If the source consists of two (or more) signals,
we know from linear system theory that the output voltage equals
the sum of the outputs produced by each signal alone. In short,
linear circuits are a special case of linear systems, and
therefore superposition applies. In particular, suppose these
component signals are complex exponentials, each of which has a
frequency different from the others. The transfer function
portrays how the circuit affects the amplitude and phase of each
component, allowing us to understand how the circuit works on a
complicated signal. Those components having a frequency less
than the cutoff frequency pass through the circuit with little
modification while those having higher frequencies are
suppressed. The circuit is said to act as a filter,
filtering the source signal based on the frequency of each
component complex exponential. Because low frequencies pass
through the filter, we call it a lowpass filter to
express more precisely its function.
We have also found the ease of calculating the output for
sinusoidal inputs through the use of the transfer function. Once
we find the transfer function, we can write the output directly
as indicated by
the output
of a circuit for a sinusoidal input.
Let's apply these results to a final example, in which the
input is a voltage source and the output is the inductor
current. The source voltage equals
V
in
=2cos2π60t+3
V
in
2
2
60
t
3
.
We want the circuit to pass constant (offset) voltage
essentially unaltered (save for the fact that the output is a
current rather than a voltage) and remove the 60 Hz term.
Because the input is the sum of two
sinusoids--a constant is a zero-frequency cosine--our approach
is
- find the transfer function using impedances;
- use it to find the output due to each input component;
- add the results;
- find element values that accomplish our design criteria.
Because the circuit is a series combination of elements, let's
use voltage divider to find the transfer function between
V
in
V
in
and
V
V,
then use the
v-i relation of the inductor
to find its current.
I
out
V
in
=i2πfLR+i2πfL1i2πfL=1i2πfL+R=Hf
I
out
V
in
2
f
L
R
2
f
L
1
2
f
L
1
2
f
L
R
H
f
(1)
where
voltage divider=i2πfLR+i2πfL
voltage divider
2
f
L
R
2
f
L
and
inductor admittance=1i2πfL
inductor admittance
1
2
f
L
[Do the units check?] The form of this transfer function
should be familiar; it is a lowpass filter, and it will
perform our desired function once we choose element values
properly.
The constant term is easiest to handle. The output is given by
3|H0|=3R
3
H
0
3
R
.
Thus, the value we choose for the resistance will determine
the scaling factor of how voltage is converted into current.
For the 60 Hz component signal, the output current is
2|H60|cos2π60t+∠H60
2
H
60
2
60
t
H
60
.
The total output due to our source is
i
out
=2|H60|cos2π60t+∠H60+3×H0
i
out
2
H
60
2
60
t
H
60
3
H
0
(2)
The cutoff frequency for this filter occurs when the real and
imaginary parts of the transfer function's denominator equal
each other. Thus,
2π
f
c
L=R
2
f
c
L
R
,
which gives
f
c
=R2πL
f
c
R
2
L
.
We want this cutoff frequency to be much less than 60 Hz.
Suppose we place it at, say, 10 Hz. This specification would
require the component values to be related by
RL=20π=62.8
R
L
20
62.8
.
The transfer function at 60 Hz would be
|1i2π60L+R|=1R|16i+1|=1R137≃0.16×1R
1
2
60
L
R
1
R
1
6
1
1
R
1
37
0.16
1
R
(3)
which yields an attenuation (relative to the gain at zero
frequency) of about
1/6
16,
and result in an output amplitude of
0.3R
0.3
R
relative to the constant term's amplitude of
3R
3
R
.
A factor of 10 relative size between the two components seems
reasonable. Having a 100 mH inductor would require a 6.28
Ω resistor. An easily available resistor value is 6.8
Ω; thus, this choice results in cheaply and easily
purchased parts. To make the resistance bigger would require a
proportionally larger inductor. Unfortunately, even a 1 H
inductor is physically large; consequently low cutoff
frequencies require small-valued resistors and large-valued
inductors. The choice made here represents only one
compromise.
The phase of the 60 Hz component will very nearly be
−π2
2
,
leaving it to be
0.3Rcos2π60t−π2=0.3Rsin2π60t
0.3
R
2
60
t
2
0.3
R
2
60
t
.
The waveforms for the input and output are shown in
Figure 2.
Note that the sinusoid's phase has indeed shifted; the lowpass
filter not only reduced the 60 Hz signal's amplitude, but also
shifted its phase by 90°.
"Electrical Engineering Digital Processing Systems in Braille."