Complete Response II – Sinusoidal Input

If the input voltage to the RC circuit considered previously is sinusoidal $v(t)=V_s \cos(\omega t+\psi)$, then the DE becomes

$$v_R(t)+v_C(t)=\tau\frac{d}{dt} v_c(t)+v_c(t) =\tau\dot{v}_c(t)+v_c(t)=V_s \cos(\omega t+\psi)$$ (146)

The homogeneous solution is the same as before $v_h(t)=A e^{-t/\tau}$. To find the particular solution $v_p(t)$ as the steady state response to the input $V_s\cos(\omega t+\psi)$, we first assume the input is $V_s e^{j(\omega t+\psi)}$, and the particular solution is $y_p(t)=B e^{j\omega t}$ and $\dot{y}_p(t)=j\omega B e^{j\omega t}$. The DE can now be written as

$$\tau\dot{v}_c(t)+v_c(t)=j\omega\tau Be^{j\omega t}+Be^{j\omega t} =(j\omega\tau +1) B e^{j\omega t} =V_s e^{j\omega t} e^{j\psi}$$ (147)

Solving for $B$ we get

$$B=\frac{V_s e^{j\psi}}{j\omega\tau+1}$$ (148)

Substituting $B$ back into $y_p(t)$ we get

$$y_p(t)=B e^{j\omega t}=\frac{V_s e^{j\psi}}{j\omega\tau+1} e^{j\omega t} =\frac{V_s}{\sqrt{(\omega\tau)^2+1}}e^{j(\omega t+\psi-\phi)}$$ (149)

where $\phi=\tan^{-1}\omega\tau$. Taking the real part of the above we get the particular solution as the response to the actual input $V_s\cos(\omega t+\psi)=Re(V_s e^{j(\omega t+\psi)})$

$$y_p(t)=Re\left[\frac{V_s e^{j(\omega t+\psi-\phi)}}{j\omega\tau+1}\right] =\frac{V_s}{\sqrt{(\omega\tau)^2+1}}\cos((\omega t+\psi-\phi)$$ (150)

Alternatively, we can also use the phasor method. The input can be written as

$$v(t)=V_s \cos(\omega t)=Re[\dot{V} e^{j\omega t}]$$ (151)

where $\dot{V}=V_s\; e^{j\psi}$ is the phasor form of the input voltage. The voltage across $C$ is (voltage divider):

$$\dot{V}_C = \dot{V} H(\omega)=\dot{V} \frac{Z_C}{Z_R+Z_C} =\dot{V} \frac{1/j\... ...c{\dot{V}}{j\omega \tau+1} =\frac{\dot{V}}{\sqrt{(\omega \tau)^2+1}\;e^{j\phi}}$$

$$= \frac{\dot{V}\;e^{-j\phi}}{\sqrt{(\omega \tau)^2+1}} =\frac{V_se^... ...\sqrt{(\omega \tau)^2+1}} =\frac{V_se^{j(\psi-\phi)}}{\sqrt{(\omega \tau)^2+1}}$$ (152)

where $\phi=\tan^{-1}\omega\tau$, and $H(\omega)$ is the frequency response function (FRF) of the system:

$$H(\omega)=\frac{Z_C}{Z_C+Z_R}=\frac{1/j\omega C}{R+1/j\omega C} =\frac{1}{\sqrt{1+(\omega\tau)^2}} e^{-j\phi}$$ (153)

In time domain the steady state voltage $v_p(t)$ is:

$$v_p(t) = Re\left[\dot{V}_C e^{j\omega t} \right] =Re\left[ \frac{V_s e^{j(... ...)^2+1}} \right] = \frac{V_s}{\sqrt{(\omega \tau)^2+1}} \cos(\omega t+\psi-\phi)$$

$$= \vert H(\omega)\vert V_s\cos(\omega t+\psi-\angle H(\omega))$$ (154)

Note that this steady state output is simply the input $v(t)=V_S\cos(\omega t+\psi)$ scaled by the magnitude of the FRF $\vert H(\omega)\vert=1/\sqrt{(\omega\tau)^2+1}$ with a phase shift equal to the phase of the FRF $\angle H(\omega)=-\phi$.

We get the complete solution as the sum of the homogeneous and particular solutions:

$$v_C(t)=v_h(t)+v_p(t)= \frac{V_s}{\sqrt{(\omega \tau)^2+1}}\cos(\omega t+\psi-\phi)+A e^{-t/\tau}$$ (155)

From the initial condition $v_C(0)=V_0$, we have

$$v_C(t)\bigg\vert _{t=0}=v_C(0) =V_0=\frac{V_s}{\sqrt{(\omega \tau)^2+1}}\cos(\psi-\phi)+A$$ (156)

Solving for $A$ we get

$$A=V_0-\frac{V_s}{\sqrt{(\omega \tau)^2+1}}\cos(\psi-\phi)$$ (157)

Substituting $A$ back to the expression of $v_C(t)$, we get

$$v_C(t)=\frac{V_s}{\sqrt{(\omega \tau)^2+1}}\cos(\omega t+\psi-\ph... ...left[V_0-\frac{V_s}{\sqrt{(\omega \tau)^2+1}}\cos(\psi-\phi)\right] e^{-t/\tau}$$ (158)

The same result can also be obtained using the Laplace transform method.

Again we consider a short-cut method, by generalizing the result above to

$$f(t)=f_{\infty}(t)+[f(0)-f_{\infty}(0)]e^{-t/\tau}$$ (159)

in terms of three essential components

1. $f(0)$: the initial value (same as before);
2. $f_{\infty}(t)$: the steady state response, and $f_{\infty}(0)=f_{\infty}(t)\big\vert _{t=0}$ is $f_{\infty}(t)$ evaluated at $t=0$;
3. $\tau$: the time constant of the system (same as before).

Note that when the input $V_s$ is a constant, the steady state response $f(\infty)$ is a constant, but when the input $V_s\cos(\omega t+\psi)$ is sinusoidal, the steady state response $f_{\infty}(t)$ is a function of time $t$, but its evaluation at $t=0$ $f_{\infty}(0)=f_{\infty}(t)\big\vert _{t=0}$ is still a constant. We see that the complete response is composed of the steady state response and the exponential decay of the difference between the initial value $f(0)$ and the steady state response evaluated at $t=0$.

If the initial voltage on $C$ is zero $V_0=0$, then

$$v_C(t)=\frac{V_s}{\sqrt{(\omega \tau)^2+1}}[\cos(\omega t+\psi-\phi) -\cos(\psi-\phi) e^{-t/\tau}]$$ (160)

Note that the initial magnitude of the transient component at $t=0^+$ varies depending on the angle $\psi-\phi$.

• If $\psi-\phi=\pm 90^\circ$, then $\cos(\psi-\phi)=0$, and the transient component disappears altogether.
• If $\psi-\phi=0^\circ$ or $180^\circ$, then $\cos(\psi-\phi)=\pm 1$, i.e., the magnitude of the transient component reaches maximum (either positive or negative), and if $\tau$ value is large and therefore the transient component decays slowly, the magnitude of the initial voltage $v_C(t)$ could be close to three times the peak of the steady state.
• In all other cases, the amplitude of the transient component is between 0 in the first case and the maximum value in the second case.

The three cases for $\psi-\phi$ to be $0^\circ$, $90^\circ$, and $180^\circ$ are shown below:

Example 2:

An electromagnet, modeled by a resistor $R=20\Omega$ and $L=0.3H$, is powered by sinusoidal voltage of $120V$ and $60Hz$. Find the current through the circuit when the switch is closed at $t=0$ when the phase angle happens to be $\psi=10^\circ$, i.e., $v(t)=120\sqrt{2}\; \cos(6.28\times 60 t+10^\circ)$.

• Find initial value: $i(0)=0$.
• Find impedance of circuit:

  $$Z=R+j\omega L=20+j6.28\times 60 \times 0.3=20+j113=114.8\angle{80^\circ}$$ (161)

• Find steady state value $i_\infty(t)$ by phasor method:

  $$\dot{V}=120 \angle{10^\circ},\;\;\;\;\dot{I}=\frac{\dot{V}}{Z} =\frac{120\angle{10^\circ}}{114.8\angle{80^\circ}} =1.05\angle{-70^\circ}$$ (162)

  $$i_\infty(t)=1.05\sqrt{2}\;\cos(6.28\times 60 t -70^\circ)$$ (163)

  $$i_\infty(0)=1.05\sqrt{2}\;\cos(-70^\circ) =1.05\sqrt{2}\times 0.342=0.51$$ (164)

• Find time constant $\tau=L/R=0.3/20=0.015$.
• Find current

  $$i(t)=i_{\infty}(t)+[i(0)-i_{\infty}(0)]e^{-t/\tau} =1.05\sqrt{2}\;\cos(6.28\times 60 t -70^\circ)-0.51 e^{-t/0.015}$$ (165)