For the same simulation, does Gear method requires more points than Trap method

Check chapter about transient analysis from this book

Dominik Przyborowski said:

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Description of The Designer's Guide to SPICE and Spectre

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Check chapter about transient analysis from this book

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I read that book again after you mentioned it.

"For a fixed time step, the most accurate of these methods in a local sense is the trapezoidal rule, followed by Gear2. ... but the trapezoidal rule would typically allow larger time steps and so require a shorter run time. ...
thus it is not usually the best choice when reltol is loose."

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First, I understand that it is hard to say the default reltol(1e-3) is tight or loose without having the circuit in hand, but I assume the default one for a full-chip setting is loose.

"The higher-order backward-difference formulas are efficient when tolerances are tight or when computing very smooth waveforms."

Click to expand...

Usually, Gear2 is the go to (not a higher order) and I would not say it is a higher-order method.

I tried to derive the relationship from the formula standpoint.


Gear2:\[ X^{n+1} = \frac{4}{3}X^n - \frac{1}{3}X^{n-1} + \frac{2}{3}h^n \dot{X}^{n+1} \]
Trap: \[ X^{n+1} = X^n + \frac{h^n}{2}(\dot{X}^{n+1}+\dot{X}^n) \]

F(t) = Gear2 - Trap with the assumption that \[X^n, X^{n-1}\] are the same timept under 2 methods, aiming to check who has a bigger \[X^{n+1}\] under different assumption.

\[ F(t) =/frac{1}{3} X^n -\frac{1}{3}X^{n-1} + \frac{2}{3}h^n \dot{X}^{n+1} - \frac{h^n}{2}(\dot{X}^{n+1}+\dot{X}^n) \]


Where can I go from here ?