Convolution in Frequency Domain

In summary, the Fourier transform of the signal x(t) = sin(t) -pi<=t<=pi can be found by using the FT table and the FT properties. The FT table shows that the frequency convolution is x1(t)*x2(t)==> (1/2pi)X1(\omega) CONVOLVED WITH X2(\omega), while the Fourier transform of the signal x(t) = rect(t/\tau) can be found by using the FT table and the FT properties. The FT table shows that the Fourier transform of the signal x(t) = sin(t) * rect(t/2pi)
  • #1
f00lishroy
6
0

Homework Statement


Find the Fourier transform of the following signal JUST by using the FT table and the FT properties

x(t) = sin(t) -pi<=t<=pi
0 otherwise

NOTE: I am using CONVOLVED WITH as a substitute for * (the real convolution operator) because I cannot express multiplication in any other way that I know of).

Homework Equations



Frequency Convolution : x1(t)*x2(t) ==> (1/2pi)X1([itex]\omega[/itex]) CONVOLVED WITH X2([itex]\omega[/itex])

Fourier Transforms:
sin(ω0t) ===> (pi/j)[δ(ω-ω0)-δ(ω+ω0)]

rect(t/[itex]\tau[/itex]) ===> [itex]\tau[/itex]sinc([itex]\omega[/itex][itex]\tau[/itex]/2pi)

The Attempt at a Solution



I know that x(t) can be expressed as a sine multiplied with a rectangle function

x(t) = sin(t) * rect(t/2pi)

so by the frequency convolution property, X(ω)= (1/2pi)[[itex]\frac{pi}{j}[/itex][δ(ω-1)+δ(ω+1)] CONVOLVED WITH 2pi*sinc(ω)]My problem is how to do the convolution. Is it easier than it looks? I don't know how the imaginary (j) factors into the convolution. Can someone help explain how to simplify X(ω)? Thank you
 
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  • #2
I get from the problem statement that you're not supposed to doany convolving.

How about using u(t-π) and u(t+π) along with sin(t) and forgetting about your rectangle function?
 
  • #3
My teacher actually said that using the rectangle function was the correct method of doing it, but didn't elaborate. All he said was set x(t) = sin(t)*rect(t/2pi)
 
  • #4
f00lishroy said:
My teacher actually said that using the rectangle function was the correct method of doing it, but didn't elaborate. All he said was set x(t) = sin(t)*rect(t/2pi)


EDIT:
OK, then, how about the fact that

f(ω)**δ(ω-ω0) = f(ω-ω0) where I use ** to denote convolution. You have delta functions in your transform of sin(t); these are easily eliminated by using this theorem in convolving with the transform of the rectangular time function. The rest is very messy algebra.
 
Last edited:
  • #5
.First, let's rewrite the Fourier transform of a sine function using the properties of the Fourier transform:

sin(ω0t) = (1/2j)[exp(jω0t) - exp(-jω0t)]

Now, let's rewrite the rectangle function using the properties of the Fourier transform:

rect(t/\tau) = τsinc(ωτ/2π)

Substituting these into the expression for X(ω) and using the frequency convolution property, we get:

X(ω) = (1/2π) * (1/2j)[exp(jω) - exp(-jω)] * 2πsinc(ω)

= (1/4jπ) * [exp(jω) - exp(-jω)] * sinc(ω)

= (1/4jπ) * [2j*sin(ω)] * sinc(ω)

= (1/2π) * sin(ω) * sinc(ω)

= (1/2π) * sin(ω) * (τsinc(ωτ/2π))

= (τ/2π) * sin(ω) * sinc(ωτ/2π)

So, the final Fourier transform of x(t) is given by:

X(ω) = (τ/2π) * sin(ω) * sinc(ωτ/2π)

I hope this helps!
 

1. What is convolution in the frequency domain?

Convolution in the frequency domain is a mathematical operation that combines two signals by multiplying their Fourier transforms. It is used to analyze the frequency components of a signal and is commonly used in signal processing and image processing.

2. How is convolution in the frequency domain different from convolution in the time domain?

In the time domain, convolution is performed by multiplying the two signals together and integrating over time. In the frequency domain, convolution is performed by multiplying the Fourier transforms of the two signals and then taking the inverse Fourier transform of the result. This can be a more efficient method for certain types of signals.

3. What are the advantages of using convolution in the frequency domain?

Convolution in the frequency domain can be more efficient and accurate for certain types of signals, particularly those with a lot of high frequency components. It can also make certain types of calculations, such as filtering, easier to perform.

4. What types of signals can be analyzed using convolution in the frequency domain?

Convolution in the frequency domain can be used to analyze any type of signal that can be represented in the frequency domain, such as audio signals, images, and even multidimensional data.

5. How is convolution in the frequency domain used in real-world applications?

Convolution in the frequency domain is used in a variety of real-world applications, including audio and image processing, radar and sonar systems, and medical imaging. It is also commonly used in computer vision and machine learning algorithms for tasks such as image recognition and object detection.

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