Natural Response of Series RLC Circuits

\[\begin{aligned} &-12+4 i\left(0^{+}\right)+v_{L}\left(0^{+}\right)+v\left(0^{+}\right)=0 \\ &\Rightarrow v_{L}\left(0^{+}\right)=12-8-4=0 \\ &\frac{d i\left(0^{+}\right)}{d t}=\frac{v_{L}\left(0^{+}\right)}{L}=\frac{0}{0.25}=0 \mathrm{~A} / \mathrm{s} \\ &i(\infty)=0 \mathrm{~A}, \quad v(\infty)=12 \mathrm{~V} \end{aligned}\]

• Substituting \(i = A \cdot e^{st}\) in second-order D.E. \[\begin{aligned} &\frac{d^{2} i}{d t^{2}}+\frac{R}{L} \frac{d i}{d t}+\frac{i}{L C}=0 \\ &\Rightarrow A s^{2} e^{s t}+\frac{A R}{L} s e^{s t}+\frac{A}{L C} e^{s t}=0 \\ & \Rightarrow A e^{s t}\left(s^{2}+\frac{R}{L} s+\frac{1}{L C}\right)=0 \end{aligned}\]

• Since \(i=A e^{s t}\) is the assumed solution we are trying to find, only the expression in parentheses can be zero: \[s^{2}+\frac{R}{L} s+\frac{1}{L C}=0 \Leftarrow \color{red}{\text{characteristic equation of D.E.}}\]

• Roots of the characteristic equation \[\begin{aligned} &s_{1}=-\frac{R}{2 L}+\sqrt{\left(\frac{R}{2 L}\right)^{2}-\frac{1}{L C}} \\ &s_{2}=-\frac{R}{2 L}-\sqrt{\left(\frac{R}{2 L}\right)^{2}-\frac{1}{L C}} \end{aligned}\]

• Compact form \[\begin{aligned} &\boxed{s_{1}, s_{2}=-\alpha\pm\sqrt{\alpha^{2}-\omega_{0}^{2}}} \quad \boxed{\alpha=\frac{R}{2 L}} \quad \boxed{\omega_{0}=\frac{1}{\sqrt{L C}}} \end{aligned}\]

• \(s_{1}\) and \(s_{2}\) : natural frequencies, in nepers per second \((\mathrm{Np} / \mathrm{s})\), because they are associated with the natural response of the circuit

• \(\omega_{0}\) : resonant frequency or undamped natural frequency, in radians per second \((\mathrm{rad} / \mathrm{s})\)

• \(\alpha\) : neper frequency or damping factor, in nepers per second.

• In terms of \(\alpha\) and \(\omega_{0}\) Eq. can be written as \[s^{2}+\frac{R}{L} s+\frac{1}{L C}=0 \Rightarrow \boxed{s^{2}+2 \alpha s+\omega_{0}^{2}=0}\]

• Natural response of the series RLC circuit \[i(t) = A_1\cdot e^{s_1t} + A_2\cdot e^{s_2t}\]

• Constants \({A_1,A_2}\) are determined using initial values \(i(0)\) and \(\dfrac{di(0)}{dt}\)

• Three types of solutions:

  1. If \(\alpha>\omega_{0} \Rightarrow\) over-damped case

  2. If \(\alpha=\omega_{0} \Rightarrow\) critically damped case

  3. If \(\alpha<\omega_{0} \Rightarrow\) under-damped case

Over-damped Case:

\[\begin{aligned} &\boxed{s_{1}, s_{2}=-\alpha\pm\sqrt{\alpha^{2}-\omega_{0}^{2}}} \quad \boxed{\alpha=\frac{R}{2 L}} \quad \boxed{\omega_{0}=\frac{1}{\sqrt{L C}}} \end{aligned}\]

• \(\alpha > \omega_0 \Rightarrow C > 4L/R^2\)

• both roots \(s_1\) and \(s_2\) are negative and real

• Response : \[i(t) = A_1\cdot e^{s_1t} + A_2\cdot e^{s_2t}\]

• As \(t \uparrow\) response decays and approaches zero.

Critically Damped Case: \[\begin{aligned} &\boxed{s_{1}, s_{2}=-\alpha\pm\sqrt{\alpha^{2}-\omega_{0}^{2}}} \quad \boxed{\alpha=\frac{R}{2 L}} \quad \boxed{\omega_{0}=\frac{1}{\sqrt{L C}}} \end{aligned}\]

• \(\alpha = \omega_0 \Rightarrow C = 4L/R^2\)

• Roots : \[s_1 = s_2 = -\alpha = -\dfrac{R}{2L}\]

• Response : \[i(t) = A_1\cdot e^{-\alpha t} + A_2\cdot e^{-\alpha t} = A_3 \cdot e^{-\alpha t} \quad (A_3 = A_1 + A_2)\]

• Wrong solution, because the two initial conditions cannot be satisfied with the single constant \(A_3\) .

• What then could be wrong?

• When \(\alpha = \omega_0 = R/2L\)

\[ \begin{aligned} \frac{d^{2} i}{d t^{2}}+2 \alpha \frac{d i}{d t}+\alpha^{2} i &=0 \\ \frac{d}{d t}\left(\frac{d i}{d t}+\alpha i\right)+\alpha\left(\frac{d i}{d t}+\alpha i\right) &=0 \\ \frac{d f}{d t}+\alpha f &=0 \quad \left(f=\frac{d i}{d t}+\alpha i \right) \\ \frac{d i}{d t}+\alpha i &=A_{1} e^{-\alpha t} \quad \Leftarrow (\text{solution of}~ f)\\ e^{\alpha t} \frac{d i}{d t}+e^{\alpha t} \alpha i &=A_{1} \\ \frac{d}{d t}\left(e^{\alpha t} i\right) &=A_{1} \\ e^{\alpha t} i &=A_{1} t+A_{2} \Leftarrow \text{on Integrating}\\ i=&\left(A_{1} t+A_{2}\right) e^{-\alpha t} \end{aligned} \]

\[\boxed{ i=\left(A_{1} t+A_{2}\right) e^{-\alpha t}}\]

• Under-damped case: \(\alpha < \omega_{0}\)

  \[\begin{aligned} s_{1} &=-\alpha+\sqrt{-\left(\omega_{0}^{2}-\alpha^{2}\right)}=-\alpha+j \omega_{d} \\ s_{2} &=-\alpha-\sqrt{-\left(\omega_{0}^{2}-\alpha^{2}\right)}=-\alpha-j \omega_{d} \\ \omega_{d} &=\sqrt{\omega_{0}^{2}-\alpha^{2}} \end{aligned}\]

• Response: \[\begin{aligned} i(t) &=A_{1} e^{-\left(\alpha-j \omega_{d}\right) t}+A_{2} e^{-\left(\alpha+j \omega_{d}\right) t} \\ &=e^{-\alpha t}\left(A_{1} e^{j \omega_{d} t}+A_{2} e^{-j \omega_{d} t}\right) \end{aligned}\]

• Using Euler’s identities, \[e^{j \theta}=\cos \theta+j \sin \theta, \quad e^{-j \theta}=\cos \theta-j \sin \theta\]

• we get \[\begin{aligned} i(t) &=e^{-\alpha t}\left[A_{1}\left(\cos \omega_{d} t+j \sin \omega_{d} t\right)+A_{2}\left(\cos \omega_{d} t-j \sin \omega_{d} t\right)\right] \\ &=e^{-\alpha t}\left[\left(A_{1}+A_{2}\right) \cos \omega_{d} t+j\left(A_{1}-A_{2}\right) \sin \omega_{d} t\right] \end{aligned}\]

• Replacing constants \(\left(A_{1}+A_{2}\right)\) and \(j\left(A_{1}-A_{2}\right)\) with \(B_{1}\) and \(B_{2}\), \[i(t)=e^{-\alpha t}\left(B_{1} \cos \omega_{d} t+B_{2} \sin \omega_{d} t\right)\]

• Presence of sine and cosine functions makes the natural response is exponentially damped and oscillatory in nature

• Response has time constant of \(1/\alpha\) and period \(T=2\pi/\omega_d\)

• Due to two types of storage elements oscillatory response is possible with flow of energy back and forth between the two

• Ringing: The damped oscillation ehibited by underdamped response

• Critically damped (decays fastest) is the borderline between underdamped and over-damped

• Over-damped has longest settling time i.e. to dissipate the intial stored energy

• Critically response is the best choice to led the response settle fastest without oscillation and ringing

\[\begin{aligned} s_{1,2}&=-9 \pm j 4.359 ~\Rightarrow \alpha < \omega ~\Rightarrow \text{Under-damped} \\ i(t)&=e^{-9 t}\left(A_{1} \cos 4.359 t+A_{2} \sin 4.359 t\right) \\ i(0)&=1=A_{1} \\ \left.\frac{d i}{d t}\right|_{t=0}&=-\frac{1}{L}[R i(0)+v(0)]=-2[9(1)-6]=-6 \mathrm{~A} / \mathrm{s} \\ \frac{d i}{d t}&=-9 e^{-9 t}\left(A_{1} \cos 4.359 t+A_{2} \sin 4.359 t\right) \\ &+e^{-9 t}(4.359)\left(-A_{1} \sin 4.359 t+A_{2} \cos 4.359 t\right) \\ -6&=-9\left(A_{1}+0\right)+4.359\left(-0+A_{2}\right) \\ \Rightarrow \quad A_{2}&=0.6882 \\ i(t)&=e^{-9 t}(\cos 4.359 t+0.6882 \sin 4.359 t) \mathrm{A} \end{aligned}\]