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Two Coupled LC Circuits

Consider the LC circuit pictured in Figure 17. Let $ I_1(t)$ , $ I_2(t)$ , and $ I_3(t)$ be the currents flowing in the three legs of the circuit, which meet at junctions $ A$ and $ B$ . According to Kirchhoff's first circuital law, the net current flowing into each junction is zero (Grant and Phillips 1975). It follows that $ I_3=-(I_1+I_2)$ . We deduce that this is a two degree of freedom system whose instantaneous configuration is specified by the two independent variables $ I_1(t)$ and $ I_2(t)$ . It follows that there are two independent normal modes of oscillation. The potential drops across the left, middle, and right legs of the circuit are $ Q_1/C+L\,\dot{I}_1$ , $ Q_3/C'$ , and $ Q_2/C+L\,\dot{I}_2$ , respectively, where $ \dot{Q}_1=I_1$ , $ \dot{Q}_2=I_2$ , and $ Q_3=-(Q_1+Q_2)$ . However, because the three legs are connected in parallel, the potential drops must all be equal, so that

$\displaystyle Q_1/C + L\,\dot{I}_1$ $\displaystyle = Q_3/C' = -(Q_1+Q_2)/C',$ (182)
$\displaystyle Q_2/C+L\,\dot{I}_2$ $\displaystyle = Q_3/C' = -(Q_1+Q_2)/C'.$ (183)

Differentiating with respect to $ t$ , and dividing by $ L$ , we obtain the coupled time evolution equations of the system:

$\displaystyle \ddot{I}_1 + \omega_0^{\,2}\,(1+\alpha)\,I_1+\omega_0^{\,2}\, \alpha\,I_2$ $\displaystyle =0,$ (184)
$\displaystyle \ddot{I}_2 + \omega_0^{\,2}\,(1+\alpha)\,I_2+ \omega_0^{\,2}\,\alpha\,I_1$ $\displaystyle =0,$ (185)

where $ \omega_0=1/\sqrt{L\,C}$ and $ \alpha=C/C'$ .

Figure 17: A two degree of freedom LC circuit.
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It is fairly easy to guess that the normal coordinates of the system are

$\displaystyle \eta_1$ $\displaystyle =(I_1+I_2)/2,$ (186)
$\displaystyle \eta_2$ $\displaystyle =(I_1-I_2)/2.$ (187)

Forming the sum and difference of Equations (184) and (185), we obtain the evolution equations for the two independent normal modes of oscillation,

$\displaystyle \ddot{\eta}_1+\omega_0^{\,2}\,(1+2\,\alpha)\,\eta_1$ $\displaystyle =0,$ (188)
$\displaystyle \ddot{\eta}_2+\omega_0^{\,2}\,\eta_2$ $\displaystyle =0.$ (189)

(We can be sure that we have correctly guessed the normal coordinates because the previous two equations do not couple to one another.) These equations can readily be solved to give

$\displaystyle \eta_1(t)$ $\displaystyle =\hat{\eta}_1\,\cos(\omega_1\,t-\phi_1),$ (190)
$\displaystyle \eta_2(t)$ $\displaystyle =\hat{\eta}_2\,\cos(\omega_0\,t-\phi_2),$ (191)

where $ \omega_1=(1+2 \alpha)^{1/2} \omega_0$ . Here, $ \hat{\eta}_1$ , $ \phi_1$ , $ \hat{\eta}_2$ , and $ \phi_2$ are constants determined by the initial conditions. It follows that

$\displaystyle I_1(t)$ $\displaystyle =\eta_1(t)+\eta_2(t) = \hat{\eta}_1\,\cos(\omega_1\,t-\phi_1)+\hat{\eta}_2\,\cos(\omega_0\,t-\phi_2),$ (192)
$\displaystyle I_2(t)$ $\displaystyle = \eta_1(t)-\eta_2(t) = \hat{\eta}_1\,\cos(\omega_1\,t-\phi_1)-\hat{\eta}_2\,\cos(\omega_0\,t-\phi_2).$ (193)

As an example, suppose that $ \phi_1=\phi_2=0$ and $ \hat{\eta}_1=\hat{\eta}_2= I_0/2$ . We obtain

$\displaystyle I_1(t)$ $\displaystyle = I_0\,\cos(\omega_-\,t)\,\cos(\omega_+\,t),$ (194)
$\displaystyle I_2(t)$ $\displaystyle = I_0 \sin (\omega_- t) \sin(\omega_+ t),$ (195)

where $ \omega_\pm = (\omega_0\pm \omega_1)/2$ . This solution is illustrated in Figure 18. [Here, $ T_0=2\pi/\omega_0$ and $ \alpha=0.2$ . Thus, the two normal frequencies are $ \omega_0$ and $ 1.18\,\omega_0$ .] Observe the beat modulation generated by the superposition of two normal modes with similar normal frequencies. (See Section 3.6.)

Figure 18: Coupled oscillations in a two degree of freedom $ LC$ circuit.
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We can also solve the problem in a more systematic manner by specifically searching for a normal mode of the form

$\displaystyle I_1(t)$ $\displaystyle =\hat{I}_1 \cos(\omega t-\phi),$ (196)
$\displaystyle I_2(t)$ $\displaystyle =\hat{I}_2 \cos(\omega t-\phi).$ (197)

Substitution into the time evolution equations (184) and (185) yields the matrix equation

$\displaystyle \left( \begin{array}{cc} \hat{\omega}^2-(1+\alpha) & -\alpha [0...
... \end{array}\right) = \left( \begin{array}{c} 0 [0.5ex] 0 \end{array}\right),$ (198)

where $ \hat{\omega}=\omega/\omega_0$ . The normal frequencies are determined by setting the determinant of the matrix to zero. This gives

$\displaystyle \left[\hat{\omega}^2-(1+\alpha)\right]^{\,2}-\alpha^2 = 0,$ (199)

or

$\displaystyle \hat{\omega}^{ 4} - 2 (1+\alpha) \hat{\omega}^{ 2}+1+2 \alph...
...eft(\hat{\omega}^{ 2}-1\right)\left(\hat{\omega}^{ 2}-[1+2 \alpha]\right)=0.$ (200)

The roots of the preceding equation are $ \hat{\omega}=1$ and $ \hat{\omega}=(1+2 \alpha)^{1/2}$ . (Again, we have neglected the negative frequency roots, because they generate the same patterns of motion as the corresponding positive frequency roots.) Hence, the two normal frequencies are $ \omega_0$ and $ (1+2 \alpha)^{1/2} \omega_0$ . The characteristic patterns of motion associated with the normal modes can be calculated from the first row of the matrix equation (198), which can be rearranged to give

$\displaystyle \frac{\hat{I}_1}{\hat{I}_2}= \frac{\alpha}{\hat{\omega}^{ 2}-(1+\alpha)}.$ (201)

It follows that $ \hat{I}_1=-\hat{I}_2$ for the normal mode with $ \hat{\omega}=1$ , and $ \hat{I}_1=\hat{I}_2$ for the normal mode with $ \hat{\omega}=(1+2 \alpha)^{1/2}$ . We are thus led to Equations (192) and (193), where $ \hat{\eta}_1$ and $ \phi_1$ are the amplitude and phase of the higher frequency normal mode, whereas $ \hat{\eta}_2$ and $ \phi_2$ are the amplitude and phase of the lower frequency mode.


next up previous
Next: Three Spring-Coupled Masses Up: Coupled Oscillations Previous: Two Spring-Coupled Masses
Richard Fitzpatrick 2013-04-08