Double Exponential

[kmortis]

Ok, it may be that I am an idiot, or I might just have forgotten 80% of what I was taught in college about double exponentials and initial states, but I just can't figure out how to make this work. The general equation for a transient pulse (ie lightning or EMP) is:

v(t)=V0 * k * (e^(-at) - e^(-bt)) V/m

Where:
V0 = Voltage max
k = scaling factor, usually 1.03
a & b = coefficients to adjust the waveform.

The problem is, the waveforms are generally defined by time markers. For example, RTCA DO-160D defines waveform 5a as a double exponential waveform with a risetime of no more than 6.4 usec and a fall time of no more than 120 usec. (Please forgive me if I have either number wrong, my copy of DO-160 is at work.)

So, how do I determine a & b so I can use MS Excel, MathCAD or any other number crunching software to analyse a circuits performance when under the effects of this waveform? I've tried to set the initial conditions such that:
v(T1) = A*k = A * k (e^(-aT1) - e^(-bT1))
v(T2) = 0 = A * k (e^(-aT2) - e^(-bT2))
v'(T1) = 0 = A * k (-a*e^(-aT1) + b*e^(-bT1))

but all I get is crap for answers. Can anyone help?

 

[69dodge]

How does the standard define "fall time"?

The v(t) you've given never falls all the way to v=0, after t=0.

 

[kmortis]

How does the standard define "fall time"?

The v(t) you've given never falls all the way to v=0, after t=0.

It does. It's all in how you choose a&b.

I'd have to check, tho. I think that t_f is defined as the time it takes for the signal to fall from .9Vo to .1Vo.

 

[69dodge]

It does. It's all in how you choose a&b.

Are you sure?

If x differs from y, then e^x differs from e^y. So, how could the difference between the two exponentials ever be 0, except at t = 0 when they're both 1?

 

[Dilb]

It does. It's all in how you choose a&b.

I'd have to check, tho. I think that t_f is defined as the time it takes for the signal to fall from .9Vo to .1Vo.

Nope. Unless a=b (which makes for a somewhat pointless result (pun intended)) it's always some non-zero value.

Incidently, you can do superscripts and subscripts with [ sup ] and [ sub ] without the spaces.

Edit- opps, did that wrong. Let me think a bit

 

[kmortis]

Nope. Unless a=b (which makes for a somewhat pointless result (pun intended)) it's always some non-zero value.

Incidently, you can do superscripts and subscripts with [ sup ] and [ sub ] without the spaces.

Anyway, divide both sides by e^-a so you have

V/V₀k = 1 - e^(a-b)t
And solve for t with whatever conditions, e.g.
0.9 = 1 - e^(a-b)t
use first result for rise time

0.1 = 1 - e^(a-b)t
use second result for fall time

Where'd the e^at go on the LHS?

 

[Dilb]

Sorry, this is harder than I first appreciated, at least the way I'm trying to do it.

 

[Dilb]

In fact, I've totally forgotten how to deal with exponentials.

 

[kalen]

Are you assuming a and b are both postive and real? If so, dodge is right and the equations in the OP will only have the trivial solution ( A=0). I suggest plotting out your original equation for various a,b and you will see that this is the case.

As for your analysis, I'm not sure I understand what the problem is. Do you have some data? Do you want to find some kind of properties of this equation?

 

[Dilb]

I don't think I'm currently up to finding exact solutions, but as a rough estimate assume the terms are independant, so that you have the e^-bt term going to some small number at the rise time, then the e^-at term going to some small number at the fall time.

I can get an okay waveform using 0.05 as the small number, so

a = -ln(0.05)/t_fall
b = -ln(0.05)/t_rise

although this does require a k of about 1.25 if you want to actually get V₀.

The k value can be found by some calculus and plugging into equations. The maximum voltage occurs at

t_max = ln(a/b)/(a-b)

So k = 1/V(t_max)

although it's pretty close to just the rise time. It gets a less accurate if the fall time isn't much more than the rise time.

 

[T'ai Chi]

v(T1) = A*k = A * k (e^(-aT1) - e^(-bT1))
v(T2) = 0 = A * k (e^(-aT2) - e^(-bT2))
v'(T1) = 0 = A * k (-a*e^(-aT1) + b*e^(-bT1))

Doesn't this imply that a=b ?

 

[kmortis]

Doesn't this imply that a=b ?

Not in a real situation. The values we use for MIL-STD-464 Lighitning are a=~64,000 and b=~120,000; I can't recall off the top of my head what the real values are, but those are the ballparks.

Are you assuming a and b are both postive and real? If so, dodge is right and the equations in the OP will only have the trivial solution ( A=0). I suggest plotting out your original equation for various a,b and you will see that this is the case.

As for your analysis, I'm not sure I understand what the problem is. Do you have some data? Do you want to find some kind of properties of this equation?

Yes, a&b are real, and I've only ever seen them positive.

I don't think I'm currently up to finding exact solutions, but as a rough estimate assume the terms are independant, so that you have the e^-bt term going to some small number at the rise time, then the e^-at term going to some small number at the fall time.

I can get an okay waveform using 0.05 as the small number, so

a = -ln(0.05)/t_fall
b = -ln(0.05)/t_rise

although this does require a k of about 1.25 if you want to actually get V₀.

The k value can be found by some calculus and plugging into equations. The maximum voltage occurs at

t_max = ln(a/b)/(a-b)

So k = 1/V(t_max)

although it's pretty close to just the rise time. It gets a less accurate if the fall time isn't much more than the rise time.

I'll have to get a real set of values when I get back to work tomorrow.

 

[chipotle]

Take the log of your initial equation. Then it will be in a linear form where you can use a least squares estimation of a and b.

Edited to add a link to this thread where another linearization example is shown.
http://www.internationalskeptics.com/forums/showthread.php?t=58521

 

[Art Vandelay]

Take the log of your initial equation. Then it will be in a linear form where you can use a least squares estimation of a and b.

No, with two exponentials, it isn't linear.

The general equation for a transient pulse (ie lightning or EMP) is:

v(t)=V0 * k * (e^(-at) - e^(-bt)) V/m

Only if t is defined so that v(0)=0. The truly general form would be
v(t)=V0(k1e^(-at)-k2e^(-bt))

(If V0 is given in V/m, then you don't need to put in the extra V/m at the end)

In which case, if T2 is the time at which v(t)=0,
a-b=ln(k1/k2)/T2

Where'd the e^at go on the LHS?

v(T2)=V0(k1e^(-at)-k2e^(-bt))=0
Divide both sides by V0 (presumably, V0 isn't zero)
k1e^(-at)-k2e^(-bt)=0
add k2e^(-bt))V/m to both sides
k1e^(-at)=k2e^(-bt)
Divide both sides by k1e^(-bt) (presumably, k1 isn't zero)
(e^(-at))/(e^(-bt)=k2/k1
use the rules of exponents to combine the RHS into one exponential
e^(-at-bt)=k2/k1
factor out -t from the exponent
e^(-t(a-b))=k2/k1
take the reciprocal of both sides
e^(t(a-b))=k1/k2
take the log of both sides
t(a-b)=ln(k1/k2)
divide both side by t (presumably, t is not zero)
a-b=ln(k1/k2)/t
Plug T2 into t
a-b=ln(k1/k2)/T2

Note that if you have only one k, that is, k1=k2, then ln(k1/k2)=0, a-b=0, so a=b.

If you take a and b as given, then you have two unknowns (k1, and k2), and you need two data points. If, however, a, b, k1, and k2 are all unknown, then you'll need four data points. Although, if you use the fact that V0 is the maximum voltage, then your three additional data points may be sufficient.

 

[kmortis]

There is only one k.

Oh...I see now.

except, if k1=k2=k then you have (a-b)=ln(1)/T2=0/T2; a=b which just isn't true.

 

[Dilb]

There is only one k.

Oh...I see now.

except, if k1=k2=k then you have (a-b)=ln(1)/T2=0/T2; a=b which just isn't true.

Right, you have the wrong initial conditions. Using the common definition of rise and fall time, you want

t2 - t1 = rise time
V(t1) = 0.1 * Vo
V(t2) = 0.9 * Vo

and

t4 - t3 = fall time
V(t3) = 0.9 * Vo
V(t4) = 0.1 * Vo

which is much harder to do. I'm pretty sure an analytic solution isn't possible, because the exponentials interfere with each other and change when the maximum voltage occurs.

 

[kmortis]

Right, you have the wrong initial conditions. Using the common definition of rise and fall time, you want

t2 - t1 = rise time
V(t1) = 0.1 * Vo
V(t2) = 0.9 * Vo

and

t4 - t3 = fall time
V(t3) = 0.9 * Vo
V(t4) = 0.1 * Vo

which is much harder to do. I'm pretty sure an analytic solution isn't possible, because the exponentials interfere with each other and change when the maximum voltage occurs.

OK,
Now I'm at work, and here's the real stuff (not going off my memory). From MIL-STD-464A (which has a public distribution statement):

And from DO-160D ((c) RTCA 1997):

OK, so I was wrong, the fall time is to 50%. I do know that in reality, the voltage (or current) goes to zero, as capacitors are used to charge the Marx generator; but that's neither here not there as far as the math goes.

An important point to make here; notice how the two standards define the waveforms. The MS uses the a & b terms, but DO-160 uses t_r & t_f. This is how I got into this conundrum to begin with. Don't worry that one is a current waveform and the other is voltage. This general equation is used for both.

I'll retry the ICs with this.

 

[sphenisc]

...
(e^(-at))/(e^(-bt)=k2/k1
use the rules of exponents to combine the RHS into one exponential
e^(-at-bt)=k2/k1
factor out -t from the exponent
e^(-t(a-b))=k2/k1
...

[nitpick]
I think this should be e^(-at+bt)=k2/k1

But it's corrected in the next step anyway.
[/nitpick]

 

[Dilb]

OK, so I was wrong, the fall time is to 50%. I do know that in reality, the voltage (or current) goes to zero, as capacitors are used to charge the Marx generator; but that's neither here not there as far as the math goes.

Well sure, in reality. But as we all know reality has an annoying habit of not conforming to theory.

An important point to make here; notice how the two standards define the waveforms. The MS uses the a & b terms, but DO-160 uses t_r & t_f. This is how I got into this conundrum to begin with. Don't worry that one is a current waveform and the other is voltage. This general equation is used for both.

I'll retry the ICs with this.

That's a new definition for me. I guess the DO-160 was designed more for experimental measurements. Certainly would be easier to read off an oscilloscope than trying to get exponential coefficients.

 

[chipotle]

No, with two exponentials, it isn't linear.

So you are trying to say, what, that log(A-B) doesn't simplify to something nice? Sorry I misread the initial equation. But this still can be solved with linearization and least squares estimation. It just requires another step. First set b to some value, B. Now the equation is linearizable and can be solved with least squares for a, and there will be an r^2 associated with that. So now there is a function to get an r^2 for a given value of B. Plot it out and pick the r^2 maximizing B as your estimate of B, and the corresponding estimate of a as your estimate of a.