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 February 28th, 2012, 05:43 PM #1 Senior Member   Joined: Oct 2009 Posts: 895 Thanks: 1 How can I do the calculation here Hi Resolution = Vfs/2^n Resolution = 3 Vfs = 2 V find n how I can find the n ... ?
 February 28th, 2012, 06:04 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: How can I do the calculation here If I am interpreting this correctly, you have: $2^{1-n}=3$ $1-n=\frac{\log(3)}{\log(2)}$ $n=1-\frac{\log(3)}{\log(2)}\approx-0.5849625007211561$
 February 28th, 2012, 06:22 PM #3 Senior Member   Joined: Oct 2009 Posts: 895 Thanks: 1 Re: How can I do the calculation here thanks
 February 28th, 2012, 07:16 PM #4 Senior Member   Joined: Oct 2009 Posts: 895 Thanks: 1 Re: How can I do the calculation here If Resolution = 3 Vfs = 5V n = 1 - (lpg(3) / log(5)) is correct ?
 February 28th, 2012, 07:27 PM #5 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: How can I do the calculation here No, that isn't correct. We can solve for n in general: $R=\frac{V_{fs}}{2^n}$ $2^n=\frac{V_{fs}}{R}$ $n=\frac{\log$$\frac{V_{fs}}{R}$$}{\log(2)}$ Now, when given values for $R$ and $V_{fs}$ just plug them into the above formula for n.
 February 29th, 2012, 01:35 AM #6 Senior Member   Joined: Oct 2009 Posts: 895 Thanks: 1 Re: How can I do the calculation here OK .. thanks

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