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More ugly xfer function

M

Mike

Jan 1, 1970
0
I think you still have a sign error; the amplifier has a gain of k,
not -k, and with the caps shorted, Vo/Vi is k

Ugh. You're right, and that's what was obvious by inspection. I even proved
that in my notes before posting:

Vo| -kc1c2c3
--| = -------------------------------------------------------
Vi|s->inf c1c3^2 + c2c3^2 + c2^2c3 - c3(c1c2 + c2^2 + c1c3 + c2c3)

which reduces to

Vo| -kc1c2c3
--| = --------- = k
Vi|s->inf -c1c2c3

I must have sign dyslexia. After posting that the gain is -k, I went one
step further and factored the equation with unity component values in
Mathcad. The result Mathcad produced was

-s^3
--------------------
-s^3 - 4s^2 - 4s - 1

Which I dutifully transcribed, switching the signs of the denominator
coefficients as I did so, while leaving the numerator untouched.

Let's see... if I make one more sign error, I'll be up to four, and they'll
all cancel out...

-- Mike --
 
T

The Phantom

Jan 1, 1970
0
Nonsense- the most direct approach in the so-called "old days" would be
to write the circuit nodal equations in matrix form, multiply through
with appropriate value of S^N to obtain integers powers of S throughout,
then use simple matrix triangularization to solve for the output
symbolically

Please show us how it's done.

and have a determinant simple enough for even you as the
 
H

Helmut Sennewald

Jan 1, 1970
0
The Phantom said:
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

This highpass circuit has a nice numerical property for
a 3rd order Butterworth filter if k=4 is used.

1. Choose C and k=4

2. C=C1=C2=C3

3. R3 = 1/(2*pi*fg*C) fg is -3dB corner frequency
R2 = R3*2
R1 = R3/2


Some other examples for 3rd order Butterworth
---------------------------------------------

Fg=1000Hz, C1=C2=C3=10nF
k=1 R1=11.428e3 R2=4.4873e3 R3=78.614e3
k=2 R1=9.3300e3 R2=18.356e3 R3=23.540e3
k=4 R1=7.9577e3 R2=31.831e3 R3=15.915e3

Fg=1000Hz, C1=C2=C3=22nF
k=1 R1=5.1947e3 R2=2.0397e3 R3=35.734e3
k=2 R1=4.2409e3 R2=8.3434e3 R3=10.700e3
k=4 R1=3.6172e3 R2=14.469e3 R3=7.2343e3

Best regards,
Helmut
 
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