Natural Response of Series RLC Circuits

Demonstrative Video


Series RLC Circuit without Source


Second Order Circuits or RLC Circuits

  • First-order circuits containing only one energy storage element, combined with a passive network which partly determined the time taken by either the \(C\) or \(L\) to charge/discharge.

  • The D.E. resulted from analysis were always first-order.

  • A second-order circuit may have two storage elements of different type or the same type (provided elements of the same type cannot be represented by an equivalent single element).

  • The result is a second-order D.E. for any voltage or current of interest.

  • We will now need two initial conditions to solve each D.E.

  • Such circuits occur in a wide variety of applications, such as oscillators, filters, automobile suspension systems, temperature controllers, etc.

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Finding Initial and Final Values

  • Unless otherwise stated \(v\) denotes capacitor voltage and \(i\) the inductor current

  • \(v(0), i(0), \dfrac{dv(0)}{dt}, \dfrac{di(0)}{dt}, i(\infty), v(\infty)\)

  • Note: carefully handle polarity of \(v(t)\) across the capacitor and \(i(t)\) through the inductor as per passive sign convention.

  • \[v(0^+) = v(0^-) \quad \text{and} \quad i(0^+) = i(0^-)\]
    \(t=0\)Remember capacitor voltage and inductor current are continuous

Solved Problem

The switch has been closed for a long time and open at \(t=0\). Find: (a) \(i\left(0^{+}\right), v\left(0^{+}\right)\), (b) \(d i\left(0^{+}\right) / d t, d v\left(0^{+}\right) / d t\), (c) \(i(\infty), v(\infty)\).

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\[\begin{aligned} &i\left(0^{-}\right)=\frac{12}{4+2}=2 \mathrm{~A}, \quad v\left(0^{-}\right)=2 i\left(0^{-}\right)=4 \mathrm{~V} \\ &i\left(0^{+}\right)=i\left(0^{-}\right)=2 \mathrm{~A}, \quad v\left(0^{+}\right)=v\left(0^{-}\right)=4 \mathrm{~V} \\ &i_{C}\left(0^{+}\right)=i\left(0^{+}\right)=2 \mathrm{~A} \\ &\frac{d v\left(0^{+}\right)}{d t}=\frac{i_{C}\left(0^{+}\right)}{C}=\frac{2}{0.1}=20 \mathrm{~V} / \mathrm{s} \end{aligned}\]
\[\begin{aligned} &-12+4 i\left(0^{+}\right)+v_{L}\left(0^{+}\right)+v\left(0^{+}\right)=0 \\ &\Rightarrow v_{L}\left(0^{+}\right)=12-8-4=0 \\ &\frac{d i\left(0^{+}\right)}{d t}=\frac{v_{L}\left(0^{+}\right)}{L}=\frac{0}{0.25}=0 \mathrm{~A} / \mathrm{s} \\ &i(\infty)=0 \mathrm{~A}, \quad v(\infty)=12 \mathrm{~V} \end{aligned}\]

The Source-Free Series RLC Circuit

\[\begin{aligned} v(0) &=\frac{1}{C} \int_{-\infty}^{0} i d t=V_{0} \\ i(0) &=I_{0} \end{aligned}\]
  • \[\begin{aligned} &R i+L \frac{d i}{d t}+\frac{1}{C} \int_{-\infty}^{t} i d t=0 \\ &\frac{d^{2} i}{d t^{2}}+\frac{R}{L} \frac{d i}{d t}+\frac{i}{L C}=0 \quad \Leftarrow \color{red}{\text{second-order D.E.}} \end{aligned}\]
    Applying KVL and then differentiating
  • Solution has general form \(i = A \cdot e^{st}\)

  • \[\begin{aligned} &R i(0)+L \frac{d i(0)}{d t}+V_{0}=0 \\ & \Rightarrow \frac{d i(0)}{d t}=-\frac{1}{L}\left(R I_{0}+V_{0}\right) \end{aligned}\]
    More initial condition
  • \[\begin{aligned} &\frac{d^{2} i}{d t^{2}}+\frac{R}{L} \frac{d i}{d t}+\frac{i}{L C}=0 \\ &\Rightarrow A s^{2} e^{s t}+\frac{A R}{L} s e^{s t}+\frac{A}{L C} e^{s t}=0 \\ & \Rightarrow A e^{s t}\left(s^{2}+\frac{R}{L} s+\frac{1}{L C}\right)=0 \end{aligned}\]
    in second-order D.E. Substituting
  • \[s^{2}+\frac{R}{L} s+\frac{1}{L C}=0 \Leftarrow \color{red}{\text{characteristic equation of D.E.}}\]
    is the assumed solution we are trying to find, only the expression in parentheses can be zero: Since
  • \[\begin{aligned} &s_{1}=-\frac{R}{2 L}+\sqrt{\left(\frac{R}{2 L}\right)^{2}-\frac{1}{L C}} \\ &s_{2}=-\frac{R}{2 L}-\sqrt{\left(\frac{R}{2 L}\right)^{2}-\frac{1}{L C}} \end{aligned}\]
    Roots of the characteristic equation
  • \[\begin{aligned} &\boxed{s_{1}, s_{2}=-\alpha\pm\sqrt{\alpha^{2}-\omega_{0}^{2}}} \quad \boxed{\alpha=\frac{R}{2 L}} \quad \boxed{\omega_{0}=\frac{1}{\sqrt{L C}}} \end{aligned}\]
    Compact form
  • \(s_{1}\) 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.

  • \[s^{2}+\frac{R}{L} s+\frac{1}{L C}=0 \Rightarrow \boxed{s^{2}+2 \alpha s+\omega_{0}^{2}=0}\]
    Eq. can be written as and In terms of
  • \[i(t) = A_1\cdot e^{s_1t} + A_2\cdot e^{s_2t}\]
    Natural response of the series RLC circuit
  • 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

  • \[i(t) = A_1\cdot e^{s_1t} + A_2\cdot e^{s_2t}\]
    Response :
  • As \(t \uparrow\) response decays and approaches zero.

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\[\begin{aligned} &\boxed{s_{1}, s_{2}=-\alpha\pm\sqrt{\alpha^{2}-\omega_{0}^{2}}} \quad \boxed{\alpha=\frac{R}{2 L}} \quad \boxed{\omega_{0}=\frac{1}{\sqrt{L C}}} \end{aligned}\]
Critically Damped Case:
  • \(\alpha = \omega_0 \Rightarrow C = 4L/R^2\)

  • \[s_1 = s_2 = -\alpha = -\dfrac{R}{2L}\]
    Roots :
  • \[i(t) = A_1\cdot e^{-\alpha t} + A_2\cdot e^{-\alpha t} = A_3 \cdot e^{-\alpha t} \quad (A_3 = A_1 + A_2)\]
    Response :
  • Wrong solution, because the two initial conditions cannot be satisfied with the single constant \(A_3\) .

  • What then could be wrong?

  • When \(\alpha = \omega_0 = R/2L\)

\[ \begin{aligned} \frac{d^{2} i}{d t^{2}}+2 \alpha \frac{d i}{d t}+\alpha^{2} i &=0 \\ \frac{d}{d t}\left(\frac{d i}{d t}+\alpha i\right)+\alpha\left(\frac{d i}{d t}+\alpha i\right) &=0 \\ \frac{d f}{d t}+\alpha f &=0 \quad \left(f=\frac{d i}{d t}+\alpha i \right) \\ \frac{d i}{d t}+\alpha i &=A_{1} e^{-\alpha t} \quad \Leftarrow (\text{solution of}~ f)\\ e^{\alpha t} \frac{d i}{d t}+e^{\alpha t} \alpha i &=A_{1} \\ \frac{d}{d t}\left(e^{\alpha t} i\right) &=A_{1} \\ e^{\alpha t} i &=A_{1} t+A_{2} \Leftarrow \text{on Integrating}\\ i=&\left(A_{1} t+A_{2}\right) e^{-\alpha t} \end{aligned} \]
\[\boxed{ i=\left(A_{1} t+A_{2}\right) e^{-\alpha t}}\]
  • Reaches a maximum value at \(t=1/\alpha\), one time constant and then decays to zero

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

    \[\begin{aligned} s_{1} &=-\alpha+\sqrt{-\left(\omega_{0}^{2}-\alpha^{2}\right)}=-\alpha+j \omega_{d} \\ s_{2} &=-\alpha-\sqrt{-\left(\omega_{0}^{2}-\alpha^{2}\right)}=-\alpha-j \omega_{d} \\ \omega_{d} &=\sqrt{\omega_{0}^{2}-\alpha^{2}} \end{aligned}\]
  • \[\begin{aligned} i(t) &=A_{1} e^{-\left(\alpha-j \omega_{d}\right) t}+A_{2} e^{-\left(\alpha+j \omega_{d}\right) t} \\ &=e^{-\alpha t}\left(A_{1} e^{j \omega_{d} t}+A_{2} e^{-j \omega_{d} t}\right) \end{aligned}\]
    Response:
  • \[e^{j \theta}=\cos \theta+j \sin \theta, \quad e^{-j \theta}=\cos \theta-j \sin \theta\]
    Using Euler’s identities,
  • \[\begin{aligned} i(t) &=e^{-\alpha t}\left[A_{1}\left(\cos \omega_{d} t+j \sin \omega_{d} t\right)+A_{2}\left(\cos \omega_{d} t-j \sin \omega_{d} t\right)\right] \\ &=e^{-\alpha t}\left[\left(A_{1}+A_{2}\right) \cos \omega_{d} t+j\left(A_{1}-A_{2}\right) \sin \omega_{d} t\right] \end{aligned}\]
    we get
  • \[i(t)=e^{-\alpha t}\left(B_{1} \cos \omega_{d} t+B_{2} \sin \omega_{d} t\right)\]
    , and with and Replacing constants
  • 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 \(T=2\pi/\omega_d\)

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Peculiar properties of an RLC network

  • The behaviour is due to damping - gradual loss of initial stored energy represented by decrease in response amplitude

  • Damping effect is due to \(R\) and damping factor \(\alpha\) determine the rate of damping

  • For lossless circuit \(R=0~\Rightarrow \alpha =0\), means \(LC\) circuit with \(\omega_0 = 1/\sqrt{LC}\)

  • \(\alpha < \omega_0~\Rightarrow\) response undamped and oscillatory

  • Adjusting \(R~\Rightarrow\) undamped, overdamped, critically-damped, or under-damped

  • 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


Solved Problem

Find \(i(t)\) assuming the circuit has reached steady-state at \(t=0^{-}\)

  • For \(t<0~\Rightarrow\) switch closed \(\Rightarrow\) \(C \rightarrow\) OC and \(L \rightarrow\) SC

  • \[\begin{aligned} i(0)& = \dfrac{10}{4+6}=1~A\\ v(0)&=6i(0)=6~V \end{aligned}\]
    At
  • \[\begin{aligned} \alpha & = \dfrac{R}{2L} = 9 \quad \omega_0 = \dfrac{1}{\sqrt{LC}}=10\\ s_{1,2} & = -\alpha \pm \sqrt{\alpha^2-\omega_0^2}= -9 \pm j4.359 \end{aligned}\]
    source-free series RLC circuit For

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