Given (in this case) a square wave of fixed frequency, fast but non-zero rising & falling edges, 50% duty cycle, and an exact integer number of cycles (etc.), what's the narrowest possible harmonic bandwidth that ought to be expected? In other words, when is the simulation as optimized as it can be?

On a more general note, assuming artifacts of a non-ideal simulation have been minimized, does the harmonic bandwidth provide useful information, or do only the peak values matter?

开云体育

When some sort of modulation is going on, energy gets transferred from the signal at the peak, into the sidebands around it. ?So as the amount of modulation gets worse, I think I would expect to see the peak shown by the FFT to decrease, while the width increases.

-- 

AC2CL 

I do not think there is any thrill that 
can go through the human heart like that felt by the inventor as 
he sees some creation of the brain unfolding to success... 
Such emotions make a man forget food, sleep, friends, love, everything. 

- Nikola Tesla 

 

? RC and John Thanks? for the response,but I am blinded by science now as

this is my first attempt to model FT behaviour

I would like to be clear on at least one matter? " 'the value of the window"

Is this the same as the number of cycles AKA Span ?

Is it possible to display the harmonics of say a 10MHz square wave to 100Hz resolution ? My 30 year-old HP spectrum analyzer can

Can anyone illustrate? with the pulse and Tran statements using a square wave at any freqency

? RC and John Thanks? for the response,but I am blinded by science now
as

this is my first attempt to model FT behaviour


I would like to be clear on at least one matter? " 'the value of the
window"

Is this the same as the number of cycles AKA Span ?

Is it possible to display the harmonics of say a 10MHz square wave to
100Hz resolution ? My 30 year-old HP spectrum analyzer can

Working backwards through that last bit in the Help, to get 100 Hz resolution, the window width is 1/100 s = 10 milliseconds. You need to consider at least harmonics up to the ninth, 90 MHz, so the number of samples has to be at least 90 MHz/100 Hz = 90000. LTspice supports much larger numbers. But the necessary simulation time may be archaeological.

Here are some tips.

(1) ?To get finer frequency resolution in the FFT results, you need to
use longer simulation times. ?The two are directly related.

You need at least two time samples at the highest frequency of the FFT.

1/90MHz * 0.5 = 5.56 ns

Helmut's example simulates for 20 ms, which actually gives you 50 Hz
resolution, enough to see the "trough" between alternate components
that are 100 Hz apart. ?Given that, you need at least 3600000 for the
"Number of data point samples in time" to have time samples no greater
than 5.56 ns apart. ?Using Helmut's recommended 4194304 (= 2^22) gives
you that. ?That simulation's FFT goes up to (1/2) * 1/(20ms/4194304) =
104.858 MHz, which is what I see when I run it.

Using the bare minimum (3600000) probably stops exactly AT 90 MHz and
isn't enough to show you the 90 MHz component. ?So always go higher if
there is any doubt.

John Woodgate wrote:

To get x Hz resolution, you should, in practice, simulate for 2/x
seconds.

Actually, x Hz resolution needs only 1/x seconds.

Simulating for 2/x seconds gives you x/2 Hz resolution, which in
practice might be useful if you know you have discrete frequency
components at x Hz increments. ?Having those in-between components in
the FFT plot both makes it easier to see the components, and can be
used as a rough gauge of the FFT's accuracy. ?If components which
should be very low, are not low, then something is wrong and you need
to take a closer look