Proving the Form of a Homogenous LC Circuit?

In summary, the conversation discusses a circuit with an input source and an inductor and capacitor in series. The speaker has found the roots of the resulting differential equation to be complex, and is asked to show the homogeneous solution in a specific form. They receive a suggestion to convert the equation into an ordinary quadratic equation and make a substitution before solving for z.
  • #1
Rome_Leader
19
0

Homework Statement



I have a circuit with input source x(t), which contains also an inductor and a capacitor in series which I have found to be related to the output voltage y(t) (across the capacitor) like so: LC*d2y/dt2 + y(t) = x(t). I have also found its roots through the quadratic equation to be complex, and equal to +/- the imaginary unit j.

The question asks me to show its homogenous solution has the form K1ej*a*t + K2ej*b*t, where a and b are the roots I found above.

Homework Equations



Unsure. See above for my progress so far.

The Attempt at a Solution



My solution has led me here, having gotten the above equation and its roots. I just do not know how to show that I can express the solution to the differential equation in the requested form. Any help would be greatly appreciated!
 
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  • #2
Hi Rome_Leader! :smile:

If you have a homogeneous quadratic equation in y(t),

write D = d/dt, and convert the equation into an ordinary quadratic equation in D …

(D - a)(D - b)(y) = 0​

then put (D - b)(y) = z, so (D - a)(z) = 0 :wink:
 
  • #3
I follow, but that does not appear to express it in the form K1*e^(j*a*t) + K2*e^(j*b*t). I don't think it makes it any simpler either, at least not to my eyes.
 
  • #4
Rome_Leader said:
I follow, but that does not appear to express it in the form K1*e^(j*a*t) + K2*e^(j*b*t).

yes it does, try it :smile:
 
  • #5
Expanding, I get d^2y/dt^2 - (dy/dt)*b - (dy/dt)*a + ab.

Your z substitution is a little confusing, but then I get (dy/dt-a)*(dy/dt-b) = 0 again?
 
  • #6
solve (D - a)z(t) = 0 first …

z = … ? :wink:

(and then solve the non-homogeneous equation (D - b)y(t) = z(t))
 

1. What is a homogenous LC circuit?

A homogenous LC circuit is a type of electronic circuit that consists of a capacitor (C) and an inductor (L) connected in series. The circuit is called "homogenous" because the components are made of the same material and have the same physical properties.

2. How do you prove the form of a homogenous LC circuit?

The form of a homogenous LC circuit can be proven through mathematical analysis and experiments. By using Kirchhoff's laws and the equations for capacitors and inductors, we can derive the differential equation that describes the behavior of the circuit. This equation can then be solved to show the form of the circuit.

3. Why is proving the form of a homogenous LC circuit important?

Proving the form of a homogenous LC circuit is important because it helps us understand the behavior of the circuit and how it responds to different inputs. This knowledge is essential for designing and troubleshooting electronic systems that use homogenous LC circuits.

4. Can the form of a homogenous LC circuit be modified?

Yes, the form of a homogenous LC circuit can be modified by changing the values of the capacitor and inductor. This can affect the resonance frequency and the overall behavior of the circuit.

5. Are there any real-life applications of homogenous LC circuits?

Yes, homogenous LC circuits have many real-life applications. They are commonly used in electronic filters, oscillators, and frequency-selective networks. They are also used in wireless communication systems and power supply circuits.

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