Communication Math

Khaled Mahmud

System Capacity, Throughput, Spectral Efficiency

Q1.

If the SNR is 100, and the bandwidth available is 4 kHz, which is appropriate for telephone communications, then by Shannon’s Capacity theorem what is the maximum attainable data rate?

Solution

Given,

Bandwidth, BW = 4 kHz

SNR (linear) = 100

We can use Shannon's capacity Theorem here.

$$ \begin{aligned} Capacity,C&=(BW)log_2 (1+SNR) \\ &=(4000)log_2 (1+100\\ &=(4000)log_2 (101) \\ &=(4000)\frac{log_{10} (101)}{log_{10} (2)} \\ &=\frac{4000x2.004}{0.301} \\ &=26628.58\ bps \\ &=26.63\ kbps \end{aligned} $$

Q2.

For a certain channel, if the SNR is 30 dB, and the bandwidth available is 10 kHz, calculate the maximum data rate (Shannon capacity) of the channel.

Solution

Given,

$ BW = 10\ kHz \\ SNR_{dB}=30\ dB $

Let's first calculate the linear value of SNR, which is required in Shannon's capacity equaiton

$ SNR\ (linear)=(10)^{(30/10)}= 10^3=1000 $

Using the capacity theorem,

$ \begin{aligned} Capacity, C\ &=\ (10000)xlog_2 (1+1000) \\ &=(10000)x\frac{log_{10} (1001)}{log_{10} (2)} \\ &=\frac{10000x3}{0.301} \\ &=99658\ bps \\ &=99.66\ kbps \end{aligned} $

Q3.

If a system it is required to transmit at 50 Mbit/s, and a bandwidth of 5 MHz is used, then, according to Shannon’s capacity theorem, what is the minimum required SNR in dB?

Solution

Given,

$ BW=5\ MHz $

$ Required\ Data\ rate\ (C)=50\ Mbps $

Use Shannon’s capacity theorem to solve for SNR.

$ Capacity, C=BWlog_2 (1+SNR) $

$ 50=(5)\frac{log_{10} (1+SNR)}{log_{10} (2)} $

$ log_{10} (1+SNR)=10xlog_{10} (2)=3.01 $

$ (1+SNR)=10^{3.01}=1024 $

$ SNR=1023\ (linear) $

$ SNR_{dB}=30.1\ dB $

Q4.

If the symbol rate of a 16QAM transmitter is 250 ksymbol/sec, what is the bit rate? What would be bit rate, if it uses 64QAM?

Solution

For any M-ary QAM, bit per symbol, n, can be calculated using the following relation

$ M=2^n $

For 16QAM,

$ n=4 $

Also, given

$ R_S=250\ ksymbole/sec $

$ \begin{aligned} R_b&=(Symbol\ rate)x(bit\ per\ symbol)\\ &=4x250\\ &=1000\ kbps\\ &=1\ Mbps \end{aligned} $

Q5.

If the symbol rate of a 16-QAM transmitter is 1.5 Msymbol/sec, what is the bit rate? What would be bit rate, if it uses 256QAM? Show your work.

Solution

(Try youself)

Q6.

A transmitter is sending using 16-QAM modulation, at a rate of 1.2 Mbps. If it switches to 64-QAM, keeping symbol rate and channel coding same, what would be the new bit rate?

Solution

(Try yourself)

Q7.

In a wireless system, the symbol rate of the output of the digital modulator (baseband) is 12000 symbol/ sec. The system adjusts its modulation according to the available SNR. For each of the following modulation schemes, calculate the resulting bit rate.

  1. BPSK (Binary PSK)
  2. QPSK (4-QAM)
  3. 16QAM
  4. 64QAM

(Try yourself)

Q8.

For a given wireless system, spectral efficiency is defined by the ratio of average number of bits transmitted by one Hz of frequency bandwidth. If a certain wireless system uses a total of 5 MHz bandwidth, and using various techniques (e.g. modulation, channel coding, MIMO, etc.) it can achieve an average data rate of 18 Mbps, what is the spectral efficiency of the system?

Solution

Given,

$ BW = 5\ MHz $

$ R_b = 18\ Mbps $

$ \begin{aligned} Spectral\ effciency&=\frac{18 Mbps}{5\ MHz}\\ &=3.6\ bit/Hz \end{aligned} $

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