Here is a derivation that I am stuck on. You may notice that it is a transfer function of an op-amp. I have seen this solution before, but no derivation so I decided to takle it myself.

My question is how in the world is (1-b) reduced from R2K/(R2 + R1 - R1K)?

See attached.

 

How about a schematic, so we can see what we're working on?

 

one clarification required,
is the question about mathematical jugglery or about opamp?

 

If the question is how to account for the term (1-b) in the expression on the right side of the initial expression through manipulation of the expression on the left side of the equation then it appears that you have the answer already.

\((1-b)= \frac{R2}{R1+R2} \ \ where\ b = \frac{R1}{(R1+R2)}\)

As far as I can tell, you have already sorted this out in your manipulations to the right of the line you have drawn down the center of the page.

hgmjr

 

If the question is how to account for the term (1-b) in the expression on the left side of the initial expression through manipulation of the expression on the right side of the equation then it appears that you have the answer already.

\((1-b)= \frac{R2}{R1+R2} \ \ where\ b = \frac{R1}{(R1+R2)}\)

As far as I can tell, you have already sorted this out in your manipulations to the right of the line you have drawn down the center of the page.

hgmjr

thats exactly why asked whether this was abt the manipulation involved.

 

If the question is how to account for the term (1-b) in the expression on the right side of the initial expression through manipulation of the expression on the left side of the equation then it appears that you have the answer already.

\((1-b)= \frac{R2}{R1+R2} \ \ where\ b = \frac{R1}{(R1+R2)}\)

As far as I can tell, you have already sorted this out in your manipulations to the right of the line you have drawn down the center of the page.

hgmjr

These two expressions are equal. I can derive the equation on the left hand side from the circuit (post later). The explanation goes on to say that the left hand side expression is equal to (1-b)*(k/(1-kb)) when b=(R1/(R1+R2)).

What I have done is derive the transfer function along with the example to arrive at the left hand side portion.

I then can show, for myself, that these are equivalent; however...

I have no idea how in god's high holy name it can be manipulated into the right hand side expression.

That is my question.

 

These two expressions are equal. I can derive the equation on the left hand side from the circuit (post later). The explanation goes on to say that the left hand side expression is equal to (1-b)*(k/(1-kb)) when b=(R1/(R1+R2)).

What I have done is derive the transfer function along with the example to arrive at the left hand side portion.

I then can show, for myself, that these are equivalent; however...

I have no idea how in god's high holy name it can be manipulated into the right hand side expression.

That is my question.

(1 - b) = R2/(R1 + R2)

(1 - b)(R1 + R2) = R2

Multiply out left hand expression:

R1 + R2 - bR1 -bR2 = R2

R1 + R2 - b(R1 + R2) = R2

R1 + R2 = R2 + b(R1 + R2)

R1 = b(R1 + R2)

b = R1/(R1 + R2)

If you write it out as I have written on paper it will be clearer.

Dave

 

Maybe my brain is not working, but its seems to me that the equivalent expression is arbitrary:

Oh, BTW R2k/(R2+R1+R1k) = (1-b)(k/(1-kb)) if b = R1/(R2+R1).

Seeing that they are equal given the constraint of b = R1/(R2+R1) is fine, the math is easy. I can derive R2k/(R2+R1+R1k) from such a simple circuit with no problem...easy as pie.

Is this a math trick like "multiplying by 1"; i.e., (Rx/Rx) to help simplify an expression?

 

I would proceed as shown in the attachment.

 

Maybe my brain is not working, but its seems to me that the equivalent expression is arbitrary:

Oh, BTW R2k/(R2+R1+R1k) = (1-b)(k/(1-kb)) if b = R1/(R2+R1).

Seeing that they are equal given the constraint of b = R1/(R2+R1) is fine, the math is easy. I can derive R2k/(R2+R1+R1k) from such a simple circuit with no problem...easy as pie.

Is this a math trick like "multiplying by 1"; i.e., (Rx/Rx) to help simplify an expression?

Apologies, I misinterpreted what you were asking. The maths trick you looking at is multiplying the whole expression by 1 which will not change the expression) - however the trick is in your case multiply by (1/(R1+R2))/(1/(R1+R2)), which is something multiplied by itself which is 1.

The Electrician's answer shows how this is done in practice.

Dave