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9.5: Properties of the Fourier Transform

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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

ˆf(ω)=f(t)eiωtdt

or

ˆf(k)=f(x)eikxdx

ω is called the angular frequency and is related to the frequency v by ω= 2πv. 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, λ, by k=2πλ.

We explore a few basic properties of the Fourier transform and use them in examples in the next section.

  1. and

    F[af]=aF[f].

  2. Transform of a Derivative: F[dfdx]=ikˆf(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.

    F[dfdx]=dfdxeikxdx=limL[f(x)eikx]LLikf(x)eikxdx

    The limit will vanish if we assume that limx±f(x)=0. The last integral is recognized as the Fourier transform of f, proving the given property.

  3. Higher Order Derivatives: F[dnfdxn]=(ik)nˆf(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(x) and applying the last result, we have

    F[d2fdx2]=F[ddxf]=ikF[dfdx]=(ik)2ˆf(k).

    This result will be true if

    limx±f(x)=0 and limx±f(x)=0.

    The generalization to the transform of the nth derivative easily follows.

  4. Multiplication by x:F[xf(x)]=iddkˆf(k)
    This property can be shown by using the fact that ddkeikx=ixeikx and the ability to differentiate an integral with respect to a parameter.

    F[xf(x)]=xf(x)eikxdx=f(x)ddk(1ieikx)dx=iddkf(x)eikxdx=iddkˆf(k).

    This result can be generalized to F[xnf(x)] as an exercise.

  5. Here we have denoted the Fourier transform pairs using a double arrow as f(x)ˆf(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 (???) is shown by the following argument. We evaluate the Fourier transform.

    F[f(xa)]=f(xa)eikxdx.

    F[f(xa)]=f(y)eik(y+a)dy=eikaf(y)eikydy=eikaˆf(k).

    The second shift property (???) follows in a similar way.

  6. Then, the Fourier transform of the convolution is the product of the Fourier transforms of the individual functions:

    F[fg]=ˆf(k)ˆg(k).

Fourier Transform Examples

In this section we will compute the Fourier transforms of several functions.

Example 9.5.1

Find the Fourier transform of a Gaussian, f(x)=eax2/2.

Solution

This function, shown in Figure 9.5.1 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.

clipboard_ea79fe202ba0558b8e58cfa5c18c84842.png
Figure 9.5.1: Plots of the Gaussian function f(x)=eax2/2 for a=1,2,3.

We begin by applying the definition of the Fourier transform,

ˆf(k)=f(x)eikxdx=eax2/2+ikxdx.

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 eβy2dy, which we can integrate. The completion of the square follows as usual:

a2x2+ikx=a2[x22ikax]=a2[x22ikax+(ika)2(ika)2]=a2(xika)2k22a.

We now put this expression into the integral and make the substitutions y= xika and β=a2.

ˆf(k)=eax2/2+ikxdx=ek22aea2(xika)2dx=ek22aikaikaeβy2dy.

One would be tempted to absorb the ika 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 9.5.2.

clipboard_e3afd60595f73d5c1a33f23b178e03a0b.png
Figure 9.5.2: Simple horizontal contour.

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

ˆf(k)=ek22aeβy2dy.

The resulting integral is a classic integral and can be performed using a standard trick. Define I by1

I=eβy2dy.

Then,

I2=eβy2dyeβx2dx.

Note that we needed to change the integration variable so that we can write this product as a double integral:

I2=eβ(x2+y2)dxdy.

This is an integral over the entire xy-plane. We now transform to polar coordinates to obtain

I2=2π00eβr2rdrdθ=2π0eβr2rdr=πβ[eβr2]0=πβ.

The final result is gotten by taking the square root, yielding

I=πβ.

We can now insert this result to give the Fourier transform of the Gaussian function:

ˆf(k)=2πaek2/2a.

Therefore, we have shown that the Fourier transform of a Gaussian is a Gaussian.

Note

Here we show

eβy2dy=πβ.

Note that we solved the β=1 case in Example 5.4.1, so a simple variable transformation z=βy is all that is needed to get the answer. However, it cannot hurt to see this classic derivation again.

Example 9.5.2

Find the Fourier transform of the Box, or Gate, Function,

f(x)={b,|x|a0,|x|>a.

Solution

This function is called the box function, or gate function. It is shown in Figure 9.5.3.

clipboard_e9c571480752542baaed7ba89edb306bd.png
Figure 9.5.3: A plot of the box function in Example 9.5.2.

 The Fourier transform of the box function is relatively easy to compute. It is given by

ˆf(k)=f(x)eikxdx=aabeikxdx=bikeikx|aa=2bksinka.

We can rewrite this as

ˆf(k)=2absinkaka2absinc ka.

Here we introduced the sinc function

sinc x=sinxx.

A plot of this function is shown in Figure 9.5.4. This function appears often in signal analysis and it plays a role in the study of diffraction.

clipboard_e5814d0f4fd9632c5d982f2dbedb99a14.png
Figure 9.5.4: A plot of the Fourier transform of the box function in Example 9.5.2. This is the general shape of the sinc function.

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.

  1. a 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 9.5.4 with Figure 9.3.1. In fact, ˆf(k)=bDa(k). [Recall the definition of DΩ(x) in Equation (9.3.10).] So, in the limit we obtain ˆf(k)=2πbδ(k). This limit implies fact that the Fourier transform of f(x)=1 is ˆf(k)=2πδ(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

    eikx=2πδ(k).

  2. b,a0, and 2ab=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 ˆf(k)2ab=1. Thus, the Fourier transform of the Dirac delta function is one. Namely, we have

    δ(x)eikx=1.

    In this case we have that the more localized the function f(x) is, the more spread out the Fourier transform, ˆf(k), is. We will summarize these notions in the next item by relating the widths of the function and its Fourier transform.

  3. The Uncertainty Principle, ΔxΔk=4π.
    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, Δx of the box function as

    Δx=2a

    The width of the Fourier transform is a little trickier. This function actually extends along the entire k-axis. However, as ˆf(k) became more localized, the central peak in Figure 9.5.4 became narrower. So, we define the width of this function, Δk as the distance between the first zeros on either side of the main lobe as shown in Figure 9.5.5. This gives

    Δk=2πa.

    Combining these two relations, we find that

    ΔxΔk=4π

    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:

    ΔxΔkc>0.

clipboard_e1d7c8672848ef16c4c5ffc2b7de806a2.png
Figure 9.5.5: The width of the function 2absinkaka is defined as the distance between the smallest magnitude zeros.

We now turn to other examples of Fourier transforms.

Example 9.5.3

Find the Fourier transform of f(x)={eax,x00,x<0,a>0

Solution

The Fourier transform of this function is

ˆf(k)=f(x)eikxdx=0eikxaxdx=1aik.

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 Δt=tf2f2 and Δω=ωˆf2f2, respectively, where f2=|f(t)|2dt and ˆf2=12π|ˆf(ω)|2dω. Under appropriate conditions, one can prove that ΔtΔω12. Equality holds for Gaussian signals. Werner Heisenberg (1901-1976) introduced the uncertainty principle into quantum physics in 1926 , relating uncertainties in the position (Δx) and momentum (Δpx) of particles. In this case, ΔxΔpx12. Here, the uncertainties are defined as the positive square roots of the quantum mechanical variances of the position and momentum.

Example 9.5.4

Find the inverse Fourier transform of ˆf(k)=1aik.

Solution

The inverse Fourier transform of this function is

f(x)=12πˆf(k)eikxdk=12πeikxaikdk.

This integral can be evaluated using contour integral methods. We evaluate the integral

I=eixzaizdz,

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 9.5.6.

clipboard_ee41a3f7558fedff46a6fbffb3b25ba4b.png
Figure 9.5.6: Contours for inverting ˆf(k)=1aik.

The integrations along the semicircles will vanish and we will have

f(x)=12πeikxaikdk=±12πCeixzaizdz={0,x<012π2πiRes[z=ia],x<0={0,x<0eax,x>0

Note that without paying careful attention to Jordan’s Lemma one might not retrieve the function from the last example.

Example 9.5.5

Find the inverse Fourier transform of ˆf(ω)=πδ(ω+ω0)+ πδ(ωω0).

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 δ(ωω0). This is a shifted function. From the shift theorems in Equations (???)-(???) we have the Fourier transform pair

eiω0tf(t)ˆf(ωω0).

Recalling from Example 9.5.2 that

eiωtdt=2πδ(ω),

we have from the shift property that

F1[δ(ωω0)]=12πeiω0t.

The second term can be transformed similarly. Therefore, we have

F1[πδ(ω+ω0)+πδ(ωω0]=12eiω0t+12eiω0t=cosω0t.

Example 9.5.6

Find the Fourier transform of the finite wave train.

f(t)={cosω0t,|t|a0,|t|>a.

Solution

For the last example, we consider the finite wave train, which will reappear in the last chapter on signal analysis. In Figure 9.5.7 we show a plot of this function.

A straight forward computation gives

ˆf(ω)=f(t)eiωtdt=aa[cosω0t+isinω0t]eiωtdt=aacosω0tcosωtdt+iaasinω0tsinωtdt=12aa[cos((ω+ω0)t)+cos((ωω0)t)]dt=sin((ω+ω0)a)ω+ω0+sin((ωω0)a)ωω0.


This page titled 9.5: Properties of the Fourier Transform is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Russell Herman via source content that was edited to the style and standards of the LibreTexts platform.

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