# What kind of Mathematics do I need?

Discussion in 'Homework Help' started by Pulsed, Sep 30, 2012.

1. ### Pulsed Thread Starter Member

Sep 10, 2012
41
3
I have a seemingly very difficult (for me) problem involving the small signal method using differential calculus... the question is as follows:

"Consider the following circuit containing the nonlinear element N:
The i-v relation for N is given by iA=10⋅(1−e−vA/5), where iA is in Amperes and vA is in Volts.
Solve for the voltage vA when vI=5.0V. Note that this requires that you solve the equation obtained using KVL iteratively for vA.
(Hint: Use the exponential term to solve for vA as a function of the assumed value of vA, and then iterate. Taking logs on both sides may facilitate convergence.)
The voltage vA is:
???

Find the incremental change in vA for a 2% increase in vI and use it to calculate the ratio ΔvAΔvI. ???

Find the value for the incremental resistance of the nonlinear element N by linearizing the expression for iA about the operating point when vI=5.0V. ???

Where do I start? How do I learn the mathematical machinery to understand this question?
What kind of mathematics do I need to study?

I have started watching the following video sequence by the Khan academy on YouTube http://www.youtube.com/course?list=EC19E79A0638C8D449&feature=plcp So my question is: Looking at this from a completely foreign perspective, imagine I am looking at something completely alien... that I only understand basic algebra, how do I even begin to understand what the question is asking??

Last edited: Sep 30, 2012
2. ### WBahn Moderator

Mar 31, 2012
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I see two basic things (beyond algebra) that are needed to understand this problem. First, what exponentials and logarithms are and how they relate to each other. Second, the basic principles of differential calculus and, for this particular problem, how to take the derivative of expressions involving exponentials. However, if you understand the basic principles, you can actually answer the question without knowing how to take that particular derivative.

Now, I'm assuming that your algebra skills include things such as the difference between linear and nonlinear equations and how to find solutions by iteration when dealing with nonlinear equatioons.

3. ### Pulsed Thread Starter Member

Sep 10, 2012
41
3

I did know about 'iteration' before when I was learning about solving fractions. I am clueless though in how to apply it.

When looking at this question, I don't know where to start!

So my question simply is: What is the first step?

Where do I look.

Are those videos on calculus by the Khan academy good for learning what I need to learn?

Also, what is your opinion of Calculus made easy by Silvanus P. Thompson?
Many thanks for your kind response.

4. ### WBahn Moderator

Mar 31, 2012
18,085
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The first step is to set up the equations that need to be solved using KVL and/or KCL. You have two unknowns, so you need two equations. KCL isn't going to do you any good, so use KVL to get one of them. Then the relationship betwee i_A and v_A is the other.

Now combine the two equations to eliminate either i_A or v_A, whichever seems easier to do. You now have a single equation in the one that is left, say v_A for discussion's sake. Now solve it partially for v_A, meaning to get v_A alone on one side, even though you will still have v_A embedded in the other side, too.

Now pick a trial value for v_A and plug it into the right hand side. If it is the correct value, it will yield that same value of v_A as a result. If not, then it will obviously either be greater than or less than the desired value. Examine the right hand side as see whether the result will increase or decrease if you increase v_A by a little bit (this is where knowing how to take derivatives is real handy, but you can do it without this). With this in mind, pick a new value of v_A and see what the result it. After a could iterations, you should get a feel for whether you are changing v_A by too big or too small a step.

Often, a good starting point would be v_A = 0V. If that is negative (i.e., too small) then your immediate goal is to find a value of v_A that results in a value that is too large. Assuming that the nonlinear function is at least continuous, then you know that a solution exists between 0V and this last value of v_A. With that in hand, you can get to a solution very quickly by just using the value midway between them as your next guess. At each stage, decide if the correct value is in the lower or upper half of your range and then pick the midpoint of the new narrower range as the next trial value. You can set this up in a spread sheet very handily

Get that much done and then we can deal with the calc-related stuff.

mkbutan likes this.
5. ### Pulsed Thread Starter Member

Sep 10, 2012
41
3
Thank you so much WBahn! This makes things so much clearer.

Attached is my solving the first question to the problem.

I am solving for VA.

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6. ### WBahn Moderator

Mar 31, 2012
18,085
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You've done a good job. Just two things you need to do.

Second, you should carry out the iterations until the two sides match to the desired number of significant figures. If you want three sig figs, then you need to keep going just another step or three. Remember, if the leading digit is a 1, then that 1 is usually not counted as a sig fig or, said another way, you want one more sig fig than normal if the value starts with a 1.

mkbutan and Pulsed like this.
7. ### Pulsed Thread Starter Member

Sep 10, 2012
41
3
Thank you very much for your time and patience WBahn!

Now, dealing with the calculus...

I am willing to forget that last part of the question that involves calculus in favor of a more seemingly simple example...

The question is as follows...

"Although vacuum tubes are no longer commonly used in computer or consumer electronics, they still have a substantial niche in high-end audio, high-power radio transmitters, particle accelerators, and microwave ovens.
A vacuum diode, the simplest vacuum tube, is an interesting two-terminal device.
A vacuum diode's voltage-current characteristic is closely approximated by the Childs-Langmuir Law, with one parameter P called the perveance:"

"i
P = {Pv^(3/2) if vPK > 0 and 0 if vPK < 0

In this problem, we will use P ≈ 2.0mA/V^(3/2) (see, for example, the 6AL5 twin diode).

(1) - What is the current IP (in Amperes) for a bias voltage of VPK=6.0V? Answer: 0.0293A

(2) - What is the incremental resistance (in Ohms) for the bias voltage of VPK=6.0V? ????

(3) - What is the current IP (in Amperes) for a bias voltage of VPK=9.0V? Answer: 0.054A

(4) - What is the incremental resistance (in Ohms) for the bias voltage of VPK=9.0V? ?????"

So from my meager understanding of differential calculus which I have gleaned watching these videos http://www.youtube.com/watch?v=ANyVpMS3HL4&feature=BFa&list=EC19E79A0638C8D449 , we need to find the 'incremental resistance' of the vacuum diode when there is a potential of 6v across is.

This means we need to find the 'limit' of the point on the curve where the x-axis = 6v??

Or we need to find the derivative of the equation (iP = {Pv^(3/2)) at the point marked incremental resistance?? So a derivative is just the slope of a curve at a point, and the point I need is that at which Vpk = 6v and Ip = 0.0293A.

So the incremental resistance is the slope at the point Vpk=6v and Ip=0.0293A.

I have attached a graph of what I think this should look like...

Is my thinking correct thus far?

• ###### Calc Graph.JPG
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Last edited: Oct 5, 2012
8. ### Pulsed Thread Starter Member

Sep 10, 2012
41
3
Attached is the equation graphed on GraphCalc, so it is to scale.

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9. ### WBahn Moderator

Mar 31, 2012
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Your latter explanation is correct. The derivative is the slope of the curve at each point.

Forgetting calculus for a minute, you can estimate the slope of the curve at a point by calculating the slope between the point of interest on the curve and a point that is nearby.

So say we have the following function:

y(x) = 4 + 10*x^3

If we want the slope at x=A, let's find the slope between the point x1=A and x2=A+Δx where Δx is just some small incremental amount.

y1 = y(A) = 4 + 10*A^3
y2 = y(A+Δx) = 4 + 10*(A+Δx)^3

Let's define Δy to be the difference between y2 and y1, so

Δy = y2 - y1

resulting in

y1 = y(A) = 4 + 10*A^3
y1 + Δy = y(A+Δx) = 4 + 10*(A+Δx)^3

So what is the slope between these two points?

m = (y2-y1)/(x2-x1) = Δy/Δx

Since this is a function of x, we will call this function y'(x)

y'(x) ≈ Δy/Δx = [y(x+Δx) - y(x)]/Δx

All y'(x) is is a function that, when evaluated at a given value of x, yields the slope of y(x) at that particular value of x. It's only approximated by Δy/Δx because that gives us the slope between to points close to x, but by choosing Δx to be really small, we can get our approximate to be as close to the true value as we like.

So, for our function, we have

Δy/Δx = [(4 + 10*(x+Δx)^3)-(4 + 10*x^3)]/Δx

Δy/Δx = [10*(x+Δx)^3 - 10*x^3 ]/Δx

You might recall from somewhere something about the binomial expansion or Pascal's triangle (or you can just perform the multiplication).

(x+Δx)^3 = (x+Δx)(x+Δx)^2
(x+Δx)^3 = (x+Δx)(x^2+2xΔx+Δx^2)

Since we want Δx to be really small, we know that Δx^2 is going to be really, really small. So we will neglect any terms that have Δx raised to any power greater than 1.

(x+Δx)^3 ≈ (x+Δx)(x^2+2xΔx)
(x+Δx)^3 ≈ x^3 + 2x^2Δx + x^2Δx + 2xΔx^2
(x+Δx)^3 ≈ x^3 + 3x^2Δx

where we have once again neglected higher order powers of Δx.

Plugging this back into the formula for our slope, we get

y'(x) ≈ [10*(x+Δx)^3 - 10*x^3 ]/Δx

y'(x) ≈ [10*(x^3 + 3x^2Δx) - 10*x^3 ]/Δx

y'(x) ≈ [10*x^3 + 10*3x^2Δx - 10*x^3 ]/Δx

y'(x) ≈ [30x^2Δx]/Δx

y'(x) ≈ 30x^2

So this is an equation that gives us an estimate of the slope of y(x) = 10x^3 at any given x. Notice that we haven't done anything other than plain old algebra.

Now, all a derivative is is the function y'(x) in the limit that Δx approaches zero. Well, in this case, our Δx terms cancelled out, but if they hadn't, we would have just taken the limit as Δx went to zero. The result is no longer an approximation of the slope of y(x) at x, it IS the slope of y(x) at x. So, in this case, we have

y'(x) = 30x^2

We call the function y'(x) the derivative of y(x). We also write this as

y'(x) = dy/dx

Notice the similarity to Δy/Δx.

The Δ means a small change, while the d indicated an infinitesimal change. You can think of the former as being a small, but finite, change while the latter is a change so small that it can't be measured.

If you've followed this up to this point, then you have a decent grasp of the fundamental concept of differential calculus.

10. ### Pulsed Thread Starter Member

Sep 10, 2012
41
3
Thank you WBahn! That is a sizable response, and equally informing!

The explanation echoes that of the khan academy.

Attached is my attempt at plugging in the variables in my question.

I have come to a wall though, as you can see, how do you expand the expression (6+x)^(3/2)?

File size:
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Sep 10, 2012
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12. ### WBahn Moderator

Mar 31, 2012
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Let's consider the 'does the answer make sense?' test.

6^(3/2) is sqrt(6^3) which is 6*sqrt(6) which is going to be between 6*sqrt(4) and 6*sqrt(9), which are 12 and 18 respectively, so somewhere around 15.

With a calculator, you can also do a quick sanity check.

6^(3/2) = 14.6969 (right near the middle of the range we expected)

Now what about 6.01^(3/2)? That's 14.7337. So the slope a x=6 is about

Δy/Δx= 0.0368/0.01 = 3.68

Hence:

(6+Δx)^(3/2) ≈ 14.697 + 3.68Δx

Notice how no calculus has been used up to this point.

In general, and stated without proof, for m != 0,

For f(x) = x^m,

f'(x) = dy/dx = m*x^(m-1)

So, in this case,

dy/dx = (3/2)x^(1/2)

which is 3.67 at x=6.

So, at x=6, we have

(6+Δx)^(3/2) ≈ 6^(3/2) + (dy/dx)Δx

(6+Δx)^(3/2) ≈ 14.697 + 3.67Δx

You can see how close the two are.

13. ### gvnc New Member

Oct 6, 2012
1
0
Hi All,

You know that whoever Pulsed is, they posted this to get answers to a graded homework assignment on Week 4 of the MITx class 6.002x, Circuits And Electronics before the deadline of October 7th.

Just FYI.

14. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
It must be frustrating for those students who put in the hard work and find their fellow students getting disproportionate rewards by exploiting sites such as AAC. I doubt there's much can be done other than what your good self is doing by highlighting the issue.
No doubt the educators also watch this forum on occasion to see if there are students from their courses looking for some advantage.

In my city there is currently a debate about university students buying essays off the net and presenting this as their own work. A student was recently ejected from their course for this.

15. ### WBahn Moderator

Mar 31, 2012
18,085
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Well, it IS the Homework forum. The assumption should always be that students are trying to get help on current homework assignments, which is why most of us insist on the student presenting their work to date and why we try to give hints, point out errors, and ask leading questions as opposed to just offering direct answers off the bat.

Most instructors of most courses well tend to agree that this is reasonable for most homework assignments.

But, of course, we get lots of students that post here just wanting a complete answer with no effort on their part. The more creative (in their own minds, anyway) ones tend to show up in the non-Homework forums. For those that do do this, though, it is a good reminder to them when someone recognized the specific assignment and points it out since it underscores that this is a public forum and that lots of people monitor it and that blatant offenders can and do come to the attention of the responsible instructors.

For the record, I don't think Pulsed has done this and has participated pretty well -- I think this thread has stayed within reasonable bounds.

16. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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In any event the entire class probably by now has watched this thread and there won't be any embarrassing failures. One needs to pragmatic about this otherwise we may as well stop helping anyone at all - even those who do make an honest attempt.

Last edited: Oct 6, 2012
17. ### WBahn Moderator

Mar 31, 2012
18,085
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Yep, and instructors are being naive if they think that students are not going to use such resources. Lots of instructors like to think that if they don't let students keep exams, that they can then re-use the exams over and over. This seldom works, particularly on any campus that has frats and sororities, who frequently go to great lengths to build up files of assignments and exams. One tactic is to split up an upcoming exam among the people taking it and have each person memorize specific question numbers and, as soon as they leave the exam, write the question down. If possible, they have multiple people assigned to each question so that they can have a better chance of getting it remembered right. They do the same thing when (and if) the exam is reviewed after it has been graded.

This is why I not only assume that any prior assignment or exam is "out in the wild", but I even the field by posting all assignments/quizzes/exams on my website along with the answers (unless I am teaching a course for which a policy is in place that doesn't allow this). Sure, I have to make up new material each time I teach it, but that is part of the price of being an instructor.

18. ### Pulsed Thread Starter Member

Sep 10, 2012
41
3
It is the policy of the homework forum, NOT to give the person asking the question explicit answers, but rather to help them come to an understanding of the concept.

That is what education like MITx should be about, should it not? Coming to an understanding.

I posted the original question because I did not understand. And I still don't fully understand differential calculus, but is there a problem with using the internet, or an all about circuits forum, to ask for help in understanding that? How is that "exploiting" a site, especially in regard to the fact that that site has a homework help section??

As to "buying essays off the net", how does that even relate to asking for help?

If this thread is to be deleted, so be it.

If it is going to go off topic, could a moderator please split the thread.

Many thanks

19. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
I've obviously become too cynical.

With that trait unacceptable to all things wholesome I therefore bid AAC a fond & permanent farewell. Life is short and needs living well.

It's a great forum - Thanks for the fun along the way .