I am studying low and high pass RL filters for Navy schooling. I need someone explain why the frequency cutoff is $fco = \frac{R}{2\pi L}$ mathematically.

From my oscilloscope lab results I got the following. $V_{pp}$ represents the peak to peak voltage :

\begin{array}{|c|c|} \hline kHz & V_{pp} \\ \hline 0.1 & 12 \\ \hline 1 &12 \\ \hline 10 &11.73 \\ \hline 100 & 5.17 \\ \hline \end{array}

The lab asked us to to find the value of $0.707 \times 12$ to determine where the change over occurred. Conceptually, I understand that the root mean square method is used to determine AC's efficiency in comparison to DC. AC are about 70.7% as effective as DC.


1 Answer 1


This is where the 0.7 comes from:

Single pole low pass frequency responses are in written in the form: $$ A_v=\dfrac{1}{\sqrt{1+\left( \dfrac{\omega}{\omega_{co}}\right) ^2}}=\dfrac{1}{\sqrt{1+\left( \dfrac{f}{f_{co}}\right) ^2}} $$ for $\omega=2\pi f$ and $A_v=\left(\dfrac{v_{out}}{v_{in}}\right)$.

Plots of frequency responses are call Bode plots (try the wikipedia page though its not so good on this topic) and it is much simpler to use linear approximations than worry about the exact response. When the frequency (or angular frequency $\omega$) is small then the "1" on the bottom line dominates and Av simplifies to: $$A_v\approx 1$$.

When the freqency is large the bottom line becomes $\frac{\omega}{\omega_{co}}$ so $$A_v\approx\dfrac{1}{\left(\frac{\omega}{\omega_{co}}\right)}=\dfrac{1}{\left(\frac{f}{f_{co}}\right)}$$

If you plot these linear approximation curves on a log-log graph (or you plot the dB gain versus log of the frequency) they will intercept at the cut off frequency point. The exact value at this point is -3dB or $\frac{1}{\sqrt{2}}$ because: $$A_v=\dfrac{1}{\sqrt{1+\left(\frac{f_{co}}{f_{co}}\right)^2}}=\dfrac{1}{\sqrt{1+\left(1\right)^2}}=\dfrac{1}{\sqrt{2}}\approx 0.707$$

The root mean square of a sine wave is $\frac{1}{\sqrt{2}}$ as well but there is not a direct connection between the two. In fact as long as you divide the input and output then it doesn't matter if you use peak values, RMS value, or peak-peak values.

As to why the RL low pass filter works like it does, you need to use complex numbers: the relationship between the RL low pass filter: $\dfrac{v_{out}}{v_{in}} =\dfrac{R}{R+j 2\pi f L}=\dfrac{1}{1+j \dfrac{2\pi f L}{R}} $

Taking the magnitude of this complex ratio gives: $\dfrac{1}{\sqrt{1 + \left(\dfrac{2\pi f L}{R} \right)^2}}=\dfrac{1}{\sqrt{1+\left( \dfrac{f}{f_{co}}\right) ^2}} $

where $f_{co}=\dfrac{R}{2 \pi L} $


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