# fundamental question

Started by September 12, 2006
```A very basic question:

Take two properly sampled signals (more than nyquist). Now I mulitply
the two sample streams. From a continous time view point it is easy to
see that the product could have frequencies for which the initial
sampling rate wouldnt be enough.

So in any DSP system is one not supposed to multiply two signals? Has
this something to do with whether multiplication with a variable value
is a linear operation or not? Is the solution that one has to up-sample
before doing such an operation to a rate that nyquist for the product
signal is met before multiplication?

- b

```
```bulk wrote:

>  Is the solution that one has to up-sample
> before doing such an operation to a rate that nyquist for the product
> signal is met before multiplication?
>
> - b

Correct.

```
```> Take two properly sampled signals (more than nyquist). Now I mulitply
> the two sample streams. From a continous time view point it is easy to
> see that the product could have frequencies for which the initial
> sampling rate wouldnt be enough.
>
> So in any DSP system is one not supposed to multiply two signals? Has
> this something to do with whether multiplication with a variable value
> is a linear operation or not? Is the solution that one has to up-sample
> before doing such an operation to a rate that nyquist for the product
> signal is met before multiplication?

multiplication in the time domain is the same as convolution in the
frequency domain.

anyway.......

try multipling sin(2*pi*f1*t)*sin(2*pi*f2*t) and you get

1/2*cos(2*pi*[f1-f2]*t)-1/2*cos(2*pi*[f1+f2]*t)

Lets say that f2=fs/2 and f1=fs/2 where fs is the sampling frequency.

Then:

Muliplying the two sine-components yields a DC-component (0Hz) and
another frequency component at fs Hz.

```
```
Joe wrote:
>
> > Take two properly sampled signals (more than nyquist). Now I mulitply
> > the two sample streams. From a continous time view point it is easy to
> > see that the product could have frequencies for which the initial
> > sampling rate wouldnt be enough.
> >
> > So in any DSP system is one not supposed to multiply two signals? Has
> > this something to do with whether multiplication with a variable value
> > is a linear operation or not? Is the solution that one has to up-sample
> > before doing such an operation to a rate that nyquist for the product
> > signal is met before multiplication?
>
> multiplication in the time domain is the same as convolution in the
> frequency domain.
>
> anyway.......
>
> try multipling sin(2*pi*f1*t)*sin(2*pi*f2*t) and you get
>
> 1/2*cos(2*pi*[f1-f2]*t)-1/2*cos(2*pi*[f1+f2]*t)
>
> Lets say that f2=fs/2 and f1=fs/2 where fs is the sampling frequency.
>
> Then:
>
> Muliplying the two sine-components yields a DC-component (0Hz) and
> another frequency component at fs Hz.

By my calculation that gives you a lot of zeroes for an output. What was

-jim

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```
```>
> By my calculation that gives you a lot of zeroes for an output. What was

My point was to illustrate what happens in the frequency domain when you
multiply two signals
with frequency f1 and f2, respectively.

```
```"bulk" <unsettledinside@yahoo.com> wrote in message
>A very basic question:
>
> Take two properly sampled signals (more than nyquist). Now I multiply
> the two sample streams. From a continous time view point it is easy to
> see that the product could have frequencies for which the initial
> sampling rate wouldn't be enough.
>
> So in any DSP system is one not supposed to multiply two signals?
You can do whatever you want.
>Has
> this something to do with whether multiplication with a variable value
> is a linear operation or not?
I don't think so : If it was, how would that help you?
>Is the solution that one has to up-sample
> before doing such an operation to a rate that nyquist for the product
> signal is met before multiplication?

That should work in all cases.

Best of Luck - Mike

```
```
bulk wrote:

> A very basic question:
>
> Take two properly sampled signals (more than nyquist). Now I mulitply
> the two sample streams. From a continous time view point it is easy to
> see that the product could have frequencies for which the initial
> sampling rate wouldnt be enough.

This is correct.

> So in any DSP system is one not supposed to multiply two signals?

Care should be taken to avoid the problems due to the possible aliasing.

> Has
> this something to do with whether multiplication with a variable value
> is a linear operation or not?

Multiplication of two variables is bilinear.

> Is the solution that one has to up-sample
> before doing such an operation to a rate that nyquist for the product
> signal is met before multiplication?

It depends on what are you trying to accomplish.

Usually, when you are multiplying two signals, your desired result is
the "difference" product. The "sum" product can be aliased, however it
does not matter if it is suppresed in the algorithm.

DSP and Mixed Signal Design Consultant

http://www.abvolt.com

```
```
Joe wrote:

>>By my calculation that gives you a lot of zeroes for an output. What was
>
>
>
> My point was to illustrate what happens in the frequency domain when you
> multiply two signals
> with frequency f1 and f2, respectively.
>

When you multiply F1 and F2, you get F1 - F2 and F1 + F2. Usually your
goal is to obtain F1 - F2. In this case, the F1 + F2 can be aliased. As
long as the alias of F1 + F2 is outside the bandwidth of interest, it
can be filtered out.

DSP and Mixed Signal Design Consultant

http://www.abvolt.com
```
```
Joe wrote:
>
> >
> > By my calculation that gives you a lot of zeroes for an output. What was
>
> My point was to illustrate what happens in the frequency domain when you
> multiply two signals
> with frequency f1 and f2, respectively.

Well yes.. er.. maybe.... But given that you chose the one particular
example that completely side steps the question the OP asked, I was just
wondering how you expected him to draw any useful conclusions from that.
I mean, multiplying by a sin function at fs/2 is probably not something
the OP needs help with.

-jim

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```
```bulk wrote:

> A very basic question:
>
> Take two properly sampled signals (more than nyquist). Now I mulitply
> the two sample streams. From a continous time view point it is easy to
> see that the product could have frequencies for which the initial
> sampling rate wouldnt be enough.
>
> So in any DSP system is one not supposed to multiply two signals?

It's not nearly that restrictive.  But before you multiply two signals
(or run one through a nonlinear function) you should think about what
you're doing.

> Has
> this something to do with whether multiplication with a variable value
> is a linear operation or not?

No, because multiplying your signal by a time varying quantity is a
linear operation (it's just not time invariant).

> Is the solution that one has to up-sample
> before doing such an operation to a rate that nyquist for the product
> signal is met before multiplication?

If you've determined that it's going to be an issue, yes.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com