My Math Forum How to take this variable out of this equation?

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 April 19th, 2019, 04:33 AM #1 Newbie   Joined: Mar 2018 From: Yanbu Posts: 14 Thanks: 0 How to take this variable out of this equation? I'm working to develop a function in my embedded code for i2c initialization. I came up to this equation which is to calculate the speed of the i2c clock. But if I want to develop a function that takes a user popular numbers for speed then I have to change the shape of the equation to get that variable in place to receive users values for adjusting the speed. This is the equation: I want to take out TWBR instead of SCL frequency. I tried to do it on a paper but I don't know how to take it out, from the multiple PrescalerValue and the addition of 16??!
 April 19th, 2019, 05:09 AM #2 Newbie   Joined: Mar 2018 From: Yanbu Posts: 14 Thanks: 0 OK, I solved it finally
 April 19th, 2019, 06:01 AM #3 Math Team     Joined: Jul 2011 From: Texas Posts: 2,923 Thanks: 1518 $TWBR = \dfrac{(\text{CPU clock freq})- 16(\text{SCL freq})}{2(\text{SCL freq})(\text{Prescaler})}$
April 19th, 2019, 07:45 AM   #4
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Quote:
 Originally Posted by wolfrose I came up to this equation . . .
How?

April 19th, 2019, 09:14 AM   #5
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Quote:
 Originally Posted by wolfrose
Yikes! Sure hope you assigned 1 letter variables to those; like:
a = SCL freq
b = CPU clock freq
c = TWBR
d = Prescaler Value

a = b / (16 + 2cd)

Then you can easily solve for b or c or d:

a = b / (16 + 2cd)
b = a(2cd + 16)

a = b / (16 + 2cd)
a(16 + 2cd) = b
16a + 2acd = b
2acd = b - 16a ****
c = (b - 16a) / (2ad)

2acd = b - 16a ****
d = (b - 16a) / (2ac)

EDIT:

Last edited by Denis; April 19th, 2019 at 09:31 AM.

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