fourier transform - Properties of Spectral Transformations - Allocation (decomposition into even and odd part)

I am trying to understand the Allocation property of Spectral Transformations. I can't.

I know that every function can be separated into an even part and into an odd part.

My problem is understanding proof for that:

$$s(t)=s_e(t)+s_o(t)$$ where:

• $s_e(t)$ - even part of function


• $s_o(t)$ - odd part of function

The problem is that in next step at every site I looked so far, it goes something like this:

$$s_e(t) =\dfrac{s(t)+s(-t)}{2}$$

Why did they do this? Where did $\frac{1}{2}$ come from, and why is $s(t)+s(-t)$

Can you please explain it to me, or point me to a good resource where I could read more about it?

Thanks!!!

Answer

Let's say that your signal is composed of two parts: even and odd:

$$s(t)=s_e(t)+s_o(t)$$

We also know following properties of this type of functions:

• Even: $f(-x)=f(x)$



• Odd: $f(-x)=-f(x)$

Let's calculate the time inversion of your signal $s(-t)$ and apply above properties:

$$s(-t)=s_e(-t)+s_o(-t)=s_e(t)-s_o(t)$$

So now let's do the trick and add the: $s(t)$ and $s(-t)$:

$$\require{cancel} s(t)+s(-t) = \color{blue}{s_e(t)}+ \cancel{\color{red}{s_o(t)}} + \color{blue}{s_e(t)}-\cancel{\color{red}{s_o(t)}}=\color{blue}{2s_e(t)}$$

Solve it for $\color{blue}{s_e(t)}$, and you will get:

$$\boxed{\color{blue}{s_e(t)}=\dfrac{s(t)+s(-t)}{2}}$$

In the end let's subtract $s(t)$ and $s(-t)$:

$$s(t)-s(-t) = \cancel{\color{blue}{s_e(t)}} + \color{red}{s_o(t)} - \cancel{\color{blue}{s_e(t)}} + \color{red}{s_o(t)}=\color{red}{2s_o(t)}$$

Rearrange and you will get:

$$\boxed{\color{red}{s_o(t)}=\dfrac{s(t)-s(-t)}{2}}$$