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?



2 Answers 2


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


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:

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

  • $\begingroup$ Aren't you assuming that the signal $s(t)$ can be expressed as the sum of an even function $s_e(t)$ and an odd function $s_o(t)$ which is exactly what needs to be proved? $\endgroup$ Commented Jun 13, 2014 at 18:32
  • $\begingroup$ @DilipSarwate: Can't see anything wrong with this answer. OP want's to understand how come $s_e(t)=\frac{s(t)+s(-t)}{2}$, is that right? Hence simple and correct explanation. If you want proof then you can always refer for example to this post (in fact reasoning is basically the same). Most important part is to use the properties of odd and even functions - rest is only juggling with variables. $\endgroup$
    – jojeck
    Commented Jun 13, 2014 at 18:45
  • $\begingroup$ Errr no, in the proof that you link to, the author first constructs the signals $s_e(t)$ and $s_o(t)$ from the given $s(t)$, shows that their sum is $s(t)$, and then proves that $s_e(t)$ is even and $s_o(t)$ is odd. He then also proves that the decomposition of $s(t)$ into the sum of an even function and an odd function is unique. Even ignoring this last part as unnecessary nitpicking, that is a reasonable proof. You begin in the middle by assuming that $s(t)$ is the sum of an even and an odd function (which is to be proved) and then show how the two functions are related to $s(t)$. $\endgroup$ Commented Jun 13, 2014 at 18:55
  • $\begingroup$ @DilipSarwate: I would like you to answer once again for my first question. Because in my opinion it is 'no'. After all I didn't noticed any word 'proof' in my post (do you?) - it is explanation. $\endgroup$
    – jojeck
    Commented Jun 13, 2014 at 19:07

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: $$\text{For each and every value of} ~~t, ~s(t) = s_e(t)+s_o(t). ~~\text{Isn't that interesting?"}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.