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Martian

Understanding Capacitance/Inductance Formulas

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A problem I come to in electronics is I can't design certain circutes (I make simple ones) because I can't use capacitors or inductors while accurately predicting their effects. (I'm self-taught in these things)

 

The reason is this:

 

Calculus.

 

I am familiar with most concepts below calculus, but the "function" and "derivitive" things are still somewhat vague to me. Formulas containing these things are used in capacitors, transformers, etc.

 

It seems to me, if I have looked at it right, that a calculus problem is like a little machine that keeps repeating and running through a changing variable causing a gradual curve, graphing the effect being studied.

 

I was rather hoping someone wouldn't mind just helping me understand the anatomy of a typical calculus problem and the relevant symbols they use. I feel if I can get a small boost, the rest should fall together.

 

You can reply here, but a personal message would be appreciated too. I'll be printing out the helpful replies and tacking it up by my work station.

 

A detailed description of what the "F" "E-shaped" "u-shaped" and "d/dx" symbols are generally used for will mean much to me. I feel with a little help from human beings (as opposed to straight book study) I may have a decent shot at getting as good in math as I would like to be.

 

Thanks!!!:D

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Differentiation gives the gradient of a line, be it straight or curved.

 

Integration is the opposite.

 

Differentiation in one easy step.

 

if [math]f(x)=kx^n[/math]

[math]f'(x)=knx^{n-1}[/math]

 

The little dash by the function just means that differentiation has happend.

Another way of writing it is with the [math]y[/math] that you'd get in a conventional graph type thing.

 

[math]y=f(x)[/math]

[math]\frac{dy}{dx}=f'(x)[/math]

 

The [math]d[/math] means difference in, I think. There's no need to understand it at an early stage.

 

An easy examle

 

When [math]x=4[/math], what is the gradient of the line [math]y=x^3+2x^2+2x[/math]?

 

It's useful to know that differentiation is distributive wich means that when you have addition, you can do it bit by bit.

 

[math]\frac{dy}{dx}=3x^2+4x+2[/math]

 

Fill in your value of [math]x[/math]:

 

[math]=3(4^2)+4(4)+2[/math]

[math]=4(4^2)+2[/math]

[math]=66[/math]

 

And that's your awnser.

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Is this AC electronics or DC electronics? And could you give some examples of the formulae you're strugalling with, it's normally easier to work from direct examples.

 

[math] \frac{d}{dx} Some Function [/math] As stated above is the gradient, which can also be written as the rate of change of the function with respect to time. so if it was:

 

[math] \frac{d}{dt} Some Function [/math] It would be how the function changes at time (t) changes.

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A function is something whereby you can put in some variable and get a different, dependant variable out. So, if f(x)=2x-3, you can put in some value, say 6, and get f(6)=2(6)-3=9.

 

Differentiation of a function is the generation of another function for which the "y-value" (value of the dependant variable at a given "x-value," or independant variable) of the second is equal to the gradient, or slope, of the first.

 

For example, take the function y=f(x)=x^2. For any given x, there is a y that is equal to x^2. The derivative of this function happens to be f1(x)=2x, meaning that for a given point on the original curve, its slope can be represented by 2x. So, at x=4, f(x)=4^2=16, and its slope at that point, f1(x)=2(4)=8, or 8 units up for every 1 unit over.

 

The dy/dx means instantaneous change in y divided by instantaneous change in x. An explanation: Slope is measured by change in y divided by change in x. So between two points on a curve, the y-value of the second minus the y -value of the first, all divided by the x-value of the second divided by the x-value of the first, will give you the slope of the straight line between those two points, also called the secant. But we want the slope at a point, which poses some problems. How can there be any change at one point? Well, there can't, really, but what we can do is find the change between two points which are closer to one another than any finite distance. We can determine through algebra that as you make the distance between them smaller and smaller, the change in y over change in x gets closer and closer to some definite ratio, which is the "limit" as the distance between them "approaches zero." Thus, the "dy/dx" is that ratio at an infinitely small distance, thereby effectively being the slope at one point.

 

I'll say more later, but right now I have actual work to do...

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I occasionally see a symbol shaped like an uppercase "E" in front of different problems. What is that?

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If it's an upper case sigma ([math]\sum[/math]) then that means the sum of a sequence.

 

[math]\sum_{n=1}^{3}n+b=(1+b)+(2+b)+(3+b)=6+3b[/math]

 

Not really to do with calculus but meh.

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It's got everything to do with integrals. An integral is the sum of the rectangles under the curve, change in x (width) times height, the change in width approaches zero and the number of rectangles approaches infinity. Sums are where integrals come from. It's basically "the sum of all y-values."

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Thanks guys. Its funny how talking to people on specifics can be so incredibly more resourceful than textbooks. In just a few paragraphs, I've been able to overcome a huge hurdle. :D

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For AC electronics, designing circuits is easily done, using complex numbers.

 

Imagine a voltage source with a angular frequency ω and amplitude A, so as function of time you have V(t) = A*cos(ωt).

 

Now, replace this with a voltage X(t) = A*exp(ωt). Now, the real voltage can be written as the real part of X(t), being Re(X(t)) = A*cos(ωt).

 

Using this formalism, you can treat every passive linear component as a complex resistor Z. For lumped devices there are basically three types:

 

Capacitor with capacity C: Z = 1/jωC

Resistor with resistance R: Z = R

Inductor with inductance L: Z = jωL

 

Here the number j has the property j² = -1.

 

Now I'll give an example with three nodes, GND, VIN, VOUT. Between GND and VIN there is a voltage source X(t). Between VIN and VOUT there is a resistor R. Between VOUT and GND is a capacitor C. What is the output voltage as function of input voltage?

 

This now can be easily solved. We introduce a complex voltage XOUT and XIN.

 

We have a series connection of two resistors. Using basic circuitry for resistors you find

 

XOUT = XIN * (ZC / (ZC + ZR)), where ZC is the capacitor's complex resistance and ZR is the resistor's complex resistance.

 

Now XOUT = XIN *(1/jωC) / (R + (1/jωC)) = XIN / (1 + jωRC)

 

So, you have XOUT as function of XIN and the angular frequency ω.

 

The amplification as function of frequency ω can be written as 1/sqrt(1+ω²R²C²). There also is a phase shift, between input and output. That is -arg(1 + jωRC). For small ω (close to DC), the phase shift is close to 0, for high ω, the phase shift is almost 90 degrees.

 

If you understand complex numbers, then this should be easy to grasp, otherwise it indeed will be very difficult for you to determine transfer functions of capactive and inductive circuits.

 

The key to understanding these things is

 

"transfer function"

"complex arithmetic"

"bode plot"

"poles and zeros"

"laplace transform"

 

http://en.wikipedia.org/wiki/Transfer_function

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