[Garbz]

The above circuit is actually the answer but I am trying to derrive it without SPICE using phasors. Call the top one R1, C, and the resistor to ground R2. What i'm trying to find is:

H(w) = Vpre/Vin = (1+jw*z) / (1+jw*p). This is the actual simplification where Z and P are functions of R1 R2 and C. (incidently the above corresponds to a pole at 1.8k rads/s, and a zero at 590 rads/s

What I found so far is the simplification starting with a voltage divider and a parallel impedance:

H(w) = R2 / (R2 + ( (R1 * 1/jwc) / (R1 + 1/jwc)))

rearranging to give me:

H(w) = ( (jw)^2 + (jw)/(R1*C) + 0 ) / ( (jw)^2 + (jw)/(R1*C) + 1/(R1*C*C) )

Sorry about the formula but my scanner is stuffed. Can anyone factorize the above equation into the form of the top one?

 

[guzzler]

Expand out the j square terms.

H(w) = ( (jw / RC) - w^2 ) / ( (1/RC^2) + (jw/RC) - w^2 )

Multiply top and bottom by RC^2

H(w) = ( jwC - w^2*R*C^2 ) / ( jwC - w^2*R*C^2 + 1 )

Rewrite numerator as

jwC - w^2*R*C + 1 - 1

Simplifies to

1 - 1/(jwC - w^2*R*C^2 + 1)

This could be entirely wrong, I'm supposed to be writing a report but this was welcome distraction!

 

[Garbz]

Yeah it is

You missed the fact R1 and R2 are different. Also the result comes to 1 pole at 1800, and 1 zero at 590. So there needs to be a (1+jw/z) / (1+jw/p) with no constants out the front since DC gain is 0.

Plus I have no idea how you got w^2 without j^2 when all the w terms are associated with a j term.

I appreciate the attempt though.

 

[PeterR]

H(s)=R2/(R2+R1||C)

=>H(s)=R2/(R2+1/(1/R1+sC)) |*(1/R1 + sC) / (R2*C)
=>H(s)=(s+1/(R1*C)) / (s + 1/C*(1/R1+1/R2))

 

[PeterR]

Quote:

Originally Posted by Garbz DC gain is 0.

More like R2/(R1+R2)

 

[rreynol]

Quote:

Originally Posted by Garbz Plus I have no idea how you got w^2 without j^2 when all the w terms are associated with a j term.

j^2 = -1 (since j = i = sqrt(-1))

 

[guzzler]

Quote:

Originally Posted by rreynol j^2 = -1 (since j = i = sqrt(-1))

Yep, and I used your rearranged equation which doesn't contain R2. Hope that much was right even if it wasn't what you're after!

 

[Garbz]

Quote:

Originally Posted by PeterR More like R2/(R1+R2)

BAH sorry I was looking black box measurements when I posted that! If DC gain were zero it wouldn't work for me

HF gain is 0

DC gain is -11.65dB if IIRC.

Btw thanks for the reply I'll check if it's correct after breakfast

Quote:

Originally Posted by rreynol j^2 = -1 (since j = i = sqrt(-1))

I know but for the purposes of creating a transfer function in the top form none of the js are allowed to disappear like that and it's the reason when doing bode analysies that you leave the j and w together, since 1+ 2*zeta*jw/f + (jw/f)^2 indicates 2 poles or zeros at f. I should have mentioned not to simplify that too much.

Quote:

Originally Posted by guzzler Yep, and I used your rearranged equation which doesn't contain R2. Hope that much was right even if it wasn't what you're after!

And crap again. Sorry the rearranged equation did contain R2. I should stop posting early in the morning or late at night

 

[rreynol]

Quote:

Originally Posted by Garbz BAH sorry I was looking black box measurements when I posted that! If DC gain were zero it wouldn't work for me HF gain is 0 DC gain is -11.65dB if IIRC. Btw thanks for the reply I'll check if it's correct after breakfast I know but for the purposes of creating a transfer function in the top form none of the js are allowed to disappear like that and it's the reason when doing bode analysies that you leave the j and w together, since 1+ 2*zeta*jw/f + (jw/f)^2 indicates 2 poles or zeros at f. I should have mentioned not to simplify that too much. And crap again. Sorry the rearranged equation did contain R2. I should stop posting early in the morning or late at night

/EDIT: Removed comment since the irony is high with this one.

 

[Garbz]

It is taking a lot of effort to respond in a civilised fashion, but A) this is not home work, B) i already got the correct result using nodal analysis before i asked, C) it's a very very small part of a much larger problem, D) if I didn't do it several times by simplification ending up with an equation I couldn't factorise then I would not have asked, and finally) if it wern't holidays I would have asked the tutors.
It's not a matter of not being able to get an answer. It's a matter of trying to figure out if it could be done using multiple methods, and if it didn't interest me I wouldn't care, I submitted this tutorial Q already about 6 weeks ago with the correct answer, so if you aren't adding something constructive don't add anything at all.

PeterR I just confirmed it and it returned the correct result. I should have just used the sum of inverses formula rather then the product over the sum, it avoides the nasty squares. I'll ask the maths department in a few weeks if there is any easy way to factorise an equation like that.

/EDIT: edited to remove profanities.