CIS 457 Midterm Guide


Specifications:

The midterm exam will consist of 4 short answer questions (40%), 2 calculation questions (20%), and approximately 20 (multiple choice / fill in blank / true-false questions) (40%). For short answer questions, I am expecting approximately 1 paragraph (several sentences) in response. Answers that are just a couple words will generally not recieve full credit.

You will be allowed to bring and use a calculator and a single page of notes. Your laptop or phone is not an acceptable calculator (calculators are available for students to borrow from the math center and from the university library). Your page of notes may be no larger than a standard letter sized piece of paper. This page must be produced by you (you can not share a page of notes someone else has written), and will be collected at the end of the exam.

Topics

Any topic covered in lecture/discussion is valid material for the exam. The topic list below is not guaranteed to be comprehensive.

Sample calculations

Compute the checksum of 10101010, 11001100, and 11100011

 10101010
+11001100
---------
101110110

 101110110
 +11100011
----------
1001011001

Now, we need to add the carry back in

01011001
      10
--------
01011011

The checksum that actually gets sent is these bits flipped, so
10100100



Now, let's try a CRC, using generator 1101. The data we will use is 11001100.

We need to divide, but we don't care about the result, only the
remainder. Remember we need to add extra 0s to the end, in this case 3
because the generator is 4 digits.

     ---------------
1101 | 11001100 000
       1101
          1110
          1101
            110 0
            110 1
                100

So, our remainder is 100. What gets transmitted is 11001100100

The reciever calculates 
     -------------
1101 | 11001100100
       1101
          1110
          1101
            1101
            1101
               000

Since the remainder is 0, the reciever concludes there are no errors.




Now, an example for data rate calcualations. We have a binary signal on a 3 kHz channel. The
signal/noise ratio is 20dB.

Applying Nyquist's theorm, we get data rate = 2 * 3k * log2(2) bits/sec
 = 2 * 3k * 1 = 6000 bps

Applying Shannon's theorm, we get data rate = 3k * log2(1+100)
bits/sec (because 20dB is a S/N ratio of 100) 
= 3K * log2(101) = 3K * 6.658 = 199974 bps

Because the Nyquist limit is lower, it is the limiting factor in our
data rate, so the maximum data rate is 6000 bps or 6kbps.