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I'm sure there is a really simple explanation for this, but I can't seem to figure it out. I am sweeping the series resistance of the source voltage and I am trying to see the transfer function output.

When plotting the \$V(out)\$ and \$V(in)\$ plots separately, the different plots are shown for increasing values of series resistance. However, when I "add trace" as \$\frac{V(out)}{V(in)}\$, only one plot appears. How do I get plots of \$\frac{V(out)}{V(in)}\$ for all the different values of series resistance?

Alternatively, is there a way I can plot \$\frac{V(out)}{V(in)}\$ directly rather than going into the waveform viewer and using "add trace"?

6 plots of V(out) for sweep in series resistance, 6 plots for V(in), Only one plot for V(out)/V(in)

Any insight on this problem would be greatly appreciated.

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  • \$\begingroup\$ Why do you think that it's different for every step? Looks to me that there's a 6 dB difference between vout and vin for every step in the beginning, so it should just be a single line, and it also matches at the end. \$\endgroup\$
    – pipe
    Commented Jun 26, 2019 at 16:51
  • \$\begingroup\$ This might not meet your definition of 'directly', but I've just discovered bv, behavioural voltage source, and am over-using it. On the schematic, place a bv, with an expression =V(out)/V(in), then plot the output of that. \$\endgroup\$
    – Neil_UK
    Commented Jun 26, 2019 at 17:00
  • \$\begingroup\$ To compute Sensitivity on the other hand is just a derivative of dG(s)/dR \$\endgroup\$
    – D.A.S.
    Commented Jun 26, 2019 at 18:01
  • \$\begingroup\$ @pipe Nice catch. There is a consistent difference between V(out) and V(in) which is why all the plots are overlaid and it looks like there's only one plot. I should have caught that earlier. Appreciate the help. \$\endgroup\$
    – Acubed_engineer
    Commented Jun 26, 2019 at 19:11

2 Answers 2

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If you're using a voltage source as the input and if the input resistance is part of the builtin parasitic (Rser) then you don't need to plot V(out)/V(in), V(out) will suffice. Otherwise, the behavioural source suggested in the comments and answer will do.

If you want to plot the result of a single value from the .STEP command, then use the @<step_number> selector, for example V(out)@3 will plot V(out) with the results from the third step. This also works for V(out)@3/V(in)@3, and you can make combinations, V(out)@3/V(in)@2.

The @ selector will not work in the behavioural source expression for obvious reasons: the behavioural expression needs to be evaluated at runtime, the latest, so you can't use V(out)@2 before the first step has been run, for example.

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  • \$\begingroup\$ Must've been in a rush because a behavioural source involving division will not work, since it's considered a nonlinear operation and its outcome cannot be linearized (dependent on the parameters of the simulation, not the circuit, itself). \$\endgroup\$
    – a concerned citizen
    Commented Dec 9, 2022 at 10:06
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One way would be to add a b-source (bv in the device list) with the function equal to V(out)/V(in) You could then plot the voltage node of the b-source

enter image description here

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