Question

A Gaussian noise channel has a bandwidth of $4\text{ kHz}$ and a signal-to-noise power ratio (SNR) of $15$. According to the Shannon-Hartley theorem, what is the maximum capacity of this channel?

• A. $32\text{ kbps}$
• B. $16\text{ kbps}$
• C. $60\text{ kbps}$
• D. $64\text{ kbps}$

Hint:

Always double-check whether the SNR is given as a linear ratio or in decibels ($\text{dB}$). If the problem states $\text{SNR} = 15\text{ dB}$, you must first convert it via $10^{\frac{15}{10}}$ before applying the Shannon formula. Here, it is given as a raw ratio value of 15, so we use it directly.

Answer 1

The Correct Option is B

Concept: The Shannon-Hartley theorem sets the theoretical upper bound on the maximum error-free information transfer rate (channel capacity, $C$) that can be achieved across a communication link given a specific transmission bandwidth and noise level. The fundamental equation is expressed as: $$C = B \log_2\left(1 + \text{SNR}\right)$$ Where:
• $C$ = Channel capacity in bits per second ($\text{bps}$).
• $B$ = Channel bandwidth in Hertz ($\text{Hz}$).
• $\text{SNR}$ = Signal-to-Noise Power Ratio (expressed as a linear ratio value, not in $\text{dB}$). Step-by-step Mathematical Calculation:
•

Step 1: Extract the numerical values provided inside the question statement: B &= 4 kHz = 4 \times 10^3 Hz
SNR &= 15 (linear ratio value)
•

Step 2: Substitute these values directly into the Shannon capacity formula: $$C = 4000 \times \log_2(1 + 15)$$
•

Step 3: Simplify the logarithmic argument expression: $$C = 4000 \times \log_2(16)$$
•

Step 4: Convert 16 into a base-2 exponent format: $$16 = 2^4 \quad \Rightarrow \quad \log_2(2^4) = 4 \times \log_2(2) = 4$$
•

Step 5: Multiply the terms out to arrive at the final quantitative result: $$C = 4000 \times 4 = 16000\text{ bps}$$ $$C = 16\text{ kbps}$$ The calculation gives exactly $16\text{ kbps}$, corresponding to Option (B).