ECE 311: Circuits II
More Circuits I Recap
$\bold{V}(t)=3cos(100t+25deg)+jsin(100t+25deg)=3e^{j(100t+25deg}$
Complex Numbers Representation
Rectangular Coordinates
$\bold{D} = a + jb$
Polar Coordinates
$\bold{D} = re^{j\phi} = r\angle\phi$
$r = \sqrt{a^2 + b^2}$
$\phi = \angle\bold{D} = arctan(b/a), a$
Phasor Analysis
$v(t) = V_p cos(\omega t + \theta) \rightarrow \bold{v}(t) = V_p e^{j(\omega t + \theta)}$
Phasor representation: $\bold{V} = V_p e^{j \theta}$
$\bold{V} = V_p e^{j \theta} e^{j \omega t}$
$V_p * e^{j \theta}$ is the phasor
$\bold{v}(t) = \bold{V}e^{j\omega t}$
$\bold{i}(t) = \bold{I}e^{j\omega t}$
Because a circuit is linear, if you drive a voltage at 100 rad/sec, all systems will have a frequency of 100 rad/sec.
$\frac{d}{dt}\bold{v}(t)=j\omega \bold{v}(t)$ $\frac{d}{dt}\bold{i}(t)=j\omega \bold{i}(t)$
Time differentiation â multiplication by $j\omega$
Application of Phasor
Sum of phasors:
$v(t) = 3cos(2\pi 60t + \pi / 7) + 2cos(2\pi 60t + \pi /4)$
Each of these waves have the same frequency, 60HZ.
Calculate a more simplified equation for $v(t)$:
$\bold{V}_1 = 3\angle\pi / 7$
$\bold{V}_2 = 2\angle\pi / 4$
$\bold{V} = \bold{V}_1 + \bold{V}_2 = 4.93\angle0.583$
$v(t) = 4.93cos(2\pi60t + 0.583)$
There are 2 main transforms, into phasor to do the addition, then back to cartesian to match the given equation.
Inductive Load
$\bold{i}(t) = I_p e^{j(\omega t + \phi)}$
$\bold{v}(t) = L\frac{d\bold{i}(t)}{dt} = j\omega LI_p e^{j(\omega t + \phi)}$
$\bold{v}(t) = j\omega LI_p e^{j(\omega t + \phi + \pi/2)}$
$\bold{I} = I_pe^{j\phi}$
$\bold{V} = V_pe^{j\theta}=j\omega LI_pe^{j\phi}$
Impedance is defined as $Z(\omega)=\frac{\bold{V}}{\bold{I}}$
$Z_L(\omega)=j\omega L$
Notice how the impedance is dependent on the frequency!
$j$ is the current lagging by 90 degrees
Capacitive Load
$\bold{v}(t) = V_p e^{j(\omega t + \theta)}$
$\bold{i}(t) = C\frac{d\bold{v}(t)}{dt} = j\omega CV_p e^{j(\omega t + \theta)}$
$\bold{V} = V_pe^{j\theta}$
$\bold{I} = I_pe^{j\phi}=j\omega CV_pe^{j\theta}$
Impedance is defined as $Z(\omega)=\frac{\bold{V}}{\bold{I}}$
$Z_C(\omega)=\frac{1}{j\omega C}$
Note that $\phi = \theta + \pi / 2$ so the current leads the voltage by 90 degrees
Impedance
$Z_R(\omega)=R$
$Z_L(\omega)=j\omega L$
$Z_C(\omega)=\frac{1}{j\omega C}$
However: This is only applicable to one harmonic frequency.
How do you do analysis of saw-tooth waves?
Decompose it to a bunch of sinusoidal waves!