Let's find the Thévenin and Mayer-Norton equivalent circuits
for Figure 2. The open-circuit voltage and
short-circuit current techniques still work, except we use
impedances and complex amplitudes. The open-circuit voltage
corresponds to the transfer function we have already
found. When we short the terminals, the capacitor no longer
has any effect on the circuit, and the short-circuit current
I
sc
I
sc
equals
V
out
R
V
out
R
.
The equivalent impedance can be found by setting the source to
zero, and finding the impedance using series and parallel
combination rules. In our case, the resistor and capacitor are
in parallel once the voltage source is removed (setting it to
zero amounts to replacing it with a short-circuit). Thus,
Z
eq
=R∥1i2πfC=R1+i2πfRC
Z
eq
∥
R
1
2
f
C
R
1
2
f
R
C
.
Consequently, we have
V
eq
=11+i2πfRC
V
in
V
eq
1
1
2
f
R
C
V
in
I
eq
=1R
V
in
I
eq
1
R
V
in
Z
eq
=R1+i2πfRC
Z
eq
R
1
2
f
R
C
Again, we should check the units of our answer. Note in
particular that
i2πfRC
2
f
R
C
must be dimensionless. Is it?

Comments:"Electrical Engineering Digital Processing Systems in Braille."