ELEC105L Fundamentals of Electrical Engineering Spring 2010ELEC105L Fundamentals of Electrical Engineering Spring 2010Lab #13: A Filter for the Electronic OrganWhy is this important?Capacitors and inductors have “resistances” to alternating currents that depend on frequency. We use a different term, impedance, to specify the ratio of voltage to current in AC circuits. The fact that impedancevaries with frequency allows us to design circuits called filters that respond differently to electronic signals at different frequencies. If your stereo or car radio has an equalizer (or just a tone control), it is actually a set of adjustable filters. Filters can be quite complex, but today we will build a simple filter composed of a resistor and a capacitor and measure some of its characteristics. But its main purpose will be to improve the tone of the electronic organ you have been developing over several lab sessions.GradingThis lab is a “familiarization exercise” and therefore will be weighted 50 points. To receive full credit, e-mail to your instructor the screen images captured in Step 7 and the data and comments requested in the same step. Add your circuits to those of the other groups to form a 13-note electronic organ. The screen images, data, and comments are due at 4 pm on the next day after the lab session. Only one submission perlab group is required; however, each member of the group must contribute to its production.BackgroundIn general, filters “pass” some frequencies and block others. In today’s exercise, a capacitor and a resistor will make a low-pass filter. By rearranging the circuit, we could also make a high-pass filter. Band-pass and band-stop filters can also be fabricated. We want to design a filter to attenuate (weaken) the high-frequency components of the output signals from the 555 timer circuits that make up the electronic organ. The unfiltered output has a rather unpleasant raspy sound because it is comprised of a signal at the fundamental frequency and many high-level harmonics, or signals at integer multiples of the fundamental frequency. Using a filter, we can attenuate the harmonic signals so that the output signal is closer to a pure sinusoid, which has a more pleasant tone.The schematic diagram shown in Figure 1 depicts a very simple filter based on a single resistor and a singlecapacitor. An application of sinusoidal steady state analysis using phasors will highlight the primary application of the circuit. In our case, input voltage vin(t) represents the output signal of the audio mixer you built in Lab #11. We assume here that the output (Thévenin) resistance of the summing amplifier circuit is negligible, which is valid for almost all op-amp circuits that have negative feedback. If we were to include the output resistance, we could simply add it to R since the output resistance and R would be in series. As Figure 1 shows, the input voltage can be represented by phasor Vin, and the output voltage by phasor Vo.The impedance of the resistor (ZR) is simply equal to the resistance itself; thus, ZR = R. The impedance of the capacitor (ZC) is defined asCjZC1.1Figure 1. A simple low-pass filter circuit.Using the voltage divider formula, the output voltage is related to the input voltage byCjRCjZZZinCRCino11 VVV.We can multiply the numerator and denominator by jC to obtain the more compact formRCjino11VV.Clearly, the output voltage’s magnitude and phase relative to those of the input voltage depend on the frequency. For example, suppose that we apply seven different sinusoidal input signals (Vin) to the filter circuit. All have the same magnitude (1 V) and phase (0°), but the frequencies are 250 Hz, 500 Hz, 1 kHz,2 kHz, 5 kHz, 10 kHz, and 20 kHz. If we substitute R = 10 k, C = 0.01 F, Vin = 1, and  = 2f, where f takes on each of the seven given values, into the expression for Vo, the output voltages at the various frequencies would have the magnitudes and phases shown in Table 1.Table 1. Output voltage magnitudes and phases at five different frequencies for the circuit shown in Figure1 with Vin = 1 in each case.Frequency (Hz)Output VoltageMagnitude (V)Output Voltage Phase(deg.)250 0.998.9500 0.95171000 0.85322000 0.62515000 0.307210,000 0.168120,000 0.07985Examination of Table 1 shows that as the frequency increases, the output voltage decreases in magnitude for a given input voltage magnitude. In other words, the RC circuit “passes” low-frequency signals (below 1000 Hz or so) with little attenuation (weakening), but it attenuates higher-frequency signals significantly, R = 10 kvin(t)  VinC0.01 F+vo(t)  Vo−+−2with progressively more attenuation as the frequency increases. There is also a phase shift that occurs as frequency increases; that is, the sinusoidal output voltage reaches its peaks later and later in time relative tothe sinusoidal input voltage. However, in audio applications phase shift is rarely distinguishable by the human ear.To determine quantitatively how much attenuation takes place at various frequencies, we can examine moreclosely the ratio of the output voltage phasor to the input voltage phasor:RCjino11VV.We are primarily concerned with how the magnitude of the voltage is affected as the signal passes through the filter. That is, we really just want to know the magnitude of the ratio. The quantity on the right-hand side is just a complex value. It has a magnitude and a phase that both vary with frequency. What we need to do is to isolate the magnitude. We can do that by multiplying the right-hand side by its own complex conjugate, which is formed simply by replacing j everywhere it appears with –j. Thus, the magnitude of thevoltage ratio becomes*22111111RCjRCjRCjinoVV,where the asterisk indicates complex conjugation. Continuing, we obtain       .111111111111122222RCjRCjRCjRCjRCjRCjRCjRCjRCjinoVVApplying the fact that j2 = 1 yields the result  222221111RCRCinoinoinoVVVVVV.We can now see that if  < 1/(RC), then the magnitude ratio will be close to one, which means that the magnitude of Vo will not be much different from the magnitude of