Comm Notes
Complete guide to Signal-to-Noise Ratio covering definitions, calculations, dB representation, SNR improvement techniques, and applications in analog and digital communication systems.
Signal-to-Noise Ratio is the most important single measure of communication system quality. It quantifies the strength of the desired signal relative to the background noise and directly determines the achievable bit error rate, voice quality, or image clarity.
Definition
SNR = Signal Power / Noise Power = P_s / P_n
In decibels:
SNR(dB) = 10 × log₁₀(P_s / P_n) dB
Or equivalently using voltages (for equal impedance):
SNR(dB) = 20 × log₁₀(V_s / V_n) dB
SNR in Different Contexts
| Context | Symbol | Definition | Application |
|---|---|---|---|
| Carrier-to-Noise | C/N | Carrier power / noise power | RF link budgets |
| Eb/N₀ | E_b/N₀ | Energy per bit / noise PSD | Digital systems |
| Es/N₀ | E_s/N₀ | Energy per symbol / noise PSD | M-ary modulation |
| SINR | SINR | Signal / (interference + noise) | Cellular systems |
| SQNR | SQNR | Signal / quantization noise | ADC performance |
SNR Calculations
Basic SNR
For a signal s(t) with noise n(t), received as r(t) = s(t) + n(t):
P_s = (1/T)∫s²(t)dt (signal power) P_n = (1/T)∫n²(t)dt = N₀B (noise power in bandwidth B)
SNR = P_s / (N₀B)
Relationship Between SNR and Eb/N₀
SNR = (E_b/N₀) × (R_b/B)
Where:
- E_b = energy per bit = P_s × T_b = P_s/R_b
- N₀ = noise power spectral density
- R_b = bit rate
- B = noise bandwidth
Or: E_b/N₀ = SNR × (B/R_b)
SNR in dB Arithmetic
| Operation | Linear | dB |
|---|---|---|
| SNR = S/N | Division | Subtraction: S(dBm) - N(dBm) |
| Double signal | ×2 | +3 dB |
| Half noise | ÷2 | +3 dB |
| 10× signal | ×10 | +10 dB |
SNR Requirements for Various Systems
| Application | Required SNR | Quality Level |
|---|---|---|
| Voice (telephone) | 30 dB | Acceptable |
| FM broadcast | 50 dB | High quality |
| Analog TV | 40-45 dB | Good quality |
| CD audio | 96 dB | 16-bit digital |
| Wi-Fi (64-QAM) | 25 dB | Required minimum |
| 5G (256-QAM) | 30 dB | Required minimum |
| Deep space comm. | -1 to 3 dB (Eb/N₀) | With coding |
SNR Improvement Techniques
1. Increase Transmit Power
- Linear improvement: doubling power adds 3 dB to SNR
- Limited by regulations, power consumption, interference
2. Reduce Bandwidth (Filtering)
- Matched filter achieves optimal SNR
- Narrower bandwidth = less noise power
- Trade-off: reduced bandwidth limits data rate
3. Use Better Antennas
- Higher gain concentrates energy directionally
- Array antennas provide additional gain
- Every 3 dB antenna gain improves SNR by 3 dB
4. Lower Receiver Noise Figure
- Use LNA at input (first stage dominates)
- Cool the receiver (cryogenic for radio astronomy)
- Better semiconductor technology
5. Error Correction Coding
- Coding gain: typically 3-10 dB
- Turbo/LDPC codes approach Shannon limit
- Trade-off: increased bandwidth or reduced data rate
6. Diversity Techniques
- Spatial diversity: multiple antennas
- Frequency diversity: transmit on multiple frequencies
- Time diversity: repeat transmission
SNR in Link Budgets
A typical link budget calculation:
| Transmit Power (dBm) | +30 |
| Transmit Antenna Gain (dBi) | +10 |
| Free Space Path Loss (dB) | -120 |
| Receive Antenna Gain (dBi) | +3 |
| Received Signal Power | -77 dBm |
| For B = 1 MHz | -174 + 60 = -114 dBm |
| Receiver NF | +4 dB |
| Total Noise Power | -110 dBm |
Solved Example
Problem: A satellite communication system has the following parameters: Transmit EIRP = 45 dBW, path loss = 200 dB, receive antenna gain = 40 dBi, system noise temperature = 150 K, bandwidth = 36 MHz. Calculate (a) received signal power, (b) noise power, (c) C/N ratio.
Solution:
(a) Received signal power: C = EIRP - Path Loss + G_r C = 45 - 200 + 40 = -115 dBW
(b) Noise power: N = k × T_sys × B N = 1.38×10⁻²³ × 150 × 36×10⁶ N = 7.45 × 10⁻¹⁴ W
N(dBW) = 10×log₁₀(7.45×10⁻¹⁴) = -131.3 dBW
(c) C/N ratio: C/N = -115 - (-131.3) = 16.3 dB
Solved Example 2
Problem: A BPSK system requires BER = 10⁻⁵. Determine the required Eb/N₀ and SNR if the system bandwidth is twice the bit rate.
Solution:
For BPSK: P_e = Q(√(2Eb/N₀)) = 10⁻⁵
From Q-function tables: Q(4.26) ≈ 10⁻⁵ Therefore: √(2Eb/N₀) = 4.26 2Eb/N₀ = 18.15 Eb/N₀ = 9.07 = 9.6 dB
SNR = (Eb/N₀) × (Rb/B) = 9.07 × (Rb/2Rb) = 9.07/2 = 4.54 = 6.6 dB
Interview Questions
Q1: Why do we use dB scale for SNR instead of linear?
The dB scale is used because: (1) signal powers can span many orders of magnitude (received power might be 10⁻¹² W while transmitted is 10 W — a ratio of 10¹³), (2) cascaded gains and losses become simple addition/subtraction, (3) human perception of loudness/quality is approximately logarithmic, (4) standard engineering tables and curves use dB.
Q2: What is the difference between SNR and Eb/N₀?
SNR is the ratio of total signal power to total noise power in the system bandwidth. Eb/N₀ is the ratio of energy per bit to noise spectral density — it's a normalized measure independent of bandwidth and data rate. They are related by: SNR = (Eb/N₀)(Rb/B). Eb/N₀ allows fair comparison between different modulation schemes regardless of bandwidth.
Q3: Can a communication system work with negative SNR (in dB)?
Yes! Spread spectrum and coded systems can operate below 0 dB SNR. GPS signals arrive at approximately -20 dB SNR but work because the processing gain (spreading factor) recovers the signal. Similarly, turbo-coded deep space links operate at Eb/N₀ ≈ 0 dB. The key is that Eb/N₀ must be sufficient even if the instantaneous SNR is negative.
Q4: How does the matched filter achieve optimum SNR?
A matched filter has impulse response h(t) = s(T-t) — the time-reversed, delayed version of the signal. It maximizes output SNR because it passes all signal energy while minimizing noise. The maximum output SNR equals 2E/N₀ (where E = signal energy), independent of signal shape. This is the theoretical optimum for any linear filter.
Exam Focus
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Interview Use
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