Given the feedback system in the figure, find $H(\omega)$ and $h(n)$.

%\begin{center}
%\includegraphics{images/feedback/feedbackdis8.ps}
%\end{center}

\begin{center}
\begin{pspicture}(0,-4)(8,2)

\rput(0,0){$x$}
\rput(7.2,0){$y$}
\rput(3.75,0){$f(n) = \alpha^n(n)$}
\rput(3.75,-2){$g(n) = \alpha\delta(n)$}
\rput(1.5,-2.5){$H$}

\psframe(1,-3)(6,1)
\psframe(2.5,-0.5)(5,0.5)
\psframe(2.5,-2.5)(5,-1.5)

% plus or minus for adder
\rput(2.1,-0.5){$-$} % input from loop
\rput(1.45,0.35){$+$} % input signal

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We know from previous examples that
$$f(n) = \alpha^nu(n) \iff F(\omega) = \frac{1}{1-e^{-i\omega}} \mbox{,}$$
and that 
$$g(n) = \alpha \delta(n) \iff G(\omega) = \alpha$$

By Black's formula, we know that $H(\omega) = \frac{F(\omega)}{1+F(\omega)G(\omega)}$. Now we can solve for $H(\omega)$:

\begin{align*}
H(\omega) &= \frac{F(\omega)}{1+F(\omega)G(\omega)} \\
H(\omega) &= \frac{ \frac{1}{1-e^{-i\omega}}}{1+ \frac{1}{1-e^{-i\omega}}\alpha} \\
H(\omega) &= \frac{1}{1 + \alpha - \alpha e^{-i\omega}} \\
H(\omega) &= \frac{1}{1+\alpha} \cdot \frac{1}{1-\frac{\alpha}{1+\alpha}e^{-i\omega}} \quad \mbox{factor out} \frac{1}{1+\alpha} \\
\end{align*}

Now we know that 
\begin{align*}
H(\omega) &= \frac{1}{1+\alpha} \cdot \frac{1}{1-\lr{\frac{\alpha}{1+\alpha}}e^{-i\omega}} \iff h(n) = \frac{1}{1+\alpha} \left( \frac{\alpha}{1+\alpha}\right)^nu(n) 
\end{align*}

We can simplify the expression to $h(t) = \frac{\alpha^n}{\left(1+\alpha\right)^{n+1}}u(n)$.

\begin{example}
Given the system in the figure, find $H(\omega)$.

%\begin{center}
%\includegraphics{images/feedback/feedback2dis8.ps}
%\end{center}

\begin{center}
\begin{pspicture}(-2,-4)(8,2)

\rput(-0.5,0){$x$}
\rput(7.7,0){$y$}
\rput(3.75,0){$F$}
\rput(3.75,-1){$G$}
\rput(0.5,-2.5){$H$}
\rput(2.65,-2){$P$}
\rput(4.85,-2){$Q$}

\psframe(0,-3)(7,1)
% for F
\psframe(3,-0.25)(4.5,0.25)
% for G
\psframe(3,-1.25)(4.5,-0.75)

% for P and Q
\psframe(2,-2.25)(3.25,-1.75)
\psframe(4.25,-2.25)(5.5,-1.75)


% plus or minus for adder
\rput(2.1,-0.5){$+$} % input from loop
\rput(1.45,0.35){$+$} % input signal

% plus or minus for adder
\rput(1.1,-0.5){$-$} % input from loop
\rput(0.45,0.35){$+$} % input signal

% input 
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% to plant
\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(2,0)(3,0)
% from plant
\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(4.5,0)(7.5,0)
% to control
\psline[linewidth=1.25 pt](5.5,0)(5.5,-1)
\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(5.5,-1)(4.5,-1)
% from control
\psline[linewidth=1.25 pt](3,-1)(1.75,-1)
% to adder from control
\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(1.75,-1)(1.75,-0.25)


% to control
\psline[linewidth=1.25 pt](6.5,0)(6.5,-2)
\psline[linewidth=1.25 pt, arrowscale=1.5]{<-}(3.25,-2)(4.25,-2)
\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(6.5,-2)(5.5,-2)
% from control
\psline[linewidth=1.25 pt](2,-2)(0.75,-2)
% to adder from control
\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(0.75,-2)(0.75,-0.25)


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\psline(1.5,0)(2,0)
\psline(1.75,0.25)(1.75,-0.25)

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\psline(0.5,0)(1,0)
\psline(0.75,0.25)(0.75,-0.25)


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\end{center}

The best approach to this is to interpret this as a two feedback systems recursively. In other words, let the overall system have a plant defined by the inner feedback system, and controller defined the the cascade of the inner feedback system with $P$ and $Q$. More abstractly,
\begin{align*}
Plant(\omega) &= \frac{F(\omega)}{1-F(\omega)G(\omega)} \\
Controller(\omega) &= P(\omega)Q(\omega)Plant(\omega) \\
\end{align*}

Thus we can solve for the overall system 
\begin{align*}
H(\omega) &= \frac{Plant(\omega)}{1+Controller(\omega)} \\
H(\omega) &= \frac{\frac{F(\omega)}{1-F(\omega)G(\omega)}}{1+P(\omega)Q(\omega)\lr{\frac{F(\omega)}{1-F(\omega)G(\omega)}}} \\
H(\omega) &= \frac{F(\omega)}{1-F(\omega)G(\omega)+F(\omega)P(\omega)Q(\omega)}
\end{align*}
\end{example}

Discussion