Calculating Low-Pass Filter Corner Frequency for LR Circuit

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 5K views
VinnyCee
Messages
486
Reaction score
0

Homework Statement



Show that a series LR circuit is a Low Pass filter if the output is taken across the resistor. Calculate the corner frequency, [itex]f_c[/itex], if L = 2mH and R = 10k[itex]\Omega[/itex].

Homework Equations



[tex]H(\omega)\,=\,\frac{V_0(t)}{V_i(t)}[/tex]

[tex]H\left(\omega_c\right)\,=\,\frac{1}{\sqrt{2}}[/tex]

[tex]f\,=\,\frac{\omega}{2\pi}[/tex]

The Attempt at a Solution



[tex]H(\omega)\,=\,\frac{R}{R\,+\,j\omega L}[/tex]

[tex]H(0)\,=\,1[/tex]

[tex]H(\infty)\,=\,0[/tex]

That does the "show" part. But now I don't know how to get the corner frequency.

[tex]\frac{1}{\sqrt{2}}\,=\,\frac{R}{R\,+\,j\omega L}[/tex]
 
Physics news on Phys.org
Never mind! It's 796 kHz.
 
Per your post on Bode plots, this transfer function would be written as:

1/(1+jwL/R), you might want to plot this before tackling that much more complicated function, if you haven't done simple ones yet.