RC Filter Networks
Evan Sheridan, Tom Power, Chris Kervick
11367741
April 1st 2013
Abstract
Both high-pass and low-pass RC circuits were setup using a signal generator, oscilloscope,
resistor and capacitor. Analysis showed that the half-power point for the high pass filter occurred
at a frequency f = 1047 ± 100Hz with phase difference 200 ± 50 microseconds corresponding
to what was theoretically predicted. It was found that the frequency of the half-power point of
the low-pass filter to be f = 1106 ± 50Hz and the phase difference of 200 ± 50 microseconds.
Finally, the response of the RC filter circuits to a square wave was analysed which led to the
calculation of the time constant τ to be0.0003876 ± 0.0005s .
1
1
Aims
•To find the half-power point and phase shift for a High-pass filter.
•To find the half-power point and phase shift for a Low-pass filter.
•Analyse the square wave response of filter circuits and determine the time constant of the circuit.
2
Backround and Theory
The High-Pass Filter circuit is given by
In analysing the response of the high-pass filter we define a function H(ω) =
expression determines it’s response
Vout
Vin
and the following
Vout
R
=
1
Vin
R + iωC
using the expression of capacitance and Kirchoff’s loop law. One can see when taking limits why
this is a high-pass filter, i.e the kind of frequencies it lets through.
Using this expression one can theoretically predict the values of the half-power point which occurs
when
Vin
Vout = √
2
And the Low-Pass Filter circuit is given by
Similarly, applying the same analysis that we did for the high-pass filter the expression for the
response is given by
Vout
1
=
Vin
1 + iωRC
1
In taking limits again we find why this is a low pass filter, i.e low frequencies are the only transmitted , as opposed to the high-pass filter.
In general, plotting log-log plots makes modelling certain phenomena a lot easier. By this, if we
have an exponential relationship on a normal scale and transfer it to a log scale we will be plotting
a straight line. Also, reading the half-power point is easier to read off the log-log plot than a normal
arithmetic scale.
In analysing the response of the RC circuit to the square wave we apply Kirchoff’s laws to the the
circuit to obtain the following expression:
dv
R
dt
where V is the voltage input, v(t) is the voltage across the capacitor, C is the capacitance of the
capacitor and R is the resistance of the resistor. Thus integrating we get the following expression:
−t
v(t) = V + (v0 − V ) exp
RC
V = v(t) + C
where v0 is the initial voltage across the capacitor which is zero at time zero.
Vout
−t
= 1 − exp
Vin
RC
Thus if τ = RC we can figure out the time constant of the circuit by
τ=
−t
ln 1 −
Vout
Vin
and determine the time constant both theoretically and experimentally.
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Experimental Method
The High-Pass Filter
Set up the high-pass RC filter circuit as in the diagram. Ensure the connections are earthed. Using
the oscilloscope and signal generator begin with a frequency of 68 Hz and take 20 readings moving
the frequency up in 250 Hz intervals each time. In each interval measure the peak-to-peak value
of both the input and output voltages using the oscilloscope. The respective voltages are given by
half the peak-to-peak values. If Vin is kept constant it simplifies things.
The Low-Pass Filter
Set up the low-pass RC filter circuit as in the diagram. This time measure the half-power point
and the phase shift at this frequency by adjusting the frequency until Vout = V√in2 . Comprehensive
results are not needed here.
The Square Wave Response of RC Filters
Using both the low-pass and high pass filter one applies a square wave from the signal generator
to both circuits. For given values of R and C, use three frequency values such that : T RC ,
T ≈ RC and T RC where f = T1 and R, C are the Resistance and the Capacitance. In each case,
there will be 6, print out the resulting waveforms. Measuring Vout and Vin for each case calculate
τ the time constant of the RC circuit.
2
4
Results and Analysis
For the high-pass filter we plotted log10 ( VVout
) against log10 (f ) which gave the following plot:
in
The half-power point of the filter can be found when :
Vin
Vout = √
2
which corresponds to when :
log10
Vout
Vin
= −0.15
Thus intersecting the curve at this point (given by the blue line) we find that f = 1047 ± 100Hz is
the frequency such that we get the half-power point. Given that we can use the equation and solve
for ω :
Vout
R
1
=
=√
1
Vin
R + ıωC
2
with R = 6.8kΩ and C = 0.022µF we get a theoretical value of the frequency at half-power point
1
to be f = 1063.87Hz using f = 2πRC
. Thus we a have a confirmation of results.
Plotting ∆t against log10 (f ) we get:
3
At the half-power point frequency we have that log10 (f ) = 3.02 so we get a value of 200 ± 50
microseconds for the phase difference.
For the low-pass filter it was observed that as we increased the frequency that the output voltage
decreased. Since the function of the low-pass filter is to not let high frequencies through this is
exactly as we expect.
It was found that the half-power point corresponds to f = 1106 ± 50Hz and the phase difference
at this point was found to be 120 ± 20 microseconds. Again , this corresponds with the theory.
Finally, analysing the response of the high-pass filter and low-pass filter to the square wave for the
frequencies 100, 13368 and 20220 Hz was successful. The 6 waveforms are in the appendix.
Using the equation :
Vout
= 1 − exp
Vin
it is confirmed that
Vout
Vin
−t
τ
= 0.63 when . Using the equation it is found that :
τ=
−t
ln 1 −
Vout
Vin
giving a value of τ to be 0.0003876 ± 0.0005s . Given that RC = 0.0004864s we are almost within
error but the results aren’t concrete.
The High-Pass Filter is interpreted as an integrating circuit if we consider the following argument
s
1 2
2
Vin = R +
ωC
but since we are dealing with the low-pass circuit we have that ωC expression for Vout is given by
4
1
R
so Vin ≈ IR. The
Vout
q
1
=
≈
C
C
Z
Vin
dt
R
and thus
Vout
1
≈
RC
Z
Vin dt
hence the reason that the High-Pass filter is known as an integrating circuit, as a result of this
expression.
Considering a similar argument for the Low-Pass filter we have that R 1
we have that Vin ≈ VC . Thus
For ω RC
Vout = iR = R
1
ωC
and thus Vin ≈
I
ωC
.
dq
d
= RC VC
dt
dt
and therefore
d
Vin
dt
hence the reason the Low-Pass filter is known as a differentiating circuit.
Vout ≈ RC
Finally, if we consider a square wave with period of
Series we get
1
f
we find that if we represent it as a Fourier
∞
2 X sin(2πnxf )
πf
n
nodd
thus we have a summation of an infinite number of terms of different frequencies. Since the low-pass
filter let’s low frequencies through and the high pass let’s only high frequencies through and what
terms in the Fourier decomposition are transmitted will determine what filter we are dealing with.
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Conclusions
•Both the operation of the high-pass and low-pass filters was successfully confirmed showing that
the high-pass filter only let’s high-frequencies pass and the low-pass filter letting only low frequencies
through. The frequency at half power point of the high-pass filter was found to be f = 1047±100Hz
with phase difference 200 ± 50 microseconds and was found from graphical analysis. From using the
apparatus the frequency at half-power point of the low-pass filter was found to be f = 1106 ± 50Hz
and the phase difference of 200±50 microseconds. Corresponding to what we predicted theoretically.
•Using a signal generator we calculated the time constant τ to be0.0003876 ± 0.0005s . Since
RC = 0.0004864s and τ ≈ RC we have a suggestion that this relation could be true but the result
is not within experimental error. Perhaps measuring the response of the circuit more accurately
would lead to the concrete verification of this statement.
•Further mathematical analysis of the action of the respective filters revealed why it is that the
Low-Pass filter is known as a differentiating circuit and the High-Pass filter an integrating circuit.
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6
Appendix
High-Pass
6
Low-Pass
7
8