Question

Following the method of equations (10.8)to (10.12), show that ${\mathbf{f}}_{\mathbf{1}}{\mathbf{f}}_{\mathbf{2}}$and ${\mathbf{g}}_{\mathbf{1}}\mathbf{*}{\mathbf{g}}_{\mathbf{2}}$are a pair of Fourier transforms.

Short answer

The convolution of the two functions $g_1$and $g_2$are a pair transform

Step-by-step solution

Given information

Equations (10.8) to (10.12) ,

Fourier transformation

The Inverse Fourier transform of the function gis defined by the equation

$f\left(x\right)={\int }_{-\infty }^{\infty }g\left(\propto \right)e^{iax}d\propto ........\left(a\right)$

And the Fourier transform for the function f(x), is given by

$g\left(\propto \right)=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }f\left(x\right)e^{-iax}dx.......\left(b\right)$

And, the convolution integral of any two function that are variable in the same variable, say$h_1\left(z\right)$and$h_2\left(z\right)$, both functions variable in z , the convolution is defined by the following

$h_1*h_2={\int }_{-\infty }^{\infty }{h}_1\left(z-x\right)h_2\left(x\right)dx$

Explanation

Required to show that $g_1*g_2$and $f_1.f_2$are Fourier pair transform, which means $g_1*g_2=FT\left\{f_1.f_2\right\}$

Or, to show that the Laplace inverse of the convolution integral of the two function $g_1$and $g_2$are equal to the product of the two functions $f_1$and $f_2$

$$\begin{gathered} F{T}^{-1}\left\{g_1*g_2\right\}=F{T}^{-1}\left\{FT\left\{f_1.f_2\right\}\right\} \\ =f_1.f_2 \end{gathered}$$

Replace Z in function $h_1$with z-x , and every variable z in function $h_2$with x , such that on integration with respect to a dummy variable x , in short x is a dummy constant it can be of any symbol, but the two functions themselves are variables in z .

Thus, one must keep in mind the proper substitution for the variables for the functions in the convolution integral.

Another example for different two functions $c_1\left(y\right)$and $c_2\left(y\right)$with a different variable say y, and choose a dummy variable say k, , then

$c_1*c_2={\int }_{-\infty }^{\infty }{c}_1\left(y-k\right)c_2\left(k\right)dk$

Through this problem, the function f is the inverse of Fourier transform of the function g , and the function g is the Fourier transform of the function f .

Product of the two functions f1 and f2

There is many approach to such problem, It is proved that the inverse Fourier transform of the convolution of the two function is$g_1$and$g_2$are equal to the product of the two function$f_1$and$f_2$"which are defined by equation (a)", proved that both the convolution of the two functions$g_1$and$g_2$ , the product of the two functions$f_1$and$f_2$are a pair Fourier transform.

Thus, evaluate the right-hand side, of equation ($†$) thus

$f_1\left(X\right).f_2\left(X\right)={\int }_{-\infty }^{\infty }{g}_1\left(u\right)e^{ixu}du.{\int }_{-\infty }^{\infty }{g}_2\left(v\right)e^{ixv}dv$

Where, u and v are dummy constants, and hence

$={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }{e}^{ix\left(u+v\right)}{g}_1\left(u\right)g_2\left(v\right)dudv$

And, let$u+v=\propto$, and hence$u=\propto -v$ , and$du=d\propto$thus

$={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }{e}^{ix\left(\propto \right)}{g}_1\left(\propto -v\right)g_2\left(v\right)d\propto dv$

Change, order of integration

$={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }{e}^{ix\left(\propto \right)}{g}_1\left(\propto -v\right)g_2\left(v\right)dvd\propto$

And, as $e^{ix\propto }$is constant with respect to the integration of dv , hence factor it out of the integration, hence

$={\int }_{-\infty }^{\infty }{e}^{ix\left(\propto \right)}{\int }_{-\infty }^{\infty }{g}_1\left(\propto -v\right)g_2\left(v\right)dvd\propto$

And, from the definition of the convolution,

$={\int }_{-\infty }^{\infty }{e}^{ix\left(\propto \right)}\left(g_1\left(\propto \right)*g_2\left(\propto \right)\right)d\propto$

Step 5: Product of f1.f2 and the convolution of the two functions g1 and g2are a pair transform

And, from the inverse Fourier definition "equation (b)", regard the convolution as one function $g_3\left(\propto \right)$being inversely transformed to a function

$f_3\left(x\right)$, such that

$f_3\left(x\right)=f_1\left(x\right).f_2\left(x\right)$

And,

$g_3\left(\propto \right)=g_1\left(\propto \right)*g_2\left(\propto \right)$

And, hence compare equation (*) , and from the definition of the inverse Fourier transform,

$$\begin{gathered} f_3\left(x\right)={\int }_{-\infty }^{\infty }{g}_3\left(\propto \right)e^{ix\propto }d\propto \\ =F{T}^{-1}\left\{g_3\left(\propto \right)\right\} \end{gathered}$$

And hence

$f_1.f_2=F{T}^{-1}\left\{g_1\left(\propto \right)*g_2\left(\propto \right)\right\}$

And hence both the product of $f_1.f_2$and the convolution of the two functions $g_1$and $g_2$are a pair transform, also take the Fourier transform of both sides,

$$\begin{gathered} FT\left\{f_1.f_2\right\}=F{T}^{-1}\left\{g_1\left(\propto \right)*g_2\left(\propto \right)\right\} \\ {g}_1*g_2=FT\left\{f_1.f_2\right\} \end{gathered}$$