# Simple High Pass Filter Problem

• SlideMan
In summary, the conversation discusses finding the value of C for a simple RC high pass filter to achieve a specific output voltage shown on a scope capture. The input voltage is given as 2.5 + 10.01cos(\omega t) and the output voltage is 3.46cos(\omega t - 60^\circ). The equations used are V_O = \frac {V_I}{1 + \frac {1}{j\omega RC}} and \frac {3.46}{(628.32)(5k)C}\angle-30 = 8.81\angle-19.9. The challenge is accounting for the DC component of the input voltage while using phasor notation.
SlideMan

## Homework Statement

Given a simple RC high pass filter, I need to find the value of C that will give the output as shown on a scope capture.

Input Voltage: $$2.5 + 10.01cos(\omega t)$$ (10.01V wave shifted up 2.5V)
Output Voltage: $$3.46cos(\omega t - 60^\circ)$$
R = 5k
$$\omega = 2\pi 100$$

## Homework Equations

$$V_O = \frac {V_I}{1 + \frac {1}{j\omega RC}}$$

## The Attempt at a Solution

I'm not sure how to take into account the DC component of the input voltage. I know it will be filtered out in the end, but I would think that it would be needed for the computations. Ignoring the DC component and using phasor notation, I come up with a phase angle of -19.9 equal to a phase angle of -30:

$$3.46\angle60 + \frac {3.46}{(628.32)(5k)C}\angle-30 = 10.01\angle0$$

$$\frac {3.46}{(628.32)(5k)C}\angle-30 = 8.81\angle-19.9$$

Where'd I go wrong?

I'm assuming it's the initial step of ignoring the DC component, but I don't know what to do with it.

In order to solve this problem, you need to use the full equation for the output voltage of a high pass filter, which includes both the input voltage and the DC component. The equation should be V_O = V_I - \frac {V_I}{1 + \frac {1}{j\omega RC}}. This takes into account the DC component of the input voltage and will give you the correct phase angle of -19.9 degrees. From there, you can solve for C using the given values for R, \omega, and the desired output voltage.

## 1. What is a Simple High Pass Filter?

A Simple High Pass Filter is an electronic circuit that allows high-frequency signals to pass through while blocking low-frequency signals. It is commonly used in audio equipment to remove unwanted low-frequency noise.

## 2. How does a Simple High Pass Filter work?

A Simple High Pass Filter works by using a capacitor and a resistor in series. The capacitor allows high-frequency signals to pass through, while the resistor blocks low-frequency signals. This creates a high-pass effect, where the high-frequency signals are allowed to pass through and the low-frequency signals are attenuated.

## 3. What are the applications of a Simple High Pass Filter?

A Simple High Pass Filter is commonly used in audio equipment, such as speakers and microphones, to filter out unwanted low-frequency noise. It is also used in radio frequency circuits to block unwanted signals and in electronic crossovers to separate high and low-frequency signals for different speakers.

## 4. How do I calculate the cutoff frequency of a Simple High Pass Filter?

The cutoff frequency of a Simple High Pass Filter is determined by the values of the capacitor and resistor in the circuit. The formula for calculating the cutoff frequency is: fc = 1 / (2πRC), where fc is the cutoff frequency in Hertz, R is the resistance in Ohms, and C is the capacitance in Farads.

## 5. What are the advantages of using a Simple High Pass Filter?

The advantages of using a Simple High Pass Filter include the ability to remove unwanted low-frequency noise, improve signal clarity and quality, and protect sensitive audio equipment from damage caused by low-frequency signals. It is also a relatively simple and inexpensive circuit to implement.

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