Example 4

Simple example - Input impedance of gyrator loaded with capacitance

###.. image:: gyrator.*

Find symbolic expression of input impedance of gyrator loaded with gounded capacitor:

import numpy, pylab
from pycircuit.circuit import *
from pycircuit.post.functions import *
from sympy import Symbol, simplify, ratsimp, sympify, factor, limit, solve, pprint, fraction, collect

C1, gm1 = [Symbol(symname, real=True , positive=True ) for symname in 'C1,gm1'.split(',')]
s = Symbol('s', complex = True)

## Create circuit object
cir = SubCircuit( toolkit=symbolic)

## Add nodes to circuit
n1, n2 = cir.add_nodes('1', '2')

## Add circuit elements
cir['cap']  = C(n2, gnd, c = C1)
# Gyrator
cir['Gyrator'] = Gyrator(n1, gnd, n2, gnd, gm = gm1, toolkit=symbolic)
# Current source for AC stimuli
cir['ISource'] = IS(gnd,n1, iac=1)

## Run symbolic AC analysis
ac = AC(cir)
result = ac.solve(freqs=s, complexfreq=True)

# Input impedance
impedance = simplify(result.v(n1, gnd))
impedance

\[\frac{C_{1} s}{gm_{1}^{2}}\]

ABCD matrix:

# Delet current source used to calculate input impedance

del cir['ISource']
del cir['cap']

## Run symbolic 2-port analysis
twoport_ana = TwoPortAnalysis(cir, Node('1'), gnd, Node('2'), gnd)
result = twoport_ana.solve(freqs=s, complexfreq=True)

## Print ABCD parameter matrix
ABCD = Matrix(result['twoport'].A)
ABCD.simplify()
ABCD

\[\begin{split}\left[\begin{matrix}0 & \frac{1}{gm_{1}}\\gm_{1} & 0\end{matrix}\right]\end{split}\]

G matrix:

cir.G(np.zeros(cir.n))

[[0 gm1 -gm1]

[-gm1 0 gm1] [gm1 -gm1 0]]

C matrix:

cir.C(np.zeros(cir.n))

[[0 0 0]

[0 0 0] [0 0 0]]

G matrix again:

del cir['Gyrator']
n3, n4 = cir.add_nodes('3', '4')
cir['Gyrator'] = Gyrator(n1, n2, n3, n4, gm = gm1, toolkit=symbolic)
cir.G(np.zeros(cir.n))

[[0 0 -gm1 gm1]

[0 0 gm1 -gm1] [gm1 -gm1 0 0] [-gm1 gm1 0 0]]