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Hi Richard,
??
Aside from the form Helmut gave you, you can set any resistor to as many changes as you need. The format, which is entered in the resistance window in place of a single value is :
??????
???????? R=table(time,0,res1,time1a,res1,time1b,res2.....,last res)
??
Example : R=table(time,0,10K,3m,10K,3.00001m,100K,6m,100K,6.00001m,1)
??
This starts with the resistance value at 10K until 3mS after start, then steps to 100K until 6mS and then to 1 ohm for the rest of the run. The run duration sets the max time end point. Careful to place comma's between each entry, and no spaces.
??
If you want a slope change instead of a step, for example a changing value of 100 ohms to 10 ohms over a 10mS time and back to 100 ohms over the next 10mS,??the form would be:
??
???????? R=table(time,0,100,10m,10,20m,100)
??
Lastly, always check your entries for pairs. There must be a time/value pair for each entry. This listing approach is invaluable for all sorts of applications which require either a slope or step change such as load testing etc.
??
- Cordially - RC
??


________________________________
From: Richard Norman <rnorman3@...>
To: "ltspice@..." <ltspice@...>
Sent: Sunday, April 14, 2013 2:30 AM
Subject: [LTspice] Step a resistor over time

??

Hi All, is it possible to step a resistor over time? Right now if I use the step command like this:

.step param R 100 1000 100
.tran 0 1ms 0 1

I
end up with a bunch of parallel horizontal lines. What I want is a
bunch of points (maybe even connected) over time. So in other words,
over the 1ms, I want the resistance to step from 100 to 1000. Is it
possible?

Thanks!
Richard

[Non-text portions of this message have been removed]




[Non-text portions of this message have been removed]

MOHAMMAD A MAKTOOMI <amaktoomamu@...> wrote:

Andy, how could you guess so quickly that waveform was missing some
peaks and the step-size be reduced in the setup

< ()?

I ask this because this will help me start thinking the way an experienced
user thinks.

When I looked at the waveforms, I noticed that the positive peaks of
the 1 MHz output were irregular, even when zoomed in. It looked to me
as if it suffered from aliasing, being sampled but not with a high
enough sampling rate. That led me to check for the plotwinsize=0
option, and then to add the max timestep parameter.

Andy

Hello,

You can only use .STEP to change the values of components in
the .AC simulation.

Best regards,
Helmut



--- In LTspice@..., "MOHAMMAD A MAKTOOMI" <amaktoomamu@> wrote:

Thank you, Andy for your hints.
But, in '.AC' we need to have an AC voltage or current source. Here, I don't have any such thing (as I wish to vary the frequency of R2, NOT that of any source), so how do I proceed?

--- In LTspice@..., Andy <Andrew.Ingraham@> wrote:

MOHAMMAD A MAKTOOMI <amaktoomamu@> wrote:

...

Note: In your circuit, you may need to decrease the maximum timestep
down to about 10ns. (.tran 0 0.1m 0 10n) The output waveform is
highly distorted at the highest frequency and LTspice was missing the
narrow peak before doing that, making the amplitude look a lot less.

It probably also means the result is unrealistic at that frequency,
but that's another matter.

Andy

Greetings,

If we make some assumptions then the math becomes easy enough that even I can do it. First, imagine that you are taking a pure sinewave and subtracting a small signal from it to create the waveform you want. The pure sinewave has no distortion, while the small signal looks roughly like a tiny triangle starting at 0V and reaching some maximum voltage before dropping back to 0V, symmetrically at each zero crossing. That maximum voltage is sine(2 pi D / T), for small values of D/T this is about linear.

If we assume that your 'dent' is small enough that it does not appreciably affect the energy at the fundamental frequency (i.e. the distortion is 'small'), then the total distortion is roughly the ratio of the RMS value of small pulse divided by the RMS voltage of the fundamental.

It's simple to show that the RMS value of a triangular pulse is proportional to (D/T)^1.5 as Helmut stated in the other thread, so that should be what you observe in THD also. The reason this disagrees with what you found has to do with the way THD is calculated by the .FOUR function. Notice that the error log shows Fourier components up to 9th order, or 9kHz in your example. Note also that the magnitude of the distortion components is the SAME for each one (since your signal is symmetric there will only be odd components, ignore the even ones as they are noise), it has not started to drop off by the 9th. The calculated THD is based on only those four frequencies, it should be easy to imagine that if you calculated more components the THD would keep going up.

How many do you need? To get an idea, look at the width of your distortion pulse, 10us in your example. It's a good bet that the 'missing' pulse has components up to at least 100kHz. So you may need to go past the 99th harmonic to get anything accurate. Or, remove the fundamental and measure the RMS of what remains, much easier.

Once this is understood it's easy to see why your observation came out too low (not enough coefficients included in the measurement) and also why the observed exponent is too high (as D increases the number of important coefficients will be reduced, so the THD will not increase as fast as you observed, exponent of 1.5 instead of 2).

HTH,
Fred

--- In LTspice@..., "Helmut" <helmutsennewald@> wrote:

Hello Echidna,

Crossover Distortion
--------------------

I calculated the formula with Fourier series and approximation.

sin(x) = x-1/3*x^3

1-x^2 = 1-2*x

THD = 100%*2*pi*sqrt(1/3*(D/T)^3)
=================================

Files > Temp > crossover_distortion.asc

You can check the distortion calculated by .FOUR with
View -> SPICE Error log

Best regards,
Helmut


--- In LTspice@..., "Echidna" <mchambin@> wrote:

Hello.
How to calculate, using maths and definitions, the THD of this signal: A sine with a small dent at the crossovers.

S(t) = 0 for 0 < t < D where D is much smaller than T
S(t) = sin omega*t for D < t < T/2 where omega is 2*pi/T
S(t) = 0 for T/2 < t < T/2 + D
S(t) = sin omega*t for T/2 + D < t < T

Using LTspice simulations I find THD = 25*( D/T )^2
In this simulation I used a 1000 HZ SINE combined with a PULSE with values of D like 0.1u 0.2u 0.5u 1u 2u 5u 10u 20u 50u
These simulations gave me a THD that perfectly fits with 25*( D/T )^2

Here is the issue.
I was enable to prove this result with maths
(instead of simulations ).
I started computing the RMS value of the error signal (the dents at the crossovers)
With sin omega*t = omega*t ( valid for t << T )
I don' t get the 25*( D/T )^2 result, I do get a fonction of D/T, so far so good, but I don't get the right exponant. It seems my approach is wrong.

Is there a signal theory / maths guru who can give the proof that for such a signal S(t) ( with D << T ) we have THD = 25*( D/T )^2

In message <kkf3is+3juc@...>, dated Sun, 14 Apr 2013, Helmut
<helmutsennewald@...> writes:

N= 1.570213683
+ RS= 0.762136823
+ IKF= 392.8188163

Not Helmut's responsibility, but isn't it disquieting that these are
given to 10 significant figures? Doesn't inspire confidence. Can we sue
if IKF turns out to be 392.8188162?(;-)
--
OOO - Own Opinions Only. See www.jmwa.demon.co.uk
Things to do today: Cut a hole in reality and push a turnip through.
John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK

Hello Echidna,

Crossover Distortion
--------------------

I calculated the formula with Fourier series and approximation.

sin(x) = x-1/3*x^3

1-x^2 = 1-2*x

THD = 100%*2*pi*sqrt(1/3*(D/T)^3)
=================================

Files > Temp > crossover_distortion.asc

You can check the distortion calculated by .FOUR with
View -> SPICE Error log

Best regards,
Helmut


--- In LTspice@..., "Echidna" <mchambin@> wrote:

Hello.
How to calculate, using maths and definitions, the THD of this signal: A sine with a small dent at the crossovers.

S(t) = 0 for 0 < t < D where D is much smaller than T
S(t) = sin omega*t for D < t < T/2 where omega is 2*pi/T
S(t) = 0 for T/2 < t < T/2 + D
S(t) = sin omega*t for T/2 + D < t < T

Using LTspice simulations I find THD = 25*( D/T )^2
In this simulation I used a 1000 HZ SINE combined with a PULSE with values of D like 0.1u 0.2u 0.5u 1u 2u 5u 10u 20u 50u
These simulations gave me a THD that perfectly fits with 25*( D/T )^2

Here is the issue.
I was enable to prove this result with maths
(instead of simulations ).
I started computing the RMS value of the error signal (the dents at the crossovers)
With sin omega*t = omega*t ( valid for t << T )
I don' t get the 25*( D/T )^2 result, I do get a fonction of D/T, so far so good, but I don't get the right exponant. It seems my approach is wrong.

Is there a signal theory / maths guru who can give the proof that for such a signal S(t) ( with D << T ) we have THD = 25*( D/T )^2