Chapter 2 - Signal Properties

Symmetry

Worked Example 2:

Find the even and odd components of the unit pulse, and simplify.

\[ f(t) = u(t) - u(t - 1) \]

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The even and odd components are given by:

\[ \begin{cases} f_{e}(t) &= \frac{u(t + 1) - u(t - 1)}{2} \\ f_{o}(t) &= \frac{-u(t + 1) + 2u(t) - u(t - 1)}{2} \\ \end{cases} \]

where

\[ f(t) = f_{e}(t) + f_{o}(t) \]

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We'll solve this directly using the formula.

For the even component, we have:

\[ \begin{align*} f_{e}(t) &= \tfrac{1}{2}\left[f(t) + f(-t)\right] \\ &= \tfrac{1}{2}\left[u(t) - u(t - 1) + u(-t) - u(-t-1)\right] \\ \end{align*} \]

We know that \(u(t) + u(-t) = 1\).

\[ \begin{align*} f_{e}(t) &= \tfrac{1}{2}\left[1 - u(-t-1) - u(t - 1)\right] \\ \end{align*} \]

We also know that:

\[ 1 - u(-(t - a)) = u(t - a) \]

So we have

\[ \begin{align*} f_{e}(t) &= \tfrac{1}{2}\left[1 - u(-t-1) - u(t - 1)\right] \\ &= \tfrac{1}{2}\left[\left(1 - u(-(t + 1))\right) - u(t - 1)\right] \\ &= \tfrac{1}{2}\left[u(t + 1) - u(t - 1)\right] \end{align*} \]

Now for the odd component:

\[ \begin{align*} f_{o}(t) &= \tfrac{1}{2}\left[f(t) - f(-t)\right] \\ &= \tfrac{1}{2}\left[\left(u(t) - u(t - 1)\right) - \left(u(-t) - u(-t-1)\right)\right] \\ &= \tfrac{1}{2}\left[u(t) - u(t - 1) - u(-t) + u(-t - 1)\right] \\ \end{align*} \]

We know that \(u(t) - u(-t) = 2u(t) - 1\)

\[ \begin{align*} f_{o}(t) &= \tfrac{1}{2}\left[2u(t) - 1 - u(t - 1) + u(-t - 1)\right] \\ \end{align*} \]

Then using \(1 - u(-(t - 1)) = u(t - 1)\) from before:

\[ \begin{align*} f_{o}(t) &= \tfrac{1}{2}\left[2u(t) - 1 - u(t - 1) + u(-t - 1)\right] \\ &= \tfrac{1}{2}\left[- \left(1 - u(-(t + 1))\right) + 2u(t) - u(t - 1) \right] \\ &= \tfrac{1}{2}\left[- u(t + 1) + 2u(t) - u(t - 1)\right] \end{align*} \]