Resonance problem in LC circuit!

Discussion in 'Homework Help' started by cdummie, Jan 15, 2018.

  1. cdummie

    Thread Starter Member

    Feb 6, 2015
    OK, i've been struggling with this for quite some time, so here it goes:

    So, let's say we have a series LC circuit with generator ug(t)=Um*cos(ωt)

    Then the output signal would be (if we consider the current in the circuit as the output signal) :


    Now, basically i can understand what happens up to this moment, where i don't understand why angle is 90 degrees if frequency of the circuit is lower than the one of the generator and -90 degree otherwise. Now, there's a special case after this. What happens when frequency of the circuit is the same as the frequency of the generator? It then says that we cannot find the solution using complex numbers representations of the values in circuit since transfer function (i believe that's the correct term for the "output over input" function) reaches infinity in this case, so instead, particular solution to the differential equation of this circuit has looks something like this:

    Why would the response to excitation look like this (since i know for the three types only, and none of them fits this)?

    So i don't know how to find the response in this circuit in case of resonance. Any help appreciated!
  2. WBahn


    Mar 31, 2012
    What if you just had an impedance of jX (i.e., purely reactive, no resistance)?

    What is the phase angle of the current relative to the voltage if X is positive (i.e., inductive)?

    What is the phase angle of the current relative to the voltage if X is negative (i.e., capacitive)?

    For a series combination of a capacitor and an inductor, under what conditions is the net reactance positive?

    For a series combination of a capacitor and an inductor, under what conditions is the net reactance negative?

    As for the case at resonance, it is little more than a mathematical curiosity.

    From a complex impedance standpoint, the impedance goes to zero so the current is "infinite". That's the base we can do. But we know that, even from a purely theoretical case, that is not possible because because of the conservation of energy -- the energy that is bounced back and forth between the capacitor and the inductor can't materialize instantaneously. We can't even resort to claiming that there is an impulse of current to instantly charge things up, because the current in the inductor has to remain continuous at all time.

    So what they are doing is running home to mama and going for what the math says when you solve the differential equation directly. In doing so, you have to work with the most general form of the solution, which has those two terms in it, as well as the terms without the factor of 't'. Any place other than resonance, these two terms can't satisfy the differential equation unless K1 and K2 are identically zero. But at resonance, the 1/LC term is ω_o² and allows these terms to survive (and now the other two terms must vanish).

    While this looks nice and fine on paper, it has a really big problem - at least at first glance. The forcing function is a sinusoid that extents in both the positive and negative 't' direction. But the solution says that the amplitude is a function of time, meaning that there is something particularly special about t=0, namely that the amplitude of the signal vanishes. Well, there IS something special about t = 0 since it is the time at which we apply our initial conditions.
    cdummie likes this.
  3. cdummie

    Thread Starter Member

    Feb 6, 2015

    Well, i do have reactive impedance, since this is LC circuit, there's no resistor in the circuit, but i don't know what does it means for this circuit.

    For completely inductive impedance current lags Pi/2 relative to the voltage.

    For completely capacitive impedance voltage lags Pi/2 relative to the current.

    However, impedance of this circuit is not completely inductive nor completely capacitive, meaning that for the frequencies above the ω_0 phase should be somewhere in between 0 and -π/2, and between 0 and π/2 if it's lower than ω_0.[This is how i intuitively understand this.]

    Now, in order for reactance to be positive, in this case, ωL-1/(ωC) >0 meaning ω>1/sqrt(LC), meaning ω>ω_0, which means that when reactance is positive then the phase angle is π/2 since the phase of the impedance dictates the phase difference between the current given as the input signal (i believe that this is the part that's been missing for me to understand this part of the problem).[This is what i got after i answered your questions, mathematically, and obviously it doesn't match with how i thought it should be in the first place .]
    But, i actually got it now, since i was holding a picture of phase diagrams with vectors being in 1st and 4th quadrant which is impossible in this case since it can happen only when i have a resistor (I haven't deleted non of this on purpose, you helped me to understand how my reasoning was wrong by asking non-direct questions so i want to keep it all here as a reminder, to know how i got from completely wrong reasoning to the point where i understood what happened, thanks! )
    However, here i got that when frequency is "greater than" then it's π/2 and not the other way around as you can see in my picture. What's the deal here?

    And, for the other part of my question, what does it actually means? Does it means that if i try to solve DE of this circuit at resonant frequency that i would got the answer i got in my second picture?