Particular Solution

• RC circuit:

  When the input to the system represented by the term on the right-hand side of the DE is non-zero, the DE can be solved to find its particular or steady state solution. We first a constant input:

  $$\tau\dot{v}_C(t)+v_C(t)=v_s(t)=V_s$$ (84)

  With a constant input, the particular solution $v_C(t)=c$ must be constant as well, and the DE becomes $c=V_s$, i.e., the solution is trivially

  $$v_C(t)=V_s$$ (85)

  We next consider a sinusoidal input $\cos(\omega t)$. We first assume the input is a complex exponential input:

  $$\tau\dot{v}_C(t)+v_C(t)=v_s(t)=e^{j\omega t}=\cos(\omega t)+j\sin(\omega t)$$ (86)

  and assume

  $$v_C(t)=A\;e^{j\omega t},\;\;\;\;\;\frac{d}{dt}v_C(t)=j\omega A\;e^{j\omega t}$$ (87)

  and substitute them back into the DE to get

  $$j\omega \tau A e^{j\omega t}+A e^{j\omega t} =(j\omega \tau+1)A\; e^{j\omega t}=e^{j\omega t}$$ (88)

  Solving for $A$ we ge

  $$A=\frac{1}{j\omega\tau+1}=\vert A\vert e^{-j\phi},\;\;\;\; \;\;\;\;\;\;\vert A\vert=\frac{1}{\sqrt{\omega^2 \tau^2+1}},\;\;\;\;\; \phi=\angle A=\tan^{-1}\omega\tau$$ (89)

  and the solution is:

  $$A\;e^{j\omega t}=\frac{1}{\sqrt{\omega^2 \tau^2+1}} e^{j(\omega t-\phi)}$$ (90)

  As the system is linear therefore superposition applies, we can find the output corrresponding to a sinusoidal input $v_s(t)=\cos(\omega t)=Re(e^{j\omega t})$, by taking the real part of the solution above:

  $$v_C(t)=Re\left[\frac{1}{\sqrt{\omega^2 \tau^2+1}} e^{j(\omega t-\phi)}\right] =\frac{1}{\sqrt{\omega^2 \tau^2+1}} \cos(\omega t-\phi)$$ (91)

  Alternatively, this RC circuit can also be solved more conveniently by the phasor method, if only the steady state solution (the particular solution) is of interest. The phasor of the input voltage $v(t)=\cos(\omega t)$ is simply 1, and the phasor of the voltage across $C$ can be found by voltage divider:

  $$V_C=V\frac{Z_C}{Z_R+Z_C}=\frac{1/j\omega C}{R+1/j\omega C} =\frac... ...ga RC+1}=\frac{1}{j\omega \tau+1} =\frac{1}{\sqrt{\omega^2\tau^2+1}} e^{-j\phi}$$ (92)

  where $\phi=\tan^{-1}\omega\tau$. Converting this phasor to time function we get

  $$v_C(t)=Re[V_Ce^{j\omega t}] =Re\left[\frac{1}{\sqrt{\omega^2\tau^... ...{j(\omega t-\phi)}\right] =\frac{1}{\sqrt{\omega^2\tau^2+1}}\cos(\omega t-\phi)$$ (93)

• RL circuit:

  Given an RL circuit consisted of a resistor and an inductor in series connected to an AC voltage source $v(t)=\cos(\omega t)$, we want to find the current $i(t)$. The governing DE describing the circuit can be obtained based on KVL:

  $$v_L(t)+v_R(t)=L\frac{d}{dt}i(t)+R\;i(t)=v(t)=\cos(\omega t)$$ (94)

  We first consider an complex exponential input:

  $$L\frac{d}{dt}i(t)+Ri(t)=e^{j\omega t}$$ (95)

  We assume

  $$i(t)=A\;e^{j\omega t},\;\;\;\;\;\frac{d}{dt}i(t)=j\omega A\;e^{j\omega t}$$ (96)

  and substitute them back into the DE to get

  $$j\omega L A e^{j\omega t}+RA e^{j\omega t} =(R+j\omega L)A\; e^{j\omega t}=e^{j\omega t}$$ (97)

  i.e.

  $$A=\frac{1}{R+j\omega L}=\vert A\vert e^{-j\phi},\;\;\;\; \;\;\;\;\;\;\vert A\vert=\frac{1}{\sqrt{R^2+\omega^2 L^2}},\;\;\;\;\; \phi=\angle A=\tan^{-1}\left(\frac{\omega L}{R}\right)$$ (98)

  and the solution is:

  $$A\;e^{j\omega t}=\frac{1}{\sqrt{R^2+\omega^2 L^2}} e^{j(\omega t-\phi)}$$ (99)

  The solution for the DE with input $v(t)=\cos \omega t$ is

  $$i(t)=Re\left[\frac{1}{\sqrt{R^2+\omega^2 L^2}} e^{j(\omega t-\phi)}\right] =\frac{1}{\sqrt{R^2+\omega^2 L^2}} \cos(\omega t-\phi)$$ (100)

  Alternatively, this RL circuit can also be resolved by the phasor method. Now the current $i(t)$ can be much more easily found by generalized Ohm's law. The phasor representation of the input voltage $v(t)=\cos(\omega t)$ is simply 1, and the overall complex impedance of the two elements in series is:

  $$Z=Z_R+Z_L=R+j\omega L=\vert Z\vert e^{j\angle Z}=\sqrt{R^2+\omega^2L^2}\;e^{j \phi}$$ (101)

  where $\phi=\tan^{-1}(\omega L/R)$. By generalized Ohm's law, the phasor representation of the current can be found as

  $$I=\frac{1}{\sqrt{R^2+\omega^2L^2}\;e^{j\phi}} =\frac{1}{\sqrt{R^2+\omega^2L^2}} e^{-j\phi}$$ (102)

  and the real current is

  $$i(t)=Re\left[\frac{1}{\sqrt{R^2+\omega^2 L^2}}e^{-j\phi)}e^{j\omega t}\right] =\frac{1}{\sqrt{R^2+\omega^2 L^2}}\cos(\omega t-\phi)$$ (103)

In general, the frequency response function (FRF) of the system is defined as the ratio of the output to the input of the system, both represented as complex exponentials. In this specific case, we have

$$H=\frac{{\bf I}}{{\bf V}}=\frac{1}{Z}$$ (104)

with

$$\vert H\vert=\frac{1}{\sqrt{R^2+\omega^2L^2}},\;\;\;\;\;\angle H=-\phi$$ (105)

Therefore the steady state output can be found to be:

$$i(t)=\vert H\vert \cos(\omega t+\angle H) =\frac{1}{\sqrt{R^2+\omega^2 L^2}}\cos(\omega t-\phi)$$ (106)

The second method, much easier than the first one, is actually a short cut representation of the first DE method. This is the justification of the complex variable or phasor method for analyzing AC circuits. However, note that the phasor method can only find the steady state solution. The homogeneous differential equation will have to be solved to obtain the transient solution.