Question

A first-order high-pass RC filter has cutoff frequency \(\text{f}_\text{c}=1\text{ kHz}\). If the magnitude of the transfer function is measured at \(\text{f}=500\text{ Hz}\), the gain magnitude is

• A. \(0.25 \)
• B. \(0.45 \)
• C. \(0.71 \)
• D. \(0.89 \)

Hint:

For a first-order filter: - High Pass Filter Magnitude: \( \frac{1}{\sqrt{1 + (f_c/f)^2}} \) - Low Pass Filter Magnitude: \( \frac{1}{\sqrt{1 + (f/f_c)^2}} \) Since the operating frequency ($500\text{ Hz}$) is exactly half of the cutoff frequency ($1\text{ kHz}$), it must experience significant attenuation (more than $3\text{ dB}$, meaning less than $0.707$). This immediately eliminates $0.71$ and $0.89$, allowing you to quickly choose between $0.25$ and $0.45$.

Answer 1

The Correct Option is B

Concept: A first-order passive High-Pass Filter (HPF) configured using a single resistor and capacitor passes signals with frequencies higher than its specific cutoff frequency and attenuates components that lie below it. The complex transfer function $H(j\omega)$ or $H(jf)$ of a first-order high-pass filter is represented mathematically as: \[ H(jf) = \frac{j\left(\frac{f}{f_c}\right)}{1 + j\left(\frac{f}{f_c}\right)} \] Taking the absolute value gives the expression for the voltage gain magnitude: \[ |H(jf)| = \frac{\frac{f}{f_c}}{\sqrt{1 + \left(\frac{f}{f_c}\right)^2}} = \frac{1}{\sqrt{1 + \left(\frac{f_c}{f}\right)^2}} \] where $f$ is the operating measurement frequency and $f_c$ is the specified $-3\text{ dB}$ cutoff frequency.

Step 1: Identifying the parameters given in the problem statement.
We are given:
• Cutoff frequency, \( f_c = 1\text{ kHz} = 1000\text{ Hz} \)
• Measurement frequency, \( f = 500\text{ Hz} \)

Step 2: Evaluating the ratio of the frequencies.
Let's compute the ratio of the cutoff frequency to the operating frequency to make our substitution step cleaner: \[ \frac{f_c}{f} = \frac{1000\text{ Hz}}{500\text{ Hz}} = 2 \]

Step 3: Calculating the total gain magnitude.
Substitute this calculated ratio into our simplified magnitude equation: \[ |H(jf)| = \frac{1}{\sqrt{1 + \left(2\right)^2}} \] \[ |H(jf)| = \frac{1}{\sqrt{1 + 4}} = \frac{1}{\sqrt{5}} \] We know that the square root of 5 is approximately \(2.236\). Let's evaluate the fraction: \[ |H(jf)| = \frac{1}{2.23607} \approx 0.4472 \] Rounding this decimal value to two decimal places yields \(0.45\). Hence, the correct choice is option (2).