $ results in
$$\int_{-\infty}^{\infty} e^{j \omega t} d t=2 \pi \delta(\omega) \tag{1.5}$$
Using this result, the Fourier transform of the given function is
$$\mathcal{F}\left[e^{j \omega_{0} t}\right]=\int_{-\infty}^{\infty} e^{j \omega_{0} t} e^{-j \omega t} d t=\int_{-\infty}^{\infty} e^{j\left(\omega_{0}-\omega\right)} d t=2 \pi \delta\left(\omega_{0}-\omega\right)$$
Since the impulse function is an even function, with $ \delta\left(\omega_{0}-\omega\right)=\delta(\omega- \left.\omega_{0}\right) $,
$$\mathcal{F}\left[e^{j \omega_{0} t}\right]=2 \pi \delta\left(\omega-\omega_{0}\right) \tag{1.6}$$
By simply changing the sign of $ \omega_{0} $, we readily obtain
$$\mathcal{F}\left[e^{-j \omega_{0} t}\right]=2 \pi \delta\left(\omega+\omega_{0}\right) \tag{1.7}$$
Also, by setting $ \omega_{0}=0 $, $$\mathcal{F}[1]=2 \pi \delta(\omega) \tag{1.8}$$ (c) By using the result in Eqs. (1.6) and (1.7), we get
$$\begin{aligned}\mathcal{F}\left[\cos \omega_{0} t\right] &=\mathcal{F}\left[\frac{e^{j \omega_{0} t}+e^{-j \omega_{0} t}}{2}\right] \\&=\frac{1}{2} \mathcal{F}\left[e^{j \omega_{0} t}\right]+\frac{1}{2} \mathcal{F}\left[e^{-j \omega_{0} t}\right] \\&=\pi \delta\left(\omega-\omega_{0}\right)+\pi \delta\left(\omega+\omega_{0}\right)\end{aligned}$$
The Fourier transform of the cosine signal is shown in Fig. 3.
Fig. 3: Fourier transform of $f (t) = \cos ω_0 t$.

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