Shannon’s and Nyquist’s equations. 1) (10 marks) Shannon equation gives the channel capacity of a communication system with AWGN. The equation is (Shannon channel capacity) C = B log2(1 + SNR) [in bps] where B is the channel bandwidth (in Hz] and SNR is the received signal-to-noise as a power ratio. Suppose the bandwidth B is 1 MHz, transmit power P is 3 Watt, the channel gain \h|2 is 10-3, the noise power spectral density No is 10-9 Watt/Hz. Please calculate the AWGN channel capacity. 2) (10 marks) Nyquist formula gives the channel capacity as a function of the number of levels per symbol. While Shannon’s equation tells us the maximum data rate possible in the presence of noise, Nyquist’s equation tells us the data rate C as a function of bandwidth B and the number of signal levels per symbol M we can achieve. The equation is (Nyquist channel capacity) C = 2B log2 M [in bps] where M is the number of signal levels per symbol. Please calculate the Nyquist data rate given the bandwidth as 1 MHz and assume 8 signal levels per symbol.
The Correct Answer and Explanation is:
1) Shannon Channel Capacity Calculation
Formula: C=Blog2(1+SNR)[bps]C = B \log_2(1 + \text{SNR}) \quad \text{[bps]}
Given:
- Bandwidth B=1 MHz=106 HzB = 1 \, \text{MHz} = 10^6 \, \text{Hz}
- Transmit Power P=3 WattP = 3 \, \text{Watt}
- Channel Gain ∣h∣2=10−3|h|^2 = 10^{-3}
- Noise Spectral Density N0=10−9 W/HzN_0 = 10^{-9} \, \text{W/Hz}
Step 1: Calculate SNR SNR=Received powerNoise power=P⋅∣h∣2N0⋅B=3⋅10−310−9⋅106=3⋅10−310−3=3\text{SNR} = \frac{\text{Received power}}{\text{Noise power}} = \frac{P \cdot |h|^2}{N_0 \cdot B} = \frac{3 \cdot 10^{-3}}{10^{-9} \cdot 10^6} = \frac{3 \cdot 10^{-3}}{10^{-3}} = 3
Step 2: Plug into Shannon formula C=106⋅log2(1+3)=106⋅log2(4)=106⋅2=2,000,000 bpsC = 10^6 \cdot \log_2(1 + 3) = 10^6 \cdot \log_2(4) = 10^6 \cdot 2 = 2,000,000 \, \text{bps}
✅ Shannon capacity = 2 Mbps
2) Nyquist Channel Capacity Calculation
Formula: C=2Blog2M[bps]C = 2B \log_2 M \quad \text{[bps]}
Given:
- Bandwidth B=1 MHzB = 1 \, \text{MHz}
- M=8M = 8 signal levels
C=2⋅106⋅log28=2⋅106⋅3=6,000,000 bpsC = 2 \cdot 10^6 \cdot \log_2 8 = 2 \cdot 10^6 \cdot 3 = 6,000,000 \, \text{bps}
✅ Nyquist data rate = 6 Mbps
Explanation
In digital communications, Shannon’s and Nyquist’s equations represent two fundamental limits of data transmission.
Shannon’s equation gives the maximum possible data rate (channel capacity) for a given channel bandwidth in the presence of Additive White Gaussian Noise (AWGN). It accounts for the effects of noise by using the Signal-to-Noise Ratio (SNR). In our example, the channel had a 1 MHz bandwidth and an SNR of 3. Plugging these into Shannon’s formula, we found the theoretical maximum data rate to be 2 Mbps. This represents the upper limit for reliable transmission using ideal coding schemes.
On the other hand, Nyquist’s equation considers the symbol rate and signal levels, assuming no noise (ideal conditions). It states that for a noiseless channel, the maximum data rate is proportional to the bandwidth and the logarithm of the number of distinct signal levels (M). Given 8 levels (i.e., 3 bits per symbol) and 1 MHz bandwidth, the Nyquist rate comes out to be 6 Mbps.
However, in real systems, noise limits the number of distinguishable signal levels. That’s why Nyquist gives the upper bound for noiseless systems, while Shannon gives the true limit in noisy conditions. In practice, achievable data rates lie below Shannon’s limit unless advanced coding techniques are used.
In summary:
- Shannon capacity reflects real-world noisy channels.
- Nyquist capacity assumes a perfect channel but shows how symbol levels affect rate.
Understanding both is key to designing efficient communication systems.
