Single Time-Constant Networks: Passive Filters
The most basic type of electronic signal filter is the single time-constant network
(STCN). The simplest implementation of a STCN is formed using one resistor and one
capacitor. Depending on the relative orientation of the resistor and capacitor, the STCN
can operate as either a high-pass (HP) or low-pass (LP) filter. Hopefully, you are all
familiar with these circuits. Figures 1(a) and 1(b) show the “topology” of high-pass and
low-pass filters respectively.
vi
vi
vo
vo
R
C
C
R
Figure 1(a).
Figure 1(b).
In order to calculate the “behavior” of either of these filters, we will express the
input and output voltages as complex quantities:
v~i  complex input voltage
v~o  complex output voltage
and we will express both the capacitance and resistance in terms of complex impedances:
~
Z R R
~
ZC 
Eq. 1(a) and 1(b)
1
jC
We can now re-draw figures 1(a) and 1(b) using the concept of complex
impedance.
vi
vo
ZC
ZR
Figure 2(a).
vi
vo
ZR
ZC
Figure 2(b).
By looking at figures 2(a) and 2(b) we see that the two components form a
voltage divider. We can calculate the output voltage as a function of the input voltage
using a simple voltage divider equation, provided we allow the voltages and impedances
to be complex. The resulting equations are:
~
ZC
~
~
LP  vo  vi ~
~
ZC  Z R
~
ZR
~
~
HP  vo  vi ~
~
Z R  ZC
Eq. 2(a) and 2(b)
At this point, it is helpful to define a “transfer function”. The transfer function is
analogous to “gain”, except that the transfer function is complex, and therefore includes
information about both amplitude and phase. We normally define gain as “voltage out
over voltage in”. The transfer function is therefore defined as ratio of complex output
voltage to complex input voltage:
v~
~
~
T s   T  j   ~o
vi
Eq. 3
We can now use equations 1(b) and 2(b) to write out the transfer function for a
low pass filter explicitly.
1
~
v~o
ZC
jC
~
TLP  j   ~  ~
~  1
vi Z C  Z R
R
jC
Eq. 4(b)
Equation 4(b) can be rearranged into a more “pleasant” form by multiplying both
numerator and denominator by jωC. Similar algebraic rearranging allows equation 2(b)
to be re-written in a similar form. The resulting equations are:
~
THP 
1
~
TLP 

1   1
 jRC 
1
1  jRC
Eq. 5(a) and 5(b)
In order to simplify equations 5(a) and 5(b), let us define:
o 
1
RC
Eq. 6
Notice that the units of resistance times capacitance (Ohms X Farads) are
seconds, and that sec-1 are units of angular frequency. Equations 5(a) and 5(b) are now
re-written as:
~
THP 
1
 
1  j o 
 
~
TLP 
1

1  j
 o
Eq. 7(a) and 7(b)



Before we continue, perhaps we should stop and consider what these equations
mean. First of all, ωo is a “characteristic frequency”. It tells us at what frequency the
filter “transitions” from “pass” to “reject”. Secondly, consider that the transfer function
is complex. In some sense, it has no meaning in the real world. It relates a complex
output voltage to a complex input voltage, but there is no such thing as a “complex
oscilloscope” or a “complex multimeter”. The complex transfer function itself tells us
nothing about the performance of the circuit in the real world; however, if we calculate
the magnitude and/or phase of the transfer function, these are both “pure real” quantities,
and therefore can be measured with an oscilloscope of multimeter.
The magnitude of the complex transfer function is what I generally refer to as the
“transmission” of the filter (although I believe that “magnitude response” is actually the
correct term). This quantity is analogous to gain: it tells us how much or how little of a
harmonic signal gets through the filter as a function of the signals frequency. We all
(hopefully) know how to find the magnitude of a complex quantity. We will multiply by
the complex conjugate and take the square root. For the low pass we get:



1
1
~
~ ~
TLP    TLP  j   TLP TLP*   

 
 1  j   o   1  j   o   
1
1
2
TLP    1    o  2 
1   2  o2



1
2
 1  j   o 1  j   o 

Similar algebraic manipulations applied to equation 7(a) yield:

THP    1  o  
2

1
2


1
1  o2  2

Presumably, we all know how to find the phase of a complex quantity. We must
rearrange the complex quantity so that it can be expressed in two parts; one pure real and
one pure imaginary. Then we calculate the arctangent of the imaginary divided by the
real. For the low pass filter, we multiply the numerator and denominator of the transfer
function by the complex conjugate of the denominator.
~
TLP  j  
   o  
1  j   o   1  j   o 
1


j



2
2 
1  j   o  1  j   o   1   2  o2
1   2  o2
1    o 
1
 

 
~
~
~
TLP  j   Re TLP  j Im TLP





1
2