A low pass filter eliminates high frequency signals from an input signal.

There are two main types of passive low pass filters, a **RC LowPass Filter**
Is made from a capacitor and a resistor while a **RL LowPass Filter**
Is made from an inductor and a resistor. 

![[rc-rl-lp-filter.png]]

We can also see that for the RL LowPass Filter the positions of the resistor and the reactive component are switched. This is because the inductor works in the opposite way of the capacitor, allowing low frequencies to pass and attenuating high ones.

**Example:**

Lets design a RC LowPass Filter with a [[Resources/electricity/glossary#Cutoff Frequency|Cutoff Frequency]] of $15.9kHz$. The Formular for calculating the cutoff frequency is the following:

![[formulas#Cutoff Frequency for RC Filters]]

So, we now have the following formula:

$\displaystyle 15900 = \frac{1}{2\pi RC}$

Now we still have two unknown variables to fill in, for my personal taste that is one variable to much. Usually when designing a low pass filter we first choose the value of $R$. 
We do not want $R$ to be too high, as that will limit the current flow. We also do not want it to be *too* low because then we would need a big capacitor. For now we will choose:

$\displaystyle R =1k\ohm$

With this our formula now looks something like this:

$\displaystyle 15900 = \frac{1}{2*\pi*1000*C}$

Now we only have one unknown variable, which is $C$ so we can solve the equation:

$$
\begin{flalign}
&15900 = \frac{1}{2*\pi*1000*C}& |& \textit{ lets simplify}\\\
&15900 = \frac{1}{2000*pi*C} & |&* 2000*pi*C \\
&31800000\pi C = 1 & |& \div318000000\pi \\
&C = \frac{1}{31800000\pi}\\
\\
&C = 0.00000001F = \textbf{10nF}

\end{flalign}
$$

Lets put those numbers into the simulator:

```circuitjs
$ 64 1e-8 30.13683688681966 50 5 50 5e-11
r 112 256 224 256 0 1000
c 224 256 224 368 0 1e-8 6.664641465475588 0.001
g 224 368 224 400 0 0
O 224 256 336 256 0 0
170 112 256 32 256 3 5000 44100 20 0.001
403 32 288 160 352 0 4_512_0_x8741c_0.0001_0.0001_-1_1_0.000049999999999999996_0
o 4 64 0 28686 19.99999999596653 0.0001 0 2 3 0
```


Or we can plot it

<iframe src="https://www.desmos.com/calculator/pesoftpl4g?embed" width="500" height="500" style="border: 1px solid #ccc" frameborder=0></iframe>


# Second Order Low Pass Filter

If we place two LPF's in series