# Show that a series LR circuit is an LP filter if the output is taken across the resis

1. Apr 11, 2007

### VinnyCee

1. The problem statement, all variables and given/known data

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

2. Relevant equations

$$H(\omega)\,=\,\frac{V_0(t)}{V_i(t)}$$

$$H\left(\omega_c\right)\,=\,\frac{1}{\sqrt{2}}$$

$$f\,=\,\frac{\omega}{2\pi}$$

3. The attempt at a solution

$$H(\omega)\,=\,\frac{R}{R\,+\,j\omega L}$$

$$H(0)\,=\,1$$

$$H(\infty)\,=\,0$$

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

$$\frac{1}{\sqrt{2}}\,=\,\frac{R}{R\,+\,j\omega L}$$

2. Apr 11, 2007

### VinnyCee

Never mind! It's 796 kHz.

3. Apr 11, 2007

### denverdoc

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.