The s-domain, transfer functions and plots

FD

Choose FD and the answers come back as functions of s, initial conditions included.

The RC circuit of Capacitors, inductors and TR with its source left as a symbol we call vs, a 1 kΩ resistor and a 1 µF capacitor:

e,1,0,vs
r,1,2,1'k
c,2,0,1'u

Set Analysis to FD — complex frequency domain and run:

Leave the source symbolic and the answer is the transfer function: divide by vs and you have H(s) = 1000/(s + 1000), a first-order low pass with its pole at −1000. Nothing is labelled "transfer function" because nothing needs to be — it is the answer with the input left as a symbol.

The pz mini-tool reads its poles and zeros. Give it the ratio:

vc/vs

It returns the poles, the roots of the denominator, and the zeros, the roots of the numerator: here one pole at -1000 and no zero. A factor the numerator and the denominator share is cancelled first, and a repeated root is written once with its multiplicity, as in -4 ×2.

t2s and s2t

Two shortcuts convert an expression between the domains:

t2s(10*exp(-1000*t))
s2t(1000/(s + 1000))

Curly brackets are shorthand for the same thing inside a value, so a source may be written in t and used in an FD run.

Limits

Evaluate takes a limit over the answers as limit(expression, variable, point). The initial- and final-value theorems are two of them:

limit(s*vo, s, oo)
limit(s*vo, s, 0)

A line in Conditions at infinity is read as a limit too, so omega = oo gives a gain at high frequency and, in TR, t = oo a final value.

The four plots

The Plot card offers four, and which ones are available depends on the analysis:

The third needs no circuit at all: type 1000/(s+1000) and get its magnitude and phase. Useful when the transfer function came from somewhere else, or when you are checking a hand derivation.

The fourth is a DC sweep — one answer plotted against another as a value varies, which is how a load line or a maximum-power curve is drawn. Give it the two names and a range.

Resonance

There is no resonance tool. Resonance is the frequency that clears the reactance, so it is a Solve problem: take the impedance answer, set its imaginary part to zero, and solve for ω. The same works for half-power points, which makes bandwidth and Q two more lines of algebra rather than a feature.