The frequency is correct. You've missed out the 2π (2pi). So it should be 1/(2pi*sqrt(LC)) - which is 1.587MHz.

As for the amplitude, I think that's just the rforce having an influence. If I set the rforce option to 100 (the default is 1 ohm). The impedance of the capacitor and inductor is 100ohms at this frequency.

Regards,

Andrew.

The initial condition problem is discussed in my book The Designer's Guide to SPICE and Spectre.

See section 4.3.5 on initial conditions and section 4.4.2 on oscillators.

-Ken

If you set the write parameter on the transient analysis to write out the DC solution at the beginning of the transient (with rforce at the default value), you'll see that it shows:

# CHECKPOINT_VERSION 1
# Generated by spectre (mode: Spectre) from circuit file `dg1.scs' during analys
is tran.
# 10:37:49 AM, Thur Feb 23, 2017
# Number of equations = 3
L0:1    1
osc     0

What is going on is that the inductor is a short at DC (and the capacitor is open). The initial condition is implemented as the norton equivalent of the desired voltage on the capacitor in series with 1 ohm (the rforce value). This is 1 amp in parallel with 1 ohm. So that's why you end up with 1Amp through the inductor.

Now, because at the resonant frequency, the impedance is 100 ohms, that will (V=IR) generate 100V at the resonant frequency.

If you set rforce=100, the norton equivalent is 0.01A with 100ohm in parallel to give 1V across the inductor. Again, because it's shorted at DC, you just get 0.01A through the inductor. So the checkpoint file contains:

# CHECKPOINT_VERSION 1
# Generated by spectre (mode: Spectre) from circuit file `dg1.scs' during analys
is tran.
# 11:26:30 AM, Thur Feb 23, 2017
# Number of equations = 3
L0:1    0.01
osc     0

Again, V=IR at the resonant frequency would then result in 1V at the resonant frequency (which is what you want).

Regards,

Andrew.