A Sound Mathematical Basis For Sampling - Lesson 3

A Sound Mathematical Basis For Sampling - Lesson 3


Good Morning, once again, Boys and Girls!

I'm sorry that I got called away yesterday; SWMBO, indeed, MBO!

Today I'll derive for you the mathematics of sampling, based on an analysis of the circuits that we actually use, rather than on some dubious mathematics (I will show later why it is dubious) that most authors use, and which they seem to crib from each other without checking! (Perhaps the unsoundness of their mathematics is why they emphasize so much the importance to them of "hands on" experience? If so, that would seem to render them as technicians rather than engineers, would it not? A disturbing thought when so many or our life-support systems depend on the results that the former pupils (that these experts have "taught") produce!)


Take the Unit Step function, U(t) which has the spectrum of 1/s from our Laplace knowledge, together with another Unit Step that has been delayed for a time T. (If we were to be discussing land-line telephony where samples are taken

8000 times per second, then T would be 125 uSecs). That second Unit Step is represented by U(t - T) and has a spectrum of (1/s) * e^(-sT).

(I will cover in my notes at the end why a delayed signal is a function of (t - T), and also the effect of a delay causing the spectrum to be multiplied by e^(-sT). In respect of this latter, remember that e^(-sT) is a complex number that represents a phase shift of all constituent parts of the spectra)

Now, if we take the first Unit Step, (which you remember rises at zero and continues thereafter to eternity) and subtract from it the second, we will get a single short pulse lasting for T, which will be 125 uSecs in the land-line telephony case.

We are relying on the Principle Of Linear Superposition here, in order to determine the spectrum of two signals added together.

Our combined signal, creating the single short pulse, is given by U(t) - U(t - T) and its spectrum is given by 1/s - (1/s) * e^(-sT) which comes down to ...

(1 - e^(-sT)) / s


Now, let us create a second pulse which starts as the first one finishes, and then goes on for a further 125 uSecs.

We can create this with a Unit Step that is delayed by T and then subtract from it another Unit Step that has been delayed by 2T... U(t - T) - U(t - 2T)

This second pulse will clearly have the same frequency content as the first, but with a different set of phases so that the infinite spectral components of the Laplace Transform now add up constructively at a different point in time to give the delayed pulse, and cancel out every where else. (Remember that the effect on the frequency spectrum of a delay of T is to multiply the spectrum by e^(-sT), a complex number having a modulus of 1 thus retaining the amplitude of the spectrum, but shifting the phase of all components linearly)

The spectrum of our second pulse is given by...

(e^(-sT) - e^(-s2T))/s

which simplifies to ... e^(-sT) * (1 - e^(-sT)) / s

This is the same as the spectrum of the first pulse except that all the phases have been modified by the delay factor e(-sT)!


In the same way, a third such delayed pulse would be ... U(t - 2T) - U(t - 3T)

with a spectrum of ....

(e^(-s2T) - e^(-s3T))/s

which simplifies to ... e^(-s2T) * (1 - e^(-sT)) / s

and so on, and so on, ad infinitum.


Our real-world sampling circuits take a sample of the input circuit, and hold it steady so that it can be digitised. Then, at the sample interval (Yes, our T of 125 uSecs above, you're getting ahead of me, well done!) take another sample, then another, and another and so on, and so on.

What we end up with is a series of pulses of differing heights determined by the value of the input signal at the start of the sampling interval.


If our analogue input signal could be termed f(t), then we end up with a series of samples, f(0), f(T), f(2T), f(3T) and so on and so on.


Now, each sample looks like the pulses that we created just now, but with each one of an amplitude governed by the input waveform at the start of the pulse ....

We end up with the pulse sequence ...

f(0) * [ U(t) - U(t - T) ] + f(T) * [ U(t - T) - U(t - 2T) ] + f(2T) * [ U(t - 2T) - U(t - 3T) ] + f(3T) * [ U(t - 3T) - U(t - 4T) ] + ...

with the resultant frequency spectrum given by ...

f(0) * [ (1 - e^(-sT)) / s ] + f(T) * [ (e^(-sT) - e^(-s2T))/s ] + f(2T) * [ (e^(-s2T) - e^(-s3T))/s ] + f(3T) * [ (e^(-s3T) - e^(-s4T))/s ] + ... Quite a mouthful!


We now take out the common term (1 - e^(-sT)) / s , which, as you will remember from above, is the Laplace Transform of the first pulse, and we get ....

[ (1 - e^(-sT)) / s ] * [ f(0) + f(T) * e^(-sT) + f(2T) * e^(-s2T) + f(3T) * e^(-s3T) + ... ]

Still quite a mouthful for the output of our sampler ready to be presented to the input of our DSP's!!


We do two things to reduce the seeming complexity of the above expression.

  1. In the frequency domain, you will remember that multiplication is commutative, A * B giving the same result as B * A, and so we take the factor [ (1 - e^(-sT)) / s ], (what might be thought of as what gives us the pulse shapes), and ignore it until the output stages of our DSP. This leaves us with a sequence ...
[ f(0) + f(T) * e^(-sT) + f(2T) * e^(-s2T) + f(3T) * e^(-s3T) + ... ]



Remember from yesterday's lecture that the sampling circuit is, as far as Laplacian analysis goes, a SIGNAL GENERATOR that cannot be analysed any deeper!

  1. We make a parameter substition using the relationship ...

z = e^(sT)

so that e^(-sT) becomes z^(-1) or 1/z

so giving us ...

[ f(0) + f(T) * z^(-1) + f(2T) * z^(-2) + f(3T) * z^(-3) + ... ]


So, we have arrived with the same mathematical expression for a sampled input sequence as other authors, but we have not had to appeal to Dirac's Delta Function and neither have we had to invent mythical properties for that function in order to justify the claims made for that approach!


In my next lecture, I will attack the misuse of Dirac's Delta Function as it is usually presented by other authors and I will show the fallacies that they propagate - they are good examples of the faulty memes described by Richard Dawkins!

Reply to
Airy R. Bean
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Reply to
Airy R. Bean

You only needed to apologise for coming back, retard. When will it penetrate your thick skull that there are real engineers of a status you can only dream of holding a door open for on this group, and that these real engineers have no wish to be lectured to by a mental defective with an over inflated ego. The simple fact that you start your rubbish with the words "Good Morning" on a post timed at 17:29 speaks volumes. David

Reply to
David Sausins

He/She probably failed to get into a university so he's trying to preach some text-book stuff to make himself/herself seem intelligent. Maybe if he/she had a job or a real profession, we wouldn't get this garbage posted here.

Obviously no one in the real world would listen.

....Poor thing.

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