# Electronics/Expanded Edition Resonance

## Resonance

Simple resonant circuit, description.

## Filters

### Ideal Filters

(Discussion of how ideal filters provide the signal until the break frequency/cies and then provide total attenuation. That is they look like a $u(\omega )-uu(\omega -\omega _{c})$  for low pass filter, where u(w) is the unit step function or Heaviside function. That is they have infinite drop off at the cutoff frequency. How this is not possible. Pretty diagrams of all the filters Low Pass, High Pass, Band Pass, Band Stop.)

This section introduces first order butterworth low pass and high pass filters. An understanding of Laplace Transforms or at least Laplace Transforms of capacitors, inductors and resistors.

### Low Pass

Transforming the Resistor and Capacitor to the Laplace domain we get:

R and ${\frac {1}{Cs}}$ .

Expressing $V_{out}(s)$  using terms of $V_{in}(s)$ .

$V_{out}(s)={\frac {V_{in}(s){\frac {1}{Cs}}}{R+{\frac {1}{Cs}}}}$
$V_{out}(s)={\frac {V_{in}(s)}{RCs+1}}$

The transfer function is

$H(s)={\frac {V_{out}(s)}{V_{in}(s)}}$

So

$H(s)={\frac {1}{RCs+1}}$

For the Frequency Domain we put $s=j\omega$

$H(\omega )={\frac {1}{jRC\omega +1}}$

The magnitude is

$|H(\omega )|={\frac {1}{\sqrt {R^{2}C^{2}\omega ^{2}+1}}}$

and the angle is

$

As $\omega$  increases $|H(j\omega )|$  decreases so this circuit must represent low pass filter.

Using the -3 dB definition of band width. $|H(j\omega )|={\frac {1}{\sqrt {2}}}$

${\frac {1}{\sqrt {2}}}={\frac {1}{\sqrt {R^{2}C^{2}\omega ^{2}+1}}}$

Therefore

$2=R^{2}C^{2}\omega ^{2}+1$
$\omega ={\frac {1}{RC}}$

Which gives the general form of a low pass butterworth filter as:

$|H(\omega )|={\frac {1}{\sqrt {1+({\frac {\omega }{\omega _{c}}})^{2k}}}}$

, where k is the order of the filter and $\omega _{c}$  is the cut-off frequency.

### High Pass

(Image of a first order RL high pass filter)

If all the component of the circuit are transformed into the Laplace Domain. The resistor becomes $R$  and the inductor becomes $Ls$ . Using voltage divider rule $H(s)$  below is reached.

$H(s)={\frac {\frac {Rs}{L}}{1+{\frac {Rs}{L}}}}$

If $H(s)$  is transformed into the frequency domain by putting $s=j\omega$ .

$H(\omega )={\frac {\frac {j\omega R}{L}}{1+{\frac {j\omega R}{L}}}}$

Which has a magnitude of

$|H(\omega )|={\frac {\sqrt {\frac {\omega ^{2}R^{2}}{L^{2}}}}{\sqrt {1+{\frac {\omega R^{2}}{L^{2}}}}}}$

and an angle of

$
(cut-off frequency is w (R/L)^0.5)