approximation formula proof

[anhnha]

Hi.
Please help me prove the approximation formula below given in my book. Thanks.

**broken link removed**

 

[albbg]

Please find my answer in the attached file.

 

[anhnha]

Well, thank you very much. Your proof is interesting. I tried to plot two expressions to compare. What was surprising is that the two expressions doesn't come close to each other as one time constant is much larger than the other.

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I just find an error in the expression after "rearranging".

 

[albbg]

I just find an error in the expression after "rearranging".

I think the derivation I did is correct but not as accurate as in the book. Sorry, but I didn't read carefully your attachement. I'll try again.

 

[anhnha]

Thank you. I am eager to read your proof.
Here is the mistake in the proof. I I had problem with internet connection and couldn't make it clear.

 

[albbg]

You are rigth.

 

[albbg]

I tried to develop a proof based on MacLaurin expansion and another based on LMS. Both of them have failed (at least, up to now) due to the excessive complexity of the resulting expressions.

I just developed a calculation to approximate the function and its first derivative in t=0. I hope this time there are no errors.

 

[zorro]

Hi,

Unfortunately there is a mistake in the derivation for f'(t). This derivative at t=0 is 0, as corresponds to the step response of a system with two zeroes at infinity.

But suppose that we want to approximate the response f(t) (4.11) by a single-pole waveform, i.e. of the form g(t)=H*exp(-t/\[\tau\]) with these conditions:

a) f(0)=g(0)
b) the "areas under the curves" are the same, i.e. \[\int\]f(t)dt=\[\int\]g(t)dt where the integrals cover from 0 to \[\infty\].

Then, a calculation similar to albbg's gives H=1 and \[\tau\]=\[\tau\]₁+\[\tau\]₂.

Regards

Z

 

[albbg]

Yes, you are right. f'(t) and g'(t) are instead the integral of f(t) and g(t), not the derivatives.

As I said, on the paper I made different trying. Among these compare the two integrals (from 0 to inf). I don't know why when, after some times, I translated it into a doc file I "converted" them into derivatives. Probably because on my messy paper I omitted the \[\int\] sign.
I'm sorry for my unaccuracy. I should check better before to post.

Thank you zorro to pointed it out.

 

[zorro]

OK.
This "equal area" criterium looks right for some circumstances. For example, if the waveform is a current we find an "equivalent" simple waveform with the same charge.
But the attachment to the first post mentions that the parameter of interest is the propagation delay. I find interesting that it can lead to the same result.

Anhnha: could you upload Figure 4.11 ?
Thanks

Z

 

[Ratch]

anhnha,
As you can see in the attachment, the book is wrong, and the values are different depending on whether the two different Tau's are large or small in relation to each other. "k" is the ratio of one Tau to the other, and the final value of the term depends on whether k goes to infinity or zero.


Ratch

 

[zorro]

As you can see in the attachment, the book is wrong, and the values are different depending on whether the two different Tau's are large or small in relation to each other. "k" is the ratio of one Tau to the other, and the final value of the term depends on whether k goes to infinity or zero.

Sorry, this is wrong.
It's easy to see that if Tau1 and Tau2 are both positive and finite, and Tau1!=Tau2, then the final value is 0.

Regards

Z

 

[Ratch]

zorro,
Please explain what you mean. I aver that when tau1 >> tau2, then the expression is approximated by exp(-t/tau1), which is what the book says it is under those conditions. Of course, when "t" increases, then the expression becomes 0. When tau1 << tau2, the expression is close to 1 at any reasonable value of "t". In both cases, tau1 does not equal tau2.

Ratch

 

[albbg]

If tau1, tau2 are positive and tau1 != tau2, the limits of the function are

lim f(t)=1
t->0

lim f(t)=0
t->inf

regardless wich one of tau1 or tau2 is greater

Furthermore if I swap tau1 and tau2 the shape of the function is still the same

 

[zorro]

I mean that

the values are different depending on whether the two different Tau's are large or small in relation to each other.

is not right.
The final value (i.e. for t->Inf) of the expression is 0 for tau1<tau2 as well as for tau1>tau2, both >0 and finite.

Regards

Z

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Oh, I had cross-posting with albbg!

The two tau's don't need to be very different. The limits given by albbg are valid anyway (with the only restrictions said in the previous posts).

 

[Ratch]

If tau1, tau2 are positive and tau1 != tau2, the limits of the function are

lim f(t)=1
t->0

lim f(t)=0
t->inf

regardless wich one of tau1 or tau2 is greater

Furthermore if I swap tau1 and tau2 the shape of the function is still the same

The book is not asking for what happens when "t" goes to infinity or zero. It is asking what the term is when the the Tau's are radically different. Therefore you have to find the limit as "k" goes to infinity and zero, where Tau1 = k*Tau2. As I show in post #11, only one of the terms at k->0 agrees with the book. If you could show where the term at k->infinity agrees with the book, I sure would appreciate it.

Ratch

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I mean that


is not right.
The final value (i.e. for t->Inf) of the expression is 0 for tau1<tau2 as well as for tau1>tau2, both >0 and finite.

Regards

Z

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Oh, I had cross-posting with albbg!

The two tau's don't need to be very different. The limits given by albbg are valid anyway (with the only restrictions said in the previous posts).

I say the same thing as I said in regards to what albbg posted. We are looking for an approximation when "k" is at one extreme or the other. We are not trying evaluate the approximation when t goes to zero or infinity.

Ratch

 

[zorro]

Ratch,

Then, now I see that when you say "final value" you don't mean the value for t->inf, like many of us are used (see for example https://en.wikipedia.org/wiki/Final_value_theorem). You mean another thing, right? No problem.

It's clear that when tau1>>tau2 the expression approaches exp(-t/tau1) .
You get that result considering tau2=k*tau1 and looking the limit for k->0+ .

Now, let's see what happens when tau1<<tau2 .
To that end, please consider tau1=k*tau2 and consider the limit for k->0+ .
Obviously, the result is exp(-t/tau2) .

As pointed out by albbg in post #14, "if I swap tau1 and tau2 the shape of the function is still the same".

Regards

Z

 

[Ratch]

Ratch,

Then, now I see that when you say "final value" you don't mean the value for t->inf, like many of us are used (see for example https://en.wikipedia.org/wiki/Final_value_theorem). You mean another thing, right? No problem.

It's clear that when tau1>>tau2 the expression approaches exp(-t/tau1) .
You get that result considering tau2=k*tau1 and looking the limit for k->0+ .

Now, let's see what happens when tau1<<tau2 .
To that end, please consider tau1=k*tau2 and consider the limit for k->0+ .
Obviously, the result is exp(-t/tau2) .

As pointed out by albbg in post #14, "if I swap tau1 and tau2 the shape of the function is still the same".

Regards

Z

Looking over the problem again, I think we are both wrong. Here's why. I gave my proof in terms of Tau1. You submited your proof in terms of Tau2. The book gives the expression in terms of Tau. It further defines Tau = Tau1 + Tau2. So we have to find the expression in terms of Tau, not Tau1 or Tau2, and then see what it will be when Tau1 is vastly different from Tau2. I will try to work on it some more tomorrow.

Ratch

 

[anhnha]

Thank you, everyone.
Sorry for the late reply.

And as Ratch said, the result should be the expression in terms of Tau. I also tried to plot two expressions, one of the left and one on the right hand side of that equation. However, I don't see two curves come close together as one time constant is much larger than the other.
BTW, this is figure 4.11.

 

[Ratch]

anhnha,
Plug into the given formula, Tau1 = k*Tau and Tau2 = (1-k)*Tau. Evaluate at k--->0 and k--->1. I think that will give you the result you want.

Ratch