Question

A particular RC low-pass filter has a breakpoint frequency of \(200 .\) Hz. At what frequency will the output voltage divided by the input voltage be \(0.100 ?\)

Short answer

Answer: The frequency at which the output voltage divided by the input voltage will be 0.100 for the given RC low-pass filter is approximately 628.9 Hz.

Step-by-step solution

Understand the RC low-pass filter characteristics

An RC low-pass filter is an electronic circuit that allows low-frequency signals to pass through while attenuating the high-frequency signals. The breakpoint frequency is the frequency at which the output voltage becomes 0.707 times the input voltage.

Define the transfer function for the RC low-pass filter

The transfer function, \(H(f)\), for an RC low-pass filter can be given as: $$H(f) = \frac{1}{\sqrt{1 + (\frac{f}{f_c})^2}} $$ Here, \(H(f)\) is the output voltage divided by the input voltage as a function of frequency \(f\), and \(f_c\) is the breakpoint frequency.

Plug in the given values

We're given that the breakpoint frequency, \(f_c\), is 200 Hz, and we need to find the frequency at which the output voltage divided by the input voltage will be 0.100. So, let's plug in the values in the transfer function equation: $$0.100 = \frac{1}{\sqrt{1 +(\frac{f}{200})^2}}$$

Solve for the frequency 'f'

Now, let's solve the equation for \(f\): First, take the reciprocal of both sides: $$\frac{1}{0.100} = \sqrt{1 + (\frac{f}{200})^2}$$ Then, square both sides: $$(\frac{1}{0.100})^2 = 1 + (\frac{f}{200})^2$$ Subtract 1 from both sides: $$(\frac{1}{0.100})^2 - 1 = (\frac{f}{200})^2$$ Take the square root of both sides: $$\sqrt{(\frac{1}{0.100})^2 - 1} = \frac{f}{200}$$ Finally, multiply both sides by 200 to get the desired frequency, \(f\): $$f = 200 \times \sqrt{(\frac{1}{0.100})^2 - 1}$$ Upon calculating, we find that the frequency at which the output voltage divided by the input voltage will be 0.100 is approximately 628.9 Hz.