Connect with us

LC circuit in the time domain

Discussion in 'General Electronics Discussion' started by Joe Bachi, Nov 7, 2014.

Scroll to continue with content
  1. Joe Bachi

    Joe Bachi

    21
    0
    Feb 3, 2013
    Hi everyone,
    When analyzing an LC series circuit, we usually work in the frequency domain which simplifies the equations.
    Z(L)=jLw
    Z(C)=1/(jCw)

    V(L) = V * Z(L)/[Z(L)+Z(C)]
    = V * LCw^2/(LCw^2 - 1)

    V(C) = V * Z(C)/[Z(L)+Z(C)]
    = V * 1/(1 - LCw^2)

    Assuming V is a simple sine wave generator,
    If we measure the voltage drop V(L) or V(C) using an oscilloscope, we get a function that resembles sin(a)*sin(b) at resonance (in the time domain)
    Why is that?
    shouldn't the output be in the form of [sin(a)]^2 since the generator and the inductor (or capacitor) have the same w?

    Thank you in advance
     
  2. Laplace

    Laplace

    1,252
    184
    Apr 4, 2010
    The simple answer is that capacitors and inductors are treated as linear devices so that an LC combination is a linear circuit that can only produce a linear output.

    A circuit with an input of sin(ωt) and an output in the form of sin²(ωt) would be a non-linear circuit; therefore, your assumption that an LC circuit can produce a non-linear output is wrong.

    Note that there is a proper method to shift from the steady-state frequency domain to the time domain that does not lead to this impossible result.
     
    Harald Kapp likes this.
  3. Joe Bachi

    Joe Bachi

    21
    0
    Feb 3, 2013
    can you elaborate please?
     
  4. Harald Kapp

    Harald Kapp Moderator Moderator

    9,540
    1,967
    Nov 17, 2011
    In the time domain:
    Voltage across an inductor: V=L*di/dt
    Current through a capacitor: I=C*dv/dt
    You set up the node equations using kirchhoff's laws and solve the resulting differential equation.
     
  5. Joe Bachi

    Joe Bachi

    21
    0
    Feb 3, 2013
    Yes I am aware of that, but i wanted to see if there was a way to solve the whole thing in the frequency domain and then do an inverse transform (Laplace, or Fourier) to the result so that i can get it in the time domain and thus avoiding all the messy differential equations.

    V(L) = V * Z(L)/[Z(L)+Z(C)]
    = V * LCw^2/(LCw^2 - 1)

    Is V supposed to be a constant here? or is it the Fourier Transform of sin(wt)?
    v = Asin(wt) in the time domain
    So if I do an inverse Fourier Transform, could i say that v(L) = convolution product (v , inverse FT of the impedance)?

    Thank you :)
     
  6. LvW

    LvW

    604
    145
    Apr 12, 2014
    Perhaps it helps to remember that the denominator of the transfer function (frequency domain) is identical to the characteristic polynominal of the corresponding diff. equation (time domain)?
     
  7. Joe Bachi

    Joe Bachi

    21
    0
    Feb 3, 2013
    How can that help?
     
  8. LvW

    LvW

    604
    145
    Apr 12, 2014
    I`ve got the impression that it is your desire to find a way back from the frequency into the time domain.
    And - solving the diff. equation (time domain) you arrive at the characteristic equation which must be solved. And this equation can be derived also from the transfer function.
    That`s all I wanted to point out.
     
  9. Joe Bachi

    Joe Bachi

    21
    0
    Feb 3, 2013
    Alright
    But in the frequency domain, what would be the expression of V, given that it's a sine wave in the time domain?
     
  10. Harald Kapp

    Harald Kapp Moderator Moderator

    9,540
    1,967
    Nov 17, 2011
  11. Laplace

    Laplace

    1,252
    184
    Apr 4, 2010
    One should first clarify whether the interest here is in the complex frequency domain, where the complex frequency variable s=σ+jω, or in the steady-state frequency domain where σ=0 and s=jω.

    In the complex frequency domain it is possible to analyze how a circuit reacts to transient events from the time domain by using simple frequency domain methods (at least simpler than using differential equations). If the time domain response is needed then one can take the inverse transform back to the time domain.

    In the steady-state frequency domain there are no transient events from the time domain. The single-frequency excitation signal is assumed to have been present since the beginning of time. In practical terms, it will be far enough removed from any transient event (such as turning on the excitation signal) that any transient response has long since died out. There may be multiple frequencies present in the circuit, but each one is analyzed individually using superposition. Also, a string of separate frequencies can be individually analyzed for the purposes of generating a Bode plot for the circuit.

    Assuming the question here concerns the steady-state frequency domain, a linear RLC circuit can be represented by a complex-value transfer function, T(C,ω) where C represents the circuit configuration & component values and ω is the single excitation frequency. Generally, the highest exponent of ω^n in the transfer function will be n= # of reactive components in the circuit. Since T(C,ω) is a complex expression evaluated at a single frequency at any one time, the transfer function can be evaluated at that frequency to obtain the magnitude and phase of the complex expression (in polar form) M(ω)@ф(ω) (or evaluated as a string of frequencies for a Bode plot). In the steady-state frequency domain, at the given excitation frequency, a linear circuit will change the magnitude and phase of the excitation signal appearing at the output.

    In the time domain, an excitation signal E∙sin(ωt) appears in the frequency domain as E. When the excitation signal is acted upon by the transfer function, the result is E∙[email protected]ф which appears in the time domain as E∙M∙sin(ωt+ф). This is the simplicity of the steady-state frequency domain -- it is all just magnitude and phase.
     
  12. Joe Bachi

    Joe Bachi

    21
    0
    Feb 3, 2013
    Thank you for the detailed answer it really helped :)
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-