# Change of variables from one set of coordinates to another in Fourier

1. Jun 7, 2013

### anhnha

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I am curious to know why we have to multiply with $$e^{-j\omega t}$$ in Fourier transform? What is the purpose of this? I have heard somewhere that the transform is merely a change of variables from one set of coordinates to another.
I would like to know more about this.
Can you help me?

2. Jun 7, 2013

### MisterX

Have you studied linear algebra? It is like representing the same vector in two different basis. We can consider functions in $\mathbb{R}$ or $\mathbb{C}$ to be vectors too, although they have an infinite amount of coordinates.

Consider the vector to be $f(x)$. We can also represent this vector in the basis of functions $e^{j\omega t}$. The coordinates in this basis are $F(\omega)$, which is the Fourier transform of f(x).

$$f(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} F(\omega)e^{j\omega t} \mathop{d\omega}$$

So, why do we multiply by $e^{-j\omega t}$ to get $F(\omega)$? If I had the vector $\begin{bmatrix} 7 \\ 3 \end{bmatrix}$ and I wanted to get the x component of this vector, I could take the dot product with the x basis vector.

$$\hat{x} \cdot \begin{bmatrix} 7 \\ 3 \end{bmatrix} = \begin{bmatrix}1 & 0\end{bmatrix}\begin{bmatrix} 7 \\ 3 \end{bmatrix} = 7 + 0$$

What were are doing is taking the component of the basis vector and multiplying it by the component of the second vector and then adding this up for each component. For functions, we still multiply each component (a value of $f(x)$) by a corresponding component of the basis, but instead of adding we integrate.

But why do we multiply by $e^{-j\omega t}$ when getting the Fourier transform instead of $e^{+j\omega t}$? Well this has to do with it being a complex vector space.
Consider what happens here:

$$\begin{bmatrix}\frac{1}{\sqrt{2}} & \frac{i}{\sqrt{2}}\end{bmatrix}\begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{i}{\sqrt{2}}\end{bmatrix} = 1/2 - 1/2 = 0$$

This is a problem, since we want every vector to be some number times a unit vector in that direction. When we take the dot product of the unit vector with the vector, we should get the number. This will not work if the dot product of some (non-zero) vectors with themselves is zero.

The way to fix this is to take the complex conjugate.
$$\left(\begin{bmatrix}\frac{1}{\sqrt{2}} & \frac{i}{\sqrt{2}}\end{bmatrix}\right)^* \begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{i}{\sqrt{2}}\end{bmatrix} = \begin{bmatrix}\frac{1}{\sqrt{2}} & \frac{-i}{\sqrt{2}}\end{bmatrix}\begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{i}{\sqrt{2}}\end{bmatrix}= 1/2 + 1/2 = 1$$

3. Jun 7, 2013

### anhnha

Well, thanks a lot!
I am starting to understand it.

4. Jun 7, 2013

### mathman

The integral of a function multiplied by $$e^{-j\omega t}$$ is a definition of Fourier transform.

5. Jun 8, 2013

### anhnha

Hi, can you explain more about this? May be a link?
I intended to think about it more but I got stuck.

6. Jun 8, 2013

### MisterX

Basically if the components of vectors are complex numbers, it is necessary to use a modified version of the inner product so that it works well with concepts like length, orthogonality, and projection.

$\vec{v} = \begin{bmatrix} a_1\\ a_2 \end {bmatrix}$

$(\vec{v}^T)^* \vec{v} = a_1^*a_1 + a_2^*a_2 \geq 0$

$\left\| \vec{v}\right\| = \sqrt{a_1^*a_1 + a_2^*a_2}$

If we omitted the conjurgation, there would be more than one vector whose product with itself is zero. This is bad; we want to connect a zero product to orthogonality, and a non-zero vector should not be orthogonal to itself.

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