LTSPICE Laplace simulation discrepancy ??

I am not sure what I am doing wrong??

I have implemented a very simple L R C circuit just to test laplace

L in series with R & C in parallel as the output (see below)

I placed both the L R C schematic and laplace equation in the same circuit

laplace equation "V=V(Si) Laplace=(1/({L}*{C}))*(1/(s**2+s/({R}*{C})+1/({L}*{C})))"

I added a BV voltage source in front of the L R C circuit to isolate both circuits

I use the same source signal for both circuits

The AC response for both are very similar but not exactly the same

but the transient responses ares completly different ???

see below screen shots

I appreciate any comments in advance

Thank you very much

Fausto bartra

  • 0
    •  Super User 
    on Jun 6, 2019 6:02 PM

    Hello Fausto,

    There are some problems with your circuit.

    1. AC Simulation
    The difference in the .AC simulation is caused by a series resistance of the inductor.
    An inductor has a series resistance of 1mOhm by default. You can change this value to 0.
    Therefore right-click on the inductor and enter 0 in "Series Resistance" of the dialog window, OK.
    You can make this visible with CTRL and right-click on the inductor. This will show you all the attributes. Mark "Visibility" for the line with Rser=0.
    Now the results from the .AC-analysis will perfectly agree with the results from the Laplace formula.

    By the way you have chosen a quality factor Q of 1million for this resonator circuit.
    Only the very best crystal oscillators have a Q of 1million.

    The AC-analysis should show a peak of 120dB, but it only shows 80dB in the simulation.
    The reason is that LTspice doesn't exactly hit the rersonance frequency.
    I had to change the ac-setting as below to get the peak with 120dB (1million).

    .ac lin 30001 15.9k 15.93k

    2. Transient Simulation
    It's generally difficult to get good results with Laplace equations in time domain simulations.
    You should only use it for demonstration, if you already know in advance the correct result. Then you can play with NFFT and WINDOW to get a more useful result.
    I tried my best with the following options, but the result is still far off and the simulation speed is badly slow.

    V= V(in) Laplace=1/(1 +s*L/R + s**2*L*C) nfft=16384 window=20m

    By the way your circuit with Q=1million has a settling time of about 100s. This means you have to simulate 100s if you want see steady state.
    At the beginning of the simulation, you see a 10kHz sine plus a sine with the resonance frequency. In steady state you will only see a pure sine wave with 10kHz.

    Learning from this example
    Laplace is fully OK with the .AC-analysis, but you may be mostly disappointed in the .TRAN simulation.