Passive Filters¶

Impedance Model¶

Impedance $Z$ is defined as $Z=R+jX$ , where $R$ is the resistance of the circuit element and $X$ is the reactance of the circuit element
Extending Ohm's Law to complex numbers, we get the relationship

$V = IZ$

where $V$ is the potential difference between any two points in the circuit, $I$ is the current through the conductor and $Z$ is the impedance of the circuit element.

Tip

Impedance of a resistor: $Z_{R} = R$
Impedance of a capacitor: $Z_{C} = \frac{1}{j\omega C}$
Impedance of an inductor: $Z_{L} = j\omega L$

Impedance Across A Resistor¶

Let's say we have a time-varying potential difference across a resistor $v(t)$ and a time-varying current $i(t)$ flowing through the resistor.

We know that there is no phase difference between the voltage and current waveforms.

For simplicity's sake, we let

$v(t) = V \sin(\omega t)\\ i(t) = I \sin(\omega t)$

Therefore, using Ohm's Law the impedance of the resistor is simply

$\begin{aligned} Z_{resistor} &= \frac{v(t)}{i(t)}\\ &= \frac{V \sin(\omega t)}{I \sin(\omega t)}\\ &= R \end{aligned}$

Impedance Across A Capacitor¶

Let's say we have a time-varying potential difference across a capacitor $v(t)$ and a time-varying current $i(t)$ flowing through the capacitor.

Capacitors have a unique property - they can store electric charges. The amount of electric charge $Q$ that a capacitor stores can be calculated using

$Q = CV$

where $C$ is the capacitance of the capacitor and $V$ is the potential difference across the capacitor.

To find out the rate of change of electric charge in a capacitor, we can take the derivative of the equation:

$\begin{aligned} \frac{dQ(t)}{dt} &= C \frac{dV(t)}{dt}\\ \because \frac{dQ(t)}{dt} &= I(t), \\ \therefore I(t) & = C\frac{dV(t)}{dt} \end{aligned}$

We now have the relationship between $i(t)$ and $v(t)$ . If we let

$\begin{aligned} v(t) = V \sin(\omega t)\\ \Rightarrow i(t) = \omega CV \cos(\omega t) \end{aligned}$

Using Ohm's Law we can find the impedance of the capacitor:

$\begin{aligned} Z_{capacitor} &= \frac{v(t)}{i(t)}\\ &= \frac{V \sin(\omega t)}{\omega CV \cos(\omega t)}\\ &= \frac{V \sin(\omega t)}{\omega CV \sin(\omega t + \frac{\pi}{2})}\\ &= \operatorname{Im}\left(\frac{{Ve^{j\omega t}}}{\omega CVe^{j(\omega t+\frac{\pi}{2})}}\right)\\ &= \operatorname{Im}(\frac{1}{\omega C}e^{-j\frac{\pi}{2}})\\ &= -\frac{j}{\omega C}\\ & = \frac{1}{j\omega C} \end{aligned}$

Impedance Across An Inductor¶

Let's say we have a time-varying potential difference across an inductor $v(t)$ and a time-varying current $i(t)$ flowing through the inductor.

From Lenz's Law we know that

$\mathcal{E} = -\frac{\partial \Phi_{B}}{\partial t}$

where $\Phi_{B}$ is the magnetic flux and $\mathcal{E}$ is the induced electromotive force.

Inductors have a quantifiable property known as inductance $L$ , which is simply the ratio of the magnetic flux linkage $\Phi_{B}$ to its time-varying current $i(t)$ flowing through the inductor.

$L = \frac{\Phi_{B}}{i(t)}$

By separation of variables we can combine the equations together:

$\begin{aligned} \mathcal{E} &= -\frac{\partial \Phi_{B}}{\partial t}\\ &= -\frac{\partial \Phi_{B}}{\partial i(t)}\cdot \frac{\partial i(t)}{\partial t} \end{aligned}$

The induced electromotive force $\mathcal{E}$ varies with time so we replace it with $v(t)$ . $\frac{\partial \Phi_{B}}{dt}$ is simply $L$ and can be replaced.

Thus the magnitude of the time-varying potential difference across an inductor $v(t)$ is related to the current $i(t)$ passing through it by the following relationship:

$\begin{aligned} v(t) &= L\frac{di(t)}{dt} \end{aligned}$

We now have the relationship between $i(t)$ and $v(t)$ . If we let

$\begin{aligned} i(t) &= I \sin(\omega t)\\ \Rightarrow v(t) &= \omega IL \cos(\omega t) \end{aligned}$

Using Ohm's Law we can find the impedance of the capacitor:

$\begin{aligned} Z_{inductor} &= \frac{v(t)}{i(t)}\\ &= \frac{\omega IL \cos(\omega t)}{I \sin(\omega t)}\\ &= \frac{\omega I L \sin(\omega t +\frac{\pi}{2})}{I \sin(\omega t)}\\ &= \operatorname{Im}\left(\frac{\omega IL e^{j(\omega t + \frac{\pi}{2})}}{Ie^{j\omega t}}\right)\\ &= \operatorname{Im}\left(\omega Le^{j\frac{\pi}{2}}\right)\\ &= j \omega L \end{aligned}$