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 November 11th, 2015, 10:54 AM #1 Newbie   Joined: Sep 2015 From: England Posts: 5 Thanks: 1 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.
 November 11th, 2015, 12:20 PM #2 Math Team     Joined: May 2013 From: The Astral plane Posts: 2,156 Thanks: 877 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
 November 12th, 2015, 04:09 AM #3 Newbie   Joined: Sep 2015 From: England Posts: 5 Thanks: 1 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? Thanks from topsquark
November 12th, 2015, 09:11 AM   #4
Math Team

Joined: May 2013
From: The Astral plane

Posts: 2,156
Thanks: 877

Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
 Originally Posted by sgtsixpack 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

 November 12th, 2015, 08:05 PM #5 Newbie   Joined: Sep 2015 From: England Posts: 5 Thanks: 1 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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