Simple High Pass Filter Problem

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.
  • #1
SlideMan
42
0

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

Homework Equations



[tex] V_O = \frac {V_I}{1 + \frac {1}{j\omega RC}}[/tex]

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:

[tex]3.46\angle60 + \frac {3.46}{(628.32)(5k)C}\angle-30 = 10.01\angle0[/tex]

[tex]\frac {3.46}{(628.32)(5k)C}\angle-30 = 8.81\angle-19.9[/tex]

Where'd I go wrong?
 
Physics news on Phys.org
  • #2
I'm assuming it's the initial step of ignoring the DC component, but I don't know what to do with it.
 
  • #3




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.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
16
Views
948
  • Engineering and Comp Sci Homework Help
Replies
6
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
13
Views
4K
  • Engineering and Comp Sci Homework Help
Replies
11
Views
3K
  • Electrical Engineering
Replies
15
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
10
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
9
Views
34K
  • Engineering and Comp Sci Homework Help
Replies
23
Views
2K
  • Electrical Engineering
Replies
6
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
4
Views
3K
Back
Top