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We now return to the Fourier transform. Before actually computing the Fourier transform of some functions, we prove a few of the properties of the Fourier transform.
First we note that there are several forms that one may encounter for the Fourier transform. In applications functions can either be functions of time, \(f(t)\), or space, \(f(x)\). The corresponding Fourier transforms are then written as \[\hat{f}(\omega)=\int_{-\infty}^{\infty} f(t) e^{i \omega t} d t\label{eq:1}\] or \[\hat{f}(k)=\int_{-\infty}^{\infty} f(x) e^{i k x} d x\label{eq:2}\] \(\omega\) is called the angular frequency and is related to the frequency \(v\) by \(\omega=\) \(2 \pi v\). The units of frequency are typically given in Hertz (Hz). Sometimes the frequency is denoted by \(f\) when there is no confusion. \(k\) is called the wavenumber. It has units of inverse length and is related to the wavelength, \(\lambda\), by \(k=\frac{2 \pi}{\lambda}\).
We explore a few basic properties of the Fourier transform and use them in examples in the next section.
- Linearity: For any functions \(f(x)\) and \(g(x)\) for which the Fourier transform exists and constant \(a\), we have \[F[f+g]=F[f]+F[g]\nonumber \] and \[F[a f]=a F[f] .\nonumber \] These simply follow from the properties of integration and establish the linearity of the Fourier transform.
- Transform of a Derivative: \(F\left[\frac{d f}{d x}\right]=-i k \hat{f}(k)\)
Here we compute the Fourier transform (9.3.5) of the derivative by inserting the derivative in the Fourier integral and using integration by parts. \[\begin{align} F\left[\frac{d f}{d x}\right] &=\int_{-\infty}^{\infty} \frac{d f}{d x} e^{i k x} d x\nonumber \\ &=\lim _{L \rightarrow \infty}\left[f(x) e^{i k x}\right]_{-L}^{L}-i k \int_{-\infty}^{\infty} f(x) e^{i k x} d x\label{eq:3} \end{align}\] The limit will vanish if we assume that \(\lim _{x \rightarrow \pm \infty} f(x)=0\). The last integral is recognized as the Fourier transform of \(f\), proving the given property. - Higher Order Derivatives: \(F\left[\frac{d^{n} f}{d x^{n}}\right]=(-i k)^{n} \hat{f}(k)\)
The proof of this property follows from the last result, or doing several integration by parts. We will consider the case when \(n=2\). Noting that the second derivative is the derivative of \(f^{\prime}(x)\) and applying the last result, we have \[\begin{align} F\left[\frac{d^{2} f}{d x^{2}}\right] &=F\left[\frac{d}{d x} f^{\prime}\right]\nonumber \\ &=-i k F\left[\frac{d f}{d x}\right]=(-i k)^{2} \hat{f}(k) .\label{eq:4} \end{align}\] This result will be true if \[\lim _{x \rightarrow \pm \infty} f(x)=0 \text { and } \lim _{x \rightarrow \pm \infty} f^{\prime}(x)=0 .\nonumber \] The generalization to the transform of the \(n\)th derivative easily follows. - Multiplication by \(x: F[x f(x)]=-i \frac{d}{d k} \hat{f}(k)\)
This property can be shown by using the fact that \(\frac{d}{d k} e^{i k x}=i x e^{i k x}\) and the ability to differentiate an integral with respect to a parameter. \[\begin{align} F[x f(x)] &=\int_{-\infty}^{\infty} x f(x) e^{i k x} d x\nonumber \\ &=\int_{-\infty}^{\infty} f(x) \frac{d}{d k}\left(\frac{1}{i} e^{i k x}\right) d x\nonumber \\ &=-i \frac{d}{d k} \int_{-\infty}^{\infty} f(x) e^{i k x} d x\nonumber \\ &=-i \frac{d}{d k} \hat{f}(k) .\label{eq:5} \end{align}\] This result can be generalized to \(F\left[x^{n} f(x)\right]\) as an exercise. - Shifting Properties: For constant \(a\), we have the following shifting properties: \[\begin{align} f(x-a) & \leftrightarrow e^{i k a} \hat{f}(k),\label{eq:6} \\ f(x) e^{-i a x} & \leftrightarrow \hat{f}(k-a) .\label{eq:7} \end{align}\] Here we have denoted the Fourier transform pairs using a double arrow as \(f(x) \leftrightarrow \hat{f}(k)\). These are easily proven by inserting the desired forms into the definition of the Fourier transform (9.3.5), or inverse Fourier transform (9.3.6). The first shift property \(\eqref{eq:6}\) is shown by the following argument. We evaluate the Fourier transform. \[F[f(x-a)]=\int_{-\infty}^{\infty} f(x-a) e^{i k x} d x .\nonumber \] Now perform the substitution \(y=x-a\). Then, \[\begin{align} F[f(x-a)] &=\int_{-\infty}^{\infty} f(y) e^{i k(y+a)} d y\nonumber \\ &=e^{i k a} \int_{-\infty}^{\infty} f(y) e^{i k y} d y\nonumber \\ &=e^{i k a} \hat{f}(k) .\label{eq:8} \end{align}\] The second shift property \(\eqref{eq:7}\) follows in a similar way.
- Convolution of Functions: We define the convolution of two functions \(f(x)\) and \(g(x)\) as \[(f\ast g)(x)=\int_{-\infty}^\infty f(t)g(x-t)dx.\label{eq:9}\] Then, the Fourier transform of the convolution is the product of the Fourier transforms of the individual functions: \[F[f\ast g]=\hat{f}(k)\hat{g}(k).\label{eq:10}\] We will return to the proof of this property in Section 9.6.
Fourier Transform Examples
In this section we will compute the Fourier transforms of several functions.
Example \(\PageIndex{1}\)
Find the Fourier transform of a Gaussian, \(f(x)=e^{-a x^{2} / 2}\).
Solution
This function, shown in Figure \(\PageIndex{1}\) is called the Gaussian function. It has many applications in areas such as quantum mechanics, molecular theory, probability and heat diffusion. We will compute the Fourier transform of this function and show that the Fourier transform of a Gaussian is a Gaussian. In the derivation we will introduce classic techniques for computing such integrals.
We begin by applying the definition of the Fourier transform, \[\hat{f}(k)=\int_{-\infty}^{\infty} f(x) e^{i k x} d x=\int_{-\infty}^{\infty} e^{-a x^{2} / 2+i k x} d x .\label{eq:11}\]
The first step in computing this integral is to complete the square in the argument of the exponential. Our goal is to rewrite this integral so that a simple substitution will lead to a classic integral of the form \(\int_{-\infty}^{\infty} e^{\beta y^{2}} d y\), which we can integrate. The completion of the square follows as usual: \[\begin{align} -\frac{a}{2} x^{2}+i k x &=-\frac{a}{2}\left[x^{2}-\frac{2 i k}{a} x\right]\nonumber \\ &=-\frac{a}{2}\left[x^{2}-\frac{2 i k}{a} x+\left(-\frac{i k}{a}\right)^{2}-\left(-\frac{i k}{a}\right)^{2}\right]\nonumber \\ &=-\frac{a}{2}\left(x-\frac{i k}{a}\right)^{2}-\frac{k^{2}}{2 a} .\label{eq:12} \end{align}\]
We now put this expression into the integral and make the substitutions \(y=\) \(x-\frac{i k}{a}\) and \(\beta=\frac{a}{2}\). \[\begin{align} \hat{f}(k) &=\int_{-\infty}^{\infty} e^{-a x^{2} / 2+i k x} d x\nonumber \\ &=e^{-\frac{k^{2}}{2 a}} \int_{-\infty}^{\infty} e^{-\frac{a}{2}\left(x-\frac{i k}{a}\right)^{2}} d x\nonumber \\ &=e^{-\frac{k^{2}}{2 a}} \int_{-\infty-\frac{i k}{a}}^{\infty-\frac{i k}{a}} e^{-\beta y^{2}} d y .\label{eq:13} \end{align}\]
One would be tempted to absorb the \(-\frac{i k}{a}\) terms in the limits of integration. However, we know from our previous study that the integration takes place over a contour in the complex plane as shown in Figure \(\PageIndex{2}\).
In this case we can deform this horizontal contour to a contour along the real axis since we will not cross any singularities of the integrand. So, we now safely write \[\hat{f}(k)=e^{-\frac{k^{2}}{2 a}} \int_{-\infty}^{\infty} e^{-\beta y^{2}} d y .\nonumber \]
The resulting integral is a classic integral and can be performed using a standard trick. Define \(I\) by\(^{1}\) \[I=\int_{-\infty}^{\infty} e^{-\beta y^{2}} d y .\nonumber \] Then, \[I^{2}=\int_{-\infty}^{\infty} e^{-\beta y^{2}} d y \int_{-\infty}^{\infty} e^{-\beta x^{2}} d x .\nonumber \] Note that we needed to change the integration variable so that we can write this product as a double integral: \[I^{2}=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\beta\left(x^{2}+y^{2}\right)} d x d y .\nonumber \] This is an integral over the entire \(x y\)-plane. We now transform to polar coordinates to obtain \[\begin{align} I^{2} &=\int_{0}^{2 \pi} \int_{0}^{\infty} e^{-\beta r^{2}} r d r d \theta\nonumber \\ &=2 \pi \int_{0}^{\infty} e^{-\beta r^{2}} r d r\nonumber \\ &=-\frac{\pi}{\beta}\left[e^{-\beta r^{2}}\right]_{0}^{\infty}=\frac{\pi}{\beta} .\label{eq:14} \end{align}\] The final result is gotten by taking the square root, yielding \[I=\sqrt{\frac{\pi}{\beta}} .\nonumber \]
We can now insert this result to give the Fourier transform of the Gaussian function: \[\hat{f}(k)=\sqrt{\frac{2 \pi}{a}} e^{-k^{2} / 2 a} .\label{eq:15}\] Therefore, we have shown that the Fourier transform of a Gaussian is a Gaussian.
Note
Here we show \[\int_{-\infty}^{\infty} e^{-\beta y^{2}} d y=\sqrt{\frac{\pi}{\beta}} .\nonumber \] Note that we solved the \(\beta=1\) case in Example 5.4.1, so a simple variable transformation \(z=\sqrt{\beta} y\) is all that is needed to get the answer. However, it cannot hurt to see this classic derivation again.
Example \(\PageIndex{2}\)
Find the Fourier transform of the Box, or Gate, Function, \[f(x)=\left\{\begin{array}{ll}b,&|x|\leq a \\ 0,&|x|>a\end{array}\right. .\nonumber\]
Solution
This function is called the box function, or gate function. It is shown in Figure \(\PageIndex{3}\). The Fourier transform of the box function is relatively easy to compute. It is given by \[\begin{align}\hat{f}(k)&=\int_{-\infty}^\infty f(x)e^{ikx}dx\nonumber \\ &=\int_{-a}^a be^{ikx}dx\nonumber \\ &=\left.\frac{b}{ik}e^{ikx}\right|_{-a}^a\nonumber \\ &=\frac{2b}{k}\sin ka.\label{eq:16}\end{align}\] We can rewrite this as \[\hat{f}(k)=2ab\frac{\sin ka}{ka}≡2ab\text{sinc }ka.\nonumber\] Here we introduced the sinc function \[\text{sinc }x=\frac{sin x}{x}.\nonumber\] A plot of this function is shown in Figure \(\PageIndex{4}\). This function appears often in signal analysis and it plays a role in the study of diffraction.
We will now consider special limiting values for the box function and its transform. This will lead us to the Uncertainty Principle for signals, connecting the relationship between the localization properties of a signal and its transform.
- \(a \rightarrow \infty\) and \(b\) fixed
In this case, as a gets large the box function approaches the constant function \(f(x)=b\). At the same time, we see that the Fourier transform approaches a Dirac delta function. We had seen this function earlier when we first defined the Dirac delta function. Compare Figure \(\PageIndex{4}\) with Figure 9.3.1. In fact, \(\hat{f}(k)=b D_{a}(k)\). [Recall the definition of \(D_{\Omega}(x)\) in Equation (9.3.10).] So, in the limit we obtain \(\hat{f}(k)=2 \pi b \delta(k)\). This limit implies fact that the Fourier transform of \(f(x)=1\) is \(\hat{f}(k)=2 \pi \delta(k)\). As the width of the box becomes wider, the Fourier transform becomes more localized. In fact, we have arrived at the important result that \[\int_{-\infty}^\infty e^{ikx}=2\pi\delta (k).\label{eq:17}\] - \(b \rightarrow \infty, a \rightarrow 0\), and \(2 a b=1\).
In this case the box narrows and becomes steeper while maintaining a constant area of one. This is the way we had found a representation of the Dirac delta function previously. The Fourier transform approaches a constant in this limit. As a approaches zero, the sinc function approaches one, leaving \(\hat{f}(k) \rightarrow 2 a b=1\). Thus, the Fourier transform of the Dirac delta function is one. Namely, we have \[\int_{-\infty}^\infty\delta (x)e^{ikx}=1.\label{eq:18}\] In this case we have that the more localized the function \(f(x)\) is, the more spread out the Fourier transform, \(\hat{f}(k)\), is. We will summarize these notions in the next item by relating the widths of the function and its Fourier transform. - The Uncertainty Principle, \(\Delta x \Delta k=4 \pi\).
The widths of the box function and its Fourier transform are related as we have seen in the last two limiting cases. It is natural to define the width, \(\Delta x\) of the box function as \[\Delta x=2 a \text {. }\nonumber \] The width of the Fourier transform is a little trickier. This function actually extends along the entire \(k\)-axis. However, as \(\hat{f}(k)\) became more localized, the central peak in Figure \(\PageIndex{4}\) became narrower. So, we define the width of this function, \(\Delta k\) as the distance between the first zeros on either side of the main lobe as shown in Figure \(\PageIndex{5}\). This gives \[\Delta k=\frac{2 \pi}{a} .\nonumber \] Combining these two relations, we find that \[\Delta x \Delta k=4 \pi \text {. }\nonumber \] Thus, the more localized a signal, the less localized its transform and vice versa. This notion is referred to as the Uncertainty Principle. For general signals, one needs to define the effective widths more carefully, but the main idea holds: \[\Delta x \Delta k \geq c>0 .\nonumber \]
We now turn to other examples of Fourier transforms.
Example \(\PageIndex{3}\)
Find the Fourier transform of \(f(x)=\left\{\begin{array}{cc}e^{-a x}, & x \geq 0 \\ 0, & x<0\end{array}, a>0\right.\)
Solution
The Fourier transform of this function is \[\begin{align} \hat{f}(k) &=\int_{-\infty}^{\infty} f(x) e^{i k x} d x\nonumber \\ &=\int_{0}^{\infty} e^{i k x-a x} d x\nonumber \\ &=\frac{1}{a-i k} .\label{eq:19} \end{align}\]
Next, we will compute the inverse Fourier transform of this result and recover the original function.
Note
More formally, the uncertainty principle for signals is about the relation between duration and bandwidth, which are defined by \(\Delta t=\frac{\|t f\|_{2}}{\|f\|_{2}}\) and \(\Delta \omega=\frac{\|\omega \hat{f}\|_{2}}{\|f\|_{2}}\), respectively, where \(\|f\|_{2}=\int_{-\infty}^{\infty}|f(t)|^{2} d t\) and \(\|\hat{f}\|_{2}=\frac{1}{2 \pi} \int_{-\infty}^{\infty}|\hat{f}(\omega)|^{2} d \omega\). Under appropriate conditions, one can prove that \(\Delta t \Delta \omega \geq \frac{1}{2}\). Equality holds for Gaussian signals. Werner Heisenberg (1901-1976) introduced the uncertainty principle into quantum physics in 1926 , relating uncertainties in the position \((\Delta x)\) and momentum \(\left(\Delta p_{x}\right)\) of particles. In this case, \(\Delta x \Delta p_{x} \geq \frac{1}{2} \hbar .\) Here, the uncertainties are defined as the positive square roots of the quantum mechanical variances of the position and momentum.
Example \(\PageIndex{4}\)
Find the inverse Fourier transform of \(\hat{f}(k)=\frac{1}{a-i k}\).
Solution
The inverse Fourier transform of this function is \[f(x)=\frac{1}{2\pi}\int_{-\infty}^\infty \hat{f}(k)e^{-ikx}dk=\frac{1}{2\pi}\int_{-\infty}^\infty\frac{e^{-ikx}}{a-ik}dk.\nonumber\]
This integral can be evaluated using contour integral methods. We evaluate the integral \[I=\int_{-\infty}^\infty \frac{e^{-ixz}}{a-iz}dz,\nonumber\] using Jordan’s Lemma from Section 8.5.8. According to Jordan’s Lemma, we need to enclose the contour with a semicircle in the upper half plane for \(x < 0\) and in the lower half plane for \(x > 0\) as shown in Figure \(\PageIndex{6}\).
The integrations along the semicircles will vanish and we will have \[\begin{align} f(x) &=\frac{1}{2 \pi} \int_{-\infty}^{\infty} \frac{e^{-i k x}}{a-i k} d k\nonumber \\ &=\pm \frac{1}{2 \pi} \oint_{C} \frac{e^{-i x z}}{a-i z} d z\nonumber \\ &=\left\{\begin{array}{cc}0,& x<0 \\-\frac{1}{2 \pi} 2 \pi i \operatorname{Res}[z=-i a], & x<0\end{array}\right.\nonumber \\ &=\left\{\begin{array}{cc} 0, & x<0 \\ e^{-a x}, & x>0 \end{array}\right.\label{eq:20} \end{align}\]
Note that without paying careful attention to Jordan’s Lemma one might not retrieve the function from the last example.
Example \(\PageIndex{5}\)
Find the inverse Fourier transform of \(\hat{f}(\omega)=\pi \delta\left(\omega+\omega_{0}\right)+\) \(\pi \delta\left(\omega-\omega_{0}\right)\).
Solution
We would like to find the inverse Fourier transform of this function. Instead of carrying out any integration, we will make use of the properties of Fourier transforms. Since the transforms of sums are the sums of transforms, we can look at each term individually. Consider \(\delta\left(\omega-\omega_{0}\right)\). This is a shifted function. From the shift theorems in Equations \(\eqref{eq:6}\)-\(\eqref{eq:7}\) we have the Fourier transform pair \[e^{i \omega_{0} t} f(t) \leftrightarrow \hat{f}\left(\omega-\omega_{0}\right) .\nonumber \] Recalling from Example \(\PageIndex{2}\) that \[\int_{-\infty}^{\infty} e^{i \omega t} d t=2 \pi \delta(\omega),\nonumber \] we have from the shift property that \[F^{-1}\left[\delta\left(\omega-\omega_{0}\right)\right]=\frac{1}{2 \pi} e^{-i \omega_{0} t} .\nonumber \]
The second term can be transformed similarly. Therefore, we have \[F^{-1}\left[\pi \delta\left(\omega+\omega_{0}\right)+\pi \delta\left(\omega-\omega_{0}\right]=\frac{1}{2} e^{i \omega_{0} t}+\frac{1}{2} e^{-i \omega_{0} t}=\cos \omega_{0} t .\right.\nonumber \]
Example \(\PageIndex{6}\)
Find the Fourier transform of the finite wave train. \[f(t)=\left\{\begin{array}{cl} \cos \omega_{0} t, & |t| \leq a \\ 0, & |t|>a \end{array} .\right.\nonumber \]
Solution
For the last example, we consider the finite wave train, which will reappear in the last chapter on signal analysis. In Figure \(\PageIndex{7}\) we show a plot of this function.
A straight forward computation gives \[\begin{align} \hat{f}(\omega)&=\int_{-\infty}^{\infty} f(t) e^{i \omega t} d t\nonumber \\ &=\int_{-a}^{a}\left[\cos \omega_{0} t+i \sin \omega_{0} t\right] e^{i \omega t} d t\nonumber \\ &=\int_{-a}^{a} \cos \omega_{0} t \cos \omega t d t+i \int_{-a}^{a} \sin \omega_{0} t \sin \omega t d t\nonumber \\ &=\frac{1}{2} \int_{-a}^{a}\left[\cos \left(\left(\omega+\omega_{0}\right) t\right)+\cos \left(\left(\omega-\omega_{0}\right) t\right)\right] d t\nonumber \\ &=\frac{\sin \left(\left(\omega+\omega_{0}\right) a\right)}{\omega+\omega_{0}}+\frac{\sin \left(\left(\omega-\omega_{0}\right) a\right)}{\omega-\omega_{0}} .\label{eq:21}\end{align}\]
