A source generates three symbols with probability 0.25, 0.25, 0.50 at a rate of 3000 symbols per second. Assuming independent generation of symbols, the most efficient source encoder would have average bit rate of
4500 bits/sec
To determine the most efficient average bit rate for a source encoder, we must first calculate the source's entropy. Entropy quantifies the average amount of information produced by the source per symbol, representing the theoretical minimum average number of bits required to encode each symbol. Once the entropy per symbol is known, we multiply it by the rate at which symbols are generated to find the average bit rate for the most efficient source encoder.
Source Information and Probabilities
The problem provides specific details about the source that generates symbols:
- Number of distinct symbols generated: 3
- Probabilities for each of these symbols are given as:
- Probability of Symbol 1 (P1): 0.25
- Probability of Symbol 2 (P2): 0.25
- Probability of Symbol 3 (P3): 0.50
- The source generates symbols at a rate of: 3000 symbols per second
- A key assumption mentioned is the independent generation of symbols, meaning the probability of one symbol appearing does not affect the probability of another.
It's important to verify that the sum of all symbol probabilities equals 1, which is $0.25 + 0.25 + 0.50 = 1.00$. This confirms the probabilities are correctly distributed.
Source Entropy Calculation
The entropy ($H$) of a discrete memoryless source measures the average information content or uncertainty associated with each symbol generated by the source. For a source with $n$ symbols, where each symbol $i$ has a probability $P_i$, Shannon's entropy formula is used:
\[ H = - \sum_{i=1}^{n} P_i \log_2(P_i) \]
In this formula:
- $P_i$ represents the probability of the i-th symbol.
- $\log_2(P_i)$ denotes the logarithm base 2 of the probability of the i-th symbol.
- $n$ is the total count of unique symbols.
Let's calculate the term $P_i \log_2(P_i)$ for each symbol given their respective probabilities:
| Symbol | Probability ($P_i$) | $\log_2(P_i)$ | $P_i \log_2(P_i)$ |
|---|---|---|---|
| 1 | 0.25 | $\log_2(0.25) = \log_2(1/4) = \log_2(2^{-2}) = -2$ | $0.25 \times (-2) = -0.5$ |
| 2 | 0.25 | $\log_2(0.25) = \log_2(1/4) = \log_2(2^{-2}) = -2$ | $0.25 \times (-2) = -0.5$ |
| 3 | 0.50 | $\log_2(0.50) = \log_2(1/2) = \log_2(2^{-1}) = -1$ | $0.50 \times (-1) = -0.5$ |
Now, we sum up all the $P_i \log_2(P_i)$ values:
\[ \sum P_i \log_2(P_i) = (-0.5) + (-0.5) + (-0.5) = -1.5 \]
Finally, we calculate the source entropy $H$ by taking the negative of this sum:
\[ H = - (-1.5) = 1.5 \text{ bits/symbol} \]
Thus, the average information content per symbol generated by this source is 1.5 bits.
Average Bit Rate Determination
The average bit rate of the most efficient source encoder represents the minimum average number of bits that must be transmitted per second to convey the information from the source. It is calculated by multiplying the source entropy (measured in bits per symbol) by the symbol generation rate (measured in symbols per second).
We have the following values:
- Calculated Source Entropy ($H$): 1.5 bits/symbol
- Given Symbol Generation Rate ($R$): 3000 symbols/second
The average bit rate (ABR) can be calculated using the formula:
\[ \text{ABR} = H \times R \]
Substituting the values:
\[ \text{ABR} = 1.5 \text{ bits/symbol} \times 3000 \text{ symbols/second} \]
\[ \text{ABR} = 4500 \text{ bits/second} \]
Therefore, the most efficient source encoder for this source would require an average bit rate of 4500 bits per second to transmit the information.
Important Questions from Information Content of a Discrete Memoryless System
_______ is defined as the rate of change of signal on transmission medium after encoding and modulation have occurred.
A source emits bit 0 with probability 1/3 and bit 1 with probability 2/3. The emitted bits are communicated to the receiver. The receiver decides for either 0 or 1 based on the received value R. It is given that the conditional density functions of R are as
\({{\rm{f}}_{\left( {{\rm{R}}/0} \right)}}\left( {\rm{x}} \right) = \left\{ {\frac{1}{4}, - 3 \le {\rm{x}} \le 1} \right.{\rm{and\;}}{{\rm{f}}_{\left( {{\rm{R}}/1} \right)}}\left( {\rm{x}} \right) = \left\{ {\frac{1}{6}, - 1 \le {\rm{x}} \le 5} \right.\)
The minimum decision error probability isAn analog signal is bandlimited to 4 KHz, sampled at Nyquist rate and the samples are quantized into 4 levels. The quantized levels are assumed to be independent and equally probable. If we transmit two quantized samples per sec, the information rate is
