Skip to main content

Symbol rate, entropy and information rate (6 symbols per 10 μs)

Short answer

Symbol rate r=1/(10 μs)=100 ksym/s. Entropy H = Σ p_i log2(1/p_i) = 2.375 bits/symbol. Information rate R = rH = 237.5 kbps.

A discrete source emits one of 6 possible symbols per 10 μs in statistically independent manner. The symbol probabilities are 1/4, 1/4, 1/4, 1/8, 1/16 and 1/16 respectively. Calculate symbol rate, entropy and information rate.

Given: one symbol per Ts=10 μT_s = 10\ \mus; P={14,14,14,18,116,116}P = \{\tfrac14, \tfrac14, \tfrac14, \tfrac18, \tfrac1{16}, \tfrac1{16}\} (sum = 1).

Symbol rate

r=1Ts=110×10−6=105 symbols/s=100 kbaudr = \frac{1}{T_s} = \frac{1}{10\times10^{-6}} = 10^5\ \text{symbols/s} = 100\ \text{kbaud}

Entropy

Symbolpip_ilog⁡2(1/pi)\log_2(1/p_i)pilog⁡2(1/pi)p_i\log_2(1/p_i)
x10.252.00000.5000
x20.252.00000.5000
x30.252.00000.5000
x40.1253.00000.3750
x50.06254.00000.2500
x60.06254.00000.2500
Total12.3750
H=3(14log⁡24)+18log⁡28+2(116log⁡216)=3(0.5)+0.375+2(0.25)=1.5+0.375+0.5=2.375 bits/symbol\begin{aligned} H &= 3\left(\tfrac14\log_2 4\right) + \tfrac18\log_2 8 + 2\left(\tfrac1{16}\log_2 16\right) \\ &= 3(0.5) + 0.375 + 2(0.25) \\ &= 1.5 + 0.375 + 0.5 = 2.375\ \text{bits/symbol} \end{aligned}

(Maximum possible for 6 symbols: log⁡26=2.585\log_2 6 = 2.585 bits/symbol.)

Information rate

R=rH=105×2.375=2.375×105 bits/s=237.5 kbpsR = rH = 10^5 \times 2.375 = 2.375\times10^5\ \text{bits/s} = 237.5\ \text{kbps}

Answer: r=105r = 10^5 symbols/s, H=2.375H = 2.375 bits/symbol, R=237.5R = 237.5 kbps.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Communication Systems chapter 6 questions →