2

I'm working on a system that includes a bunch of elements and I arrived at the following general expression for the output:

$$\mathcal{F}\left\{ T\cdot\left(\left[T\cdot\mathcal{F}\left\{ E_{1}\right\} \right]* h\right)\right\} $$

where \$ mathcal{F}\$ is the Fourier transform of whatever is inside of it.

From the way it looks, it's begging for me to employ the convolution theorem, and so I did: $$\mathcal{F}\left\{ T\cdot\left(\left[T\cdot\mathcal{F}\left\{ E_{1}\right\} \right]* h\right)\right\} =\mathcal{F}\left\{ T\right\} *\left(\mathcal{F}\left\{ \left[T\cdot\mathcal{F}\left\{ E_{1}\right\} \right]* h\right\} \right)=$$ $$\mathcal{F}\left\{ T\right\} *\left[\mathcal{F}\left\{ T\cdot\mathcal{F}\left\{ E_{1}\right\} \right\} \cdot\mathcal{F}\left\{ h\right\} \right]$$ $$=\mathcal{F}\left\{ T\right\} *\left[\left(\mathcal{F}\left\{ T\right\} *\mathcal{F}\left\{ \mathcal{F}\left\{ E_{1}\right\} \right\} \right)\cdot\mathcal{F}\left\{ h\right\} \right]$$

Assuming I did these steps right, I'm not sure how to carry on from here? can I make it simpler?

Thanks in advance and hope you're all safe and healthy!

FlyGuy
  • 21
  • 2
  • First, this seems like a question meant for Math Stack Exchange. Second, how would you apply twice a Fourier transform in \$ \mathcal{F}\left\{ \mathcal{F}\left\{ E_{1}\right\} \right\} \$? – jDAQ Mar 28 '20 at 19:05
  • Isn't it when the Fourier transform applied twice we get the same function $$E_1$$ but with minus the coordinate? – FlyGuy Mar 28 '20 at 19:09
  • Also, are you using \$\otimes\$ as the convolution operator? Usually represent it by \$ *\$. – jDAQ Mar 28 '20 at 19:10
  • Yes. Sorry in the book I'm reading from it uses $$\otimes$$ – FlyGuy Mar 28 '20 at 19:10
  • You are right about the result of applying twice the transform, I just had never seen it being used. https://dsp.stackexchange.com/questions/31285/repeated-fourier-transform-what-happens. The fact you are only using symbols like \$T,E,h\$ without point if they are scalars or matrices, if they are dependent on \$E(t)\$ or \$E(f)\$ might also lead to some confusion. – jDAQ Mar 28 '20 at 19:31

0 Answers0