Example of linear convolution given, solve analytically by finding the regions of interests. These will be partial overlapping (both left and right end as needed) and full overlap. Example was \(x\left ( n\right ) =u\left ( n\right ) -u\left ( n-5\right ) \) and \(h\left ( h\right ) =\alpha ^{n}u\left ( n\right ) \) for \(\left \vert \alpha \right \vert <0\)
Circular convolution question will be on final exam
More definitions given: Casual system: System can’t predict input and react to it before it occurs. If system is LTI and casual, then \(h\left ( n\right ) =0\) for \(n<0\). Casual systems work in read time.
Definition of stability. BIBO. When checking for BIBO, remember to take the limit as \(n\rightarrow \infty \) or \(t\rightarrow \infty \) all the time. To check for BIBO, given a bounded input, say \(\left \vert x\left ( n\right ) \right \vert \leq M\), and use this in the convolution to find \(y\left ( n\right ) \) and see if \(y\left ( n\right ) \) will be bounded as \(n\rightarrow \infty \)
Main theory: LTI system is stable iff \[ S={\sum \limits _{n=-\infty }^{\infty }}\left \vert h\left ( n\right ) \right \vert <\infty \]
Proof was given. Need to check for both directions here as this is an iff. Proof will not be on exam.
Examples given to check for stability: \(y\left ( n\right ) ={\sum \limits _{k=0}^{n}}x\left ( k\right ) \), to solve this, let \(\left \vert x\left ( n\right ) \right \vert \leq M\), then
\begin{align*} \left \vert y\left ( n\right ) \right \vert & =\left \vert{\sum \limits _{k=0}^{n}}x\left ( k\right ) \right \vert \\ & \leq{\sum \limits _{k=0}^{n}}\left \vert x\left ( k\right ) \right \vert \\ & ={\sum \limits _{k=0}^{n}}M\\ & =\left ( n+1\right ) M \end{align*}
Now, remember to take the limit \(n\rightarrow \infty \), so we see that \(y\left ( \infty \right ) \rightarrow \infty \) hence this is not stable system, since the input was bounded, but the output is not. Another example given is \[ y\left ( n\right ) ={\sum \limits _{k=0}^{n}}a^{k}x\left ( k\right ) \]
Follow the same steps as above
\begin{align*} \left \vert y\left ( n\right ) \right \vert & =\left \vert{\sum \limits _{k=0}^{n}}a^{k}x\left ( k\right ) \right \vert \\ & \leq{\sum \limits _{k=0}^{n}}\left \vert a^{k}x\left ( k\right ) \right \vert \\ & ={\sum \limits _{k=0}^{n}}\left \vert a^{k}\right \vert \left \vert x\left ( k\right ) \right \vert \\ & \leq{\sum \limits _{k=0}^{n}}\left \vert a^{k}\right \vert M\\ & =M{\sum \limits _{k=0}^{n}}\left \vert a^{k}\right \vert \\ & =M\left ( \frac{1-a^{n+1}}{1-a}\right ) \end{align*}
Hence, only for \(\left \vert a\right \vert <1\) will the above becomes \(\frac{M}{1-a}\), ie. bounded. Hence system is table only for \(\left \vert a\right \vert <1\)
Next, introduce difference equations as a way to describe LTI discrete system, an \(N\) order LTI system is
\[{\sum \limits _{k=0}^{N}}a_{k}y\left ( n-k\right ) ={\sum \limits _{r=0}^{M}}b_{k}y\left ( n-r\right ) \]
Examples: \(y\left ( n\right ) =3x\left ( n\right ) \), \(y\left ( n\right ) =3x\left ( n\right ) +2x\left ( n-3\right ) \) and \(y\left ( n\right ) +y\left ( n-1\right ) =x\left ( n\right ) \), this last one is different since \(y\) shows twice on the LHS. To solve this last one, let \(x\left ( n\right ) =\delta \left ( n\right ) \) and find \(h\left ( n\right ) \) (which will be \(y\left ( n\right ) \) in this special case). Then go back and find \(y\left ( n\right ) \) using convolution But remember, when letting \(x\left ( n\right ) =\delta \left ( n\right ) \), we need to check for 2 cases, when \(n=0\) and when \(n\neq 0\), and we have to assume values for \(y\left ( -1\right ) \) and \(y\left ( 0\right ) \). We need additional condition to finally find \(h\left ( n\right ) \).
Another way, is to solve the difference equation using \(Z\) transform. Which is what we will probably end up doing.
End of 3rd lecture.