# What is right

1. Oct 8, 2011

### matematikuvol

What is right definition?

$$(f*g)(x)=\frac{1}{\sqrt{2\pi}}\int^{\infty}_{-\infty}f(x-\xi)g(\xi)d\xi$$

or

$$(f*g)(x)=\frac{1}{2\pi}\int^{\infty}_{-\infty}f(x-\xi)g(\xi)d\xi$$

or

$$(f*g)(x)=\int^{\infty}_{-\infty}f(x-\xi)g(\xi)d\xi$$

this is for me huge problem. For example

$$f*\delta=f$$

or

$$f*2\pi\delta=f$$

or

$$f*\sqrt{2\pi}\delta=f$$

?

?

Last edited: Oct 8, 2011
2. Oct 8, 2011

### mathman

Re: Convolution

I believe it is simply a matter of convention. I am used to the without any coefficient, but as long as you are consistent, it won't matter.

3. Oct 8, 2011

### matematikuvol

Re: Convolution

Tnx for the answer. Just one more question is definition of convolution in some bond with definition of Fourier transform? So if I say

$$\mathcal{F}\{f(x)\}=\frac{1}{\sqrt{2\pi}}\int^{+\infty}_{-\infty}f(x)e^{-ikx}dx$$

and

$$\mathcal{F}\{g(x)\}=\frac{1}{\sqrt{2\pi}}\int^{+\infty}_{-\infty}g(x)e^{-ikx}dx$$

then

$$(f*g)(x)=?$$

and if I define

$$\mathcal{F}\{f(x)\}=\int^{+\infty}_{-\infty}f(x)e^{-ikx}dx$$

$$\mathcal{F}\{g(x)\}=\int^{+\infty}_{-\infty}g(x)e^{-ikx}dx$$

then

$$(f*g)(x)=?$$

4. Oct 9, 2011

### mathman

Re: Convolution

The connection to Fourier transform is essentially the Fourier transform of a convolution is the product of the Fourier transforms of the individual functions. There are three different ways to define things as you noted. If you want to avoid 2π or its square root as a coefficient for the integral, then the exponent of e must be -2πikx for the transform and 2πikx for the inverse transform.

When you leave out 2π in the exponent, then you need 1/(2π) as a coefficient for either the transform or its inverse. If you prefer symmetry, in this case, use 1/√(2π) for both.

Last edited: Oct 9, 2011
5. Oct 14, 2011

### matematikuvol

Re: Convolution

I know that. My question is does definition of convulution depends of which definition of Fourier transform is choosen?

6. Oct 14, 2011

### mathman

Re: Convolution

I suggest you try a simple example. I believe that if you use any of the definitions of transform that I described, i.e. coefficient in the transform, then you don't need any in the convolution.

7. Oct 14, 2011

### matematikuvol

Re: Convolution

Ok. But whole story for me has problems. For example

Take function $$f(x)=e^{-ax^2}$$

What is Fourier transform?

If I define

$$\mathcal{F}\{f(x)\}=\frac{1}{2\pi}\int^{\infty}_{-\infty}f(x)e^{-ikx}dx$$

then

$$\mathcal{F}\{e^{-ax^2}\}=\frac{1}{2\sqrt{\pi a}}e^{-\frac{k^2}{4a}}$$

and if I define

$$\mathcal{F}\{f(x)\}=\frac{1}{\sqrt{2\pi}} \int^{\infty}_{-\infty}f(x)e^{-ikx}dx$$

$$\mathcal{F}\{e^{-ax^2}\}=\frac{1}{\sqrt{2 a}}e^{-\frac{k^2}{4a}}$$

How is that posible? That aren't same functions. I have trouble with that. Can you please help me with some explanation?

8. Oct 15, 2011

### mathman

Re: Convolution

Your example involves two possible definitions of Fourier transform, each of which requires a consistent definition of inverse transform. In your two examples the first one needs a coefficient of 1 and the other 1/√(2π).

All that matters is that you and your professor or anyone else you communicate with agree on which definition to use.

My specialty is in probability theory. Here we put 2π in the exponent so the coefficient then is 1 for both forward and back transforms.

Last edited: Oct 15, 2011
9. Oct 15, 2011

### matematikuvol

Re: Convolution

Ok. But that means that Fourier transform of some function depends of definition. If you use one definition, and me the other our results aren't the same. I'm not happy because of that.

10. Oct 16, 2011

### mathman

Re: Convolution

I can't help that. All this means is that, when you communicate any result, make sure you describe the definition you are using.

11. Oct 16, 2011

### rbj

Re: Convolution

not all conventions are equal. electrical engineers, particularly those that worry about communications, control systems, and signal processing most commonly use these definitions (that were missed in this thread) for the Fourier transform and inverse:

$$\mathcal{F}\{x(t)\} \triangleq X(f) = \int^{+\infty}_{-\infty} x(t) \ e^{-j 2 \pi f t} \ dt$$

$$\mathcal{F}^{-1}\{X(f)\} \triangleq x(t) = \int^{+\infty}_{-\infty} X(f) \ e^{+j 2 \pi f t} \ df$$

(note the change of notation.) frequency is in Hz instead of rad/sec. this way you loose all these nasty scaling factors regarding $2 \pi$ or square root thereof. but you have to remember to put it into the exponential. the theorems for convolution,
$$(x*y)(t) \triangleq \int^{\infty}_{-\infty} x(t-u) \ y(u) \ du = \int^{\infty}_{-\infty} x(u) \ y(t-u) \ du$$
$$\mathcal{F}\{(x*y)(t)\} = X(f) \cdot Y(f)$$

Parsevals,
$$\int_{-\infty}^{\infty} |x(t)|^2 \ dt = \int_{-\infty}^{\infty} |X(f)|^2 \ df$$

area under the curve,
$$X(0) = \int^{+\infty}_{-\infty} x(t) \ dt$$
$$x(0) = \int^{+\infty}_{-\infty} X(f) \ df$$

duality,
$$\mathcal{F}\{g(t)\} = G(f) \quad \iff \quad \mathcal{F}\{G(t)\} = g(-f)$$

all come out to be quite clean and devoid of nasty scaling factors.
all this is soooo simple by using the right convention. not all conventions are of equal utility.

Last edited: Oct 16, 2011