approximation formula proof

Ratch,

When one of the two Tau goes to 0 (and the other doesn't) the result is seen immediately from the expression itself: the terms containing the Tau that vanishes disappear from the numerator and from the denominator. Whether it is Tau1=0 or Tau2=0 is irrelevant.

Te book states "This approximation works best when one time constant is significantly bigger that the other". Then shows that it works better for ratio 6.5 (error <7%) than for ratio 1 (error <15%).
The case in which one of the Tau's is 0 is not of interest.

I tried to show that your assertion that the situations tau1<<tau2 and tau2<<tau1 are different is wrong.
Your assertion that the book is wrong is unfair too.

Ratch said:

Looking over the problem again, I think we are both wrong.

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I don't understant why you say that I'm wrong too. Please explain. Thanks

Z

zorro,

When one of the two Tau goes to 0 (and the other doesn't) the result is seen immediately from the expression itself: the terms containing the Tau that vanishes disappear from the numerator and from the denominator. Whether it is Tau1=0 or Tau2=0 is irrelevant.

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Yes, that is true.

Te book states "This approximation works best when one time constant is significantly bigger that the other". Then shows that it works better for ratio 6.5 (error <7%) than for ratio 1 (error <15%).
The case in which one of the Tau's is 0 is not of interest.

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So it does, but that is not what we wanted to prove.

I tried to show that your assertion that the situations tau1<<tau2 and tau2<<tau1 are different is wrong.

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Yes, you proved that the reduction formula was not related to Tau1 or Tau2 alone, but at that time, you did not state that the relationship was related to the sum of Tau1 and Tau2 or Tau.

Your assertion that the book is wrong is unfair too.

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My assertion about the book was wrong.

I don't understant why you say that I'm wrong too. Please explain. Thanks

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Sure, I will be glad to. At the time neither of us noticed that the book stated the reduced expression in terms of Tau, not Tau1 or Tau2. On that point, we were both wrong. When I realized I was wrong, I carefully read again what the book said about the relationship being in terms of Tau, not Tau1 or Tau2.

Ratch

Sorry, but probably I didn't understand exactly what you are saying. In my post #7 I found a single time constant approximation of the second order system that agrees with the book. The time constant was tau=tau1+tau2 so the approximation is
exp[-t/(tau1+tau2)];
In the doc I just called tau=z, tau1=x and tau2=y for simplicity of writing.
However I've plotted the curves: Original (second order system) and approximated (with tau=tau1+tau2) for K=1.02, K=2, K=5 and K=20

albbg,
I looked at the attachment in post #7 and found it too cryptic to follow. Also, I do not see a summary statement declaring the reduction formula to be exp(t/Tau)

Ratch

In the "doc" file of post #7 I simply called f(t) the original function and g(t) the approximated one. As said for simplicity of writing I renamed tau1 with x and tau2 with y.

The function to be used to approximate f(t), (i.e. g(t)) was H*exp(-t/z), with H and z unknown.

First I proved that H=1 because in zero we want g(0)=f(0). As a result g(t) was reduced to exp(-t/z).
After that I've integrated (not derived as I've wrongly written in the doc) with limit (0, inf) both f(t) and g(t).

Then I equated the results of the integrals of the two functions and I found z=x+y that is z=tau+tau2
So we can rename z with tau, so tau=tau1+tau2

I hope now it's more clear.

albbg,
Yes, it is somewhat clearer. But why do calculus operations when the reduction formula can be discerned by algebraic operations and limits calculatons? Also, the book already gave the relationship Tau = Tau1 + Tau2.

Ratch

Ratch,

Ratch said:

So it does, but that is not what we wanted to prove.

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What we wanted to prove (or even better, to deduce) is that a good approximation to the second-order response (4.11) is a single-pole response with tau=tau1+tau2.

Ratch said:

Yes, you proved that the reduction formula was not related to Tau1 or Tau2 alone, but at that time, you did not state that the relationship was related to the sum of Tau1 and Tau2 or Tau.

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Please read my post #8.

Ratch said:

At the time neither of us noticed that the book stated the reduced expression in terms of Tau, not Tau1 or Tau2. On that point, we were both wrong.

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Please read my post #8.

Ratch said:

But why do calculus operations when the reduction formula can be discerned by algebraic operations and limits calculatons?

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It was not shown in this thread (nor in the page of the book) a rationale for the approximation Tau=Tau1+Tau2 using algebraic operations and limits calculatons.
Do you dislike calculus?

Ratch said:

Also, the book already gave the relationship Tau = Tau1 + Tau2.
Ratch

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But this is precisely what we try to prove!!!
Please see post #1.

Regards

Z

zorro,

What we wanted to prove (or even better, to deduce) is that a good approximation to the second-order response (4.11) is a single-pole response with tau=tau1+tau2.

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I will agree with that.

Please read my post #8.

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I did. It shows that Tau = Tau1+Tau2, which the book already states is true. But, the formula with Tau only was not derived.

It was not shown in this thread (nor in the page of the book) a rationale for the approximation Tau=Tau1+Tau2 using algebraic operations and limits calculatons.
Do you dislike calculus?

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That is true, but I will take a given. Neither the book or the OP asked to derive Tau. I don't dislike calculus, but if algebra works, then I will use that first.

Originally Posted by Ratch

Also, the book already gave the relationship Tau = Tau1 + Tau2.
Ratch
But this is precisely what we try to prove!!!
Please see post #1.

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No, the request was to prove the reduction formula, not Tau = Tau1 + Tau2.

Ratch

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anhnha said:

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

**broken link removed**

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I sure feel like one dumb bunny. I should have seen this earlier. This problem is almost solvable by inspection. The book does all the heavy work. No need to get involved with the Maclaurin series, calculus or any other high octane math. Look at equation 4.12 in the book. It shows Tau1 and Tau2 in terms of a constant added or subtracted by a fearsome looking radical. However, when Tau1 and Tau2 are added together, then the radical disappears, and only the constant remains. The book calls this constant Tau, which is based on the circuit RC values, and proves that Tau1 and Tau2 are equal to that constant. Knowing that, it is a simple matter to do an easy substitution as I outlined in post #20 to get the desired simplified formula.

Ratch

Ratch,

I did. It shows that Tau = Tau1+Tau2, which the book already states is true.

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The book does not show why.

But, the formula with Tau only was not derived.

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No, the request was to prove the reduction formula, not Tau = Tau1 + Tau2.

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Do you call "reduction formula" the relationship (4.11)\[\approx\] VDD*exp(-t/Tau) ?

It doesn't need to be proved.
The book approximates the second-order circuit as a first-order system with a single time constant. Its response has the form VDD*exp(-t/Tau), and he fact that it is a suitable approximation is evident. This is a model with one parameter that has to be determined (Tau).

It would be possible to propose a different approximation, for instance a linear-piecewise one, or a polynomial one if its range of validity can be restricted. In that cases, we would need to estimate the parameters (slopes, coefficients, etc) in terms of Tau1 and Tau2.
(For the purposes of the book, the single-pole model is more suitable than the two above mentioned examples.)

Z

zorro,

The book does not show why.

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Yes, the book does show why it is true in equation (4.12) as I already explained in my last post. Furthermore, it shows that Tau is a easily obtained constant

Do you call "reduction formula" the relationship (4.11) VDD*exp(-t/Tau) ?

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Yes, I do.

It doesn't need to be proved.

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It has to be derived, which is about the same thing.

The book approximates the second-order circuit as a first-order system with a single time constant. Its response has the form VDD*exp(-t/Tau), and he fact that it is a suitable approximation is evident.

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Yes, that simplification is the book's achievement. It is not evident at first glance that it is a suitable approximation. That is why the OP asked about it. However, it is easily proven to be the case.

This is a model with one parameter that has to be determined (Tau).

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Yes, and the book shows and proves that Tau is a easily determined constant.

It would be possible to propose a different approximation, for instance a linear-piecewise one, or a polynomial one if its range of validity can be restricted. In that cases, we would need to estimate the parameters (slopes, coefficients, etc) in terms of Tau1 and Tau2.
(For the purposes of the book, the single-pole model is more suitable than the two above mentioned examples.)

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Why make it more difficult when it is already easy.

Ratch

Ratch said:

zorro,
Yes, the book does show why it is true in equation (4.12) as I already explained in my last post. Furthermore, it shows that Tau is a easily obtained constant
Ratch

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As stated by the book, the equation 4.12 is too complicated and it has to be simplified. If you want to reduce it to the sum of the two time constants, you should prove that the term under the square root is negligible.
I think is simpler to prove (as I did) that a second order system can be approximated with a single order system setting its time constant equal to the sum of the two constant of the second order time system. The approximation is valid under the circumstances already discussed.

I tought anhnha asked about tau=tau1+tau2, however he (or she) can clarify

albbg,

As stated by the book, the equation 4.12 is too complicated and it has to be simplified. If you want to reduce it to the sum of the two time constants, you should prove that the term under the square root is negligible.

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I don't think you understood what I said about Tau1 and Tau2 in post #27. Let's look at 4.12 again. It says that Tau1 is (X/2)*(1+sqrt(Y) and Tau2 is (X/2)*(1-sqrt(Y) . Where X and Y are constants dependent on the circuit component values. We can easily see that adding Tau1 and Tau2 gives us just X, because the sqrt(Y) cancels out no matter how large its value is. The book shows and proves that the sum is a simple easily calculated constant dependent on circuit values. Once we know that Tau1 plus Tau2 are a constant we can call Tau, then applying the method outlined in post #20 gives us the approximation formula as the first equation in the book. In summary, finding Tau1 or Tau2 is difficult, but finding their sum is easy.

Ratch

Yes, the 4.12 says that you can easily found tau1+tau2 in terms of components value, that is:

tau1+tau2=R1C1+(R1+R2)C2

despite the fact that tau1 and tau2 are quit complicated to find.

But this doesn't mean that tau1+tau2 is the time constant that approximates well the behaviour of a second oprder system, unless you can prove it. This, in my opinion, is why we need to proof that tau=tau1+tau2 is a good approximation

albbg,

But this doesn't mean that tau1+tau2 is the time constant that well approximate the behaviour of a second oprder system, unless you can prove it

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The book already did that by doing the algebra for the transfer function (4.10) and then the response to a unit step function in (4.11). Those are exact equations involving Tau1 and Tau2.

Ratch