Phase and Bode Plots: Struggling to Understand High-Pass Filters

  • Thread starter Thread starter revival
  • Start date Start date
  • Tags Tags
    Bode Phase Plots
Click For Summary
SUMMARY

This discussion focuses on understanding phase plots for high-pass filters and LCR circuits, specifically how to derive the phase from the transfer function G = jwCR/(1+jwCR). The user struggles with the phase calculation and seeks clarity on the equation involving tan(phi). The solution involves expressing both the numerator and denominator in polar form to facilitate the determination of phase in Bode plots.

PREREQUISITES
  • Understanding of transfer functions in electrical circuits
  • Familiarity with Bode plots and their components
  • Knowledge of polar form representation of complex numbers
  • Basic concepts of high-pass filters and LCR circuits
NEXT STEPS
  • Study the derivation of phase angles in transfer functions
  • Learn how to express complex numbers in polar form
  • Explore the relationship between gain and phase in Bode plots
  • Investigate the mathematical properties of the arctangent function in phase calculations
USEFUL FOR

Electrical engineering students, circuit designers, and anyone looking to deepen their understanding of phase and Bode plots in filter design.

revival
Messages
1
Reaction score
0

Homework Statement


I am really struggling to understand how to draw a phase plot for circuits such as low/high-pass filters and other LCR circuits. I understand how to find the gain and use this information to make a Bode plot but when it comes to phase I just don't even know where to begin.
My notes seem to indicate some sort of equation along the lines of tan(phi)= tani^-1[tan(something) -tan(something)] but I haven't a clue what the 'somethings' are.


Homework Equations


If I take a simple high pass filter... G= jwCR/(1+jwCR).. how can I find the phase from this, help!


The Attempt at a Solution

 
Physics news on Phys.org
Start by expressing the numerator and denominator in polar form re. Then you can easily find G in polar form.
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
5K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 10 ·
Replies
10
Views
4K