2.5 Lecture 5, Monday, Feb. 8,2010

Example of finding inverse Z transform using partial fractions. Watch out if the sequence comes out unstable. rewrite for stable and not causal i.e. \(u\left ( -n-1\right ) \) and do only for the part that is not stable., leave the other using \(u\left ( n\right ) \)

Now using one sided Z transform, the Z transform of a delayed sequence is generated. \(Z\left ( x[n-1]\right ) =z^{-1}Z\left ( x[n]\right ) +x\left ( -1\right ) \)

Next an example of a difference equation is given, with an initial condition, and the Z transform is used to solve it. Do not use \(u\left ( n\right ) \) in the answer, since initial condition is at \(n=-1\), just write \(n\geq 0\)

Next, we are given an LSI system with \(e^{j\omega n}\) as input, and proofed that output \(y\left ( n\right ) =x\left ( n\right ) H\left ( j\omega \right ) \) where \(H\left ( j\omega \right ) \) is the frequency response. it is the DTFT of \(h\left ( n\right ) \) of the system. i.e. \(H\left ( j\omega \right ) ={\sum \limits _{n=-\infty }^{\infty }}h\left ( n\right ) e^{-j\omega n}\).

\(H\left ( j\omega \right ) \) is periodic of period \(2\pi \) and we normally use \(\omega =-\pi \cdots \pi \) as the range. Note \(\omega \) is continuous and has units of radians (I thought it was in units radians/sample).

Next, we learned how to express \(H\left ( j\omega \right ) \) as \(\left \vert H\left ( j\omega \right ) \right \vert e^{j\arg \left ( H\left ( j\omega \right ) \right ) }\) to make it easier to draw the magnitude and phase diagrams of the frequency response of the system. The trick to use is to factor out \(e^{ja\omega }\) out and to end up with expression as \(A\left ( \omega \right ) e^{j\arg \left ( H\right ) }\). We are shown 2 sequences \(x\left ( n\right ) \) and asked to find its \(DTFT\) and put the result in this form. Next, the magnitude and phase diagrams are plotted. Remember the following: Since we are using absolute value on the magnitude, when \(\omega <0\) and we get a negative value for the absolute, we multiply it by -1 to get +ve, then for that region of \(\omega \) remember to add a \(\pi \) when doing the phase diagram.

HW1 was given.

This was the end of this lecture.