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 December 31st, 2016, 07:08 AM #1 Newbie   Joined: Mar 2016 From: UK Posts: 20 Thanks: 1 Rearranging Equation Where Unknown Appears Twice Hello, I'm struggling to rearrange an equation where I have an unknown appear twice. The example says: $\displaystyle \sin \left ( \phi _m \right )=\frac{\alpha-1}{\alpha+1}$ $\displaystyle \therefore \alpha =\frac{1+\sin \left ( \phi_m \right )}{1-\sin \left ( \phi_m \right )}$ When I rearrange I get: $\displaystyle \sin \left ( \phi _m \right )=\frac{\alpha-1}{\alpha+1}$ $\displaystyle \sin \left ( \phi _m \right )\left ( \alpha+1 \right )=\alpha-1$ Multiply out LHS: $\displaystyle \alpha\sin \left ( \phi _m\right )+\sin\left(\phi _m\right)=\alpha-1$ Move $\displaystyle \alpha$ over to LHS: $\displaystyle \alpha\sin \left ( \phi _m\right )+\sin\left(\phi _m\right)-\alpha=-1$ Move $\displaystyle \sin\left(\phi_m\right)$ over to RHS: $\displaystyle \alpha\sin \left ( \phi _m\right )-\alpha=-\sin\left(\phi _m\right)-1$ Now the $\displaystyle \alpha$ terms can be factorised: $\displaystyle \alpha\left(\sin\left(\phi_m\right)-1\right)=-\sin\left(\phi_m\right)-1$ And finally, divide LHS by $\displaystyle \sin\left(\phi_m\right)$: $\displaystyle \alpha=\frac{-\sin\left(\phi_m\right)-1}{\sin\left(\phi_m\right)-1}$ So I suppose I'm asking, how is: $\displaystyle \frac{1+\sin \left ( \phi_m \right )}{1-\sin \left ( \phi_m \right )}\equiv\frac{-\sin\left(\phi_m\right)-1}{\sin\left(\phi_m\right)-1}$ If I plug $\displaystyle \phi_m=45$, I get the same for each one. I look forward to hearing from you. Kind Regards Edward
 December 31st, 2016, 07:35 AM #2 Senior Member   Joined: May 2016 From: USA Posts: 466 Thanks: 198 $GIVEN:\ \alpha = \dfrac{-\ \sin( \theta_m) - 1}{\sin( \theta_m ) - 1} \implies$ $\alpha = \dfrac{(- 1) * \{ \sin( \theta_m) + 1\}}{\sin( \theta_m ) - 1} \implies$ $\alpha = \dfrac{(- 1) * \{ \sin( \theta_m) + 1\}}{(-\ 1) * \{-\ \sin( \theta_m ) + 1\}} \implies$ $\alpha = \dfrac{\cancel {(-\ 1)} * \{ \sin( \theta_m) + 1\}}{\cancel {(-\ 1)} * \{-\ \sin( \theta_m ) + 1\}} \implies$ $\alpha = \dfrac{\sin( \theta_m) + 1}{-\ \sin( \theta_m ) + 1} \implies$ $\alpha = \dfrac{1 + \sin( \theta_m)}{1 -\ \sin( \theta_m )}.$ Last edited by skipjack; December 31st, 2016 at 06:00 PM.
 December 31st, 2016, 08:09 AM #3 Newbie   Joined: Mar 2016 From: UK Posts: 20 Thanks: 1 @JeffM1 Thank you Jeff, I forgot that: $\displaystyle \frac{-1}{-1}=1$ And therefore, by multiplying the expression I had by it, you could arrive at the same equation that the example had. Kind Regards Edward
 December 31st, 2016, 04:46 PM #4 Senior Member   Joined: Feb 2016 From: Australia Posts: 764 Thanks: 285 Math Focus: Yet to find out. If this is from Ogata (5E), the examples on page 507 and 558 might help.
January 2nd, 2017, 02:54 PM   #5
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Quote:
 Originally Posted by Joppy If this is from Ogata (5E), the examples on page 507 and 558 might help.
Thanks, it's in some Control Systems notes I have on a Plase Lead Compensation procedure... although I may look for the reference as my algebra has never been a strong point!

EDIT: I've been able to find a copy of the eBook and the solutions on Academia.edu.

Kind Regards

Edward

Last edited by edwardholmes91; January 2nd, 2017 at 03:10 PM.

January 2nd, 2017, 05:33 PM   #6
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Math Focus: Yet to find out.
Quote:
 Originally Posted by edwardholmes91 Thanks, it's in some Control Systems notes I have on a Plase Lead Compensation procedure... although I may look for the reference as my algebra has never been a strong point! EDIT: I've been able to find a copy of the eBook and the solutions on Academia.edu. Kind Regards Edward
In that case, i would recommend having a look at Dorf. Ogata tends to brush over some things which seem important when first learning, both mathematical and theoretical. Especially in the later chapters.

Dorf also uses some different, albeit simpler methods for designing lead/lag compensators. At the time it really helped me to switch between these two books for understanding the design process better.

 January 2nd, 2017, 11:49 PM #7 Newbie   Joined: Mar 2016 From: UK Posts: 20 Thanks: 1 @Joppy, thank you for the other reference, I really appreciate it

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