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Algebra help!

  1. Aug 20, 2013 #1
    Can someone show some mathematical steps on how we go from step 1 to step 2 please.

    Step#1
    vout = (vcm + vd/2)(Ro/Rb + Ro) (1+Rf/Ra)- (vcm - vd/2)(Rf/Ra)

    Step#2
    vout = vcm [(Ro/Rb + Ro) (1+ Rf/Ra) -(Rf/Ra)] + vd(1/2) [(Ro/Rb + Ro) (1+ Rf/Ra) + (Rf/Ra)]


    thank you in advance!
     
  2. jcsd
  3. Aug 21, 2013 #2
    Is this what you wrote?

    [itex]v_{out}=(v_{cm}+\frac{v_d}{2})(\frac{R_0}{R_b}+R_0)(1+\frac{R_f}{R_a})-(v_{cm}-\frac{v_d}{2})(\frac{R_f}{R_a})[/itex]

    [itex]\Rightarrow v_{out}=v_{cm}[(\frac{R_0}{R_b}+R_0)(1+\frac{R_f}{R_a})-\frac{R_f}{R_a}]+\frac{v_d}{2}[(\frac{R_0}{R_b}+R_0)(1+\frac{R_f}{R_a})+\frac{R_f}{R_a}][/itex]

    If so, you should write it out and do it yourself. It is just two uses of the distributive property and then a slight rearrangement and the distributive property again (but backwards). When all is said and done, it should be no more than 6 lines (probably less, but it depends on your brain's current math RAM). Or maybe I misread what you wrote and it is completely different.
     
  4. Aug 21, 2013 #3
    "If so, you should write it out and do it yourself. "

    thanks
     
    Last edited: Aug 21, 2013
  5. Aug 21, 2013 #4
    Are you able to figure it out, or you need some help with that?
     
  6. Aug 21, 2013 #5
    I checked out your other posts and there is nothing here that you weren't able to figure out before. I also outlined the steps. I apologize if you find that to be a rude statement, there is no point in people feeding you answers. Sorry if that is what you wanted.
     
  7. Aug 21, 2013 #6
    Hello Electric Red,

    I am totally baffled and don't even know where
    to start... I attempted to get vcm out of the brackets like so:

    vcm(1+vd/2) and vcm(1-vd/2)

    but then don't know where to go from there. I tried going further by dividing vcm on both sides but I have to be honest here .... I am totally lost.... I tried looking up some math tutorials but could not relate them to a resolution in respect to my problem.

    A step by step illustration would surely help getting me started and would really be appreciated.

    Thanks with sincere regards
     
  8. Aug 21, 2013 #7
    That is not correct. From that you just wrote ##v_{cm}(1+\frac{v_d}{2})=(v_{cm}+\frac{v_{cm}v_d}{2})## which does not appear anywhere. As I said earlier, but more explicitly now, distribute over the first and fourth parentheses and then rearrange to isolate ##v_{cm}## and ##v_d##.

    It might be easier if you define new variables for yourself. Let
    ##A=(\frac{R_0}{R_b}+R_0)## and
    ##B=(1+\frac{R_f}{R_a})##
     
  9. Aug 21, 2013 #8

    Mark44

    Staff: Mentor

    Neither of these is correct, so I can see why you're having some problems.

    If you factor vcm out of vcm + vd/2, you don't get vcm(1 + vd/2). As a check, expanding the last result here gives vcm + vcmvd/2), which isn't what you started with.


    This is not valid either. You don't really have "both sides." All you are doing is simplifying an expression, so you are very limited in what you can do, unlike when you have an equation to work with.
     
  10. Aug 22, 2013 #9
    Try looking at it like this.

    ##v_{out}=(v_{cm}+\frac{v_d}{2})(\frac{R_0}{R_b}+R_0)(1+\frac{R_f}{R_a})-(v_{cm}-\frac{v_d}{2})(\frac{R_f}{R_a})##

    is the same as

    ##v_{out}=(v_{cm}+\frac{v_d}{2})A-(v_{cm}-\frac{v_d}{2})B##

    where

    ##A=(\frac{R_0}{R_b}+R_0)(1+\frac{R_f}{R_a})##
    and
    ##B=\frac{R_f}{R_a}##

    With these new definitions (not the ones I made before), you are trying to show that

    ##v_{out}=(v_{cm}+\frac{v_d}{2})A-(v_{cm}-\frac{v_d}{2})B=v_{cm}[A-B]+\frac{v_d}{2}[A+B]##
     
  11. Aug 23, 2013 #10
    Hello DrewD,

    totally swamped here... I will get back as soon as I have a bit of free time...
    Thanks
     
  12. Sep 12, 2013 #11
    Ahhh!

    The light from the sun hit my head!

    I see it now....

    thanks DrewD
     
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