Simple alegbra has me stumped!

J

jsbeckton

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Oct 24, 2005
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174
Well I am in my last semester of mechanical engineering and am working on a radar tracking system for an automatic controls class. I have developed equations in the s-plane for my electro-mechanical system and not I need to form transfer functions to develop relationship between the input and output of a voltage signal. Long story short, I am having troubkle converting this equation to the form of I/u=....


\(\displaystyle i = \frac{1}{{sL + R}}(u + kw)\)

I need this in the form:


\(\displaystyle \frac{i}{u} = ??????\)

I have tried to get it but can'y eem to simplify to that form for some reason. Anybody see how to solve this in that form? It woudl be greatly appreciated.

Thanks in advance.
 
jsbeckton said:
Well I am in my last semester of mechanical engineering and am working on a radar tracking system for an automatic controls class. I have developed equations in the s-plane for my electro-mechanical system and not I need to form transfer functions to develop relationship between the input and output of a voltage signal. Long story short, I am having troubkle converting this equation to the form of I/u=....


\(\displaystyle i = \frac{1}{{sL + R}}(u + kw)\)

I need this in the form:


\(\displaystyle \frac{i}{u} = ??????\)

I have tried to get it but can'y eem to simplify to that form for some reason. Anybody see how to solve this in that form? It woudl be greatly appreciated.

Thanks in advance.
Click to expand...
If you simply divide both sides by 'u' - you would get 'i/u' on the left-hand-side
 
Yes, but then I would have a 2 "u's" on the right side. This is a transfer function where I want the "i" and "u" only on the left side, and only in the form i/u. Can you see an algebraic manipulation that will give me this?

Thanks
 
Your "complicated looking" equation can be looked at as a = (b + c) / (d + e) : thus less headaches in manipulating!

I'm no expert, but it's clear to me that a's value DEPENDS partly on b's value;
like, you can get a value for a ONLY if b,c,d,e's values are given:
make 'em 12, 4, 1, 3 : then a = 4; only then can you get a/b = 4/12
 
Denis said:
Your "complicated looking" equation can be looked at as a = (b + c) / (d + e) : thus less headaches in manipulating!

I'm no expert, but it's clear to me that a's value DEPENDS partly on b's value;
like, you can get a value for a ONLY if b,c,d,e's values are given:
make 'em 12, 4, 1, 3 : then a = 4; only then can you get a/b = 4/12
Click to expand...

Ok, say

b=1910
k=6.67
R=50
L=66.7

"s" stands for derivitave w/respect to time so it doesn't have a value. We still have a=(b+c)/(d+e) equation that nees to be manipulated to the form a/b. I don't see how the actual value of b helps if you don't have u or s. Do you see any way to manipulate the equation?

Thanks
 
I can't figure out what you're asking...
You're first saying s has no value, then that it does with your statement "...if you don't have u or s".

What I did was:
let a = i
let b = u
let c = kw
let d = sL
let e = R .... making the equation a = (b + c) / (d + e)

Let's start fresh and pretend someone posted this (forget YOUR equation entirely):
Take equation a = (b + c) / (d + e)
Is it possible to move b to the left such that a/b = some expression in terms of c,d,e ?

You're a "tutor" and give the student asking this the solution...
We're awaiting... :wink:
 
That IS the question that I am asking.

If I knew the answer, I wouldn't have posted the question!

:D

Do you see how it can be done?
 
jsbeckton said:
Do you see how it can be done?
Click to expand...
I "see" it can't be done! Perhaps someone else will: my $5 against your $1 that it's impossible...
 
What's wrong with Dennis' first solution? i/u = (1/(sL+R))*(1+kw/u) If you know all the values, where's the problem?

You could do this...

u = i(sL + R) - kw

i/u = i/(i(sL + R) - kw)

then you could plot i/u = f(i) and the other parameters.

Or, I like this version

\(\displaystyle \frac{i}{u}\;=\;\frac{-1}{sL+R} \sum_{n=1}^{\infty}\left(\frac{i \cdot (sL+R)}{kw}\right)^{n}\)

From there, of course, you are free to pick any approximation you like.
 
I understood it that a transfer function was a relationship of the output over the input "on the left", and the "gain" on the right. All of the transfer functions that we have done involved developing this relationship w/o the input (u) or the output (i) being on the right hand side in any form.

If this is not possible, how then would I develop a transfer function of the original equation relating input (u) and output (i)?

Thanks for everyones help!
 
The very task you describe is why i printed that last one, that may have been perceived as over-kill. It was not.

The question is, just how fast does the summation converge?

A first approximation to your transfer function is tf0 = -1/(sL+R), just figuring the summation is close to unity. This may be a reasonbale approximation, depending on what you are doing.

A second approximation might be taking just one term, producing tf1 = -i/kw. I realize it still has that pesky 'i' on the RHS, but perhaps this can tell you something.

This causes me to be VERY interested in how close these first two approximations are. tf0 = tf1?

A third approximation might be, taking two terms, tf2 = -1/(sL+R)[i(sL+R)/kw + i^2(sL+R)^2/(kw^2)] = tf1 - (sL+R)*(i/(kw))^2

If you can put bounds on the value of 'i', you just might be in business.
 
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