Chapter 2 - Signal Properties¶
Symmetry¶
Worked Example 3:¶
Find the even and odd components a half-wave rectified sine wave, and simplify.
\[
f(t) =
\begin{cases}
\sin{\left(\tfrac{2\pi t}{T}\right)},\quad & 0\leq t \leq \frac{T}{2} \\
0,\quad & \frac{T}{2} < t < T
\end{cases}
\qquad f(t + T) = f(t)
\]
Show answer
The even and odd components are given by:
\[
\begin{cases}
f_{e}(t) &= \frac{1}{2} \left|\sin{\left(\frac{2\pi t}{T}\right)}\right| \\
f_{o}(t) &= \frac{1}{2} \sin{\left(\frac{2\pi t}{T}\right)} \\
\end{cases}
\]
where
\[
f(t) = f_{e}(t) + f_{o}(t)
\]
Show solution
We'll solve this directly using the formula.
\[
\begin{align*}
f_{e}(t) &= \tfrac{1}{2}\left[f(t) + f(-t)\right] \\
f_{o}(t) &= \tfrac{1}{2}\left[f(t) + f(-t)\right] \\
\end{align*}
\]
Let's break this into the two cases:
Case 1: \(0 < t < \tfrac{T}{2}\)¶
\[
\Rightarrow f(t) = \sin{\left(\tfrac{2\pi t}{T}\right)},\quad f(-t) = 0
\]
For the even component
\[
\begin{align*}
f_{e}(t) &= \tfrac{1}{2}\left[f(t) + f(-t)\right] \\
&= \tfrac{1}{2}\left[\sin{\left(\tfrac{2\pi t}{T}\right)} + 0 \right] \\
&= \tfrac{1}{2}\sin{\left(\tfrac{2\pi t}{T}\right)}
\end{align*}
\]
For the odd component
\[
\begin{align*}
f_{o}(t) &= \tfrac{1}{2}\left[f(t) - f(-t)\right] \\
&= \tfrac{1}{2}\left[\sin{\left(\tfrac{2\pi t}{T}\right)} - 0 \right] \\
&= \tfrac{1}{2}\sin{\left(\tfrac{2\pi t}{T}\right)}
\end{align*}
\]
Case 2: \(\tfrac{T}{2} < t < T\)¶
\[
\Rightarrow f(t) = 0, \quad f(-t) = \sin{\left(-\tfrac{2\pi t}{T}\right)} = -\sin{\left(\tfrac{2\pi t}{T}\right)}
\]
For the even component
\[
\begin{align*}
f_{e}(t) &= \frac{1}{2}\left[f(t) + f(-t)\right] \\
&= \frac{1}{2}\left[0-\sin{\left(\tfrac{2\pi t}{T}\right)}\right] \\
&= -\frac{1}{2}\sin{\left(\tfrac{2\pi t}{T}\right)}
\end{align*}
\]
For the odd component
\[
\begin{align*}
f_{o}(t) &= \frac{1}{2}\left[f(t) - f(-t)\right] \\
&= \frac{1}{2}\left[0 + \sin{\left(\tfrac{2\pi t}{T}\right)}\right] \\
&= \frac{1}{2}\sin{\left(\tfrac{2\pi t}{T}\right)}
\end{align*}
\]
Combining Cases¶
Now, we can combine the two cases together:
For the even component
\[
f_{e}(t) =
\begin{cases}
\frac{1}{2}\sin{\left(\tfrac{2\pi t}{T}\right)},\quad & 0 < t < \tfrac{T}{2} \\
-\frac{1}{2}\sin{\left(\tfrac{2\pi t}{T}\right)},\quad & \tfrac{T}{2} < t < T \\
\end{cases}
\qquad f_{e}(t + T) = f_{e}(t)
\]
This can be written as:
\[
f_{e}(t) = \frac{1}{2}\left| \sin{\left(\frac{2\pi t}{T}\right)}\right|
\]
For the odd component
\[
f_{o}(t) =
\begin{cases}
\frac{1}{2}\sin{\left(\tfrac{2\pi t}{T}\right)},\quad & 0 < t < \tfrac{T}{2} \\
\frac{1}{2}\sin{\left(\tfrac{2\pi t}{T}\right)},\quad & \tfrac{T}{2} < t < T \\
\end{cases}
\qquad f_{e}(t + T) = f_{e}(t)
\]
Clearly this is
\[
f_{o}(t) = \frac{1}{2}\sin{\left(\frac{2\pi t}{T}\right)}
\]
So to summarise, we have:
\[
\begin{cases}
f_{e}(t) &= \frac{1}{2} \left|\sin{\left(\frac{2\pi t}{T}\right)}\right| \\
f_{o}(t) &= \frac{1}{2} \sin{\left(\frac{2\pi t}{T}\right)} \\
\end{cases}
\]