# First Order RL Filter

To complete out the triad of passive components, we’ll next be looking at the inductor to help us make a high pass filter. Clearly serving as the complement to the low-pass filter, HPFs allow us to remove ugly DC noise from our circuits. This is critical for anything wireless, like GPS or WiFi or BlueTooth. It’s also useful lower frequency items like speakers or headphones, since we all hate hearing the humming noise of batteries or wall outlets in our music. While we can simply swap the order of the components in our past RC Filter, it’s always nice to have more options. Let’s start by defining the impedance equation for an inductor:

$$ Z_L = jwL \text{, where } j = \sqrt{-1}\text{, } w_{rad}=2\pi f_{Hz} $$

Immediately we can see it differs from the capacitor due to the relationship between frequency and impedance. The capacitor had an inverse relationship and here we have a linear one. Thinking again in limits, we can see at DC the inductor is a short, and at high frequencies it represents an open circuit. Just to define our intuition in a slightly more formal manner:

$$ \lim_{w \rightarrow 0} Z_L = jL\cdot 0 = 0 $$

$$ \lim_{w \rightarrow \infty} Z_L = jL\cdot \infty = \infty $$

I always found it easier to remember inductor formulas, because I visualize the frequency as the loops. Low frequency means less loops means less distance, so we have less inductance and therefore impedance. High frequency is lots of loops and consequently lots of impedance. Let’s sketch out a quick high pass filter below:

Now calculating the transfer function from input to output voltage to get the 3dB point:

$$ \dfrac{V_{out}}{V_{in}} = \dfrac{jwL}{jwL+R} = \dfrac{1}{1+\frac{R}{jwL}} = \dfrac{1}{1-\frac{jR}{wL}} $$

$$ \left| \dfrac{1}{1-\dfrac{jR}{wL}} \right| = \left| \dfrac{1}{1-\dfrac{jR}{wL}} \cdot \dfrac{1+\dfrac{jR}{wL}}{1+\dfrac{jR}{wL}} \right| = \left| \dfrac{1+\dfrac{jR}{wL}}{1+\dfrac{R^2}{(wL)^2}} \right| $$

$$ \dfrac{1}{\sqrt{2}} = \sqrt{ \left(\dfrac{1}{1+\dfrac{R^2}{(wL)^2}}\right)^2 + \left(\dfrac{\dfrac{R}{wL}}{1+\dfrac{R^2}{(wL)^2}}\right)^2} = \sqrt{ \dfrac{1+\left(\dfrac{R}{wL}\right)^2}{\left(1+\frac{R^2}{(wL)^2}\right)^2}} $$

$$\dfrac{1}{2} = \dfrac{1}{1+\dfrac{R}{wL}}$$

$$ 2 = 1 + \dfrac{R}{wL} $$

$$ 1 = \dfrac{R}{wL} $$

$$ w = \dfrac{R}{L} $$

Now we’ll simulate it with a 10kΩ resistor and a 1mH inductor:

```
* Simple RL Filter
r1 n1 n0 10k
l1 n1 0 1mH
Vin n0 0 ac 1v
.ac DEC 10 1 1E8
.print ac v(n1) db(v(n1))
.end
```

Now it serves as an exact mirror of our plot from before. We start off at very, very low levels, just as expected. As we ramp up the frequency, we get more and more power across the inductor, finally leveling off. From our calculations prior, we know we’ll get half the power at the $R/L=10k\Omega/1mH = $ 10MHz point. After that, we effectively take in all the power, so there’s nothing more to see!