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DERIVE-NEWS  March 2000

DERIVE-NEWS March 2000

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Subject:

HELP!!!!!!!!!!!!!!!!!!!!!!

From:

Jesús Ramírez <[log in to unmask]>

Reply-To:

Jesús Ramírez <[log in to unmask]>

Date:

Sun, 19 Mar 2000 14:34:40 PST

Content-Type:

multipart/mixed

Parts/Attachments:

Parts/Attachments

text/plain (77 lines) , files.zip (77 lines)

Hi Derivers!!!!

I have a little problem but I can't find where the mistake is. I hope you 
can help me on this...

I'm working with a heating system closed-loop transfer function and I'm 
trying to find its Inverse Laplace Transform (I know Derive doesn't do that, 
but wait a little bit). I'm writing down every step between the frecuency 
domain and the time domain because this is part a of a Control Course 
homework that I have to do. The way I'm working is the following one:

- I make this substitution inside the transfer function --> Te(s) = 10/s so 
I get this:

	numerator = 100 + 10/s

	denominator = 60 s + (1.71) (100 + 10/s) + 1

- I simplify the expression via Derive and I get this:

	numerator = 100 (10 s +1)

	denominator = 600 s^2 + 1720 s +171

- In order to find the Inverse Laplace Transform, first I have to find the 
roots on the denominator so I can name them by a symbol (let's say "a" & 
"b") and expand the transfer function by partial fractions ("expand" command 
on Derive) and finally look up at a Laplace Transforms Table (remember that 
this is a homework and, although I can use the computer, I must show the 
intermediate results)...

- Finally I use the Student Edition of MATLAB 4.0 (which has a symbolic 
toolbox with access to the Maple Kernel) in order to check what I got and I 
find the Inverse Laplace Transform with "invlaplace" (using the first 
expression, where I just made the substitution Te(s) = 10/s).

Well, the problem is that I came up with two completely different answers:

- Using the first expression (without simplification and all that stuff) I 
get the right answer (I know that's the right answer because I'm designing 
an automatic control closed-loop and that's the answer that less disturbs 
the right operation of the heating system), which is the following one:

		T(t) = 1.6686 e^(-2.7635 t) - 0.0019 e^(-0.1031 t)

- Using the expression with the factorized roots, that is

	numerator = 100 + 10/s

	denominator = (s - a)(s - b)

I get what I find to be a wrong answer, that is:

		T(t) = 1001.17 e^(-2.7635) - 1.17 e^(-0.1031)

I found again the roots using MATLAB and there's no error in Derive 
regarding this aspect....

I'm enclosing a zip file containing the text file of the MATLAB outputs as 
well as a jpg image of the same computations made with Mathematica 4.0. 
Don't be afraid to download it, it doesn't contain virus...

I hope you can lend me a hand with this problem. Maybe I'm making a mistake 
over and over and the one that is in a "closed loop" is me!!

Thanks in advance!!!!!!!!!!!!

Greetings from Guadalajara, Jalisco, México!!!!!!!!!!!!!!!

Atte:
        Jesús Ramírez

PD: By the way, I'm using Derive 4.13 for DOS
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