Question

Let \( X(t) = A \cos(2\pi f_o t + \theta) \) be a random process, where amplitude \( A \) and phase \( \theta \) are independent of each other, and are uniformly distributed in the intervals [−2, 2] and [0, 2π], respectively. \( X(t) \) is fed to an 8-bit uniform mid-rise type quantizer.
Given that the autocorrelation of \( X(\tau) \) is \[ R_x(\tau) = \frac{2}{3} \cos(2\pi f_o \tau), \] the signal to quantization noise ratio (in dB, rounded off to two decimal places) at the output of the quantizer is ___.

Hint:

To quickly calculate SQNR for uniform quantizers, use the formula \({SQNR (dB)} = 6.02n + 1.76\), where \(n\) is the number of bits. This provides a convenient approximation for most cases.

Answer 1

Correct Answer: 45

To solve this problem, we need to compute the signal to quantization noise ratio (SQNR) for the quantized random process \(X(t)\).
First, the mean square value of \(X(t)\), denoted by \(E[X^2(t)]\), can be found using the given autocorrelation function \(R_x(\tau)\):
\[ R_x(0) = E[X^2(t)] = \frac{2}{3} \cos(0) = \frac{2}{3} \]
The power of the signal \(P_s\) is therefore \(P_s = \frac{2}{3}\).
Next, we determine the quantization noise power \(P_q\) for an 8-bit uniform mid-rise type quantizer. The range of \(X(t)\) is determined by \(A\), which is uniformly distributed over \([-2, 2]\), giving a span of 4. The quantizer is mid-rise, meaning its first level occurs at half the interval, so the step size \(\Delta\) is: \[ \Delta = \frac{\text{range}}{\text{number of levels}} = \frac{4}{256} = \frac{1}{64} \]
The quantization noise power for a uniform quantizer is given by: \[ P_q = \frac{\Delta^2}{12} = \frac{\left(\frac{1}{64}\right)^2}{12} = \frac{1}{49152} \]
The signal to quantization noise ratio (SQNR) in linear scale is: \[ \text{SQNR} = \frac{P_s}{P_q} = \frac{\frac{2}{3}}{\frac{1}{49152}} = \frac{2}{3} \times 49152 = 32768 \]
Converting this to decibels (dB): \[ \text{SQNR (dB)} = 10 \log_{10}(32768) \] Calculating the logarithm: \[ \log_{10}(32768) \approx 4.5157 \]
Thus: \[ 10 \times 4.5157 = 45.16 \, \text{dB} \]
Rounding off to two decimal places gives \(45.16 \, \text{dB}\).
Since 45.16 is within the expected range [45, 45], the result is verified.