Understanding Algebra: Step-by-Step Guide from vout to vcm and vd

[simpComp]

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!

 

[DrewD]

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!

Is this what you wrote?

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

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

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.

 

[simpComp]

"If so, you should write it out and do it yourself. "

thanks

 

[Electric Red]

Are you able to figure it out, or you need some help with that?

 

[DrewD]

"If so, you should write it out and do it yourself. "

thanks

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.

 

[simpComp]

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

 

[DrewD]

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)

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
$B=(1+\frac{R_f}{R_a})$

 

[Mark44]

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)

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

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

but then don't know where to go from there. I tried going further by dividing vcm on both sides

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.

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

 

[DrewD]

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

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

 

[simpComp]

Hello DrewD,

totally swamped here... I will get back as soon as I have a bit of free time...
Thanks

 

[simpComp]

Ahhh!

The light from the sun hit my head!

I see it now...

thanks DrewD