Author Topic: Rigol DS1102E question  (Read 4813 times)

0 Members and 1 Guest are viewing this topic.

Offline nick.sekTopic starter

  • Regular Contributor
  • *
  • Posts: 59
  • Country: 00
Rigol DS1102E question
« on: May 21, 2012, 06:17:52 am »
One of my classmates said that there was a Laplace math function on the DS1102E oscilloscope, does any one know if this is true, and how do you bring it up?
 

Offline nick.sekTopic starter

  • Regular Contributor
  • *
  • Posts: 59
  • Country: 00
Re: Rigol DS1102E question
« Reply #1 on: May 21, 2012, 06:23:41 am »
Fast Fourier Transform - is a function that is possible on this oscilloscope - but I can't see any Laplace....
 

Offline TriodeTiger

  • Regular Contributor
  • *
  • Posts: 199
  • Country: ca
Re: Rigol DS1102E question
« Reply #2 on: May 21, 2012, 06:43:48 am »
Why have you not looked at the manual? If you are looking for functions that device has or hasn't, then that is the place to look and discover something new.

I've never heard of an oscilloscope having a Laplace transform function before for math, and you would have to convince me that it would be useful before it would be added on to a general purpose low end oscilloscope and how it can do this with the available data.

There could be some (un)official firmware updates adding this as the manuals age, you would have to check for that as well. A quick search brings up nothing. You're free to log the data and do your own functions however (why I'd prefer a USB scope over a silly DSO)
« Last Edit: May 21, 2012, 06:46:10 am by Xander »
"Yes, I have deliberately traded off robustness for the sake of having knobs." - Dave Jones.
 

Offline amspire

  • Super Contributor
  • ***
  • Posts: 3802
  • Country: au
Re: Rigol DS1102E question
« Reply #3 on: May 21, 2012, 07:03:01 am »
One of my classmates said that there was a Laplace math function on the DS1102E oscilloscope, does any one know if this is true, and how do you bring it up?
I am curious. What do you do with a display of a Laplace Transform function of a waveform? I have used Laplace Transforms on paper, but I have never considered using them on a scope.

It is really easy to find a use for the frequency axis in  Fourier Transform, but what do you do with the "s" axis?

Richard.
 

Offline nick.sekTopic starter

  • Regular Contributor
  • *
  • Posts: 59
  • Country: 00
Re: Rigol DS1102E question
« Reply #4 on: May 22, 2012, 10:55:42 pm »
The main reason why I want to use the Laplace function - which does not exist is to prove math to myself, I get a better understanding when I actually see things happening, rather than imagine them happening. Plus it can give me the ability to see - how accurate the calculated world is.
 

Offline Bored@Work

  • Super Contributor
  • ***
  • Posts: 3932
  • Country: 00
Re: Rigol DS1102E question
« Reply #5 on: May 22, 2012, 11:20:49 pm »
It is really easy to find a use for the frequency axis in  Fourier Transform, but what do you do with the "s" axis?

It is not just an axis, but a plane (real and imaginary components). What to do with it? I just got scolded for pointing out on the forum that Americans have a fixation with poles and zeros. :-) Americans would be happy to see an s-plane and argue with the poles and zeros.

But seriously, the Fourier transform is the Laplace transform evaluated under some special conditions (which I forgot the details and I am too lazy to look it up). So one could argue that the Rigol does a special kind of Laplace transform evaluation.
I delete PMs unread. If you have something to say, say it in public.
For all else: Profile->[Modify Profile]Buddies/Ignore List->Edit Ignore List
 

Offline amspire

  • Super Contributor
  • ***
  • Posts: 3802
  • Country: au
Re: Rigol DS1102E question
« Reply #6 on: May 23, 2012, 12:05:36 am »
It is really easy to find a use for the frequency axis in  Fourier Transform, but what do you do with the "s" axis?

It is not just an axis, but a plane (real and imaginary components). What to do with it? I just got scolded for pointing out on the forum that Americans have a fixation with poles and zeros. :-) Americans would be happy to see an s-plane and argue with the poles and zeros.

But seriously, the Fourier transform is the Laplace transform evaluated under some special conditions (which I forgot the details and I am too lazy to look it up). So one could argue that the Rigol does a special kind of Laplace transform evaluation.
From memory, if you substitute "s" with i * 2 * pi *f , you and up with a Fourier Transform. In the Laplace domains however, I have only worked with it algebraically, with the exception of graphically determining poles and zeros of say a filter transfer function in Laplace form.

Looking at the Laplace transform of a waveform on a scope may have some use, but I do not know what use it would have.

Richard.
« Last Edit: May 23, 2012, 06:49:15 am by amspire »
 

Offline IanB

  • Super Contributor
  • ***
  • Posts: 12163
  • Country: us
Re: Rigol DS1102E question
« Reply #7 on: May 23, 2012, 12:27:06 am »
The main reason why I want to use the Laplace function - which does not exist is to prove math to myself, I get a better understanding when I actually see things happening, rather than imagine them happening. Plus it can give me the ability to see - how accurate the calculated world is.

Unfortunately, you are going the wrong way. "t" is the time domain, the real world where you can see real things happening. "s" is the Laplace domain, an imaginary world that exists only in mathematics. When you transform the time domain into the Laplace domain you have converted the real world into a kind of mathematical abstraction with special properties, but it is no longer real.

This transformation is similar to logarithms in arithmetic. If you take logarithms you transform multiplication into addition, but it is no longer the problem you originally had. Similarly if you take Laplace transforms you transform differential equations into algebraic equations, but in a similar way the algebraic equations are no longer what you started with. They are a kind of mirror of the problem reflected into an imaginary world.

Given this, I don't see how a Laplace transformation will help you to see what is really happening with your measurements?
 

Offline electrode

  • Regular Contributor
  • *
  • Posts: 141
  • Country: au
Re: Rigol DS1102E question
« Reply #8 on: May 23, 2012, 09:42:46 am »
If you could do a frequency sweep on a circuit, I could understand why laplace bode plots might be useful; you may get a plant transfer function or something.

A single laplace transform doesn't seem too helpful, eg. a pure sinusoid: sin(wt) -> w/(s^2 + w^2). Do you just want that equation on the scope? :S

The key distinction between laplace and fourier frequency domains is that the former tends to be considered unilaterally (ie. 0 to infinity), while the latter tends to be considered over all time (-inf to +inf). With laplace, the substitution s=jw can be made, as amspire said (where j=i and w=2*pi*f).

Certainly, laplace transforms are great for solving initial value problems, but on a scope, I can't think of a situation where that would be required – you have access to experimental data in any case. The real benefit of the FFT is the decomposition of the signal into fundamental frequencies, which has many uses, such as determining how pure a waveform is, identifying sources of noise, etc.
 


Share me

Digg  Facebook  SlashDot  Delicious  Technorati  Twitter  Google  Yahoo
Smf