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August 28th, 2017, 05:30 AM   #1
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How to obtain an RLC circuit from this differential?

(D^2 + 4D + 13)x(t) = (4D + 2)y(t).

I know the term that is squared related to the inductance, the 4D term is related to the capacitance and the term with no D is the resistance. But I am unsure whether I have to solve for y(t) as x(t) is the input and y(t) is the output.

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August 28th, 2017, 06:13 AM   #2
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What instructions came with the equation?
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August 28th, 2017, 06:45 AM   #3
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The instructions were to model an electric circuit using the differential equation above. So as far as I understand, it would be to divide both sides by (4D +2) to get y(t) by itself and then solve from there.
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August 28th, 2017, 07:08 AM   #4
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Quote:
Originally Posted by Damiene View Post
(D^2 + 4D + 13)x(t) = (4D + 2)y(t).

I know the term that is squared related to the inductance, the 4D term is related to the capacitance and the term with no D is the resistance. But I am unsure whether I have to solve for y(t) as x(t) is the input and y(t) is the output.

Thanks
Your notation is rather odd, but this should help...

Consider a circuit with an ohmic resistor, an inductor and a capacitor. The voltage, $\displaystyle V=V(t)$, for each of these components relates to the charge, $\displaystyle q = q(t)$, using the following equations:

Resistor: $\displaystyle V_R = R \frac{dq}{dt}$

Inductor: $\displaystyle V_L = L \frac{d^2q}{dt^2}$

Capacitor: $\displaystyle V_C = \frac{q}{C}$

In a series RLC circuit,

$\displaystyle V_{total} = V_R + V_L + V_C$
$\displaystyle = L \frac{d^2q}{dt^2} + R \frac{dq}{dt} + \frac{q}{C}$

If we introduce a new operator 'D' such that $\displaystyle Dq = \frac{dq}{dt}$ and $\displaystyle D^2q = \frac{d^2q}{dt^2}$, then

$\displaystyle V_{total} = (LD^2 + RD + C)q$

Therefore, by comparison:

L = 1 H
R = 4 $\displaystyle \Omega$
C = 13 F
$\displaystyle V_{total} = (4D + 2) y(t)$

To solve, consider dividing through by L and then investigate solutions to second order differential equations
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August 28th, 2017, 07:18 AM   #5
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Thank you for the detailed reply Benit13. Howcome you suggest to divide by L? I stuck on having to solve it as I did Laplace transforms many moons ago.
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