Communication Systems Tutoring

A communications course asks you to follow a message from a microphone to a loudspeaker, or from a bit stream to a radio wave and back. Each block has its own math: spectra and filters, modulation, noise, sampling, and probability. Students handle each block alone and then get lost when an exam problem chains several together. Sessions keep the block diagram on the whiteboard the whole time. For each block we ask what it does to bandwidth, power, and noise, and then do the arithmetic.

Analog topics include amplitude modulation, DSB-SC, SSB, frequency and phase modulation, Carson's rule, and receiver structures. Digital topics include sampling, quantization, PCM, line codes, ASK, FSK, PSK, QAM, matched filters, bit error rate, and an introduction to information theory and channel capacity. Sessions follow your textbook, such as Haykin, Lathi and Ding, Proakis and Salehi, or Couch, along with your professor's notes and the formulas your exams provide.

A typical AM problem, worked

An AM transmitter sends $$s(t) = 10 \, [1 + 0.5 \cos(2 \pi \times 1000 \, t)] \cos(2 \pi \times 100000 \, t)$$ volts into a 1 Ω load. Find the modulation index, the spectrum, the total power, and the efficiency.

Compare with the standard form $A_c [1 + \mu \cos(2 \pi f_m t)] \cos(2 \pi f_c t)$. Then $A_c = 10\ \text{V}$, $\mu = 0.5$, $f_m = 1\ \text{kHz}$, and $f_c = 100\ \text{kHz}$.

Expanding the product gives three spectral lines: a carrier of 10 V at 100 kHz, and two sidebands of $\mu A_c / 2 = 2.5\ \text{V}$ each, at 99 kHz and 101 kHz. The bandwidth is $2 f_m = 2\ \text{kHz}$.

Power into 1 Ω is amplitude squared over 2. The carrier carries $10^{2} / 2 = 50\ \text{W}$ and each sideband carries $2.5^{2} / 2 = 3.125\ \text{W}$, so the total is 56.25 W. Check: $P_c (1 + \mu^{2} / 2) = 50 \times 1.125 = 56.25\ \text{W}$.

Efficiency is sideband power over total power: $\dfrac{6.25}{56.25} = 11.1\%$. Only the sidebands carry the message, and that low number is one reason DSB-SC and SSB exist.

Where students usually get stuck

• Mixing up decibels and ratios. An SNR of 30 dB is a ratio of 1000, and capacity formulas need the ratio.
• Treating Carson's rule as exact. It estimates FM bandwidth as $2 (\Delta f + f_m)$.
• Putting $E_b/N_0$ in decibels straight into the $Q$ function instead of converting to a ratio first
• Confusing bit rate with symbol rate. An M-ary scheme carries $\log_2 M$ bits per symbol.
• Forgetting that content above half the sampling rate aliases unless a filter removes it first