However it is easy to see that $\hat{y}[n] \neq y_1[n]$ (in fact, $\hat{y}[n] = \ldots, 0, 0, \underline{4}, 0, 1, 1, 0, 0, \ldots$). As a consequence, $\mathcal{H}$ is not linear.
\item $\delta[n] = x_2[n+3]$ so that, because of time invariance, $\mathcal{H}\{\delta[n]\} = y_2[n+3] = \ldots, 0, 0, \underline{2}, 1, 0, 0, \ldots$
From this we can write the input/output relation in the $z$ domain
\[
Y(z) = D(z)[X(z) + \alpha z^{-1} Y(z)]
\]
from which we obtain the transfer function
\[
H(z) = \frac{D(z)}{1-\alpha z^{-1}D(z)}
\]
If we write $D(z)$ as a ratio of polynomials, i.e. $D(z) = B(z)/A(z)$, we finally obtain
\[
H(z) = \frac{B(z)}{B(z)-\alpha z^{-1}A(z)}
\]
From the figure, it is immediate to see that $D(z)$ is an (incomplete) second order section in direct form II, incomplete since it has a single zero. Its transfer function is therefore
\item There is a zero in $z = 5/4$ and two poles in $z = \pm \sqrt{1/2}$
\begin{center}
\begin{dspPZPlot}[width=5cm,clabel={$1$}]{1.8}
\dspPZ[type=zero,label=none]{1.25,0}
\dspPZ[label=none]{-.7,0}
\dspPZ[label=none]{0.7,0}
\end{dspPZPlot}
\end{center}
\item since pole and zero on the positive real axis are almost equidistant from one, their effects cancel each other out; the pole in $z = -\sqrt{1/2}$ brings the magnitude of the frequency response up to create a highpass characteristic:
\item The inverse transfer function is not stable because the zero of $H(z)$ is outside the unit circle. By choosing
\[
G(z) = \frac{1 -(1/2)z^{-2}}{(5/4) - z^{-1}}
\]
the product G(z)H(z) is the allpass term $(1 - (5/4)z^{-1})/((5/4) - z^{-1})$ whose frequency response magnitude is one.
\item If we remove the feedback branch, the transfer function becomes $H(z) = D(z)$. The poles of $D(z)$ are larger than one in magnitude ($|z_{1,2}| = |1 \pm j| = \sqrt{2}$) and so the system would not be stable.
\end{enumerate}
\end{exercise}
\begin{exercise}{}
\begin{enumerate}
\item The easiest way to prove the result is to invoke the convolution theorem. In the $z$-transform domain, the filter's output can be expressed as $Y(z) = H(z)X(z)$. If $x[n]$ is finite-support and $h[n]$ is FIR, then both $X(z)$ and $H(z)$ are finite-degree polynomials in $z$ and so their product is also finite-degree.
Alternatively, you can consider the convolution sum for a FIR with support over $[M_1,M_2]$:
If $x[n]$ is zero outside of $[N_1,N_2]$, then all the terms in the sum will be zero for $n> N_2+M_2$ and for $n<N_1+M_1$.
\item The simplest example is to have an input that exactly cancels the filter's poles. As an IIR filter, take a simple leaky integrator with transfer function
\[
H(z) = \frac{1}{1-\lambda z^{-1}}
\]
and consider the finite-support input $x[n] = \ldots, 0, 0, \underline{1}, -\lambda, 0, 0, \ldots$. We have that $Y(z) = H(z)X(z) = 1$ so that $y[n] = \delta[n]$, which is finite-support.
\item Just as in oversampling, the idea is to send the signal more slowly as to occupy less bandwidth. If the signal has a smaller bandwidth, we can increase its amplitude without exceeding the power constraint, which will allow us to have a better SNR over the band of interest.
the downsampler stretches the signal to full band again but reduces the spectral ``amplitude'' by a factor of 3, which means the psd gets divided by 9:
Whenever we have a sum of the form $\sum_{n=-\infty}^{\infty} x[n]$ it is always useful to see if we can compute the DTFT of $x[n]$ since the sum is equal to the value of the DTFT in $\omega = 0$