T
The Phantom
- Jan 1, 1970
- 0
A modern active filter building block is the GIC (generalized
immittance converter). Descriptions can be found in many textbooks
with expressions for the relevant properties, but these usually assume
the amplifier gains are infinite.
For the circuit shown, find the input impedance for finite amplifier
gains. There are two amplifiers of gain A1 and A2, as shown. There
are five impedances, Z1 through Z5. Assume the nodes are numbered 1
through 5, with node 1 being the input, node 2 between Z1 and Z2, node
3 between Z2 and Z3, etc.
Obtain a final result in the form of a fraction with the numerator
consisting of a sum of products with no fractions, and the denominator
likewise. Leave the numerator and denominator unfactored so that
comparison with other people's results is facilitated. The numerator
and denominator should look *something* like this:
Z1 Z2 Z3 A1 + Z5 A1 A2 + Z2 Z4 Z5 A2 + Z3 Z4 Z5 A1 + ...
---------------------------------------------------------------------
Z2 Z4 Z5 A2 + Z1 Z2 Z3 Z4 A1 A2 + Z2 Z3 Z4 + Z1 Z5 + A1 Z1 Z2 +...
Since you have the general expression with finite amplifier gains, you
can then substitute an expression for amplifier gain with rolloff and
see what the effect of amplifier imperfection is.
Or, let the amplifier gains go to infinity, and find the limiting
input impedance for that case.
You might use a different method to solve this problem if you have
access to a computer algebra system (CAS) such as Mathcad, than if you
do it all by hand. That would be an interesting topic for discussion:
how best to solve this kind of problem given modern computer
mathematical assistants.
Vin ----o-----------.
| |
.-. |
Z1 | | |
| | |
'-' .-----. |
.---------| | | |
| | | | |
| .-. | .-----.
| Z2 | | | \- +/
| | | | \ / A1
| '-' | V
A2 /^\ .---|---' |
/+ -\ | | |
'-----' | .-. |
| | | | | Z3 |
| | | | | |
| '----' '-' |
| |----------'
| |
| .-.
| Z4 | |
| | |
| '-'
'----------|
|
.-.
Z5 | |
| |
'-'
|
GND ---
-
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de
immittance converter). Descriptions can be found in many textbooks
with expressions for the relevant properties, but these usually assume
the amplifier gains are infinite.
For the circuit shown, find the input impedance for finite amplifier
gains. There are two amplifiers of gain A1 and A2, as shown. There
are five impedances, Z1 through Z5. Assume the nodes are numbered 1
through 5, with node 1 being the input, node 2 between Z1 and Z2, node
3 between Z2 and Z3, etc.
Obtain a final result in the form of a fraction with the numerator
consisting of a sum of products with no fractions, and the denominator
likewise. Leave the numerator and denominator unfactored so that
comparison with other people's results is facilitated. The numerator
and denominator should look *something* like this:
Z1 Z2 Z3 A1 + Z5 A1 A2 + Z2 Z4 Z5 A2 + Z3 Z4 Z5 A1 + ...
---------------------------------------------------------------------
Z2 Z4 Z5 A2 + Z1 Z2 Z3 Z4 A1 A2 + Z2 Z3 Z4 + Z1 Z5 + A1 Z1 Z2 +...
Since you have the general expression with finite amplifier gains, you
can then substitute an expression for amplifier gain with rolloff and
see what the effect of amplifier imperfection is.
Or, let the amplifier gains go to infinity, and find the limiting
input impedance for that case.
You might use a different method to solve this problem if you have
access to a computer algebra system (CAS) such as Mathcad, than if you
do it all by hand. That would be an interesting topic for discussion:
how best to solve this kind of problem given modern computer
mathematical assistants.
Vin ----o-----------.
| |
.-. |
Z1 | | |
| | |
'-' .-----. |
.---------| | | |
| | | | |
| .-. | .-----.
| Z2 | | | \- +/
| | | | \ / A1
| '-' | V
A2 /^\ .---|---' |
/+ -\ | | |
'-----' | .-. |
| | | | | Z3 |
| | | | | |
| '----' '-' |
| |----------'
| |
| .-.
| Z4 | |
| | |
| '-'
'----------|
|
.-.
Z5 | |
| |
'-'
|
GND ---
-
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de