Consider a feedback system where our plant is the system $F(\omega) = \frac{1}{1-\frac{1}{2}e^{-i\omega}}$, and our control is defined by $G(\omega) = z^{-1}$. Also, before the adder, there is a multiplier of $\alpha$. Note that $z^{-1}$ represents a system such that:

\begin{align*}
q(t) \to &\fbox{$z^{-1}$} \to q(n-1) \\
q(t) \to &\fbox{$D_1$} \to q(n-1) \quad \mbox{basic delay}\\
\end{align*}

So in order to determine what happens, lets pass $e^{i\omega n}$ through $z^{-1}$:
$$ e^{i\omega n}  \to \fbox{$z^{-1}$} \to e^{i\omega (n-1)} = e^{-i\omega} \cdot e^{i\omega n} $$
Therefore we know the frequency response, thus $G(\omega) =  e^{-i\omega}$. Since $\alpha$ is simply multiplied before the system finishes the loop, we can just bundle that with $G(\omega)$, so $\alpha  e^{-i\omega}$ can be our $G(\omega)$.


Now we can simply apply Black's equation, 

\begin{align*}
H(\omega) &= \frac{F(\omega)}{1+G(\omega)F(\omega)} \\
H(\omega) &= \frac{ \frac{1}{1-\frac{1}{2}e^{-i\omega}}}{1-\alpha e^{-i\omega} \frac{1}{1-\frac{1}{2}e^{-i\omega}}} \\
H(\omega) &= \frac{1}{1-(\alpha + \frac{1}{2})e^{-i\omega}} \\
\end{align*}

Notice that we know have the ability to create a low-pass or high-pass filter using this feedback system. Originally, we had only a low pass filter. The system could use $\alpha$ as a parameter to determine whether the system represents a low or high pass filter. We know that $\abs{\alpha + \frac{1}{2}} \lt 1$ in order for the sum $\sum_{\R_+}\lr{\lr{\alpha + \frac{1}{2}}e^{-i\omega}}^n$ to converge (since $h(t) = \lr{\alpha + \frac{1}{2}}^n\cdot u(n)$). This tells us the range for $\alpha$ as a parameter: $\alpha \in (-\frac{3}{2},\frac{1}{2})$ for a high pass filter, and $\alpha \in (0,\frac{1}{2})$ for a low pass filter.

Feedback Systems