# complex convolution

Started by May 11, 2005
```Rune Allnor wrote:
(snip)

> While I suspect the conjugate/no conjugate issue could be
> a matter of convention, an appealing argument in favour of
> no conjugates is that the frequency-domain expression
> becomes

>     Y(w) = X(w)H(w)

> where Y(w), X(w) and H(w) are Fourier transforms of complex-
> valued y(t), x(t) and h(t), respectively.

Well, it could also be written as Y(w) = X(w)H*(w)

My guess is that in the end it will be just as obvious as the sign
rule for multiply and divide, though I can't follow the argument
all the way through right now.

-- glen

```
```Thanks for all replies. Is it mean that convolution of two complex
sequences processes is really similar with the convolution of two real
sequences? If so, might I use matrix reprentation of complex
convolution as Toeplitz matrix as well as representation matrix of real
convolution?

Many thanks.

Regard,
Estdev

```
```estdev wrote:
> Thanks for all replies. Is it mean that convolution of two complex
> sequences processes is really similar with the convolution of two
real
> sequences?

Yes.

> If so, might I use matrix reprentation of complex
> convolution as Toeplitz matrix as well as representation matrix of
real
> convolution?

Yes.

Rune

```
```Hi Glen,

You wrote that we can write complex convolution as Y(w) = X(w)H*(w). It
means H(w) and X(w) are complex number. So we convolve two complex
sequences. Is it correct?

Then how to calculate those convolution? I mean we have real part and
imaginary part. please explain to me?

Thanks.

regard,
estdev

```
```Hi Brad,

Do you mean we can calculate real and imaginary part convolution
separately? So what is the convolution result/output will be? real form
or also in complex form?

Thanks.

regard,
estdev

```
```estdev wrote:

> You wrote that we can write complex convolution as Y(w) = X(w)H*(w). It
> means H(w) and X(w) are complex number. So we convolve two complex
> sequences. Is it correct?

> Then how to calculate those convolution? I mean we have real part and
> imaginary part. please explain to me?

For real convolution the Fourier transform of the convolution is the
product of the transforms of the two functions being convolved.

The question, then, is what happens for complex convolution.  Someone
suggested that if the product still applies that would define the
complex convolution.  I didn't go all the way through the math, but
suggested that it might be the product of one with the conjugate of
the other.  It seems that it might have the right symmetry, but
I didn't test it.

-- glen

```
```On 12 May 2005 06:51:45 -0400, Randy Yates
<randy.yates@sonyericsson.com> wrote:

>"Matt Timmermans" <mt0000@sympatico.nospam-remove.ca> writes:
>
>> "Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message
>> news:9qKdnap81OQ9Ox_fRVn-oQ@centurytel.net...
>> > Tim,
>> >
>> > I think convolution is the simplest form with no conjugates.
>> > Correlation uses the conjugate of one of the arguments.
>>
>> Yes, but now the OP needs an authoritative reference:
>>
>> http://mathworld.wolfram.com/Cross-Correlation.html
>
>A good, authoritative hardcopy reference on the subject is
>
>@BOOK{bracewell,
>  title = "{The Fourier Transform and Its Applications}",
>  author = "{Ronald~N.~Bracewell}",
>  publisher = "McGraw-Hill",
>  edition = "second, revised",
>  year = "1986"}

Ah yes,
Bracewell's book is a bit hard to read,
but there's much info there if you keep
"digging away" at it.  And that book has the most
beautiful artwork I have ever seen in a book.

See Ya',
[-Rick-]

```