# transform signal

Hello everyone I need help solving a Fourier transform for the given signal, I know it will be a frequency convolution for the first function it will be a window function and for the second function I do not know how to convert it (I do not know how to turn it into a sinc) I would love if someone could help me with this And within the boundaries of the integration of the convolution between them what will be the boundaries? This is the exercise

$$x_6(t) = \frac{\sin(\pi t)}{\pi t}\cdot \frac{\sin(2\pi (t-1))}{\pi (t-1)}$$

I know the sinc is $$\operatorname{sinc}(t):=\frac{\sin(\pi t)}{\pi t}$$.

This is what I wrote I want to know how to find X 2

• "I can't convert it to a sinc" uh, can you write down the formula of a sinc in your question, please? Apr 22 at 9:40
• sinc(x)=sin(pix)/pix Apr 22 at 12:47
• that's not what I meant with "in your question", so I quickly added the formula to your question. OK; so, I think you need to express all that you can in your $x_6$ with sinc(t) and modifications of that yourself first. That really doesn't seem very hard, so please do that, or explain where the hardness is! Apr 22 at 13:50
• Hi Orel! (Do) you know the multiplication-convolution property of the Fourier transform? Have you tried it? It will turn out to be a convolution integral in frequency, but have you actually written it? Which part is hard for you? Please write down your steps either in handwriting (picture) or in better LaTex. Apr 22 at 13:57
• Hi thank you very much for the answer I understand the exercise, I have a multiplication between functions while it is a convolution at this frequency I understand I know that the transformation of the first function would actually be a rectangle but regarding the second transformation I can not convert it I would be happy to help I ask for help Apr 22 at 14:40

First you need to remember these 3 facts about Fourier Transform:

1. FT of sinc is rect function

$$FT[sinc(t)]=rect(w)$$

1. FT of a shifted signal is the FT of the original signal multiplied by exponent:

$$FT[x(t-t_0)]=FT[x(t)]e^{-2jwt_0\pi}$$

1. FT of a scaled signal is FT of the original divided by the scaling:

$$FT[x(at)]=\frac{1}{a}FT[x(\frac{w}{a})]$$

Now just use 1 and 3 to find the FT of $$x_1(t)$$, where $$a=\pi$$ in this case.

Then use 1, 2 and 3 to find the FT of $$x_2(t)$$, where $$a=2\pi$$ and $$t_0=1$$ in this case.

Finally, substitute them in the convolution expression you wrote in the first line: $$\frac{1}{2\pi}FT[x_1(t)]*FT[x_2(t)]$$

• Thank you very much! I really owed the feature 2 you helped me a lot. Apr 25 at 6:18