Solve Fourier Transform of f(x+6,y), f(x,-y), f(2x+6,y) - Help Me

In summary, the Fourier transform of the given functions can be found using a standard technique of variable change, which shows how shifts and scales in the time-domain affect the frequency-domain.
  • #1
hanafnaf
8
0
fourier transform please help me

if the Fourier transform of f(x,y) is f(u,v) what is the Fourier transform of the following:

f(x+6,y)
f(x,-y)
f(2x+6,y)

please solve it and help me
 
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  • #2


Just try substituting these into the definition

[tex]F[f(x,y)](u,v)=\int^{\infty}_{-\infty}\int^{\infty}_{-\infty}f(x,y)e^{-i(ux+vy)}dxdy[/tex]

f.e. let's do the first one:

f(x+6,y):

[tex]F[f(x+6,y)]=\int^{\infty}_{-\infty}\int^{\infty}_{-\infty}f(x+6,y)e^{-i(ux+vy)}dxdy[/tex]

I'm going to make the next change of variables:
[tex]x->\tilde{x}-6[/tex]

The boundaries of the integral remains unchanged, and so does the differential, but we get:

[tex]\int^{\infty}_{-\infty}\int^{\infty}_{-\infty}f(\tilde{x},y)e^{-i(u\tilde{x}-6u+vy)}d\tilde{x}dy[/tex]

Notice that [tex]e^{i6u}[/tex] is constant wrt to the integration, so we pull it out and what we have left of the integral is simply the Fourier transform of f(x,y) [because x* is just a dummy variable]:

[tex]e^{i6u}\int^{\infty}_{-\infty}\int^{\infty}_{-\infty}f(\tilde{x},y)e^{-i(u\tilde{x}+vy)}d\tilde{x}dy=e^{i6u}f(u,v)[/tex]

This technique of variable change is the standard technique to observe how shift & scale of the time-domain (the original function) affects the frequency-domain (its Fourier transform).
 
  • #3


i can't understand, please explain more
thanks for your attention
 
  • #4


You obviously don't understand simple substitution in integrals, maybe you should brush up on that first.
 

1. What is a Fourier Transform?

A Fourier Transform is a mathematical tool used to convert a function in the time or spatial domain to its corresponding representation in the frequency domain. It decomposes a function into its constituent frequencies and provides information about the amplitude and phase of each frequency component.

2. How do I solve a Fourier Transform?

To solve a Fourier Transform, you need to first identify the function and its variables. Then, you can use a set of mathematical equations to convert the function from the time or spatial domain to the frequency domain. This process involves integrating the function over a specific range and using complex numbers to represent the amplitude and phase of each frequency component.

3. What is the purpose of using the Fourier Transform?

The Fourier Transform is used in many fields of science and engineering to analyze and manipulate signals and data. It is particularly useful in image and signal processing, as well as in solving differential equations. It allows us to understand the frequency components of a function and how they contribute to the overall behavior of the system.

4. Can you provide an example of solving a Fourier Transform?

Sure, let's say we have a function f(x) = cos(x). To solve its Fourier Transform, we would first need to determine the range of the function, which is from 0 to 2π. Then, we would use the equation F(ω) = ∫f(x)e^(-iωx)dx to integrate the function over this range. After solving the integral, we would get F(ω) = π(δ(ω + 1) + δ(ω - 1)), where δ represents the Dirac delta function. This result tells us that the function f(x) has two frequency components, ω = 1 and ω = -1, with equal amplitudes of π.

5. How can I apply the Fourier Transform to solve practical problems?

The Fourier Transform has many practical applications, such as in image and sound processing, signal filtering, and solving differential equations. For example, in image processing, the Fourier Transform can be used to remove noise from an image by filtering out high-frequency components. In signal processing, it can be used to analyze and manipulate audio signals, such as in equalization or compression. In solving differential equations, it can be used to convert a differential equation in the time domain to an algebraic equation in the frequency domain, which is often easier to solve.

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