As shown earlier, the \emph{convolution sum} can be defined as 

\begin{nicebox}
\begin{align*}
y(n) &= \sum \limits _{m \in \Z} x(m)h(n-m) = (x * h)(n)
\end{align*}
\end{nicebox}

or using the commutative property of convolution we can write 

\begin{nicebox}
\begin{align*}
y(n) &=  \sum \limits _{k \in \Z} h(k)x(n-k) = (h * x)(n)
\end{align*}
\end{nicebox}

where $h(n)$ denotes the \emph{impulse response} of systems $H$ to the input signal $x(n)$, defined as 
$$x(n) =\cdots + x(-1)\delta(n+1) + x(0)\delta(n) + x(1)\delta(n-1) + \cdots \mbox{,}$$
and the output signal of the LTI system $y(n)$ is defined as 
$$ y(n) =\cdots + x(-1)h(n+1) + x(0)h(n) + x(1)h(n-1) + \cdots \mbox{.} $$

Note that $x * \delta = \delta * x = x$. In other words, $\delta(n)$ is the identity element of the convolution sum.

\begin{problem}
Determine the output signal of the following:
%\begin{figure}[htp]
%\centering
%\includegraphics{images/conv50.ps}
%\caption{ $h(n) = \delta(n-50)$, $x(n)=\delta(n) + 2\delta(n-1)$ }
%\label{ fig:conv50 }
%\end{figure}

$h(n) = \delta(n-50)$, $x(n)=\delta(n) + 2\delta(n-1)$

\begin{center}
\begin{pspicture}(-3,-1)(3,4.3)
	% line
\psline{-*}(0,0)(0,1)
%  value
\rput(0.25,1){(1)}
%  position
\rput(0,-0.4){50}
% axes
\psline{->}(-3,0)(3,0)

\rput(3,-0.3){ $n$ }
\rput(2.5,4){ $x(n)$ }
\rput(3,1.7){ $n$ }
\rput(2.5,1.2){ $h(n)$ }


\psline{-*}(0.000,2)(0.000,3)
\psline{-*}(1.000,2)(1.000,4)
%  value
\rput(0.25,3){(1)}
\rput(1.25,4){(2)}
%  position
\rput(0,1.6){0}
\rput(1,1.6){1}
% axes
\psline{->}(-3,2)(3,2)

	
\end{pspicture}
\end{center}

$h(n) = \delta(n-50)$, $\quad x(n)=\delta(n) + 2\delta(n-1)$.

\end{problem}

\begin{solution}
There are many ways to look at this, here we go over an algebraic approach:
\item $h(n) = \delta(n-50)$, $\quad x(n)=\delta(n) + 2\delta(n-1)$.


\begin{align*}
y(n) &= \sum \limits _{m \in \Z} x(m)h(n-m) = (x * h)(n) \\
y(n) &= x(0)h(n) + x(1)h(n-1) \\
y(n) &= 1\cdot h(n) + 2\cdot h(n-1) \\
y(n) &= \delta(n-50) + 2\delta(n-51) 
\end{align*}


This shows us that $y(n) = x(n-50) = \delta(n-50) + 2\delta(n-51)$. Note that this could have been done with one term if we use the formula $y(n) = (h * x)(n) = \sum_{k\in\Z} h(k)x(n-k)$. In that form we could have done $h(50)x(n-50) = \delta(n-50) + 2\delta(n-51)$. Here the system can be represented as $ x(n) \to \fbox{ $\delta(n-50)$} \to x(n-50) = y(n) $.

%\begin{figure}[htp]
%\centering
%\includegraphics{images/conv50ans.ps}
%\caption{ $y(n) = x(n-50)$ }
%\label{ fig:conv50ans }
%\end{figure}

\begin{center}
\begin{pspicture}(-1,-1)(3,3)
	% line
\psline{-*}(0.000,0)(0.000,1)
%  value
\rput(0.25,1){(1)}
%  position
\rput(0.000,-0.4){50}
% line
\psline{-*}(1.000,0)(1.000,2)
%  value
\rput(1.25,2){(2)}
%  position
\rput(1.000,-0.4){51}
% axes
\psline{->}(-1.5,0)(3,0)
\rput(2.5,-0.3){ $n$ }
\rput(2,2){ $y(n)$ }
	
\end{pspicture}
\end{center}

\end{solution}

\begin{problem}
$h(n) = \delta(n) + \delta(n-1)$, $\quad x(n)=\delta(n) + 2\delta(n-1)$.
\end{problem}

\begin{solution}
Here is our system: $$ x(n) \to \fbox{ $\delta(n) + \delta(n-1)$} \to x(n) + x(n-1) = y(n) $$

\begin{align*}
y(n) &= \sum \limits _{m \in \Z} x(m)h(n-m) = (x * h)(n) \\
y(n) &= x(0)h(n) + x(1)h(n-1) \\
y(n) &= 1\cdot h(n) + 2\cdot h(n-1) \\
y(n) &= \left[ \delta(n) + \delta(n-1) \right] + 2 \left[ \delta(n-1) + \delta(n-2) \right] \\
y(n) &= \delta(n) + 3\delta(n-1) + 2\delta(n-2)
\end{align*}
\end{solution}

Here is a cool method for solving a convolution problem in discrete time:

Let the impulse response $h(n) = 5\delta(n) + 3\delta(n-1) +5\delta(n-2) + 2\delta(n-3)$ be represented as a vector, 
$
\left[
\begin{array}{c}
5 \\
3 \\
5 \\
2
\end{array}
\right]
$.

Now, let the input signal $x(n) = 3\delta(n) + 2\delta(n-1) + \delta(n-2)$ be represented as a matrix:
$
\left[
\begin{array}{cccc}
3 & 0 & 0 & 0 \\
2 & 3 & 0 & 0 \\
1 & 2 & 3 & 0 \\
0 & 1 & 2 & 3 \\
0 & 0 & 1 & 2 \\
0 & 0 & 0 & 1 
\end{array}
\right]
$. 

Then multiply the matrices:

$
\left[
\begin{array}{cccc}
3 & 0 & 0 & 0 \\
2 & 3 & 0 & 0 \\
1 & 2 & 3 & 0 \\
0 & 1 & 2 & 3 \\
0 & 0 & 1 & 2 \\
0 & 0 & 0 & 1 
\end{array}
\right]
\cdot
\left[
\begin{array}{c}
5 \\
3 \\
5 \\
2
\end{array}
\right]
=
\left[
\begin{array}{c}
15 \\
19 \\
26 \\
19 \\
9 \\
2
\end{array}
\right]
= 15\delta(n) + 19\delta(n-1) + 26\delta(n-2) + 19\delta(n-3) + 9\delta(n-4) + 2\delta(n-5)
$


Convolution