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November 11th, 2015, 10:54 AM   #1
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Using brackets & squared constant

Kd = constant (stands for differential I believe).

S(squared) = is S x10 to the power 2

Kd.S(squared)

It can be written as

(Kd + 1) S(squared)

I want to know how to go between these two ways of writing the same result.

I thought of writing it as Kd . S . S (1)

(Kd + 1) . S (2)

Going from (1) to (2) doesn't make sense to me; can you explain?

If oyu want the background to this question, it's here:

Last edited by skipjack; November 12th, 2015 at 04:30 AM.
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November 11th, 2015, 12:20 PM   #2
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Math Focus: Wibbly wobbly timey-wimey stuff.
The problem in the denominator is $\displaystyle s(s - 2) + K_ds^2 + K_ps + K_i$. Collecting the s^2 terms: $\displaystyle s^2 + K_ds^2 = s^2(1 + K_d)$.

I want to make it clear: $\displaystyle K_d s^2 \neq (K_d + 1)s^2$ unless s = 0.
-Dan
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November 12th, 2015, 04:09 AM   #3
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I have just seen her tap the board and recognize now that she is combining 2 terms and that (1) is an incomplete unit of the denominator. Could we agree on this?
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November 12th, 2015, 09:11 AM   #4
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Quote:
Originally Posted by sgtsixpack View Post
I have just seen her tap the board and recognize now that she is combining 2 terms and that (1) is an incomplete unit of the denominator. Could we agree on this?
Seeing as that's what I'm trying to tell you I'm happy to agree.

-Dan
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November 12th, 2015, 08:05 PM   #5
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Something is troubling me. @ 4:34 she is looking at the "S's".

i.e. -2s + Kp.s

which she evaluates to (Kp -2s)

whereas I would evaluate it to (Kp -2)s

The objective is to simplify the terms and use brackets to signify that both terms are multiplied by s. Am I correct?
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