# LTspice Question

A

#### amdx

Jan 1, 1970
0
Is it possible to graph R and X of a circuit?
Such as looking into a filter, instead of amplitude
and phase, could I graph R and X.
If yes, how?

Thanks, Mike

K

#### Ken Smith

Jan 1, 1970
0
Is it possible to graph R and X of a circuit?
Such as looking into a filter, instead of amplitude
and phase, could I graph R and X.
If yes, how?

Get the plot showing amplitude and phase.

Bring the cursor over the lets say V(n002) at the top of the plot

Right click and change what is plotted.

re(v(n002)) plots the in phase part of a voltage

To get both, you need to get the raw signal displayed again and edit the
second one.

Next you double (left) click the scale down the right side and tell it not
to plot phase.

A

#### amdx

Jan 1, 1970
0
Hi Ken,
I don't quite have it,
I'm plotting (v(in)) which is the voltage out of the signal generator
(with internal resistance) feeding a filter. The plot shows voltage
amplitude
and phase. If I right click and change (v(in)) to re(v(in)) then I still
need to do
something with im(v(in)). I either have improper syntax or I'm putting it in
the
wrong place.
If I get it right, what is the label on the left y axis? It seems to be
voltage I was
looking for ohms.
Maybe I need to plot something different from the start?
Thanks for working me through this,
Mike

T

#### Tom Bruhns

Jan 1, 1970
0
Is it possible to graph R and X of a circuit?
Such as looking into a filter, instead of amplitude
and phase, could I graph R and X.
If yes, how?

Thanks, Mike

Yes... OK, there are many ways you could do this. Here's just one.

I'm assuming you will do an AC analysis, to get the response versus
frequency. Excite the circuit with a current source, say I1. It's
convenient to set it to AC 1 amp, zero phase, but not necessary. Have
that current flowing INTO the circuit. Then the voltage at the input
divided by the current will be a complex quantity equal to the net
impedance. If you made the current = 1A, you don't even have to
divide by it, but there's an advantage to do so... The real part of
that will be resistance, and the imaginary part will be reactance.
Let's say you labeled the input node "in" Then to start, plot V(in).
The default is a dB plot. Now right-click on the label at the top. A
box will come up; change the plotted quantity to re(V(in)/I(I1)), and
"ok" that. Now put your cursor over the plot, and right click.
Select "manual limits". Set the left vertical axis to linear. You
should now have a plot of the real part of the input impedance,
displayed in ohms. But wait! If you change the plotted quantity back
to just V(in)/I(I1), it will plot the real part (resistance) on the
left axis, and the imaginary part (reactance) on the right axis.

Dat help any?

Cheers,
Tom

A

#### amdx

Jan 1, 1970
0
Thanks Tom,
That info got me almost all the way there.
The last step, has no effect.
("But wait! If you change the plotted quantity back
to just V(in)/I(I1), it will plot the real part (resistance) on the
left axis, and the imaginary part (reactance) on the right axis.")
For the last step I did as follows, Right click on the label at the top,
changed (V(in)/I(I1)) to V(in)/I(I1) "hmm" (is that correct?) and then
click run, nothing changes.
The curves don't change, I don't get ohms on the right side.
What did I miss?
Thanks again, Mike

Q

#### qrk

Jan 1, 1970
0
Is it possible to graph R and X of a circuit?
Such as looking into a filter, instead of amplitude
and phase, could I graph R and X.
If yes, how?

Thanks, Mike

Left double-click on the Y-axis numbers on the graph. The
"Representation" selecto box that shows Bode, change that to
Cartesian. You will see your results in real an imaginary components.
You can get you plot to display |Z| and phase if you select Linear for
the Bode plot.

T

#### Tom Bruhns

Jan 1, 1970
0
Thanks Tom,
That info got me almost all the way there.
The last step, has no effect.>("But wait! If you change the plotted quantity back

For the last step I did as follows, Right click on the label at the top,
changed (V(in)/I(I1)) to V(in)/I(I1) "hmm" (is that correct?) and then
click run, nothing changes.
The curves don't change, I don't get ohms on the right side.
What did I miss?
Thanks again, Mike
Notice that what I said to plot at first was re(V(in)/I(I1)). That
plots just the resistance; the dashed line, corresponding to the right
vertical axis, is zero. Drop the "re" part (and the parens if you
wish) and you display the resistance on the left axis, with the solid
line, and the reactance on the right, with the dashed line. Cool
thing is that the units are displayed as ohms; it knows that voltage
divided by current is ohms. And there's an "i" in the label for the
reactance, as there should be.

Cheers,
Tom

K

#### Ken Smith

Jan 1, 1970
0
Hi Ken,
I don't quite have it,
I'm plotting (v(in)) which is the voltage out of the signal generator
(with internal resistance) feeding a filter. The plot shows voltage
amplitude
and phase. If I right click and change (v(in)) to re(v(in))

Stop at this point and admire the trace.

Now click on the same node again.

Look at the two traces. You now have the "re(V(in))" and just "V(in)" in
the top of the graph.

Now, move the mouse over the "V(in)" label at the top of the graph. Click
and edit like you did before.

then I still
need to do
something with im(v(in)). I either have improper syntax or I'm putting it in
the
wrong place.

You missed a step.
If I get it right, what is the label on the left y axis? It seems to be
voltage I was
looking for ohms.

You will only be able to plot voltages or currents. If you force a
current of 1A through an impedance, you get a voltage equal to that
impedance so this is how you can get impedances to plot.

Z = V/I

If you want you can use 1mA instead and divide by the 1mA. You could use
the 1uA or 1pA or 1fA or 1aA or ......

T

#### Tom Bruhns

Jan 1, 1970
0
Stop at this point and admire the trace.

Now click on the same node again.

Look at the two traces. You now have the "re(V(in))" and just "V(in)" in
the top of the graph.

Now, move the mouse over the "V(in)" label at the top of the graph. Click
and edit like you did before.

You missed a step.

You will only be able to plot voltages or currents. If you force a
current of 1A through an impedance, you get a voltage equal to that
impedance so this is how you can get impedances to plot.

Not true! If you plot V(in)/I(source) -- that is, a function which is
a voltage divided by a current -- it will display in ohms, complete
with the Omega symbol. That way, you can use any current you want
(not that it makes a difference for an AC analysis, which is taken to
be linear around an operating point -- but it's nice for transient
analyses), and properly display the result in ohms. This works for
quite a few other units, too: volts*current is power, etc.

Cheers,
Tom

A

#### amdx

Jan 1, 1970
0
Thanks guys,
I got the graph that I was looking for, Real ohms on the left ( solid
line)
imaginary ohms on right (dotted line).
At this point I got as far as I could with Tom's info then used qrk's
cartesian
change to finish the graph.
I'll reread Tom's third post and see where I went wrong.
Thanks for all the help,
Mike

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