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

Question

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 1

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:

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

Answer 2

Suppose that you have available to you $s(t)$ as a known function of time. What this is means is that if I ask you what if the value of $s(t)$ when, say, $t = 1.23$, you can provide me with the value, say $4.56$ which is conventionally written as $s(1.23) = 4.56$. But there is nothing magical about $1.23$. I can choose any real number value for $t$ and you are able to tell me what the value of $s$ is at that point in time.

So, now I ask you for the values of $s(1.23)$ and $s(-1.23)$, and the values of $s(4.56)$ and $(-4.56)$ etc. until you are sick and tired of answering my questions, and build for myself two functions that I name as $s_e(t)$ and $s_o(t)$. Their values are as follows: $$\begin{align} s_e(1.23) &= \frac{s(1.23) + s(-1.23)}{2}\\ s_e(4.56) &= \frac{s(4.56) + s(-4.56)}{2}\\ \end{align}$$ and, more generally $\displaystyle \qquad\qquad \qquad \quad s_e(t) = \frac{s(t) + s(-t)}{2}\quad$ for each and every value of $t$. Similarly,$\displaystyle \qquad\qquad\qquad\qquad \qquad \quad s_o(t) = \frac{s(t) - s(-t)}{2}\quad$ for each and every value of $t$.
That bold-faced stuff is important: I can apply the definition even when $t$ is a negative number: there is nothing in the definition that requires that $t$ be a positive number because there is no $-$ sign in front of it in the formula. It is perfectly OK for $t$ to be $-1.23$ and $-t = 1.23$ in the formula.

These functions that I have built for myself, and which you could build for yourself too, have the following interesting properties. I know that $$s_e(1.23) = \frac{s(1.23) + s(-1.23)}{2}.$$ But, $$s_e(-1.23) = \frac{s(-1.23) + s(1.23)}{2} = s_e(1.23)$$ and more generally, $s_e(t) = s_e(-t)$ for all values of $t$. In other words, $s_e(t)$ is an even function of $t$. Similarly, it can be shown that $s_o(t)$ is an odd function of $t$.

Now that I have built my functions, I tell you about them and say
"Hey, lookit, these functions that I just made up have a very special property: $$\text{For each and every value of} ~~t, ~s(t) = s_e(t)+s_o(t). ~~\text{Isn't that interesting?"}$$