Output Signal-to-Noise Ratio Through a Lossy Channel: 25-Comp-B5 Computer Communications, December 2016, Q4

Question 4: Output Signal-to-Noise Ratio Through a Lossy Channel (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Check: the noise level is read as 3 μW; the only physically sensible reading at the scale of a 1 W signal attenuated by 10 dB (giving a 0.1 W output) is 3 microwatt (μW), i.e. $3\times10^{-6}\text{ W}$ — a picowatt noise floor would make the SNR meaninglessly large for a "find the SNR in dB" exam question.

Given.

Quantity	Value
Channel loss	10 dB
Input signal power $P_{\text{in}}$	1.0 W
Output noise level $N_{\text{out}}$	$3\ \mu\text{W} = 3\times10^{-6}\text{ W}$

Find. The output signal-to-noise ratio, in dB.

Approach. Convert the 10 dB loss to a power ratio to get the output signal power, then form the ratio of output signal power to output noise power and convert that ratio to dB.

Output signal power from the channel loss. A loss of $L_{\text{dB}}$ means $P_{\text{out}} = P_{\text{in}}/10^{L_{\text{dB}}/10}$: $$P_{\text{out}} = \frac{1.0}{10^{10/10}} = \frac{1.0}{10} = \boxed{0.1\text{ W}}$$
Form the output SNR as a power ratio. $$\text{SNR}_{\text{out}} = \frac{P_{\text{out}}}{N_{\text{out}}} = \frac{0.1}{3\times10^{-6}} = 33{,}333.3$$
Convert to decibels. $$\text{SNR}_{\text{out}}(\text{dB}) = 10\log_{10}(33{,}333.3) = 10 \times 4.5229$$ $$\text{SNR}_{\text{out}}(\text{dB}) = \boxed{45.23\text{ dB}}$$

Quantity	Value
Output signal power	0.1 W
Output SNR (ratio)	≈ 33,333
Output SNR	≈ 45.23 dB

Topic: Decibels, channel loss, and signal-to-noise ratio.

Key relations: loss/gain in dB, $L_{\text{dB}} = 10\log_{10}(P_{\text{in}}/P_{\text{out}})$; SNR in dB, $\text{SNR(dB)} = 10\log_{10}(P_{\text{signal}}/P_{\text{noise}})$.

Why this works: decibels turn multiplicative chains of gains/losses into simple addition/subtraction, and because both loss and SNR are expressed the same logarithmic way, the two conversions (power ratio → dB, then dB → power ratio for the loss) compose cleanly through one intermediate linear quantity, $P_{\text{out}}$.

Common pitfall: computing the SNR from the INPUT signal power instead of the output (the noise is measured at the output, after the loss has already reduced the signal, so both quantities in the ratio must be taken at the same point in the channel); or applying the 10 dB loss to the noise instead of the signal.

Source: Stallings, Data and Computer Communications, 10th ed., Ch.3 (Data Transmission) §3.2–3.3, Transmission Impairments and Decibels.

Question 4 of 7: Output Signal-to-Noise Ratio Through a Lossy Channel

Question 4: Output Signal-to-Noise Ratio Through a Lossy Channel (20 marks)