Ugly xfer function

A

Active8

Guest
Being a true masochist, I thought I'd try deriving the Laplace xfer
function for this filter and it's getting ugly. Before I go further,
I thought I'd ask if I'm setting it up right, er, umm, ...
efficiently?



+------+-------+ _______
| | | 1 | R1+R3
| .-. | fr= ----- \|-------
--- | |R2 | 2pi*C R1*R2*R3
C --- | | |
| '-' | R2
___ | || | |\ | gain= - -----
o-|___|-+-+--||--+--|-\ | 2R1
R1 | || | >-+-o
.-. C +--|+/
| |<-+ | |/
R3| | | | 1
'-' | | Bandwidth= -------
| | | pi*R2*C
=== === ===
GND GND GND
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de
view\fixed font


I figured if I wrote the branch currents coming off the node to the
right of R1, I'd get both e_in and e_out in the equation and go from
there. So I set the inv op amp input to 0, and figured the
combination of the 2 caps and R2 in || with R3 - call that Z_ugly.
That would give me a voltage divider eq for

e_n = e_i * Z_ugly / (R1 + Z_ugly), e_n being the voltage at the
node to the right of R1.

Isn't there a better way to set this up? I'd like to know before I
try to simplify Z_ugly any further.

TIA
--
Best Regards,
Mike
 
Active8 wrote:
Being a true masochist, I thought I'd try deriving the Laplace xfer
function for this filter and it's getting ugly. Before I go further,
I thought I'd ask if I'm setting it up right, er, umm, ...
efficiently?



+------+-------+ _______
| | | 1 | R1+R3
| .-. | fr= ----- \|-------
--- | |R2 | 2pi*C R1*R2*R3
C1 --- | | |
| '-' | R2
___ vx | || | |\ | gain= - -----
Vi o-|___|-+--+--||--+--|-\ | 2R1
R1 | || | >-+-o Vo
.-. c2 +--|+/
| |<-+ | |/
R3| | | | 1
'-' | | Bandwidth= -------
| | | pi*R2*C
=== === ===
GND GND GND
Click to expand...
sometimes you can use a trick.

Vo = -Vx * R2 / (1/sC2)

I1 = current through R1 = Vx/R3 + Vx/(1/sC2) + (Vx-Vo)/(1/sC1)

Vi = Vx + R1*I1

solve the first equation for Vx
Vx = -Vo/sRC2
substitute that into the second equation
I1 = (-Vo/sRC2)/R3 + (-Vo/sRC2)/(1/sC2) + (-Vo/sRC2 -Vo)/(1/sC1)
and put that into the klast equation

Vi=Vx+R1*[ (-Vo/sRC2)/R3 +(-Vo/sRC2)/(1/sC2) +(-Vo/sRC2 -Vo)/(1/sC1)]
 
On Sat, 07 Aug 2004 15:00:55 GMT, Roy McCammon wrote:

Active8 wrote:
Being a true masochist, I thought I'd try deriving the Laplace xfer
function for this filter and it's getting ugly. Before I go further,
I thought I'd ask if I'm setting it up right, er, umm, ...
efficiently?

+------+-------+ _______
| | | 1 | R1+R3
| .-. | fr= ----- \|-------
--- | |R2 | 2pi*C R1*R2*R3
C1 --- | | |
| '-' | R2
___ vx | || | |\ | gain= - -----
Vi o-|___|-+--+--||--+--|-\ | 2R1
R1 | || | >-+-o Vo
.-. c2 +--|+/
| |<-+ | |/
R3| | | | 1
'-' | | Bandwidth= -------
| | | pi*R2*C
=== === ===
GND GND GND

sometimes you can use a trick.

Vo = -Vx * R2 / (1/sC2)

I1 = current through R1 = Vx/R3 + Vx/(1/sC2) + (Vx-Vo)/(1/sC1)

Vi = Vx + R1*I1

solve the first equation for Vx
Vx = -Vo/sRC2
substitute that into the second equation
I1 = (-Vo/sRC2)/R3 + (-Vo/sRC2)/(1/sC2) + (-Vo/sRC2 -Vo)/(1/sC1)
and put that into the klast equation

Vi=Vx+R1*[ (-Vo/sRC2)/R3 +(-Vo/sRC2)/(1/sC2) +(-Vo/sRC2 -Vo)/(1/sC1)]
Click to expand...
What's assumptions (if that's the right word) do we make to justify
using that trick?

In the mean time, I'll try simplifying the above and BTW, I'm trying
to come up with those formulas, above, one way or the other.
--
Best Regards,
Mike
 
Active8 wrote:
On Sat, 07 Aug 2004 15:00:55 GMT, Roy McCammon wrote:

Active8 wrote:
Being a true masochist, I thought I'd try deriving the Laplace xfer
function for this filter and it's getting ugly. Before I go further,
I thought I'd ask if I'm setting it up right, er, umm, ...
efficiently?

+------+-------+ _______
| | | 1 | R1+R3
| .-. | fr= ----- \|-------
--- | |R2 | 2pi*C R1*R2*R3
C1 --- | | |
| '-' | R2
___ vx | || | |\ | gain= - -----
Vi o-|___|-+--+--||--+--|-\ | 2R1
R1 | || | >-+-o Vo
.-. c2 +--|+/
| |<-+ | |/
R3| | | | 1
'-' | | Bandwidth= -------
| | | pi*R2*C
=== === ===
GND GND GND

sometimes you can use a trick.

Vo = -Vx * R2 / (1/sC2)

I1 = current through R1 = Vx/R3 + Vx/(1/sC2) + (Vx-Vo)/(1/sC1)

Vi = Vx + R1*I1

solve the first equation for Vx
Vx = -Vo/sRC2
substitute that into the second equation
I1 = (-Vo/sRC2)/R3 + (-Vo/sRC2)/(1/sC2) + (-Vo/sRC2 -Vo)/(1/sC1)
and put that into the klast equation

Vi=Vx+R1*[ (-Vo/sRC2)/R3 +(-Vo/sRC2)/(1/sC2) +(-Vo/sRC2
-Vo)/(1/sC1)]

What's assumptions (if that's the right word) do we make to justify
using that trick?

In the mean time, I'll try simplifying the above and BTW, I'm trying
to come up with those formulas, above, one way or the other.
Click to expand...
Here there is the transfer function, w=omega.

R2*R3
- -----C*wr*P
Vout R1+R3
---- = --------------------------------------
Vin 2R1*R3 R1*R2*R3
1 + ------C*wr*P + --------(C*wr*P)^2
R1+R3 R1+R3

You can then compare this with the principal bandpass equation:

Ar
--- P
Vout Q
---- = ----------------
Vin 1
1 + --- P + P^2
Q

and derive the design equations from that.
Well, not exactly s.e.b. :)
--
ciao Ban
Bordighera, Italy
 
On Sun, 08 Aug 2004 03:02:48 GMT, Ban wrote:

Active8 wrote:
On Sat, 07 Aug 2004 15:00:55 GMT, Roy McCammon wrote:

Active8 wrote:
Being a true masochist, I thought I'd try deriving the Laplace xfer
function for this filter and it's getting ugly. Before I go further,
I thought I'd ask if I'm setting it up right, er, umm, ...
efficiently?

+------+-------+ _______
| | | 1 | R1+R3
| .-. | fr= ----- \|-------
--- | |R2 | 2pi*C R1*R2*R3
C1 --- | | |
| '-' | R2
___ vx | || | |\ | gain= - -----
Vi o-|___|-+--+--||--+--|-\ | 2R1
R1 | || | >-+-o Vo
.-. c2 +--|+/
| |<-+ | |/
R3| | | | 1
'-' | | Bandwidth= -------
| | | pi*R2*C
=== === ===
GND GND GND

sometimes you can use a trick.

Vo = -Vx * R2 / (1/sC2)

I1 = current through R1 = Vx/R3 + Vx/(1/sC2) + (Vx-Vo)/(1/sC1)

Vi = Vx + R1*I1

solve the first equation for Vx
Vx = -Vo/sRC2
substitute that into the second equation
I1 = (-Vo/sRC2)/R3 + (-Vo/sRC2)/(1/sC2) + (-Vo/sRC2 -Vo)/(1/sC1)
and put that into the klast equation

Vi=Vx+R1*[ (-Vo/sRC2)/R3 +(-Vo/sRC2)/(1/sC2) +(-Vo/sRC2
-Vo)/(1/sC1)]

What's assumptions (if that's the right word) do we make to justify
using that trick?

In the mean time, I'll try simplifying the above and BTW, I'm trying
to come up with those formulas, above, one way or the other.

Here there is the transfer function, w=omega.

R2*R3
- -----C*wr*P
Vout R1+R3
---- = --------------------------------------
Vin 2R1*R3 R1*R2*R3
1 + ------C*wr*P + --------(C*wr*P)^2
R1+R3 R1+R3

You can then compare this with the principal bandpass equation:

Ar
--- P
Vout Q
---- = ----------------
Vin 1
1 + --- P + P^2
Q

and derive the design equations from that.
Well, not exactly s.e.b. :)
Click to expand...
Oh, I wasn't sure :) I'll look at that, but I was trying to figure
out if I had the problem set up right so I could derive the
*Laplace* xfer function, myself.

What is wr? 2.pi.f ? What about P and Ar?
--
Best Regards,
Mike
 
On Sun, 08 Aug 2004 03:02:48 GMT, Ban wrote:

Active8 wrote:
On Sat, 07 Aug 2004 15:00:55 GMT, Roy McCammon wrote:

Active8 wrote:
Being a true masochist, I thought I'd try deriving the Laplace xfer
function for this filter and it's getting ugly. Before I go further,
I thought I'd ask if I'm setting it up right, er, umm, ...
efficiently?

+------+-------+ _______
| | | 1 | R1+R3
| .-. | fr= ----- \|-------
--- | |R2 | 2pi*C R1*R2*R3
C1 --- | | |
| '-' | R2
___ vx | || | |\ | gain= - -----
Vi o-|___|-+--+--||--+--|-\ | 2R1
R1 | || | >-+-o Vo
.-. c2 +--|+/
| |<-+ | |/
R3| | | | 1
'-' | | Bandwidth= -------
| | | pi*R2*C
=== === ===
GND GND GND

sometimes you can use a trick.

Vo = -Vx * R2 / (1/sC2)

I1 = current through R1 = Vx/R3 + Vx/(1/sC2) + (Vx-Vo)/(1/sC1)

Vi = Vx + R1*I1

solve the first equation for Vx
Vx = -Vo/sRC2
substitute that into the second equation
I1 = (-Vo/sRC2)/R3 + (-Vo/sRC2)/(1/sC2) + (-Vo/sRC2 -Vo)/(1/sC1)
and put that into the klast equation

Vi=Vx+R1*[ (-Vo/sRC2)/R3 +(-Vo/sRC2)/(1/sC2) +(-Vo/sRC2
-Vo)/(1/sC1)]

What's assumptions (if that's the right word) do we make to justify
using that trick?

In the mean time, I'll try simplifying the above and BTW, I'm trying
to come up with those formulas, above, one way or the other.

Here there is the transfer function, w=omega.

R2*R3
- -----C*wr*P
Vout R1+R3
---- = --------------------------------------
Vin 2R1*R3 R1*R2*R3
1 + ------C*wr*P + --------(C*wr*P)^2
R1+R3 R1+R3

You can then compare this with the principal bandpass equation:

Ar
--- P
Vout Q
---- = ----------------
Vin 1
1 + --- P + P^2
Q

and derive the design equations from that.
Well, not exactly s.e.b. :)
Click to expand...
I'll repost in sed.
--
Best Regards,
Mike
 
Active8 wrote:
Here there is the transfer function, w=omega.

R2*R3
- -----C*wr*P
Vout R1+R3
---- = --------------------------------------
Vin 2R1*R3 R1*R2*R3
1 + ------C*wr*P + --------(C*wr*P)^2
R1+R3 R1+R3

You can then compare this with the principal bandpass equation:

Ar
--- P
Vout Q
---- = ----------------
Vin 1
1 + --- P + P^2
Q

and derive the design equations from that.
Well, not exactly s.e.b. :)

Oh, I wasn't sure :) I'll look at that, but I was trying to figure
out if I had the problem set up right so I could derive the
*Laplace* xfer function, myself.

What is wr? 2.pi.f ? What about P and Ar?
Click to expand...
wr = omega resonance = 2*pi*fr fr=resonance frequency
P = complex frequency variable = j(w/wr) = j (f/fr)
Ar = gain at resonance

I gave you the solution already, sorry for that. But this way you can
control better your effort to derive it from the knots. You are on the right
track. Start with the first knot, and sum up the currents. Then make a loop
and sum up the voltages. Don't forget that the -in is a virtual gnd.
--
ciao Ban
Bordighera, Italy
 
Active8 wrote:
On Sat, 07 Aug 2004 15:00:55 GMT, Roy McCammon wrote:


Active8 wrote:

Being a true masochist, I thought I'd try deriving the Laplace xfer
function for this filter and it's getting ugly. Before I go further,
I thought I'd ask if I'm setting it up right, er, umm, ...
efficiently?

+------+-------+ _______
| | | 1 | R1+R3
| .-. | fr= ----- \|-------
--- | |R2 | 2pi*C R1*R2*R3
C1 --- | | |
| '-' | R2
___ vx | || | |\ | gain= - -----
Vi o-|___|-+--+--||--+--|-\ | 2R1
R1 | || | >-+-o Vo
.-. c2 +--|+/
| |<-+ | |/
R3| | | | 1
'-' | | Bandwidth= -------
| | | pi*R2*C
=== === ===
GND GND GND

sometimes you can use a trick.

Vo = -Vx * R2 / (1/sC2)

I1 = current through R1 = Vx/R3 + Vx/(1/sC2) + (Vx-Vo)/(1/sC1)

Vi = Vx + R1*I1

solve the first equation for Vx
Vx = -Vo/sRC2
substitute that into the second equation
I1 = (-Vo/sRC2)/R3 + (-Vo/sRC2)/(1/sC2) + (-Vo/sRC2 -Vo)/(1/sC1)
and put that into the klast equation

Vi=Vx+R1*[ (-Vo/sRC2)/R3 +(-Vo/sRC2)/(1/sC2) +(-Vo/sRC2 -Vo)/(1/sC1)]


What's assumptions (if that's the right word) do we make to justify
using that trick?

In the mean time, I'll try simplifying the above and BTW, I'm trying
to come up with those formulas, above, one way or the other.
Click to expand...
sorry to take so long to get back to you, but it looks like you got
plenty of good help.

as for the trick, is a matter of recognizing that a
sub circuit composed of the opamp, R2 and c2 is just
a standard inverting amplifier and replacing
that with the gain equation for the standard inverting
amplifier. Sorry that I didn't explain that.

But as a general rule that is the way I attack
complicated circuits; try do identify known
circuit blocks and replace then with an equivalent
equation.




--
local optimization seldom leads to global optimization

my e-mail address is: <my first name> <my last name> AT mmm DOT com
 
On Tue, 10 Aug 2004 11:15:38 -0500, Roy McCammon wrote:

Active8 wrote:
On Sat, 07 Aug 2004 15:00:55 GMT, Roy McCammon wrote:

Active8 wrote:

Being a true masochist, I thought I'd try deriving the Laplace xfer
function for this filter and it's getting ugly. Before I go further,
I thought I'd ask if I'm setting it up right, er, umm, ...
efficiently?

+------+-------+ _______
| | | 1 | R1+R3
| .-. | fr= ----- \|-------
--- | |R2 | 2pi*C R1*R2*R3
C1 --- | | |
| '-' | R2
___ vx | || | |\ | gain= - -----
Vi o-|___|-+--+--||--+--|-\ | 2R1
R1 | || | >-+-o Vo
.-. c2 +--|+/
| |<-+ | |/
R3| | | | 1
'-' | | Bandwidth= -------
| | | pi*R2*C
=== === ===
GND GND GND

sometimes you can use a trick.

Vo = -Vx * R2 / (1/sC2)

I1 = current through R1 = Vx/R3 + Vx/(1/sC2) + (Vx-Vo)/(1/sC1)

Vi = Vx + R1*I1

solve the first equation for Vx
Vx = -Vo/sRC2
substitute that into the second equation
I1 = (-Vo/sRC2)/R3 + (-Vo/sRC2)/(1/sC2) + (-Vo/sRC2 -Vo)/(1/sC1)
and put that into the klast equation

Vi=Vx+R1*[ (-Vo/sRC2)/R3 +(-Vo/sRC2)/(1/sC2) +(-Vo/sRC2 -Vo)/(1/sC1)]

What's assumptions (if that's the right word) do we make to justify
using that trick?

In the mean time, I'll try simplifying the above and BTW, I'm trying
to come up with those formulas, above, one way or the other.

sorry to take so long to get back to you, but it looks like you got
plenty of good help.

as for the trick, is a matter of recognizing that a
sub circuit composed of the opamp, R2 and c2 is just
a standard inverting amplifier and replacing
that with the gain equation for the standard inverting
amplifier. Sorry that I didn't explain that.

But as a general rule that is the way I attack
complicated circuits; try do identify known
circuit blocks and replace then with an equivalent
equation.
Click to expand...
I think the "trick" wasn't a trick at all. I've worked it all out
days ago. Normally, I'd first try to divide output Z by in Z, but
this had multiple feedback paths. Your "trick", IIRC was to evaluate
the node to the right of the input R by seeing that the current in
to and through R1 (Rs) to virtual gnd also flowed through whatever R
it was that connected directly to Vout. That solved node Vx and thus
the whole deal. I was looking at it ... I over complicated things.

TNX much.
--
Best Regards,
Mike
 
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