Question

The output noise variance due to an 8-bit ADC of a first-order filter with \[ H(z) = \frac{1}{1 - 0.25z^{-1}} \] for the input signal with noise variance \( \sigma^2 \) is

• A. $1.5 \sigma^2$
• B. $1.06 \sigma^2$
• C. $0.25 \sigma^2$
• D. $3.25 \sigma^2$

Hint:

Remember that for calculating the output noise variance, you first need to get the impulse response and then apply the relevant formula.

Answer 1

The Correct Option is B

Step 1: The transfer function of the system is given by \( H(z) = \frac{1}{1 - 0.25z^{-1}} \). The noise power or variance is calculated as: \[ \sigma_o^2 = \sigma^2 \times \sum_{n=-\infty}^{\infty} |h[n]|^2 \] 
Step 2: First, find the impulse response: Since \( H(z) = \frac{1}{1 - 0.25z^{-1}} \), the h[n] becomes a decaying exponential: \( h[n] = (0.25)^n u[n] \) 
Step 3: Find the sum of the square of the impulse response: \[ \sum_{n=0}^\infty |(0.25)^n|^2 = \sum_{n=0}^\infty (0.25)^{2n} \] \[ = \frac{1}{1 - 0.25^2} = \frac{1}{1 - 0.0625} = \frac{1}{0.9375} = 1.066 \] Therefore the output noise variance is 1.066\(\sigma^2\), which is approximately 1.06 \(\sigma^2\).