Given a parallel system, $h$, with two systems $f$ and $g$, determine an expression for $h(t)$ and $H(\omega)$. A parallel system is a system that can be graphically described as a circuit with two systems on parallel sides going in the same direction, connecting with an adder before the output.

%\begin{center}
%\includegraphics{images/parallel/para.ps}
%\end{center}

\begin{center}
\begin{pspicture}(0,-3)(8,3)

\rput(0,0){$x(t)$}
\rput(4,1.5){$f(t)$}
\rput(4,-1.5){$g(t)$}
\rput(8.2,0){$y(t)$}
\rput(1.5,-2){$h(t)$}

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\end{center}

Let $x(t) = \delta(t)$
$$
h(t) = f(t) + g(t)
$$

Let $x(t) = e^{i\omega t}$
\begin{align*}
H(\omega)e^{i\omega t} &= F(\omega)e^{i\omega t} + G(\omega)e^{i\omega t} \\
H(\omega)e^{i\omega t} &= e^{i\omega t}\lr{F(\omega) + G(\omega)} \\
H(\omega) &= F(\omega) + G(\omega) \\
\end{align*}

\begin{example}
Consider the system defined in the frequency domain by the equation $$ H(\omega) = \frac{1 - e^{-i\omega T}}{i\omega + \alpha} $$
Write the system as a parallel system composed of two systems $F$ and $G$. Find $h(t)$ in terms of $f(t)$ and $g(t)$.

\begin{align*}
H(\omega) &= \frac{1 - e^{-i\omega T}}{i\omega + \alpha} \\
H(\omega) &= \frac{1}{i\omega + \alpha} - \frac{e^{-i\omega T}}{i\omega + \alpha}\\
\end{align*}

Here is a hint: 
\begin{align*}
H(\omega) = \frac{1}{i\omega + \alpha} &\iff h(t) = e^{-\alpha t}u(t) \\
h(t) = \delta(t) &\iff H(\omega) = 1
\end{align*}

Lets see if we can find a relationship:

\begin{align*}
H(\omega) &= \frac{1}{i\omega + \alpha} - \frac{e^{-i\omega T}}{i\omega + \alpha}\\
H(\omega) &= F(\omega) - e^{-i\omega T}F(\omega) \\
H(\omega) &= F(\omega) - G(\omega) \\
h(t) &= f(t) - g(t) \\
h(t) &= e^{-\alpha t}u(t) - g(t) \\
h(t) &= e^{-\alpha t}u(t) - f(t-T) \\
\end{align*}

Lets solve for $G(\omega)$ using $g(t)$ = $f(t-T)$:
\begin{align*}
G(\omega) &= \infint g(t) e^{-i\omega t}dt \\
G(\omega) &= \infint f(t-T) e^{-i\omega t}dt \\
G(\omega) &= \infint f(\lambda) e^{-i\omega (\lambda + T)}d\lambda \quad \mbox{let } \lambda = t - T\\
G(\omega) &= e^{-i\omega T}\infint f(\lambda) e^{-i\omega \lambda}d\lambda \\
G(\omega) &= e^{-i\omega T}F(\omega) \\
\end{align*}

Note that if we let $g(t) =  \delta(t-T)$ we get $e^{-i\omega T}$:
\begin{align*}
\infint g(t) e^{-i \omega t} dt = e^{-i\omega T} \iff g(t) = \delta(t-T)
\end{align*}

We can now make the following additions to the statements above:
\begin{align*}
H(\omega) = \frac{1}{i\omega + \alpha} &\iff h(t) = e^{-\alpha t}u(t) \\
h(t) = \delta(t) &\iff H(\omega) = 1 \\
h(t) = \delta(t-T) &\iff H(\omega) = e^{-\omega T} \\
g(t) = f(t-T) &\iff G(\omega) = e^{-i\omega T}F(\omega)
\end{align*}

So finally we have
\begin{align*}
h(t) &= f(t) - g(t) \\
h(t) &= e^{-\alpha t}u(t) - e^{-\alpha(t-T)}u(t-T) \\
\end{align*}

Now we can see that we could have defined the system like this:

% \begin{center}
% \includegraphics{images/parallel/para2.ps}
% \end{center}
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\begin{pspicture}(0,-3)(8,3)

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\rput(4,1.5){$f(t)$}
\rput(3,-1.5){$f(t)$}
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\rput(8.2,0){$y(t)$}
\rput(1.5,-2){$h(t)$}

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We can also describe the same system as a cascade of $f(t)$ and $g(t)$, where $g(t) = \delta(t) - \delta(t-T)$. 

\end{example}

Parallel