Transfer function algebraic manipulation

AI Thread Summary
The discussion centers on manipulating the transfer function g_p = -1.43 / [(s - 1.399)(s + 5.086)]. The original poster struggles with the algebraic manipulation and expresses frustration over not being able to simplify it correctly. A response clarifies that the manipulations attempted are correct but notes that one of the constants is negative, which may indicate an issue with the original function. Additionally, it emphasizes that the final form requires combining constants differently, specifically that k_p does not equal -1.43. The conversation highlights the importance of understanding the implications of negative constants in transfer functions.
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Homework Statement


upload_2015-11-7_14-2-6.png


Homework Equations

The Attempt at a Solution


I don't know if I am just stupid, but I can't see how to make this manipulation. I tried factoring the denominator

$$ g_{p} = \frac {-1.43}{(s-1.399)(s+5.086)} $$
Then take out -1.399 and 5.086
$$ g_{p} = \frac {-1.43}{-1.399(\frac {-1}{1.399}s + 1)5.086(\frac {1}{5.086}s+1)} $$
But I know this won't work, and I am not clever enough right now to come up with the right way to manipulate this thing.
 
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Maylis said:

Homework Statement


View attachment 91494

Homework Equations

The Attempt at a Solution


I don't know if I am just stupid, but I can't see how to make this manipulation. I tried factoring the denominator

$$ g_{p} = \frac {-1.43}{(s-1.399)(s+5.086)} $$
Then take out -1.399 and 5.086
$$ g_{p} = \frac {-1.43}{-1.399(\frac {-1}{1.399}s + 1)5.086(\frac {1}{5.086}s+1)} $$
But I know this won't work, and I am not clever enough right now to come up with the right way to manipulate this thing.

Your manipulations are correct; it is just that one of your constants ##\tau_i## is negative. If that is not allowed, there must be something wrong with the original ##g_p(s)##.
 
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The negative time constant is allowed, which is what makes it unstable. My algebra is weak, so for some reason I had it in my head that if I multiplied my result out, I would come up with something different from the original transfer function!
 
Maylis said:

Homework Statement


View attachment 91494

Homework Equations

The Attempt at a Solution


I don't know if I am just stupid, but I can't see how to make this manipulation. I tried factoring the denominator

$$ g_{p} = \frac {-1.43}{(s-1.399)(s+5.086)} $$
Then take out -1.399 and 5.086
$$ g_{p} = \frac {-1.43}{-1.399(\frac {-1}{1.399}s + 1)5.086(\frac {1}{5.086}s+1)} $$
But I know this won't work, and I am not clever enough right now to come up with the right way to manipulate this thing.
You've almost got the form of gp(s) that is specified, you just need to combine all the constants in the numerator and the denominator.

Hint: kp ≠ -1.43 in the final form.
 
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Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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