Amplitude Modulation (AM)
AM Signal Time Domain
s(t) = A_c[1 + \mu\cos(2\pi f_m t)]\cos(2\pi f_c t)
| Symbol | Description | Unit |
|---|---|---|
| A_c | Carrier amplitude | V |
| \mu | Modulation index (0 ≤ μ ≤ 1) | dimensionless |
| f_c | Carrier frequency | Hz |
| f_m | Message frequency | Hz |
Worked example
Find modulation index if A_c = 10 V, A_m = 6 V.
Given: A_c = 10 V, A_m = 6 V
- μ = A_m / A_c
- μ = 6 / 10
- μ = 0.6
Answer: μ = 0.6 (60% modulation)
AM Power Distribution
P_T = P_c\left(1 + \frac{\mu^2}{2}\right)
| Symbol | Description | Unit |
|---|---|---|
| P_T | Total AM power | W |
| P_c | Carrier power | W |
| \mu | Modulation index | dimensionless |
Worked example
AM transmitter with P_c = 1000 W, μ = 0.6. Find total power.
Given: P_c = 1000 W, μ = 0.6
- P_T = P_c × (1 + μ²/2)
- = 1000 × (1 + 0.36/2)
- = 1000 × (1 + 0.18)
- = 1000 × 1.18 = 1180 W
Answer: P_T = 1180 W (180 W in sidebands)
AM Bandwidth
BW_{AM} = 2f_m
| Symbol | Description | Unit |
|---|---|---|
| BW_{AM} | AM bandwidth | Hz |
| f_m | Maximum message frequency | Hz |
Worked example
AM radio with audio up to 5 kHz. Find channel bandwidth.
Given: f_m = 5 kHz
- BW = 2 × f_m
- = 2 × 5000
- = 10000 Hz
Answer: BW = 10 kHz
SSB Power Advantage
P_{SSB} = \frac{\mu^2 P_c}{4} = \frac{P_{USB}}{1}
| Symbol | Description | Unit |
|---|---|---|
| P_{SSB} | Single sideband power | W |
Worked example
Compare SSB power to AM for P_c = 1000 W, μ = 1.
Given: P_c = 1000 W, μ = 1
- AM total = P_c(1 + 1/2) = 1500 W
- Each sideband = P_c × μ²/4 = 1000/4 = 250 W
- SSB uses only 250 W to convey same info as 1500 W AM
Answer: SSB needs only 250 W vs 1500 W AM — 6× more efficient
Frequency Modulation (FM)
FM Modulation Index
\beta = \frac{\Delta f}{f_m} = \frac{k_f A_m}{f_m}
| Symbol | Description | Unit |
|---|---|---|
| \beta | FM modulation index | dimensionless |
| \Delta f | Maximum frequency deviation | Hz |
| f_m | Message frequency | Hz |
| k_f | Frequency sensitivity | Hz/V |
Worked example
FM: Δf = 75 kHz, f_m = 15 kHz (commercial FM). Find β.
Given: Δf = 75 kHz, f_m = 15 kHz
- β = Δf / f_m
- = 75000 / 15000
- = 5
Answer: β = 5 (wideband FM)
Carson's Rule for FM Bandwidth
BW_{FM} \approx 2(\Delta f + f_m) = 2f_m(\beta + 1)
| Symbol | Description | Unit |
|---|---|---|
| BW_{FM} | Approximate FM bandwidth | Hz |
Worked example
Find commercial FM station bandwidth with β = 5, f_m = 15 kHz.
Given: β = 5, f_m = 15 kHz
- BW = 2 × f_m × (β + 1)
- = 2 × 15000 × (5 + 1)
- = 2 × 15000 × 6
- = 180000 Hz
Answer: BW ≈ 180 kHz (commercial FM stations are 200 kHz spaced)
FM SNR Improvement
\left(\frac{SNR_o}{SNR_i}\right)_{FM} = 3\beta^2(\beta + 1)
| Symbol | Description | Unit |
|---|---|---|
| SNR_o | Output SNR | dimensionless |
| SNR_i | Input SNR (baseband reference) | dimensionless |
Worked example
Compare FM figure of merit for β = 5 vs AM (DSBSC).
Given: β = 5 for FM; figure of merit for DSBSC = 1
- FM figure = 3β²(β+1) = 3×25×6 = 450
- DSBSC figure = 1
- FM improvement = 450/1 = 450 = 26.5 dB over DSBSC
Answer: FM outperforms DSBSC by 26.5 dB in SNR for β = 5
Noise in Communication Systems
Thermal Noise Power
N = kTB
| Symbol | Description | Unit |
|---|---|---|
| N | Thermal noise power | W |
| k | Boltzmann constant = 1.38×10⁻²³ J/K | J/K |
| T | Temperature | K |
| B | Bandwidth | Hz |
Worked example
Find thermal noise power in a 1 MHz bandwidth at T = 290 K.
Given: k = 1.38e-23, T = 290, B = 1e6
- N = k × T × B
- = 1.38e-23 × 290 × 1e6
- = 1.38e-23 × 2.9e8
- = 4.002e-15 W
Answer: N ≈ 4 fW = -144 dBW = -114 dBm
Noise Figure
F = \frac{SNR_{in}}{SNR_{out}} = 1 + \frac{T_e}{T_0}
| Symbol | Description | Unit |
|---|---|---|
| F | Noise figure | dimensionless |
| T_e | Equivalent noise temperature | K |
| T_0 | Reference temperature = 290 K | K |
Worked example
LNA with noise figure NF = 2 dB. Find equivalent noise temperature.
Given: NF = 2 dB → F = 10^(2/10) = 1.585
- F = 1 + T_e/T₀
- T_e = (F-1) × T₀
- = (1.585 - 1) × 290
- = 0.585 × 290
- = 169.7 K
Answer: T_e ≈ 170 K
Friis Noise Formula (Cascaded Stages)
F_{total} = F_1 + \frac{F_2 - 1}{G_1} + \frac{F_3 - 1}{G_1 G_2} + \ldots
| Symbol | Description | Unit |
|---|---|---|
| F_1, F_2 | Noise figures of stages | dimensionless |
| G_1, G_2 | Power gains of stages | dimensionless |
Worked example
LNA: G1 = 20 dB, F1 = 1.585 (2 dB NF). Mixer: F2 = 10 (10 dB NF). Find F_total.
Given: G1 = 100 (linear), F1 = 1.585, F2 = 10
- F_total = F1 + (F2-1)/G1
- = 1.585 + (10-1)/100
- = 1.585 + 0.09
- = 1.675
Answer: F_total = 1.675 (2.24 dB) — dominated by the first stage
Digital Modulation
BPSK Probability of Error
P_e = Q\left(\sqrt{\frac{2E_b}{N_0}}\right)
| Symbol | Description | Unit |
|---|---|---|
| P_e | Bit error probability | dimensionless |
| E_b | Energy per bit | J |
| N_0 | One-sided noise PSD | W/Hz |
| Q(x) | Q-function = P(Z>x) for Z~N(0,1) | dimensionless |
Worked example
Find P_e for BPSK at E_b/N_0 = 10 dB.
Given: E_b/N_0 = 10 dB = 10 (linear)
- Argument = √(2 × E_b/N_0) = √(2 × 10) = √20 = 4.47
- P_e = Q(4.47)
- Q(4.47) ≈ 3.9 × 10⁻⁶
Answer: P_e ≈ 3.9 × 10⁻⁶
QPSK Bandwidth Efficiency
\eta = \frac{R_b}{BW} = 2 \text{ bps/Hz (ideal)}
| Symbol | Description | Unit |
|---|---|---|
| \eta | Spectral efficiency | bps/Hz |
| R_b | Bit rate | bps |
Worked example
QPSK system with 10 MHz bandwidth. Find maximum bit rate.
Given: BW = 10 MHz, η = 2 bps/Hz
- R_b = η × BW
- = 2 × 10×10⁶
- = 20×10⁶ bps
Answer: R_b = 20 Mbps
QAM Spectral Efficiency
\eta = \log_2(M) \text{ bps/Hz},\quad M = \text{constellation size}
| Symbol | Description | Unit |
|---|---|---|
| M | Number of symbols (e.g., 16 for 16-QAM) | dimensionless |
Worked example
64-QAM in 10 MHz channel. Find bit rate.
Given: M = 64, BW = 10 MHz
- η = log₂(64) = 6 bps/Hz
- R_b = 6 × 10e6
- = 60e6 bps
Answer: R_b = 60 Mbps at 6 bps/Hz efficiency
FSK Minimum Bandwidth
BW_{FSK} = (M-1)\Delta f + 2B
| Symbol | Description | Unit |
|---|---|---|
| \Delta f | Frequency spacing | Hz |
| M | Number of frequencies | dimensionless |
| B | Baseband signal bandwidth | Hz |
Worked example
Binary FSK (M=2): f1 = 1200 Hz, f2 = 2200 Hz, baud rate = 1200 baud. Find BW.
Given: Δf = 2200-1200 = 1000 Hz, B = 600 Hz (half baud rate)
- BW = (2-1)×1000 + 2×600
- = 1000 + 1200
- = 2200 Hz
Answer: BW ≈ 2.2 kHz
Information Theory
Shannon's Channel Capacity
C = B\log_2\left(1 + \frac{S}{N}\right)
| Symbol | Description | Unit |
|---|---|---|
| C | Channel capacity | bps |
| B | Bandwidth | Hz |
| S/N | Signal-to-noise ratio (linear) | dimensionless |
Worked example
Find capacity of a 3 kHz voice channel with SNR = 1000 (30 dB).
Given: B = 3000 Hz, S/N = 1000
- C = 3000 × log₂(1 + 1000)
- = 3000 × log₂(1001)
- = 3000 × 9.967
- = 29902 bps
Answer: C ≈ 29.9 kbps
Entropy of a Source
H = -\sum_{i=1}^{M} p_i \log_2 p_i \text{ bits/symbol}
| Symbol | Description | Unit |
|---|---|---|
| H | Source entropy | bits/symbol |
| p_i | Probability of symbol i | dimensionless |
Worked example
Binary source with p(0) = 0.25, p(1) = 0.75. Find entropy.
Given: p0 = 0.25, p1 = 0.75
- H = -(0.25 × log₂0.25 + 0.75 × log₂0.75)
- = -(0.25 × (-2) + 0.75 × (-0.415))
- = -((-0.5) + (-0.311))
- = -(-0.811) = 0.811 bits/symbol
Answer: H = 0.811 bits/symbol (max H = 1 bit for equal probability)
Nyquist Theorem (Noise-Free Channel Capacity)
C_{max} = 2B\log_2 M
| Symbol | Description | Unit |
|---|---|---|
| C_{max} | Maximum noiseless channel capacity | bps |
| M | Number of discrete signal levels | dimensionless |
Worked example
Noiseless channel B = 3 kHz with 8 signal levels. Find max rate.
Given: B = 3000 Hz, M = 8
- C = 2 × 3000 × log₂(8)
- = 6000 × 3
- = 18000 bps
Answer: C_max = 18 kbps
Pulse Modulation and PCM
PCM Bit Rate
R_b = f_s \cdot n
| Symbol | Description | Unit |
|---|---|---|
| R_b | PCM bit rate | bps |
| f_s | Sampling frequency | Hz |
| n | Bits per sample | bits |
Worked example
PCM for telephone: f_s = 8 kHz, 8 bits/sample. Find bit rate.
Given: f_s = 8000 Hz, n = 8
- R_b = f_s × n
- = 8000 × 8
- = 64000 bps
Answer: R_b = 64 kbps (standard G.711 PCM)
PCM Quantization Noise SNR
SNR_{PCM} \approx 6.02n + 1.76 \text{ dB}
| Symbol | Description | Unit |
|---|---|---|
| n | Number of bits per sample | bits |
Worked example
Find SNR for 8-bit PCM.
Given: n = 8
- SNR = 6.02 × 8 + 1.76
- = 48.16 + 1.76
- = 49.92 dB
Answer: SNR ≈ 50 dB for 8-bit PCM
TDM Frame Efficiency
\text{Efficiency} = \frac{n \cdot N}{n \cdot N + F} \times 100\%
| Symbol | Description | Unit |
|---|---|---|
| n | Bits per channel slot | bits |
| N | Number of channels | dimensionless |
| F | Framing bits per frame | bits |
Worked example
TDM: 24 channels, 8 bits each, 1 framing bit per frame (T1 format). Find efficiency.
Given: N = 24, n = 8, F = 1
- Total bits per frame = n×N + F = 8×24 + 1 = 193
- Data bits = 192
- Efficiency = 192/193 × 100
- = 99.48%
Answer: Efficiency = 99.48% (T1 carrier)
Spread Spectrum and Multiple Access
Processing Gain (CDMA)
PG = \frac{BW_{spread}}{R_b} = \frac{f_c}{f_m}
| Symbol | Description | Unit |
|---|---|---|
| PG | Processing gain | dimensionless |
| BW_{spread} | Spread spectrum bandwidth | Hz |
| R_b | Data rate | bps |
Worked example
CDMA with chip rate 1.2288 Mcps and data rate 9.6 kbps (IS-95). Find PG.
Given: BW = 1.2288e6 Hz, R_b = 9600 bps
- PG = BW / R_b
- = 1228800 / 9600
- = 128
Answer: PG = 128 = 21 dB (IS-95 CDMA standard)
OFDM Number of Subcarriers
N = \frac{BW}{\Delta f}, \quad \Delta f = \frac{1}{T_s}
| Symbol | Description | Unit |
|---|---|---|
| N | Number of OFDM subcarriers | dimensionless |
| \Delta f | Subcarrier spacing | Hz |
| T_s | OFDM symbol duration | s |
Worked example
LTE 20 MHz BW with 15 kHz subcarrier spacing. Find N.
Given: BW = 20e6 Hz, Δf = 15000 Hz
- N = BW / Δf
- = 20000000 / 15000
- = 1333 subcarriers (LTE uses 1200 active + guard)
Answer: N ≈ 1333 total; 1200 active in LTE 20 MHz
Quick reference
| Formula | Expression |
|---|---|
| AM Modulation Index | μ = A_m/A_c |
| AM Total Power | P_T = P_c(1 + μ²/2) |
| AM Bandwidth | BW = 2f_m |
| FM Modulation Index | β = Δf/f_m |
| FM Bandwidth (Carson) | BW = 2(Δf + f_m) |
| FM Figure of Merit | 3β²(β+1) |
| Thermal Noise | N = kTB |
| Noise Figure | F = SNR_in/SNR_out |
| Friis Formula | F = F1 + (F2-1)/G1 + ... |
| Shannon Capacity | C = B log₂(1 + S/N) |
| Source Entropy | H = -Σ p_i log₂ p_i |
| BPSK P_e | P_e = Q(√(2E_b/N_0)) |
| QAM Efficiency | η = log₂(M) bps/Hz |
| PCM Bit Rate | R_b = f_s × n |
| PCM SNR | SNR = 6.02n + 1.76 dB |
Exam tips
- GATE noise analysis problems specify whether SNR is pre-detection or post-detection — DSB-SC and SSB have figure of merit 1 while FM's figure 3β²(β+1) can far exceed this.
- For AM efficiency questions, efficiency = μ²/(2+μ²) × 100% — at 100% modulation (μ=1) only 33.3% of total power is in the sidebands (useful).
- Shannon capacity is an upper bound — actual systems need coding; examiners ask you to compare achieved bit rate to capacity to find spectral efficiency.
- In digital modulation BER problems, always convert E_b/N_0 from dB to linear before substituting into Q-function formulas.
- Friis noise formula shows that the first stage dominates total system noise — a high-gain low-noise first amplifier is critical, tested in receiver design problems.
- For PCM, each additional bit increases SNR by approximately 6 dB — GATE frequently asks how many bits are needed to achieve a target SNR.