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

Step 1: Understanding the given random process
The given signal is: \[ X(t) = A \cos(2\pi f_0 t + \theta) \] where:
\( A \) is uniformly distributed in the range \([-2, 2]\),
\( \theta \) is uniformly distributed in the range \([0, 2\pi]\).

The power of a uniformly distributed random variable \( A \) over \([-2, 2]\) is: \[ E[A^2] = \frac{1}{b-a} \int_{-2}^{2} A^2 \, dA = \frac{1}{4} \left[ \frac{A^3}{3} \Big|_{-2}^{2} \right] \] \[ = \frac{1}{4} \times \frac{8 + 8}{3} = \frac{16}{12} = \frac{4}{3} \] Thus, the average power of \( X(t) \) is: \[ P_X = \frac{2}{3} \]
Step 2: Quantization noise power calculation
For an 8-bit quantizer, the total number of quantization levels is: \[ L = 2^8 = 256 \] The quantization noise power for a uniform quantizer is: \[ P_Q = \frac{\Delta^2}{12} \] The quantization step size \( \Delta \) is: \[ \Delta = \frac{\text{max value} - \text{min value}}{L} \] Since the range of \( X(t) \) is \([-2, 2]\): \[ \Delta = \frac{4}{256} = \frac{1}{64} \] Substituting: \[ P_Q = \frac{(1/64)^2}{12} = \frac{1}{4096 \times 12} = \frac{1}{49152} \]
Step 3: Signal-to-quantization noise ratio (SQNR)
\[ \text{SQNR} = \frac{P_X}{P_Q} \] \[ \text{SQNR} = \frac{2/3}{1/49152} = \frac{2 \times 49152}{3} = 32768 \]
Step 4: Convert SQNR to decibels
\[ \text{SQNR (dB)} = 10 \log_{10} (32768) \] \[ = 10 \log_{10} (2^{15}) \] \[ = 10 \times 15 \log_{10} (2) \] \[ = 10 \times 15 \times 0.301 \] \[ = 45.15 \text{ dB} \]
Final Answer:
\[ \boxed{45.15 \text{ dB}} \]