Faculty :
Subject
FACULTY OF ELECTRICAL ENGINEERING
:
Subject Code :
INDUSTRIAL
ELECTRONIC
LABORATORY
SEE 3712/ 3722
Review
:1
Release Date
: 2003
Last Amendment
: 2003
Procedure Number
: PK-UTM-FKE-(0)-10
SEE 3712 /3722
FAKULTY OF ELECTRICAL ENGINEERING
UNIVERSITY OF TECHNOLOGY MALAYSIA
CAMPUS OF SKUDAI
JOHOR
INDUSTRIAL ELECTRONIC LABORATORY
ACTIVE FILTER
ACTIVE FILTERS
OBJECTIVES :
1. To obtain the magnitude responses of active filters.
2. To obtain the phase responses of active filters
3. To observes the characteristics of active filters based on the number of poles and Q’s factor
(i.e. the effect of Q factor on cut-off frequency, the effect of pole on the rate of roll-off and
natural frequency)
BASIC INFORMATION
A low-pass filter transmits low frequencies but stops high ones. Figure 1-1 (a) shows one
way to build a low-pass filter. At very low frequencies the inductive reactance approach 0 and the
capatitive reactance approach infinity. This is equivalent to saying the inductors appear shortcircuited and the capacitors appear open. Therefore, the output voltage equals the input voltage at
very low frequencies.
As the frequency increases, the inductive reactance increase and the capacitive reactance
decrease. At some point, the output voltage starts to decrease. For very high frequencies the,
inductors appear open and the capacitors appear short-circuited; therefore, the output voltage
approaches 0.
Figure 1-1 (b) illustrates how the voltage gain of a low-pass filter varies with frequency.
Ideally, the voltage gain equals unity at lower frequencies. As the frequency increases, the voltage
gain eventually starts to drop off. The cutoff frequency is where the voltage gain equals 0.707
(equivalent to the half-power point).
Figure 1-1 (c) is an example of high-pass filter. In this case, the low frequencies are blocked
and the high frequencies are transmitted. Figure 1-1 (d) shows the graph of voltage gain versus
frequency. Again notice the cutoff frequency, this is where the voltage gain drops to 0.707.
Fig. 1-1 (a) Low-pass filter
Fig. 1-1 (b) Low-pass response
Fig. 1-1 (b) Low-pass response
Fig. 1-1 (c) High-pass filter
Fig. 1-1 (d) High-pass response
Decibels
Voltage gain is defined as the ratio of output voltage to input voltage:
A=Vout / Vin
(1-1)
In the low-pass filter of Fig. 1-1 (a), A equals unity at low frequencies. At the cutoff frequency,
A = 0.707.
Decibels are units commonly used with filters. The decibel voltage gain is defined as
AdB =20 logA
(1-2)
where the logarithm is to the base 10. The abbreviation dB stands for "decibel" (one-tenth of a bel).
Here is an example of calculating decibel voltage gain. If the voltage gain A = 100, then the decibel
voltage gain is :
AdB =20 log 100 = 20 (2) = 40 dB
As another example, if the voltage gain equals 0.707, then
AdB = 20 log 0.707 = 20 (-0.15) = -3 dB
Active Low-Pass Filter
By using op amps and reactive elements, we built active filters. Active filters have several
advantages over passive filters. To begin with, we can eliminate the inductors, which are bulky and
expensive at low cutoff frequencies. Active filters can also have variable voltage gain, allow easy
tuning of cutoff frequency, etc.
Figure 1-2 (a) shows one way to build an active low-pass filter. Here is what it does. At low
frequencies the capacitor appears open, and the circuit acts like an inverting amplifier with a
voltage gain of R2/R1.As the frequency increases, the capacitive reactance decreases, causing the
voltage gain to drop off. As the frequency approaches infinity, the capacitor appears short-circuited
and the voltage gain approaches 0.
Figure 1-2 (b) illustrates the output response. The output signal is maximum at low
frequencies. When the frequency reaches the cutoff frequency, the output is down 3 dB. Well
beyond this frequency, the gain decreases at an ideal rate of 6 dB/octave (a factor of 2 in
frequency). For instance, if the cutoff frequency is 1 kHz, then the gain decreases approximately 6
dB when the frequency increases from 2 to 4 kHz. It will decrease another 6 dB when the frequency
increases from 4 to 8 kHz, and so on.
Fig. 1-2 (a) First-order low-pass filter circuit
Fig. 1-2 (b) First-order low-pass filter response
A decrease of 6 dB/octave is equivalent to 20 dB/decade. If the cutoff frequency is 1 kHz, then
the gain decreases 20 dB when the frequency changes from 10 to 100 kHz. It changes another 20
dB when the frequency increases from 100 kHz to 1 megahertz (MHz).
A mathematical analysis leads to this formula for the cutoff frequency:
The adjustable C of Fig. 1-2 (a) allows us to vary the cutoff frequency, and the adjustable Rl lets us
the control the gain. If a fixed response is desired, we can eliminate the adjustments and use a fixed
Rl and C. Because of the negative feedback, the output impedance approaches 0, which means the
active filter can drive low-impedance loads.
Second-Order Low-Pass Filter
The filter of Fig. 1-2(a) is called a first-order filter because the gain decreases 6dB/octave beyond
the cutoff frequency. A second-order low-pass filter is one that decreases12dB/octave beyond the
cutoff frequency.
Figure 1-3(a) shows a second-order active low-pass filter. At low frequencies both
capacitors appear open, and the circuit becomes a voltage follower. As the frequency increases, the
gain eventually starts to decrease until it is down 3 dB at the cutoff frequency. Because of the two
capacitors, the rate of decrease in gain is twice as fast as before. As a result, the gain drops off at a
rate of 12 dB/octave of 40 dB/decade.
Fig. 1-3 (a) Second-order low-pass filter circuit
Fig. 1-3 (b) Second-order low-pass filter response
Figure 1-3 (b) illustrates the gain versus frequency. First, notice the gain is down 3 dB at the cutoff
frequency, this means the ordinary voltage gain equals 0.707 times the low-frequency value.
Second, notice that the gain rolls-off (decreases) at a rate of 12dB/octave. For instance, suppose, the
cutoff frequency is 1 kHz. Then the gain ideally decreases 12 dB when the frequency changes from
2 to 4 kHz decreases another 12 dB when the frequency changes from 4 to 8 kHz, and so on. Stated
another way, the gain decreases 40 dB when the frequency changes from 2 to 20 kHz, another 40
dB when the frequency changes from 20 to 200 kHz, and so forth.
An advanced mathematical analysis shows that the cutoff frequency is given by,
Second Order High-Pass Filter
Figure 1-4 (a) is a second-order high-pass filter. At low frequencies the capacitors appear open, and
the voltage gain approaches 0. At high frequencies the capacitors appear short-circuited, and the
circuit becomes a voltage follower. Figure 1-4 (b) shows the response. The cutoff frequency is
given by Eq. (1-4).
Fig. 1-4 (a) Second-order high-pass filter circuit
Fig. 1-4 (b) Second-order high-pass filter response
SUMMARY
1.
2.
3.
4.
5.
6.
7.
8.
A low-pass filter transmits low frequencies but stops high ones.
A high-pass filter blocks the low frequencies and passes the high frequencies.
Active filters use op amps and reactive elements.
One advantage of active filters is that they eliminate inductors, which are bulky and
expensive at low frequencies.
Well above the cutoff frequency, a firs t-order active low-pass filter has a voltage gain that
decreases 6 dB/octave. This means the ordinary voltage gain decreases by a factor of 2 for
each doubling of frequency.
A decrease of 6 dB/octave is equivalent to 20 dB/decade.
Well above cutoff, a second-order active low-pass filter has voltage gain that decreases 12
dB/octave, equivalent to 40 dB/decade.
At the cutoff frequency of a first-or-second-order filter, the decibel voltage gain is down
3dB. This means the voltage gain equals 0.707 of the maximum value.
SELF TEST
Check your understanding by answering these questions.
1. At the cutoff frequency the voltage gain equals _______________ of the maximum voltage
gain.
2. In terms of decibels, the gain is down _______________ dB at the cutoff frequency.
3. If A= 10, the decibel voltage gain equals _______________ dB.
4. If AdB = 0 dB, the voltage gain equals _______________ dB.
5. If Rl = 10 kΩ and R2 =100 kΩ in Fig 1-2 (a), the voltage gain is _______________ and the
decibel voltage gain is _______________ dB.
6. If R2 = 47 kΩ and C = 500 pF in Fig. 1-2(a), the cutoff frequency equals _______________
kHz.
7. If the cutoff frequency equals 1 kHz in Fig. 1-3(a), the decibel voltage gains decreases by
_______________ dB when the frequency changes from 10 to 20 kHz.
PRELIMINARY SELF PREPARATION
1.
FIRST-ORDER LOW-PASS FILTER
Refer to Fig. 1-5,
(a)
Prove that the transfer function for the rust-order low-pass filter is
(b)
Show that the magnitude gain
phase angle θ and cut-off frequency fc can be
deduced from equation (1-5.1) to become three expressions as follows :
(c)
Calculate the rate of roll-off in dB/octave and dB/decade by using values of A (dB) at
f1 = 5fo, f2 = l0fo and f3 = 50 fo.
(d)
Plot A (dB) and θ (degree) versus frequency f for 0 f 200kHz and indicate the cut-off
frequency fc and rate of roll-off on the response curves plotted.
2.
SECOND-ORDER LOW-PASS FILTER
(a)
Prove that the transfer function for second-order low-pass filter circuit in Fig. 1-6 is:
(b)
, phase angle θ and cut-off frequency fc can be
Also show that the magnitude gains
deduced from equation (1-6.1) to become three expressions as follows :
(c)
Calculate the rate of roll-off in dB/octave and dB/decade by using values of A(dB)
calculated at f1 = 5fo, f2 = l0fo and f3 = 50 fo
(d)
Plot A (dB) and θ (degree) versus frequency f for 0 f 200kHz and indicate on the
response curves the cut-off frequency and the rate of roll-off.
3.
SECOND-ORDER HIGH-PASS FILTER
(a)
Prove that the transfer function for 2nd order high-pass filter circuit in Fig. 1-7 is:
(b)
(c)
Also show that the following expressions can be deduced from equation (1-7.1)
(c)
Calculate the rate of roll-off of
at f1 = 0.08 fo), f2 = 0.4fo and f3 = 0.8fo
in dB/octave and dB/decade by using values of
(d) Plot
(dB) and θ (degree) versus frequency f for 0 f
response curves the cut-off frequency fc and the rate of roll-off.
200kHz and indicate on the
PROCEDURE
MATERIALS REQUIRED
Power supplies
Equipment
Resistors
Op amp
Capacitors
: Two 12V
: AC generator, oscilloscope
: Two 10 kΩ, 20kΩ, ½ W
: 741 C
: Two 0.01 μF, 0.022 μF
First-Order Low-Pass Filter
Fig. 1-5
1. Connect the circuit shown in Fig. 1-5.
2. Set the ac generator at low frequency (100 Hz. or below).
3. Adjust the signal level of ac generator to get 1V peak to peak at the input of the filter (and
maintain this value throughout the course of experiments).
4. Measure and record the input voltage (peak-to-peak), output voltage (peak-to-peak) and the
relative position of output waveform with respect to input waveform (along the time axis).
5. Work out the voltage gain (Vo/Vi), the equivalent decibel gain (AdB) and the phase angle (θ).
6. Plot the decibel gain, AdB and the phase angle, θ (degree) versus frequency on a semi log
graph paper.
7. Repeat step 4, step 5 and step 6 for other frequencies. Preferably the frequency coverage is
up to 100kHz. Adequate numbers of AdB and θ are required for drawing good curves.
Second-Order Low -Pass Filter
Fig. 1-6
8. Connect the circuit shown in Fig. 1-6.
9. Repeat step 2 to 7.
Second-Order High-Pass Filter
Fig 1-7
10. Connect the circuit shown in Fig 1-7.
11. Set the ac generator at higher frequency (100kHz or higher).
12. Repeat step 3 to 7.
QUESTIONS
1. What is the theoretical voltage gain at 100Hz in Fig. 1-5 and Fig. 1-6 and the
theoretical voltage gain at 100kHz in Fig. 1-7?
2. Explain why the theoretical voltages gain may differ from the measured value.
3. What is the theoretical cutoff frequency in Fig. 1-5, Fig. 1-6 and Fig. 1-7?
Explain why this may differ from the measured cutoff frequency.
4. Well above cutoff in Fig. 1-5 and Fig. 1-6, how fast should the voltage gain
decrease? How much decreases are there between 5 and 10 kHz? Explain why this
may differ from the results obtained from experiment.
5. Well below the cutoff frequency in Fig. 1-7,how fast should the voltage gain
decrease? How much decreases is there between 200Hz and 100Hz? Explain why
this may differ from the results obtained from experiment.