- #1

binbagsss

- 1,265

- 11

**Ok so I've seen the convolution theorem written as:**

**F**(h(x)[itex]\otimes[/itex]g(x))=H(k)G(k)

*(And this is how it appears when I have a quick google).*

*My book then does a problem in which is uses:*

**F**(h(x)g(x))=H(k)[itex]\otimes[/itex]G(k)

*Where H(k)=*

and

**F**(h(x)) and similarly G(k)=**F**(g(x)),and

**F**represents a Fourier transform**My question**

- I can't see how these are equivalent at all?

**Many Thanks to anyone who can help shed some light !**