The numbers in parentheses indicate the number of marked items (1 mark
each).  Annotations in the marked PDF indicate the number of marks
deducted.  A red X "(see solutions)" indicates one mark was deducted.

                                 Quizzes

                                Quiz 1 (12)

Q1

- correct rho in dB
- correct rho in linear units
- correct probability

Q2

- correct f_m (2)
- correct tau-bar
- correct bits

Q3

- correct excess delays, dB to power conversion and normalized levels (3)
- correct mean excess delay
- correct RMS delay spread

                              Quiz 2 (9)

Q1

- use of mutual information equation
- correct joint probabilities
- correct marginal probabilities
- correct answer

Q2

- correct equation
- correct answer

Q3

- correct G(x) as bit sequence
- correct method
- correct result
  
                              Quiz 3 (12)

Q1 (3)

- sampling rate = 10 MHz
- guard time - 1000ns/100ns = 10 samples
- subcarrier spacing = 10 MHz / 256 ~= 39.1 kHz

Q2 (6)

  (lambda = 300e6/3e9 = 0.1m, 4pi d ~= 1e5)
- path loss = (1e-1/1e5)^2 = 1e-12 = 120dB
- received signal power is 30 + 15 - 120 = -75 dBm
- received noise power -174 + 2 + 62 = -110 dBm
- SNR = 35 dB
- link margin = 10 dB
- YES

Q3 (3)

- OIP3 is 40dB/2 = 20 dB above output
- need 20 dBm output
- need 40 dBm OIP3


                            Midterm Exam 1
                          (marked out of 15)

[Unlike Quiz 1, if the wrong method/equation was used or a mistake was
made in a previous part of the same question, I deducted marks for
subsequent wrong answers as well.]

Q1

- only positive frequencies
- "bathtub" shape
- correct f_m=30 Hz

Q2
- equation
- result = 0.92

Q3
- convert to threshold (-20/10)
- equation
- answer = 2.3%

Q4
- probability w/o ~0.1
- probability w/ ~0.1E-3

Q5
- equation
- wavelength & loss in linear units 2/3m, 1/4
- received power ~18nW or ~-48 dBm

Q6
- switching
- why (only receive/monitor one branch)



                            Midterm Exam 2
                          (marked out of 19)

Q1 (3)

- used AWGN capacity formula
- capacity ~= 40 Mb/s
- conclusion is "not plausible"

Q2 (3)

- n=5 k=3
- transmitted codeword = [ 1 0 1 0 1 ]
- parity check matrix = [ 1 1 1 1 0 ; 0 1 1 0 1 ]

Q3 (4)

- G = [ 1 0 0 1 ; 0 1 1 1 ]
- codewords = 0000 0111 1001 1110
- distances = 3 2 3 ; 3 2 ; 3
- minimum distance = 2

[or some other reasonable explanation for dmin = 2]

Q4 (4)

- Eb/No with coding (2) (-2 if code rate not taken into account)
- Eb/No without coding
- coding gain ~= 3 or 4.8 dB

Q5 (5)

- duration ( = 102.4 us + 2 * 10 = 122.4 us) (2)
- bps ~= 10E6 (~15 if had wrong duration)
- bandwidth ~= 6 MHz
- spectral efficiency ~= 1.7 b/s/Hz (~2.5 if had wrong duration)


                              Final Exam
Q1 (3)

- use of Rayleigh CDF (no marks deducted here if got probabilites
  "backwards" although 1 mark was deducted for wrong answer final
  answer)
- solve CDF equation for rho & compute the mean
- correct answer and units (no marks deducted if also computed correct
  RMS voltage of the envelope)

Q2 (5)

- convert distance to time
- convert to powers to linear units
- compute normalized delays
- compute excess delay spread
- OFDM guard time

Q3 (5)

- correct wavelength
- correct path loss
- correct area of circle, correct gain from Ae
- convert (or give) power to dBm
- correct power

Q4 (2)

- correct normalized weights
- correct resulting SNR

Q5 (2)

- correct capacity equation for BSC
- correct result

Q6 (2)

- a correct generator matrix
- the parity-check matrix for this generator matrix

Q7 (3)

- correct method (e.g. output level in dBm - difference/2)
- correct OIP3
- correct IIP3

Q8 (1)

- correct answer (FHSS) and explanation

Q9 (1)

- correct answer (MIMO) and explanation


                                 Labs

Lab 1 (3)

- a plot of amplitude vs time
- a histogram of the amplitude and a fit Rayleigh
- a log-log plot of path loss vs distance

[The calculation of the path loss exponent was not marked for reasons
that will be explained in a lecture.]

Lab 2 (8)

- four results (GUI and spectrum analyzer captures) (4)
  - basic
  - frequency shift
  - increased PRBS period
  - disabling the low-pass filter
- answers to questions (4):
  - one frequency equal to the sum
  - any correct differences
  - periods are 500/7 ~= 71 kHz and 500/255 ~= 2 kHz
  - no anti-alising

Lab 3 (6)

- GRC flowgraph without FEC
- GRC flowgraph with FEC
- screen capture with showing signals
- spreadsheet with data
- plot
- calculation of coding gain

Lab 4 (5)

- GRC flow graph
- frequency spectra for N=128, 32 subcarrier, 1 MHz
- " for N=256 & explanation
- " for nc=64 & explanation
- " for fs=2 MHz & explanation

Lab 5 (3)

- successful publishing of secret (2)
- screen capture of RSSI display in report (1)