From kozick@charcoal.eg.bucknell.edu Wed Sep 24 10:34 EDT 1997
Date: Wed, 24 Sep 1997 10:34:26 -0400
From: Kozick Rich  <kozick@charcoal.eg.bucknell.edu>
To: gamache@bucknell.edu, louie@bucknell.edu, dkline@bucknell.edu,
        rbrooks@bucknell.edu, vary@bucknell.edu, dseiler@bucknell.edu,
        suqi@bucknell.edu, reed@bucknell.edu, mwtaylor@bucknell.edu,
        keeney@bucknell.edu, mmccrthy@bucknell.edu, blixt@bucknell.edu,
        sgordon@bucknell.edu, pankake@bucknell.edu, ganiear@bucknell.edu,
        campbell@bucknell.edu, michael@bucknell.edu, jcarey@bucknell.edu,
        pickert@bucknell.edu, sweet@bucknell.edu, bixby@bucknell.edu,
        eneff@bucknell.edu, vallone@bucknell.edu, hajdu@bucknell.edu,
        kmehaffy@bucknell.edu, pribik@bucknell.edu, ligong@bucknell.edu,
        kozick@bucknell.edu
Subject: HW for Friday

HI,

I hope you found the lecture by Profs. Sweeney and Shrivastava
interesting.  I think it was very good.  It was a nice contrast to my
plan for class today, which was solving convolution problems!

Please continue to work on your lab reports.

For Friday, please do the following:

1.  #3 on the HW 11 assignment, which is Problem 3.1 in the text.
    A hint for this problem is given below.

2.  #5 on the HW 12 assignment.  For those of you that went directly
    to room 132 and did not receive the HW 12 sheet, I have posted
    copies on the board outside my office.

(Hand-in solutions to these on Friday.)

3.  Try #4 on HW 12, which refers to #4 and #5 on HW 11.
    This will be due on Monday.  We will do examples of discrete-time
    convolution in class on Friday.  If you read the text and try
    these problems for Friday, it might help you.

I will be in Philadelphia on Thursday, but I will be available on
Friday from 11-12 and 2-6 if you'd like to talk.

Thank you.


Rich Kozick

====================================

Subject: Problem 3.1 hint

Here is that hint for Problem 3.1 that I promised:

Note that the difference equation can be written equivalently as

y[n] - 2 y[n-1] + y[n-2] = x[n-1] - x[n-2]

(Do you see why?  It might be easier to solve the problem in this
form.  However, you can certainly answer the question without changing
to this form.)

We want the output h[n] when the input x[n] = delta[n].
Thus the difference equation for h[n] is:

h[n] = 2 h[n-1] - h[n-2] + delta[n-1] - delta[n-2]

All "initial condtions" are zero.  Thus for an input x[n]=0, the
output would be y[n] = 0 for all n.

What some of you may have tried is to express the original difference
equation as:

y[n] = -y[n+2] + 2 y[n+1] + x[n+1] - x[n]

This equation is correct.  However, if you try to find y[0] you need:

y[0] = -y[2] + 2 y[1] + x[1] - x[0]

So you need FUTURE values of the output y[n], which you don't know.
The trick is to express y[n] in terms of PAST values, so that you can
iterate and use initial conditions to get started.

Thanks.


Rich Kozick