ECE 311: Circuits II

Phasor Analysis

Circuit Analysis in Frequency Domain

Using a basic RLC circuit with all components in series.

$v{in}(t)=V{in}cos(\omega t + \theta_{in})$

Calculate the voltage $v{out}(t)$:
$P{v
{in}(t)} = \bold{V}{in} = V{in} \angle \theta {in}$
$-\bold{V}{in} + R * \bold{I} + j\omega L \bold{I} + \bold{V}{out} = 0$
$-\bold{V}_{in} + \bold{V}_R + ZL \bold{I} + \bold{V}{out} = 0$
$\bold{I} = \bold{I}_C = \frac{\bold{V}C}{\frac{1}{j\omega C}} = j\omega C \bold{V}{out}$
$-\bold{V}{in} + (j\omega R C + (j\omega)^2LC + 1)\bold{V}{out} = 0$
$\bold{V}{out} = \frac{\bold{V}{in}}{j\omega R C + (j\omega)^2LC + 1}$
$v{out}(t) = V{out}cos(\omega t + \theta_{out})$

Phasor Analysis Example

Example Circuit

Phasor of source: $P{v_1(t)} = \bold{V}_1 = 10\angle \pi / 8$

$\bold{V}2 = \frac{Z{R2}}{Z_{R2} + Z_L}\bold{V}_1$
$\bold{V}_2 = \bold{V}_1 * \frac{R_2}{R_2+j \omega L} = \frac{1}{\sqrt{2}\angle \pi / 4}\bold{V}_1$
$v_2(t) = P^{-1}{5\sqrt{2}\angle-\pi / 8} = 5\sqrt{2}cos(100t - \pi / 8)$

Impedance and Admitance

For an arbitrary passive network at angluar frequency $\omega$:
$Z = R + jX\Omega$

The association in series is a series of sums.

For an inductor, $X_L = \omega L$.
For a capacitor, $X_C = \frac{-1}{\omega C}$

The same network can be modeled with $Y = \frac{1}{Z}$.
$Y = G + jB\Omega^{-1}$

When is it convenient to use the impedance or the admittance?
The association in parallel is a series of sums.

For an inductor, $B_L = \frac{-1}{\omega L}$.

Example

$f = 100kHz$ $A = 100\Omega$

What is the equivalent impedance, $Z_{eq}$?

$\bold{Z}_{eq} = 100 || (\bold{Z}_1 + \bold{Z}_2$
$\omega = 2\pi f = 6283 rad/s$
$\bold{Z}_1 = 60||\frac{-j}{\omega C} = 60||[\frac{-j33.310^3}{\omega}] = (\frac{60 \frac{-j33.310^3}{\omega}}{60 - \frac{-j33.310^3}{\omega}}) = \bold{Z}_1 = 5.28\angle-85$
$\bold{Z}_1 = 60||j\omega L = 60||[j\omega 0.01] = \frac{60j\omega 0.01}{60j\omega 0.01} = \bold{Z}2 = 43.4\angle43.7$
$\bold{Z}
{eq} = \frac{100(5.28\angle-85+43.4\angle43.7)}{100+5.28\angle-85+43.4\angle43.7}\Omega = 30.0516\angle27.2109 \Omega$

AC Power Analysis

Instantaneous Power: $p(t) = v(t) * i(t)$
Average Power: $P = \frac{1}{t_1-t_0}\int^{t1}{t_0}p(t)dt$
For a periodic signal: $P = \frac{1}{T}\int^{t0+T}{t_0}p(t)dt$