An 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
4 bit / s
Information Rate Basics
The question asks us to determine the information rate of an analog signal under specific conditions. We are given the signal's bandwidth, the sampling method (Nyquist rate), the number of quantization levels, and the transmission rate of the quantized samples. The information rate is essentially the total number of bits transmitted per second. To find this, we need to understand how many bits each sample represents and how many such samples are transmitted per second.
Analog Signal Bandwidth and Sampling
An analog signal is specified as being bandlimited to \(4 \text{ KHz}\). This means its maximum frequency component (\(B\)) is \(4 \text{ KHz}\).
The signal is sampled at Nyquist rate. The Nyquist rate (\(f_s\)) is the minimum sampling rate required to perfectly reconstruct a bandlimited analog signal from its samples without aliasing. It is given by:
\[ f_s = 2B \]
Where \(B\) is the bandwidth of the signal.
- Given bandwidth, \(B = 4 \text{ KHz}\)
- Nyquist rate, \(f_s = 2 \times 4 \text{ KHz} = 8 \text{ KHz} = 8000 \text{ samples/second}\).
While the Nyquist rate tells us how frequently the signal should be sampled, the problem specifies a different transmission rate for the quantized samples. This distinction is crucial for calculating the actual information rate.
Quantized Levels and Bits per Sample
After sampling, the samples are quantized into 4 levels. Quantization is the process of mapping continuous-amplitude samples into a finite number of discrete amplitude levels.
We are told that the quantized levels are independent and equally probable. When there are \(M\) equally probable levels, the number of bits (\(n\)) required to represent each sample is given by the formula:
\[ M = 2^n \]
Or equivalently:
\[ n = \log_2(M) \]
- Number of quantization levels, \(M = 4\)
- Bits per sample, \(n = \log_2(4) = 2 \text{ bits/sample}\).
This means each sample, after being quantized, carries 2 bits of information.
Information Rate Computation
The problem states: "If we transmit two quantized samples per sec". This is the rate at which the quantized data is being sent over a communication channel. This rate, let's call it \(R_s\), is \(2 \text{ samples/sec}\).
The information rate (\(R\)) is calculated by multiplying the number of bits per sample by the rate at which these samples are transmitted.
\[ \text{Information Rate } (R) = (\text{Transmission rate of samples}) \times (\text{Bits per sample}) \]
Using the values we've found:
- Transmission rate of samples (\(R_s\)) = \(2 \text{ samples/sec}\)
- Bits per sample (\(n\)) = \(2 \text{ bits/sample}\)
Therefore, the information rate is:
\[ R = 2 \text{ samples/sec} \times 2 \text{ bits/sample} = 4 \text{ bits/sec} \]
Final Information Rate
Based on the analysis, transmitting two quantized samples per second, where each sample is represented by 2 bits, results in an information rate of 4 bits per second.
| Parameter | Value |
|---|---|
| Analog Signal Bandwidth (\(B\)) | \(4 \text{ KHz}\) |
| Nyquist Rate (\(f_s\)) | \(8000 \text{ samples/second}\) |
| Number of Quantization Levels (\(M\)) | \(4\) |
| Bits per Sample (\(n = \log_2 M\)) | \(2 \text{ bits/sample}\) |
| Transmission Rate of Quantized Samples (\(R_s\)) | \(2 \text{ samples/sec}\) |
| Information Rate (\(R = R_s \times n\)) | \(4 \text{ bits/sec}\) |
The calculated information rate is \(4 \text{ bits/s}\).
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 isA 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
