Passive filters
Last week we began to look at passive filters, specifically a low pass filter. Parts of that
lab are listed below, if you haven’t finished that lab don’t worry we repeat the last part of
it again today. (For the background theory sections from last week refer to this week’s
appendix).
Bode plots
Decibels and logarithmic scales
To transfer function of a two port system was introduced last week, it was given as the
ratio
Vout
,
(1)
Vin
where is it understood that Vout and Vin are in their phasor representation. If we recall
from last week that the transfer function for a simple RC circuit is given by
1
(2)
H( f ) 
1  j 2fRC
It is relatively straight forward procedure to calculate the magnitude of the transfer
function.
H( f ) 
Question (1) Calculate the magnitude of the transfer function in Eq. 2.
Phase? Pg 288.
The definition of a decibel is given as
|H(f)|dB=20 log10 |H(f)| .
(3)
The ability of a filter to strongly reject a band of frequencies is its primary purpose. By
plotting the transfer function (which is a function of frequency) using a decibel scale it
makes it possible to display very large and very small magnitudes easily on a single plot.
By using log magnitude scales and log frequency scales we build up plot known as a
bode plot (see Fig. 1). These are useful because they allow large range of measurements
to be clearly displayed on a single plot, making a scientists or engineers task or
measurement recording less mundane.
10fb
0
100fb
f
-20
-40
|H(f)|dB
Figure 1. Sample bode plot for a simple
low pass filer
This also allows us to easily calculate the transfer function of a cascaded two port system
easily using log laws.
Consider the following scheme,
H1(f)
H2(f)
Vout1=Vin2
Vin=Vin1
Vout2=Vout
Figure 2. Cascaded 2 port system
we notice that the output of the 1st block is the input to the 2nd block. This implies that if
we wish to write the transfer function of the whole system we can write,
V
V
(4)
H ( f )  out  out 2 .
Vin
Vin1
We pull the little trick that we can multiply and divide by Vout1 (i.e. unity) we find,
H( f ) 
Vout1 Vout 2
.

Vin1 Vout1
Or more succinctly
H ( f )  H1 ( f )  H 2 ( f ) .
Question 2. Comment on the significance of Eq. 6.
(5)
(6)
Laboratory:
For this part of the laboratory we shall build and evaluate the frequency response of a RC
circuit, in effect we shall construct a low pass filter. Consider the circuit in Fig. 3.
Vin
R1
+
C1
Vout
Figure
3:
Circuit
diagram for the low pass
filter.
Task 1
Using a 2.7nF capacitor and 47 kΩ resistor, measure the voltage drop across the capacitor
as a function of frequency (i.e. measure Vout). Predict using your solution to Question 1
and Eq. 3 what the -3dB point, or Half power point (also known as the "corner
frequency") is. Construct a bode plot (by plotting (log) amplitude vs (log) frequency. Set
the signal generator output amplitude to 10V peak to peak. Use Log-log scales. Measure
amplitude at 10Hz, 20Hz, 50Hz, 100Hz,...., 100kHz. ) and comment how this compares
with your measured result?
Task 2
Now we consider a slight variation of the circuit in Fig. 3. Consider the circuit in Fig. 4,
using your knowledge of capacitors and the low pass filter of task 1, predict the bode plot
shape. Using the same 2.7 nF capacitor and 47 kΩ resistor build and measure Vout as a
function of frequency and hence construct a bode plot.
Vin
C1
R1
Vout
Figure
3:
Circuit
diagram for task 2.
Task 3
Design task,
Construct a band-pass filter using the previous two filters. Measure and plot a bode plot.
Comment on the bandwidth (HINT: the difference between the half power frequencies of
the cascaded system).
Inductors
Inductors can be viewed as a kind of anti-capacitor, in that while a capacitor opposes
changes in voltage, an inductor opposes changes in current. The voltage-current
relationship for an inductor is
di
.
(7)
dt
The total inductance of an inductor network is calculated in the same fashion as a
resistor network.
v(t )  L
The impedance of an inductor is
Z=Ljω.
(8)
L1
Vin
R1
Vout
Figure
4:
Circuit
diagram for task 4.
C1
Task 4
Consider the alternate band pass filter configuration shown in Fig 4.
Question 3 Using the impedance of an inductor as given in Eq. 8 calculate
the total impedance of the circuit in Fig 4. From this we can calculate the
total phasor current through the circuit, using this obtain the voltage drop
across the resistor and hence the transfer function.
The resonant frequency is defined to be the frequency at which the impedance of the
RLC circuit is purely resistive. Thus show that.
f0 
1
2 LC
Task 5
Build the circuit and obtain the bode plot. Comment on any special features.
For today’s lab we shall investigate the capacitors behavior when the placed in an AC
circuit.
Background theory.
We begin with the equation for the charge stored on a capacitor (see Fig. 2.1),
Q(t)=CV(t),
where V(t) is the time varying voltage applied.
2.1
Figure 2.1: Current i(t)
+
flowing through the
V(t) capacitor due to the
applied voltage V(t).
i(t)
The current through the capacitor is just the change in the amount of charge stored with
respect to time,
dQ(t )
i (t ) 
.
2.2
dt
Hence the current through the capacitor is,
dV (t )
i (t )  C
.
2.3
dt
Let us now consider the case where V(t)=Vs cos(ωt+φ) such that Vs is the peak value of
the voltage wave, ω is the angular frequency of the wave in radians and φ is the phase
angle. As the cosine function completes one cycle when the angle increases by 2π we
have
and
f=1/T.
2.4
T  2
V (t)
i(t)
+
-
Figure 2.2: Current i(t)
flowing through the
capacitor due to the
applied voltage V(t).
Evaluating the current through the capacitor using Eq’s. 2.3 and 2.4 we have,
i(t )  CVs (t ) sin( t   )
We note that the current lags the voltage by 90º hence we can recast Eq. 2.5 as
i(t )  CVs (t ) cos(t    90)
Using V=IZ we evaluate the impedance of the capacitor as,
2.5
2.6
Z
cos(t   )
cos(t   )

 C sin( t   )  C cos(t    90)
2.6
Using phasors (don’t worry if you haven’t seen this yet as it will come up in class) we
find that the impedance of a capacitor is,
Z
1
.
jC
2.7
The significance of this equation is that it tells us that the capacitor has frequency
dependent impedance at low frequencies (i.e. DC current) the impedance of the capacitor
is infinitely large.
Laboratory:
For this part of the laboratory we shall build and evaluate the frequency response of a RC
circuit, in effect we shall construct a low pass filter. Consider the circuit in Fig. 2.3
Vin
R1
+
C1
Vout
Figure 2.3: Circuit
diagram for the low pass
filter.
Task 1
The transfer function for a filter is given as a ratio of Vout/Vin, calculate the transfer
function of the filter above. (Hint: The impedance of the capacitor can be though of as the
resistance of the capacitor. Knowing this use KVL.)
Once you find the transfer function we can calculate at what frequency the output voltage
will drop to half the input value. Using this fact prove that this frequency is,
1
fb 
.
2.8
2 RC
Task 2
Setup two 47kΩ resistors in series with the signal generator and measure voltage drop
across one of the resistors as a function of frequency. This is known as a Bode plot.
This is constructed by plotting (log) amplitude vs (log) frequency. Set the signal
generator output amplitude to 10V peak to peak. Use Log-log scales. Measure
amplitude at 10Hz, 20Hz, 50Hz, 100Hz,...., 100kHz. Draw a smooth curve
through the measured points.
Task 3
Change one resistor to a 2.7nF capacitor and measure the voltage drop across the
capacitor as a function of frequency (i.e. measure Vout). Predict using Eq 2.8 what the 3dB point, or Half power point also known as the "corner frequency" is and how does
this compare with your measured result?
Task 4 (optional)
Plot the R-C circuit phase vs frequency over the same range 10Hz to 100kHz.
Setup the CRO to plot input Amplitude on the X axis and Output amplitude on the Y
axis. What angle on the CRO is represented as 0 degrees? Find the frequency where the
phase is 15 degrees? 30 degrees? 45 degrees? 60 degrees? 75 degrees? Plot a smooth line
through the measured points.