If a stable system has transfer function H(s), then its frequency response is H(jw). That is, the frequency response is obtained by evaluating H(s) along the vertical axis in the complex s-plane. The following mesh plot shows the magnitude |H(s)| over the complex s-plane for a 2'nd order LPF with resonant frequency f0 = 1 kHz and zeta = 1/2. The Matlab program that was used as the basis for this and the other 3-D plots in this section is available here.
The red curve shown in front is the magnitude of the frequency response |H(jw)|, where w = 2*pi*f. Note that the plot shows both positive and negative frequencies. The two poles in the left half of the s-plane, which have been clipped at the top to obtain a reasonably scaled graph, are the clearly visible "driving force" behind the passband at low frequencies.
If one of the two zeros at infinity is moved to s = 0, then the 2'nd order LPF becomes a 2'nd order BPF as shown in the next plot.
If zeta is made smaller and the poles move closer to the imaginary axis, the peaks in the magnitude of the frequency response become more and more pointed and narrow, resulting in a BPF with a higher Q factor. The graph below shows |H(s)| over the s-plane for a BPF with f0 = 1 kHz and zeta = 0.05.
If the second zero of the LPF is also moved from infinity to s = 0, then the result is a 2'nd order HPF with the following magnitude of frequency response for f0 = 1 kHz and zeta = 1/2.
Note that in all three of the above cases it is the numerator of H(s) that determines which type of filter is obtained. The denominator, however, determines the resonant frequency and, through zeta, how much of a peak |H(jw)| has near w0.
More filter types can be obtained by linear combinations from an LPF, BPF, and HPF with the same resonant frequency w0 and the same damping factor zeta. The sum of a 2'nd order LPF and a 2'nd order HPF with equal passband gains results in a BSF which has zeros at +jw0 and at -jw0. The next graph shows |H(s)| and |H(jw)| (red line) for this case.
A 2'nd order filter of a given type (LPF, BPF, or HPF) whose main characteristics (passband, transition from passband to stopband, stopband attenuation) are determined only by w0 and zeta (i.e., by the location of the poles) is called an all-pole filter. Moving the zeros of H(s) to ±jwz for wz different from 0 or infinity (or w0) yields so-called elliptic or Cauer filters. If wz > w0, then the elliptic filter is a LPF as shown in the next mesh plot for a 2'nd order filter with wz = 1.73*w0.
When the magnitude of the frequency response of this filter is compared with an all-pole LPF, then it can be seen that the elliptic LPF has a steeper transition from the passband to the stopband, at the expense of less stopband attenuation as w >> w0. This is shown in the following figure where the red curve is |H(jw)| for an elliptic LPF and the dashed purple curve is |H(jw)| for an all-pole LPF, both with the same w0 and zeta.
The next figure shows the case of a 2'nd order elliptic HPF which is obtained if wz < w0. The example shown has wz = 0.577*w0.
As was the case for the LPF, a steeper transition from passband to stopband is obtained, but at the cost of less overall attenuation in the stopband, especially near dc, because now the two available zeros are at ±jwz rather than at zero.
Another interesting case of a 2'nd order filter is obtained by adding the outputs of a LPF and a HPF and subtracting the output from a BPF, all of 2'nd order and using the same w0 and the same zeta. This yields a 2'nd order allpass filter with |H(s)| as shown below.
The name "allpass" derives from the fact that all frequencies are passed with equal gain (but different phase). This is achieved by placing zeros in the right half of the s-plane at the mirror locations of the poles. In this way |H(jw)| = 1 as can be clearly seen in the above figure from the red line above the imginary axis of the s-plane.
E1. Build and Test SVF filter. Build the following 4 OpAmp state-variable filter on your breadboard. You will use it in all of the other experiments of this lab.
Let R = 100 kohm, Rw = 15 kohm, C = 10 nF, R' = 100 kohm, R" = 10 kohm and Rz = 68 kohm. Don't forget to use decoupling capacitors (of about \span class="math">0.1 uF) between the power supply pins and ground. Verify the basic operation of your SVF by applying a sinusoidal input of about 100...200 mVpp at frequencies in the range of about 200 to 5000 Hz and observing the LP, BP and HP outputs. How would you verify that you indeed have second order LP, BP, and HP outputs? Hint: Consider measuring the magnitude and the phase of the frequency response and using the Matlab script f_resp14 to display the result.
E2. Measure zeta, listen to step response. (a) Decreasing Rz in the SVF circuit decreases the value of the damping ratio zeta. Using a range of about 68 ohms to 200 kohms, what values of zeta can be obtained in the SVF circuit? How are you going to measure the actual value of zeta? Should the method be different for small and large values of zeta? Hint: Look at the step and frequency response formulas given in the introduction.
Note: The magnitude of the frequency response of a unity gain LPF or HPF is 1/(2*zeta) at w0. Thus, if zeta = 0.01, for example, then the input voltage vS(t) is amplified by a factor of 50 for a sinusoid at or near w0. In order not to drive the OpAmps into saturation, it is therefore necessary in this case to use an amplitude of about 0.5 Vpp or less at the input. As zeta gets smaller, the input amplitude will have to be reduced even more.
(b) The wav files signal01.wav and signal02.wav were generated using a waveform generator and the SVF filter shown in E1. Determine the settings of the waveform generator, the parameter values in the SVF circuit, and the output terminal of the SVF from which the recordings were made as accurately as possible. One feature of the SVF was changed during the second recording. Which one? Hint: Listen to the signals and plot them, e.g., in Matlab. Try to reproduce signal01.wav as accurately as possible with your SVF. How would you characterize this signal and what could it be used for?
E3. Bandstop Filter, Spectral Decomposition. (a) Use an additional OpAmp and modify your SVF circuit such that you can also obtain a BS (bandstop) output, corrsponding to the transfer function
in addition to the LP, BP, and HP outputs. Hint: Use the - vLP(t) and the - vHP(t) outputs to make a + vBS(t) output. Verify the correct operation of the BS output.
(b) When you look at the sum of the bandpass and bandstop transfer functions, you see that they add up to one:
Thus by looking at the vBP(t) and the vBS(t) signals simultaneously, the input signal vS(t) of the SVF tuned to frequency w0 can be decomposed into two signal components, one which consists of A(w0) cos(w0t + phi) at the BP output and one which consists of vS(t) - A(w0) cos(w0t + phi) at the BS output.
All periodic signals like sinusoids or a rectangular or triangular test signal generated by the waveform generator have a fundamental frequency f1 = 1/T, where T is the period of the signal. Except for the sine (or cosine), which is a "pure" tone, all other periodic signals also have harmonics at (some of the) integer multiples of f1. The frequency fk = k f1, where k > 0 is called the k-th harmonic. A dc-free rectangular signal with 50% duty cycle and period T, for instance, has frequency components at f1 = 1/T, f3 = 3/T, f5 = 5/T, f7 = 7/T, etc.
Use the waveform generator to produce a dc-free 50% duty cycle rectangular waveform at the undamped resonant frequency w0 of your SVF. Use a damping factor of about zeta = 0.1 for your SVF and record the BP output and the BS output using the soundcard probe and import the two signals into Matlab. Plot the two signals individually and also plot their sum. Check that the sum is equal to the rectangular signal. The BP output shows you the amplitude of the fundamental frequency component. Compare it to the amplitude of the rectangular signal. The BS output shows you the rectangular waveform without the fundamental frequency component. An example graph for a triangular (instead of a rectangular) waveform is shown below.
(c) Repeat the measurement in (b) for the 3-rd harmonic of the rectangular waveform. Ideally, you would leave the setting of the waveform generator the same as in (b) and tune your SVF to a new resonant frequency of 3 w0 for this. However, since it is easier to change the frequency of the waveform generator, leave the SVF at w0 and change the frequency of the rectangular signal to w0/3. Measure the amplitude of the third harmonic compared to the amplitude of the rectangular pulse. Then compare the amplitude of the fundamental frequency (from (b)) with the amplitude of the 3'rd harmonic. What is their ratio?
E4. Elliptic LPF/HPF. An elliptic or Cauer filter (in honor of network theorist Wilhelm Cauer) has zeros at ±wz in addition to the poles that an all-pole filter has. As shown in this plot in the introduction, the advantage of an elliptic filter is that it can have a steeper transition from passband to stopband than an all-pole filter. However, the attenuation at frequencies w >> w0 is better for an all-pole LPF, thus indicating a tradeoff between steepness in the transition band and attenuation in the stopband.
(a) Use the adder circuit shown below to obtain vEL(t) = -(vLP(t) + k*vHP(t)) corresponding to the following system function of an elliptic LPF with passband gain of 1:
Determine k for the two cases when (i) wz = 2*w0 and (ii) wz = 6*w0. Adjust R1, R2, and R3 such that these values of k are obtained and the passband gain (for w << w0) is one.
Set zeta = 1/2 and use the soundcard probes to measure the frequency response and verify that the zeros occur at the right frequencies. Determine the -3 dB (or half power) frequency, the frequency where an attenuation of -20 dB is reached for the first time, and the attenuation for w >> w0 (e.g., at about 10*w0). Compare this to the magnitude of the frequency response of a 2'nd order all-pole LPF with the same w0 and zeta.
(b) Repeat (a) for the two cases (i) wz = (1/2)*w0 and (ii) wz = (1/6)*w0. In this case compute vEL(t) = -(vHP(t) + k*vLP(t)) with the adder circuit. Looking at the frequency response that you recorded with the soundcard probe, what kind of filter is this?
E5. Allpass Filter. Combine -vLP(t), +vBP(t), and -vHP(t) using the following adder circuit, e.g., with R = 10 kohm,
to obtain an output vAP(t) from your SVF that corresponds to the allpass filter system function:
(a) Use the same component values as in E1, but adjust Rz for a damping ratio of approximately zeta = 0.1. Measure the amplitude and the phase of the frequency response of the allpass filter using the soundcard probe. Verify that the magnitude of the frequency response is 1 (or 0 dB) for all frequencies between about 20 Hz and 20 kHz, and that only the phase changes.
(b) To see the importance of measuring not only the magnitude but also the phase of the frequency response, use a 50% duty cycle rectangular signal with frequency 1 kHz and frequency 2 kHz as input and observe the output on the oscilloscope. Any changes in the rectangular signal that you see at the output of the allpass filter must be due to the phase response since the magnitude is not changed. How do you think an APF would affect the transmission of digital signals with rectangular shape?
Lab worksheet in PDF format: lws05.pdf
Note: Each student needs to turn in a lab worksheet. The raw measured data for each student in a lab team will be the same, but the conclusions drawn from it are individual to each partner in the team.
©2003-2007, P. Mathys. Last revised: 3-22-07, PM.
