EK307 Passive Filters and Steady State Frequency Response

Transcription

1 EK307 Passive Filters and Steady State Frequency Response Laboratory Goal: To explore the properties of passive signal-processing filters Learning Objectives: Passive filters, Frequency domain, Bode plots Suggested Tools: Function generator, oscilloscope Introduction In this lab assignment, we explore the properties of passive filters. In dealing with the concept of each circuit s frequency response, we analyze and test circuits solely in the sinusoidal steady state. As we have learned in class, in the sinusoidal steady state, we can treat resistors, capacitors, and inductors as impedances, wherein resistance comprises but one form of impedance. Using impedances, we can utilize all the various circuit techniques from resistive circuit analysis, such as series and parallel combinations, voltage division, and current division.. In this context, we can think of a filter as a circuit that operates on each sinusoidal component of the input signal. Prelab: Prelab part one: The lab instructor for EK307 is often seen with a water bottle. Previously unknown to the students, there is a digital sensor in the bottle measuring and logging the temperature of the drink at a rate of 42 times per second. Recall from physics and chemistry classes that liquid has a heat capacity. Heat capacity relates the flow of heat into a material to its temperature change. This can be modeled as a capacitor, where the capacitance is equivalent to the product of the heat capacity and the quantity of liquid in the bottle. Voltage is analogous to temperature and the flow of heat is analogous to current. The water bottle material has heat flow properties (insulation) that can be modeled as a resistance. For simplicity sake we will pretend that the heat transfer properties of the bottle do not change even if the liquid quantity does. In reality we know they will because the liquid-tobottle surface area changes. Assume the room is at standard temperature and pressure (STP), and the instructor s body temperature is typical of a human. Now for the experiment: During the course of the lab session the instructor picks up the bottle with a warm hand, sips, and puts it down at a constant frequency. In the beginning of the lab session the temperature sensor does not detect much change in temperature. As the lab session progresses the temperature fluctuations increase. What is causing the increase in temperature fluctuation magnitude? Prelab part two: In your lab notebook draw a schematic of the passive filters used in level one. Calculate the values of R and C for the low pass and high pass filters given a cutoff frequency of 1050 Hz. 11/27/2016 1

2 Level 1: a) Low Pass Filter: The circuit shown below is a passive low pass filter. It is classified as passive because it does not use power other than the input signal. Another way to think of it is there are no power supplies (as in an op amp circuit) that provide power to the circuit. You may recognize this circuit as the RC circuit from the timing lab. It is the same circuit, we are simply feeding it with sinusoids instead of a step signal. The objective of this part of the lab is to measure the magnitude and phase of V OUT relative to V IN as a function of frequency. Select the appropriate value for R and C to result in a desired breakpoint (cutoff) frequency of approximately 1050 Hz. Remember to use practical values for R and C. Practical means not too small or too large. 1kΩ < R < 220kΩ, 100pF <C < 10µF. For V IN frequencies ranging from 10 Hz to 100 khz, observe and record the change in the magnitude ratio V OUT /V IN and phase angle using a function generator and an oscilloscope. Use ten or more points. Be sure some are near the cutoff frequency. Use a sufficiently large voltage amplitude sine wave because your filter will attenuate frequencies above the cutoff. If the amplitude is small to begin with it will quickly disappear into the noise as the frequency goes up. Since we want to know the relative difference between the input (V IN ) and output (V OUT ), connect one oscilloscope probe to V IN and the other to V OUT. Remember on the oscilloscope the red wires are the signal 11/27/2016 2

3 inputs and the black wires connect to ground. If the waveforms on your scope are not sinusoidal there is a problem with your setup. You can measure the sine wave magnitude (zero to peak) on each channel of the oscilloscope. Be sure that the scope is in DC coupled mode. You can measure the phase difference by measuring the delay of the zero crossing of V OUT relative to the zero crossing of V IN (Figure 1). Record the data in a spreadsheet and make two plots, one of frequency vs. magnitude (db), the other frequency vs. phase (in degrees). Collect at least 10 different frequency points. One of those frequencies should be at the theoretical breakpoint frequency. You will find the frequency response graph is not very interesting until you are within one decade (factor of 10) of the breakpoint. Therefore it is encouraged to measure more frequency points in the breakpoint region than in the pass band. Use Decibels for your magnitude plot. Note that the decibel scale is defined by db = 20*log 10 V out /V in, so that db = 0 when V out /V in = 1. Phase (degrees) = 360 * Δt / period Or Phase (degrees) = 360 * Δt * Frequency (Hz) Figure 1. Measuring phase difference of two sinusoids of the same frequency using the zero crossing time delay. The phase difference in degrees can be calculated by dividing the time difference by the period then multiplying that quantity times 360. Be mindful of vertical offset between the two waves. Rephrased, the zero of each sine wave may not be the same. Be mindful of units when making this calculation. b) High pass filter: Repeat all the procedures performed on the low pass filter for the high pass filter. In condensed form they are: Calculate the R and C values for a cut off frequency of 1050Hz Measure magnitude and phase for frequencies ranging from 1Hz to 100kHz Plot the Magnitude and phase as a function of frequency. Use loglog scale for the magnitude and semilogx for the phase. 11/27/2016 3

4 Passive first order high pass filter. Notice if looks similar to the passive first order low pass filter. The two passive components have been transposed. The cutoff frequency is the same although this filter passes frequencies above the cutoff where the low pass filter passes frequencies below the cutoff. Level 2: (complete in addition to level 1) Design, construct, and measure the frequency response of a second order RLC low pass filter. It should have flat frequency response and a cutoff frequency of 1050Hz. Compare the slopes of the first and second order filters. Background: Passive Filters A passive filter is made solely from resistors, capacitors, and/or inductors. A passive filter can t amplify, nor can it produce power. Rather, it simply allows some frequencies of the input to pass with unchanged amplitude and phase, while attenuating and/or phase-shifting others. The typical passive filter is suitable for driving high-input-impedance or open-circuit loads only. Attaching a low impedance load to the output of a passive filter usually causes the filter s frequency-response properties to change. Consider, for example, the frequency response of the following passive filter circuit. From voltage division in the sinusoidal steady state, 11/27/2016 4

5 If we use this filter to drive a load resistance R L, then the response will change to R P R PC Z in2 where R P = R R L. Similarly, if we attempt to cascade two passive RC filters, as shown below, the net response will not just the simple product of the individual responses, because the input impedance Z in2 of the second filter will load down the first filter: Here, the simple multiplication of the individual responses does not work, because the first filter is loaded by (R 2 + 1/jC 2 ) in parallel with C 1. 11/27/2016 5