Question

Let \(X(t) = A \cos(2\pi f_0 t + \theta)\) 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 \([0, 2\pi]\), respectively. \(X(t)\) is fed to an 8-bit uniform mid-rise type quantizer. Given that the autocorrelation of \(X(t)\) is: \[ R_X(\tau) = \frac{2}{3} \cos(2\pi f_0 \tau), \] 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

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 calculated as: \[ E[A^2] = \frac{1}{b-a} \int_{-2}^{2} A^2 dA = \frac{1}{4} \left[ \frac{A^3}{3} \Big|_{-2}^{2} \right] \] \[ = \frac{1}{4} \times \frac{8 + 8}{3} = \frac{16}{12} = \frac{4}{3}. \] Thus, the average power of \( X(t) \) is: \[ P_X = \frac{2}{3}. \] \medskip 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 given by: \[ P_Q = \frac{\Delta^2}{12}, \] where the quantization step size \( \Delta \) is: \[ \Delta = \frac{{max value} - {min value}}{L}. \] Since the range of \( X(t) \) is \([-2, 2]\): \[ \Delta = \frac{4}{256} = \frac{1}{64}. \] Substituting into the formula for \(P_Q\): \[ P_Q = \frac{(1/64)^2}{12} = \frac{1}{4096 \times 12} = \frac{1}{49152}. \] \medskip Step 3: Signal-to-quantization noise ratio (SQNR) The signal-to-quantization noise ratio (SQNR) is expressed as: \[ {SQNR} = \frac{P_X}{P_Q}. \] Substituting the values of \(P_X\) and \(P_Q\): \[ {SQNR} = \frac{2/3}{1/49152} = \frac{2 \times 49152}{3} = 32768. \] \medskip Step 4: Convert SQNR to decibels (dB) The SQNR in decibels is given by: \[ {SQNR (dB)} = 10 \log_{10} (32768). \] Using properties of logarithms: \[ {SQNR (dB)} = 10 \log_{10} (2^{15}), \] \[ = 10 \times 15 \log_{10} (2), \] \[ = 10 \times 15 \times 0.301, \] \[ = 45.15 \, {dB}. \] \medskip Therefore, the signal-to-quantization noise ratio is approximately: \[ \boxed{45.00 { to } 45.30 \, {dB}}. \]