Question

A white Gaussian noise \(w(t)\) with zero mean and power spectral density \(\frac{N_0}{2}\), when applied to a first-order RC low-pass filter produces an output \(n(t)\). At a particular time \(t = t_k\), the variance of the random variable \(n(t_k)\) is:

• A. \(\frac{N_0}{4RC}\)
• B. \(\frac{N_0}{2RC}\)
• C. \(\frac{N_0}{RC}\)
• D. \(\frac{2N_0}{RC}\)

Hint:

To analyze noise in RC filters, calculate the output variance by integrating the power spectral density over all frequencies. Remember to use substitution and standard integral results for efficiency.

Answer 1

The Correct Option is A

Step 1: Derive the transfer function for the RC low-pass filter.
The transfer function of a first-order RC low-pass filter is given by: \[ H(f) = \frac{1}{1 + j2\pi fRC}. \] Step 2: Connect the power spectral density and variance.
The output power spectral density \(S_{n}(f)\) is related to the input \(S_{w}(f)\) as: \[ S_{n}(f) = |H(f)|^2 S_{w}(f). \] For white Gaussian noise, \(S_{w}(f) = \frac{N_0}{2}\). The magnitude squared of the transfer function becomes: \[ |H(f)|^2 = \frac{1}{1 + (2\pi fRC)^2}. \] Step 3: Compute the variance of the output.
The variance of the output noise \(n(t_k)\) is the integral of \(S_{n}(f)\) over all frequencies: \[ {Variance} = \int_{-\infty}^\infty S_{n}(f) df = \int_{-\infty}^\infty \frac{\frac{N_0}{2}}{1 + (2\pi fRC)^2} df. \] Using the substitution \(\omega = 2\pi f\), the integral becomes: \[ {Variance} = \frac{N_0}{2} \int_{-\infty}^\infty \frac{1}{1 + (\omega RC)^2} \frac{d\omega}{2\pi}. \] The standard integral result for \(\frac{1}{1 + x^2}\) is: \[ \int_{-\infty}^\infty \frac{1}{1 + (\omega RC)^2} d\omega = \frac{\pi}{RC}. \] Substituting this result, the variance is: \[ {Variance} = \frac{N_0}{2} \cdot \frac{\pi}{RC} \cdot \frac{1}{2\pi} = \frac{N_0}{4RC}. \] Final Answer: \[\boxed{{(1) } \frac{N_0}{4RC}}\]