University of Pennsylvania Department of Electrical and Systems Engineering ESE206 Design and characterization of a Band-pass Filter Goals: • •
Design and build an active band-pass filter Measure the frequency response (magnitude and gain), the poles and bandwidth of the filter.
Background Filters are electric circuits that selectively pass signals of certain frequencies. There are several type of filters such as “Low Pass”, “High Pass” and “Band Pass” filters. As the name implies a “Low Pass” filter is a circuit that passes low-frequency signals and blocks high-frequency ones. A “High Pass” filter on the other hand passes high-frequency signals and blocks low-frequency ones. A “Band Pass” filter passes signals whose frequency lies in a certain frequency band. Filters are very important components and are used extensively in electronic and communication systems. For audio applications for instance, a filter can be used to emphasize certain frequencies and de-emphasize others. Or, you may use a filter to block out noise, e.g. a 60 Hz signal. The principle behind filters is quite simple, although the actual implementation can become complicated, depending on the specifications of the filter. Let us discuss a very simple filter, i.e. a first-order low pass filter. We will make use of the fact that the impedance value of capacitors (and inductors) is a function of the frequency: ZC= 1/jωC
(1)
Consider now the following circuit.
Figure 1: First-order low-pass filter
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The output voltage Vo is measured over the capacitor. Since the impedance of the capacitor decreases for higher frequencies, according to the expression (1), the output voltage will decrease as well. This implies that high frequency signals will be blockedout and that only low-frequency signals pass. The actual expression of the output voltage as a function of frequency can be found by using the voltage divider expression: 1 1 1 jωC Vin = Vin = Vin Vo = 1 jωCR + 1 1 + j 2πfRC R+ jω C
(2)
Or the transfer function can be written as, G ( jω ) =
Vo 1 1 1 1 = = = = Vin 1 + jωCR 1 + jωτ 1 + jω / ω b 1 + jf / f b
(3)
in which τ=RC, called the time constant and ωb=1/RC the bandwidth or –3dB point. As one can see, from the above expression (3), the transfer function is equal to 1 for very low frequencies and goes to 0 for very high frequencies (ω >>ωb). Thus, the circuit of Fig. 1 is a low-pass filter. Figure 2 shows the frequency response of the magnitude and phase (Bode plot) of the transfer function. The bandwidth is defined as the frequency at which the gain has decreased by a factor √2 or (or 3dB).
Figure 2: Bode plot of a first order low-pass filter A high-pass filter can be constructed from a resistor and capacitor as shown in Fig. 3. In this case, the signals at very low frequencies will be blocked because the capacitor acts as 2
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an open circuit. For high frequencies on the other hand, the capacitor can be considered a short circuit, passing high frequency signals.
Figure 3: First order high-pass filter A band pass filter can in principle be constructed by combining a low and high pass filter in cascade. Figure 4 shows such a filter. The first part (C1R1) will pass high frequency signals while the second part (C2R2) will pass the low frequencies (or rejects high frequency signals). However, the filter cannot be considered as a simple cascading of a high and low pass filter since the second part loads the first part. As a result, the overall transfer function is not simply the product of the individual transfer functions of the high and low pass sections (see pre-lab).
Figure 4: A passive two-pole band pass filter. The above examples of filters are called passive filters since they do not make use of amplifiers. Among the disadvantages of such a filter is that there is no gain and that the load resistor RL influences the transfer characteristic. A better way to build filters for low to mid-frequency applications is to use operational amplifiers. Such filters are called active filters. The advantage of active filters is that one can provide amplification, and have a filter whose characteristic is independent of the load. The goal of this lab is to build a passive and an active band-pass filter and to measure the transfer characteristic. A simple active band pass filter (two pole system) is shown in Fig. 5 below. The frequency response (Bode graph) of the gain is given in Fig. 6.
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Figure 5: A two-pole active band pass filter.
Figure 6: Frequency response of a bandpass filter.
Pre-lab assignment 1. Passive band pass filter of Fig. 4: a. Derive the general expression of the transfer function (Vo/V1) of the passive band pass filter of Fig. 4. Assume that RL is not present. b. Use the following values for the resistors and capacitors: R1=2kOhm, R2=20kOhm, C1=1E-6F or 1µF (1 microFarad) and C2=0.22nF. Give the expression of the transfer function and plot the Bode diagram of the gain and phase. You can sketch it by hand or use Matlab. Indicate the values of the – 3dB points and the mid frequency gain (i.e. the gain in the passband, where you can assume that the frequency f1<
Kjω (1 +
jω
ω1
)(1 +
jω
ω2
=− )
Kj 2πf jf jf (1 + )(1 + ) f1 f2
Give the expression of K, f1, f2 and the mid-frequency gain (i.e. the gain in the passband, where you can assume that the frequency f1<
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5. Plot the bode diagram of the gain and phase of the band pass filter. You can use Matlab or Excel to plot these graphs. The frequency axis should be logarithmic going from 0.1 to 100 kHz. The gain axis should be in dB. Indicate the values of the -3dB points, the band width and the mid-frequency gain on the graphs. The pre-lab needs to be handed in at the start of the lab session. Make a copy of your prelab for use during the lab (you will need the calculated component values).
In-lab assignments Parts: • • • • •
741 Opamp Resistors and capacitors Power supply Function generator Oscilloscope
Tasks:
1. Build the passive band pass filter of Fig. 4. a. Measure the actual values of the resistors and capacitors b. Give the expression of the transfer function using the actual values of the components. 2. Test the passive filter’s characteristics. a. Apply a 1V peak-to-peak sinusoidal signal to the input and measure the output voltage with the probe of the oscilloscope. Find the mid-frequency (bandpass) gain, –3dB frequencies and the phase at mid-frequency and at the –3dB points. Do this for the following cases of load resistance: no RL, RL=4.7 kOhm and RL= 470 Ohm. Do you notice the influence of the load resistor? Can you explain what happens when you add the load resistor?
RL No load 4.7 kOhm 470 Ohm
Mid Frequency (Passband) Gain Phase
Low Freq. Cutoff (-3dB) f1 Phase
High Freq. Cut-off (-3dB) f2 Phase
3. Build the band pass filter of Fig. 5. Use capacitors and resistor values as close as possible to the one you calculated in the pre-lab (see list of available components). a. Measure the actual values. 5
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b. Based on the actual capacitor and resistor values, recalculate the values of f1, f2, and mid frequency gain. 4. Testing of the filter circuit. The goal is to measure the filter characteristic of your circuit. Apply a sinusoidal frequency of 1V peak-to-peak to the input of the circuit and measure the amplitude and phase of the output voltage, using the probe of the oscilloscope. Do not add a load resistor. Change the frequency between 10Hz and 300 kHz. Change the frequency on a logarithmic scale (10, 20, 50, 100, 200, etc.). Record the values in table form. Determine the values of the –3dB frequencies f1, f2, the mid frequency gain and the phase, and compare these with the calculated ones. Next add a 4.7 kOhm load resistor and check the mid-frequency gain and the – 3dB frequencies f1, f2 (there is no need to measure the complete transfer function) Compare these with the values measured in the case where no load is present and with the values obtained for the passive filter. Created by J. Van der Spiegel; Updated February 23, 2005
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