Marking Scheme

The items below were checked.  One mark was assigned for each item
unless otherwise indicated.

A red X indicates one mark was deducted.
                                   
                               Quizzes

                                Quiz 1

Q1:

- one mark for correct answer

Q2:

- one mark for correct answer

Q3:

- two marks if correct answer
- one mark if only one or two bits wrong

Q4:

- one mark for correct number of bytes (3)
- three marks for correct result

                                 Labs

Pre-Lab 1 (3)

- 1(a) square wave average = 0, RMS = A/2
- 1(b) sine wave average = 0, rms = 0.707 A/2
- 5(a) 12-bit A/D can resolve 4096 levels

Lab 1 (9)

- three screen captures with measurements (3)
- three histograms (3)
- table with DMM, 'scope, matlab and spreadsheet voltages for 3 rows
  (3)
- -1 if DMM noise voltage reading incorrect

Pre-Lab 2 (2)

- table with bits at correct voltages
- drawing with start and stop bits of correct polarity

Lab 2 (2)

- screen capture of octave waveform
- scope display with correct decoded value

Pre-Lab 3 (2)

- correct pin name (RxD)
- correct sample rate (8x baud rate)

Lab 3 (3)

- octave plot
- 'scope screen capture with decoding (showing part of name)
- Teraterm screen capture (showing name)

Pre-Lab 4 (2)

- 3x2 rows & columns of matrix
- transpose operator

Lab 4 (5)

- 3 scope screen captures (1)
- one plot of 3 spectra (3)
- frequency of first null for NRZ and Manchester (1)

Lab 5 (2)

- descrambler schematic
- correct waveforms

Lab 6 (4)

- script listing
- measured transfer function
- vertical gain should be V/6
- a spreadsheet was submitted

Lab 7 (5)

- a spreadsheet (in the PDF)
- two flowgraphs
- a GUI screen capture
- a BER plot with two curves
- coding gain calculations

Lab 8 (5)

- hexedit screen capture with name and all fields correct
- scope screen capture showing single-ended and differential voltages
  at start of frame
- Wireshark screen capture
- 5 MHz preamble waveform frequency
- Wireshark does not show preamble or CRC



                            Midterm Exam 1

Q1 (3)

  - symbol rate = 1 kHz or 1 MHz
  - entropy (2) 0.125*3 + 0.5*2 + .53 = 1.405 bits/message
  - information rate = 1.405 kbps or Mbps

Q2 (4)

  - correct binary representation of the code point

  - 5353 = 0101 0011 01_01 0011 => E5 8D 93
  - 96FB = 1001 0110 11_11 1011 => E9 9B BB
  
  - correct number of bytes
  - correct bytes (2)

Q3 (2)

  - DTE or DCE pin 4=>DTR=>DTE 6=>DSR=>DCE
  - direct connect to DTE? Yes.  No, null modem.

Q4 (2)

  - correct order
  - correct bits AT = 0x41 0x54 = 0100 0001 0101 0100 => 1000 0010,
    0010 1010



                            Midterm Exam 2

Q1(4):

- 1/2 mark per correct bit

Q2(4):

- -1 mark per incorrect byte

Q3(4):

- correct number of seconds or bits (2)
- correct equation for m
- correct answer

Q4(4):

- correct CRC length
- correct n, k and n-k
- correct CRC (2)

                              Final Exam

Q1 (5):

- entropy equation
- entropy result
- BSC capacity equation
- BSC capacity
- correct conclusion

Q2 (3):

- correct number of bytes
- correct value of x and y
- correct bytes including prefix

Q3 (5):

- correct baud rate
- correct parity sum and error check result
- correct bit order
- correct bit polarity
- correct result

Q4 (3):

- correct mu and sigma
- correct t
- correct probability

Q5 (2):

- differential encoding
- correct waveform

Q6 (3):

- maximum symbol rate
- bits per symbol
- maximum bit rate

Q7 (2):

- common mode voltage
- differential voltage

Q8 (5):

- correct length
- correct crc calculation method
- correct crc result
- message
- correct k and n

Q9 (5):

- distances
- minimum distance
- detectable errors
- correctable errors
- most likely word

Q10 (3):

- one per correct answer