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Fun with filters!

Note: A basic knowledge of electrical engineering principles is helpful, but not vital, for understanding this node. Knowledge of the behaviour of capacitors and inductors is also desirable.

Band pass and band stop filters, although extremely similar in construction and use, perform very different functions. A band pass filter allows frequencies or ac voltages of a specific frequency (called the resonance frequency) to go through, and blocks out or at least severely attenuates any other frequency. Band stop filters are quite the opposite: they block specific frequencies from passing, and allow all others to go through. These things are used widely in various fields of engineering, from reducing noise to metal detectors.

Theory:
The relationship between voltage and current in a capacitor is given by:

```                dv                   Ic = capacitor current
Ic = C--                    C = capacitance
dt                    v = rms voltage
```
And in an inductor:
```                                     IL = inductor current
IL =  1/L∫vdt              L = inductance
v = rms voltage
```
If the applied voltage is a perfect sinusoid, i.e. v = Vsin(ωt), then the respective currents will be:
```     Ic = ωCVcos(ωt)    and    IL = 1/ωLVcos(ωt)
```
Thus, the sine wave has produced a cosine current. This is equivalent to a phase shift of 90˚. For the capacitor this shift is +90˚, for the inductor this is -90˚. However, fortuitously we can ignore the effects of phase shifting in these filters.

But where does that get us? I'm glad you asked, Timmy. Now we have expressions for V and I, we can relate the two in accordance with Ohm's Law, which produces:
```        V/IL = ωL     and     V/Ic = 1/ωC
```
And since V/I is measured in ohms (Ω), ωL and 1/ωC indicate the impedance of an inductor or capacitor to the flow of a sinewave current which has frequency ω.

For a list of symbols used, please see the end of this node.

Basic circuit representations of filters:
Crappy ASCII art, go!

```
KEY:

|                    |                  |
\                   ---        ||       3
/  OR --/\/\/\--    ---   OR --||--     3   OR  --nnnn--
|                    |         ||       |
Resistor             Capacitor           Inductor
```

Circuit 1: Band pass filter

``` __________                             A
|          |----------/\/\/\-----------+    /|\
|          |/|\                      __|__   |
|  Input   | |                      |     |  |
|Oscillator| | Vin                 ---    3  |  Vout
|          | |                     ---    3  |
|          | |                      |_____|  |
|          | |                         |     |
|__________|---------------------------+     |
B
```

Circuit 2: Band stop filter

```                       ||
__________        +---||---+           A
|          |-------|   ||   |----------+  /|\
|          |/|\    |        |          |   |
|  Input   | |     +--nnnn--+          \   |
|Oscillator| | Vin                     /   | Vout
|          | |                         \   |
|          | |                         /   |
|          | |                         |   |
|__________|---------------------------+   |
B
```

Calculating the resonance frequency

Definition of resonance frequency: That frequency at which the capacitor||inductor pair exhibits the highest impedance.

In the band pass filter, this is the frequency allowed to pass. In the band stop filter, this is the frequency blocked.

There are two ways to calculate the resonance frequency (f0): mathematically with a formula, or through experimentation. Both are explained.. below.

Mathematically:
Both the inductance (L) and capacitance (C) need to be known. These can then be plugged into the following formula:

```       ω0 = √LC   where   ω0 = 2πf0
Therefore
f0 = 1 / 2π√LC
```

Experimentation
Ah, the simple, old-fashioned method of plugging it all up and throwing the switch. The important readings to be taken are marked on the circuit diagrams as `Vout` and `Vin`.
Get readings for `Vout` and `Vin` for a range of different frequencies, say 50 - 500 Hz. For each reading, calculate `Vout / Vin`. Now, plot these points on a graph, with Vout / Vin on the y-axis and log(frequency) on the x-axis. Something approaching the following should come out:

```
1  |
|
0.9   |                         x
|                        x|x
0.8   |                       x | x
|                       x | x
0.7   |                      x  | x
|                      x  |  x
0.6   |                     x   |   x
Vo        |                    x    |    x
--  0.5   |                   x     |     x
Vi        |                 x       |       x
0.4   |               x         |           x
|                         |
0.3   |               o         |          o
|                  o      |      o
0.2   |                     o   |   o
|                       o | o
0.1   |                        o|o
|                         o
0   +-------------------------X------------------
10                  100   f0             1000
Frequency (Hz) (log scale)

KEY:
x == band pass filter
o == band stop filter
```
The point where the peak and trough line up onto the x-axis is equal to f0. This line should be straight and vertical - if not, perhaps some experimental inaccuracy is to blame.

And that, as they say, is that. A quick note: For the band pass filter, a high-value resistor (>1kΩ) is preferable, and for the band stop filter a low-value resistor (<1kΩ, ~100Ω) gives better results.
this node is almost over

List of variables, constants and other jobbies used

```   L = inductance, measured in Henrys.
C = capacitance, measured in Farads.
Ic = capacitor current, measured in amperes.
IL = inductor current, measured in amperes.
v = r.m.s. voltage, measured in volts.
V = peak voltage, measured in volts.
Vout = output r.m.s. voltage, measured in volts.
Vin = input r.m.s. voltage, measured in volts.
ω = radian frequency, equal to 2πf.
f = frequency, measured in hertz.
f0 = resonance frequency, measured in hertz.
π = constant, approximately equal to 3.1415926535...

it's over```

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