More ugly xfer function

[The Phantom]

How about this one? The amplifier has a gain of k.

|-------------------|
| |
.-. |
R2| | |
| | |
'-' |
C1 C2 | C3 |
| |
|| || | || |\ |
Vi -||--|--||--|--||--|----|k>-----| Vo
|| | || || | |/
| |
| |
.-. .-.
R1| | R3| |
| | | |
'-' '-'
| |
-------|-------
|

 

[Jim Thompson]

How about this one? The amplifier has a gain of k.

|-------------------|
| |
.-. |
R2| | |
| | |
'-' |
C1 C2 | C3 |
| |
|| || | || |\ |
Vi -||--|--||--|--||--|----|k>-----| Vo
|| | || || | |/
| |
| |
.-. .-.
R1| | R3| |
| | | |
'-' '-'
| |
-------|-------
|
------
GND


created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

One complex pole-pair and one real pole ;-)

...Jim Thompson

 

[Active8]

One complex pole-pair and one real pole ;-)

...Jim Thompson

This should be solvable without worrying about the gain of the amp.
Another one to add to my list of shit to analyze.

 

[Active8]

How about this one? The amplifier has a gain of k.

|-------------------|
| |
.-. |
R2| | |
| | |
'-' |
C1 C2 | C3 |
| |
|| || | || |\ |
Vi -||--|--||--|--||--|----|k>-----| Vo
|| | || || | |/
| |
| |
.-. .-.
R1| | R3| |
| | | |
'-' '-'
| |
-------|-------
|

Are you asking or challenging?

 

[Active8]

<snip>

Your circuit shows the amp as an op-amp. You're on the list

The OP's net (above) can't neccessarily be considered that way so
what do you think? Treat it as

an A matrix consisting of C1, C2, R1 cascaded with the A matrix
equiv of

Y( A( C3, R3 ) * A( K ) ) + Y{ R2 ) )

where for e.g. A( C3. R3 ) is shorthand for thr A matrix of the C3,
R3 net and Y( A ) is the Y equiv of A

??

So that would be (half-assed ascii matrix shorthand follows):

A( C1, C2, R1 ) * A( Y( A( C3, R3 ) * A( K ) ) + Y{ R2 ) ) )
|---------- * -----|cascade
parallel|------------------------ + ------|

Yuck.

 

[The Phantom]

Are you asking or challenging?

Whichever makes it seem most interesting to you!

 

[john jardine]

How about this one? The amplifier has a gain of k.

|-------------------|
| |
.-. |
R2| | |
| | |
'-' |
C1 C2 | C3 |
| |
|| || | || |\ |
Vi -||--|--||--|--||--|----|k>-----| Vo
|| | || || | |/
| |
| |
.-. .-.
R1| | R3| |
| | | |
'-' '-'
| |
-------|-------
|
------
GND


created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

And another one ...

(Adjust centre frequency without disturbing the gain).
___
o-----|___|-------------, (CA=CB)
|
Vin ||CA | ||CB
o ,-||-------+-----||---,
| | || | || |
| | .-. |
GND | | |Fres |
| | |Pot |
| '-' |
| | |
| GND |
| ___ |
+--------|___|--------+
| |
| |\| |
'---------|-\ |
| >--------+--o
,--|+/ Vout
| |/| o
| |
GND GND

created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

 

[Active8]

On 21 Aug 2004 03:17:38 -0500, The Phantom wrote:

Whichever makes it seem most interesting to you!

Well, if you already have the answer, it's a challenge and you'll be
able to say whether someone gets it right. If you don't, how will
you know ou get the right answer?

 

[Mike]

One complex pole-pair and one real pole ;-)

...Jim Thompson

....and three zeros. Don't forget the zeros...

How's this:

gx = 1/rx

Vo k*s^3c1c2c3
-- = -------------------------------------------------
Vi sc3(sc3-kg2)(sc1+sc2+g1) - s^2c2^2(sc3+g3) -
(sc3+g3)(sc2+sc3+g2)(sc1+sc2+g1)

-- Mike --

 

[Jim Thompson]

How about this one? The amplifier has a gain of k.

|-------------------|
| |
.-. |
R2| | |
| | |
'-' |
C1 C2 | C3 |
| |
|| || | || |\ |
Vi -||--|--||--|--||--|----|k>-----| Vo
|| | || || | |/
| |
| |
.-. .-.
R1| | R3| |
| | | |
'-' '-'
| |
-------|-------
|
------
GND


created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

My approach to node/loop analysis for this problem is posted at...

Newsgroups: alt.binaries.schematics.electronic
Subject: Another Ugly Xfer Function - UglyXferFunction.gif (1/1)
Message-ID: <[email protected]>

...Jim Thompson

 

[Roy McCammon]

How about this one? The amplifier has a gain of k.

|-------------------|
| |
.-. |
R2| | |
| | |
'-' |
C1 C2 | C3 |
| |
|| || | || |\ |
Vi -||--|--||--|--||--|----|k>-----| Vo
|| | || Vx || | |/
| |
| |
.-. .-.
R1| | R3| |
| | | |
'-' '-'
| |
-------|-------
|

trick again. I added the node Vx.

Assume Vx known and solve for Vo. From that compute
the currents in C3 and R2 and hence the current in C2.

Work backward to compute Vi

 

[Jim Thompson]

trick again. I added the node Vx.

Assume Vx known and solve for Vo. From that compute
the currents in C3 and R2 and hence the current in C2.

Work backward to compute Vi

You don't need a "VX" for this particular problem...

My approach to node/loop analysis for this problem is posted at:

Newsgroups: alt.binaries.schematics.electronic
Subject: Another Ugly Xfer Function - UglyXferFunction.gif (1/1)
Message-ID: <[email protected]>

...Jim Thompson

 

[The Phantom]

On 21 Aug 2004 03:17:38 -0500, The Phantom wrote:



Well, if you already have the answer, it's a challenge and you'll be
able to say whether someone gets it right. If you don't, how will
you know ou get the right answer?

Well then, I guess it's a challenge. The point here is to
demonstrate how very rapidly things get VERY ugly. The network posed
recently under "Ugly xfer function" was 2nd order; this one is only
3rd order; but it is very "ugly". The problem is not so much setting
it up, but doing all the algebra without making a mistake.

I see over on alt.binaries.schematics.electronic that Jim copped
out, probably for that very reason.

I first worked it out by hand about 20 years ago, and it took me
several days to convince myself that I had got it right. I worked it
out on several sheets of paper and then set them aside. Then I did it
again, and then once more, and compared results till I found all the
mistakes.

 

[Jim Thompson]

[snip]

I see over on alt.binaries.schematics.electronic that Jim copped
out, probably for that very reason.

[snip]

Naaaah! I simply ran out of paper ;-)

You will note that I did exhibit the final node voltage before Vin.
So, with a big sheet of paper, you can get to Vin easily.

Plus I'd like to have the young-bucks here actually do the grunt work
and get proficient.

I taught Algebra and passive circuit analysis for several years, at a
tech school here in Phoenix, when I was in my mid-20s, to make an
extra few bucks

Sheesh! When I was studying at MIT, the instructor used to assign 5
or 6 loop networks to solve ;-)

...Jim Thompson

 

[The Phantom]

...and three zeros. Don't forget the zeros...

How's this:

gx = 1/rx

Vo k*s^3c1c2c3
-- = -------------------------------------------------
Vi sc3(sc3-kg2)(sc1+sc2+g1) - s^2c2^2(sc3+g3) -
(sc3+g3)(sc2+sc3+g2)(sc1+sc2+g1)

-- Mike --

This is very close, but you have made some errors in signs
somewhere along the way.

As an aid to those who actually do all the algebra, if you set
the value of all the components and the amplifier gain to unity, the
roots of the denominator are: -2.61803, -1 and -.381966

 

[The Phantom]

trick again. I added the node Vx.

Assume Vx known and solve for Vo. From that compute
the currents in C3 and R2 and hence the current in C2.

Work backward to compute Vi

You have to do the work to get full credit.

 

[The Phantom]

[snip]

I see over on alt.binaries.schematics.electronic that Jim copped
out, probably for that very reason.

[snip]

Naaaah! I simply ran out of paper ;-)

You will note that I did exhibit the final node voltage before Vin.
So, with a big sheet of paper, you can get to Vin easily.

Easily? It's just algebra, right?

Plus I'd like to have the young-bucks here actually do the grunt work
and get proficient.

Then you should wait longer before giving them the answer!

 

[Active8]

[snip]

I see over on alt.binaries.schematics.electronic that Jim copped
out, probably for that very reason.

[snip]

Naaaah! I simply ran out of paper ;-)

You will note that I did exhibit the final node voltage before Vin.
So, with a big sheet of paper, you can get to Vin easily.

Easily? It's just algebra, right?

Plus I'd like to have the young-bucks here actually do the grunt work
and get proficient.

Then you should wait longer before giving them the answer!

All one needs is the self-control to not look at the answer. I can't
read it without zooming, so all I know is that he drew current
arrows. IOW he set it up for KCL node analysis, not the matrix
solution I spotted. Now I can do both for comparison - but I'm going
to have to cheat the gain block into a matrix somehow. Maybe I
can't.

 

[Mike]

This is very close, but you have made some errors in signs
somewhere along the way.

As an aid to those who actually do all the algebra, if you set
the value of all the components and the amplifier gain to unity, the
roots of the denominator are: -2.61803, -1 and -.381966

Sure enough... inspection of the circuit reveals that the gain at infinite
frequency (short the caps) should have been -k. I did indeed miss a sign,
resulting in several sign flips by the end. Try this:

Vo -ks^3c1c2c3
-- = -------------------------------------------------
Vi sc3(sc3+kg2)(sc1+sc2+g1) + s^2c2^2(sc3+g3) -
(sc3+g3)(sc2+sc3+g2)(sc1+sc2+g1)

I approached this in almost the same way Jim Thompson did (except that I
used G instead of R), and getting the equation in the form I posted took
around 15 minutes. Checking the answer (which I didn't do until after you
pointed out the error) took a couple minutes, and finding the sign error
took another five minutes or so. The analysis took ten equations, showing
all the steps, and would have fit on one page if I hadn't drawn the
schematic at the top of the page. Putting the denominator in standard
polynomial form would have taken an extra couple of equations.

It doesn't seem particularly ugly, especially since it's easy to check the
resulting transfer function at zero and infinity.

-- Mike --

 

[Roy McCammon]

You don't need a "VX" for this particular problem...

My approach to node/loop analysis for this problem is posted at:

well duh, Jim. the trick is about avoiding your approach.