PHYSICS LAB #2 Passive Low-pass and High-pass Filter Circuits and Integrator and Differentiator Circuits

Transcription

1 PHYSICS LAB #2 Passie Low-pass and High-pass Filter Circuits and Integrator and Differentiator Circuits Objectie: Study the behaior of low-pass and high-pass filters. Study the differentiator and integrator circuits. Introduction: Resistors respond to ac sources as they do to dc. Capacitors and inductors, howeer, respond to ac sources quite differently. In this lab you will study the behaior of RC circuits to sinusoidal input signals and to input signals in the form of square pulses. Because of the frequencydependent behaior of capacitor, the output oltage depends on the frequency of the oltage sources. RC circuits are used widely in electronics to perform a ariety of functions, such as frequency filtering, differentiation and integration Low-pass Filter Circuit. Let s say you want to design a filter that will block all frequencies aboe 1 khz and pass all frequencies below 1 khz. This would be called a low pass filter. If a perfect filter could be made to achiee this task what would its response cure (gain ersus frequency) look like? In other words, what would a graph of out / in ersus frequency look like? Sketch the perfect response below. Add a linear scale to the y-axis. Gain ( out / in ) Frequency (Hz) 1

2 The RC circuit shown below is a low pass filter, but its response cure is not perfect. in R C o Fig. 2-1 Low-pass filter Think of it as a oltage diider where the resistance of one of the resistors depends on frequency. (In this case, we say the impedance of the capacitor depends on frequency.) At high frequency the capacitor behaes like a short, so the oltage across it will be a small fraction of the input oltage. The circuit is not passing, it is blocking, high frequencies. At low frequencies the capacitor behaes like an open, so the output will be a large fraction of the input oltage. The circuit is passing low frequencies. Thus the low pass filter has o in at low frequencies and o 0 at high frequencies. These are the extreme cases. The gain (and the phase shift) as a function of frequency oer the entire range of frequencies for this low-pass filter circuit are gien by: Gain = A = o in = 1 1+ R 2 C 2 Phase shift = tan -1 (RC) (That is, c lags in by (2) 1 In the following exercise you will measure the gain and phase shift as a function of frequency for a low pass filter. Not only is it a study of the properties of a low pass filter, it will deelop your skill at measuring gain and phase shift from an oscilloscope trace. These are ery common measurements. Part of the skill in making measurements is in achieing the highest precision possible when necessary. Answer these questions for a reminder on how to maximize precision with an oscilloscope. Estimate the uncertainty in measuring the peak-to-peak amplitude of a sinusoidal signal on the oscilloscope. Gie your answer as plus-or-minus so many diisions. Calculate the percent uncertainty in the measured amplitude of a 5 V p-p signal when the oscilloscope is set to display 10 V/diision. Calculate the percent uncertainty in the measured amplitude of a 5 V p-p signal when the oscilloscope is set to display 1 V/diision. 2

3 What lesson is made apparent by your answers? The same principle applies when you are measuring time (phase shifts) using the horizontal diisions and the times/diision display setting. Wire the low-pass filter circuit shown in Fig. 2-1 on the breadboard. Use R = 15 kand C = 0.01 µf. Measure the input and output oltages as a function of the sine wae frequency as well as the phase shift of the output sine wae as compared to the input sine wae on the oscilloscope. If necessary, adjust the amplitude of function generator to maintain a constant in. Use an input oltage of 10 Vp-p. Frequency 40 Hz 100 Hz 200 Hz 400 Hz 500 Hz 800 Hz 1 khz 4 khz 8 khz 10 khz 40 khz 100 khz 400 khz o (p-p) Gain Measured ( o / in ) Gain Measured (in db) Phase Measured (φ) in deg. Transfer your data to Excel/Origin. Create additional columns for calculated gain and phase using formulas (1) and (2). Plot a graph that oerlays the measured and calculated gain ersus frequency, using a log scale for the frequency axis. Do this twice, once using the simple out/in ratio and again plotting the gain in db. Determine the cut-off frequency fo (also called -3dB frequency or break point frequency). This is the frequency at which the gain is Mark it on printouts of your graphs. fo (measured) = Hz fo (calculated) = 1/(2RC) = Hz Show that the output attenuates at 20 db/decade and 6 db/octae at f >> fo. Plot the phase angle difference () between o and i ersus log frequency. Do your results agree with calculations based on Eqns. (1) and (2) for a low-pass filter? 3

4 Summer High-pass Filter Circuit. Behaior of a high-pass filter is opposite to that of a low-pass filter. The circuit below is a simple RC high-pass filter. Here the output is taken across the resistor. i C R o Fig. 2-2 High-pass filter With a sinusoidal applied ac input the output oltage across the resistor is ery small (o 0 ) at low frequencies (cap. behaes like an open) and at o in at high frequencies (cap. behae like a short). The gain and the phase as a function of frequency for the high-pass filter circuit are gien by: Gain = A = o i = RC 1+ R 2 C 2 3 Phase shift = tan -1 (1/RC) (Note that R leads in by (4) Let s say you want to design a filter that will pass all frequencies aboe 5 khz and block all frequencies below 5 khz. This would be called a high-pass filter. If a perfect filter could be made to achiee this task what would its response cure (gain ersus frequency) look like? In other words, what would a graph of out / in ersus frequency look like? Sketch the perfect response below. Add a linear scale to the y-axis. Gain ( out / in ) 4

5 The frequency response of the real RC high-pass filter is not perfect. In fact, it is just like that of the low pass filter, but reersed. With the measured low-pass frequency response from the preious actiity as a model, sketch the frequency response of an RC high pass filter that has a cut-off frequency of 1000 Hz Gain ( out / in ) Gain (db) Frequency (Hz) Frequency (Hz) Let s say there is a signal that is the sum of a 10 Vp-p 1 khz sinusoid and a 10 Vp-p 12 khz sinusoid. Design an RC high pass filter that will pass the 12 khz signal and block the 1 khz signal. There are two choices to make. First, choose the cut-off frequency. For high-pass filters, any frequency aboe the cut-off frequency is in the pass-band. Any frequency below the cut-of frequency is in the stop-band. Choose it so as to minimize attenuation of the frequency you want to pass, and maximize attenuation of the frequency you want to block. In other words, the signals are equal ( 1) at the input, but you want the 12 khz to be much larger than the 1 khz ( in, 12kHz in,1khz out, 12kHz out,1 khz asbig aspossible) at the output. Shift the frequency response transparency horizontally oer the graph below to help make your selection Hz Hz Gain (db)

6 What cut-off frequency did you choose? Your second design choice is the resistor alue. Its alue should be much greater than the output resistance of the stage that precedes the filter (the source), and much less than the input resistance of the stage that follows the filter (the load). This ensures that the source and load, when connected to the filter, do not significantly alter the cut-off frequency from the theoretical alue of f0 1 2RC. What is the output resistance of the source (in this case, the function generator)? What is the input resistance of the load (in this case, the oscilloscope)? What resistance did you choose? Now calculate the required capacitance. Show your work. You ll probably hae to adjust your choices to the nearest stock alues. Use only one capacitor. (Exact alues are not necessary. A circuit that absolutely has to hae a Ω resistor and a µf capacitor is poorly designed. Resistors are typically ±5% and capacitors are een worse, often ±20%. And eerything drifts oer time and with temperature. A well designed circuit tolerates these ariations.) C = Obtain your components. Now let s calculate how well we expect the filter to perform. Based on the alues of the actual components you will use in your filter, recalculate the cut-off frequency. f 0 = Accurately sketch the frequency response of the filter on the graph aboe. From the frequency response plot, determine the output oltage of the 12 khz and 1 khz signals. (You ll hae to conert the alue read from the graph in db to a alue in olts. Show your work.) Graphical out,12khz = 6

7 Graphical out,1khz = Next, calculate a theoretical alue for the output amplitudes using equation (3). Show your work. Theory out,12khz = Theory out,1khz = Finally, build your filter. (If your capacitor is electrolytic be careful with the polarization.) Measure the output amplitude for a 10 Vp-p input at 12 khz and at 1 khz. Measured out,12khz = Measured out,1khz = Compile all these results and calculations in the tables below. FRACTIONS out,12khz in,12khz out,1khz in,1khz out,12khz out,1khz Graphical Theory Measured db 20 log out,12khz in,12khz 20 log out,1khz in,1khz 20 log out,12khz out,1khz Graphical Theory Measured 7

8 Integrating and Differentiating Circuits The shapes of electrical signals must often be modified to be in a suitable form for operation of circuits. The simple RC circuit often plays a part in the formation of suitable waeforms. Here you will study the effects of low-pass and high-pass filter circuits on the input square oltage pulses of differing pulse widths Integrator Circuit: From theory we know that the oltage across the capacitor in an RC circuit is gien by: Charging: Discharging: t - RC c = V p (1 - e ) (5) t - RC c = V p e (6) Integrator: For RC >>, c RC t Vin dt (7) (Vin = Vp in this case and is the pulse width) This is why the RC circuit used in this configuration is called an integrator circuit. Wire the circuit shown in Fig. 2-3 with R = 10 k and C = 0.1 F. Time constant of the below circuit c = RC = 8

9 0 V p R C V o Fig. 2-3 RC Integrator Using a 5 V square wae input signal (use TTL output from the function generator on the design station), sketch the wae forms for the input and output signals for the following cases. Connect your input to CH 1 and output to CH 2 of the oscilloscope. i) RC >> f = Hz, Period (T) = sec CH 1 CH 2 Time Base s/di ii) f = 100 Hz, Period (T) = sec 9

10 CH 1 CH 2 Time Base s/di Calculate the time constant of the circuit from the figure in (ii). Recall that the time constant of the RC circuit is the time it takes for the capacitor to charge to 0.632Vp during the charging process Differentiator Circuit: From theory we know that the oltage across the resistor in an RC circuit is gien by: Charging: Discharging: - = V R p e t RC - = - V R p e t RC (8) (9) Differentiator: For RC <<, R RC dv in dt (10) (Vin = Vp in this case and is the pulse width) This is why the RC circuit used in this configuration is called a differentiator circuit. Wire the circuit shown in Fig. 2-4 with R = 10 k and C = 0.1 F. Time constant of the below circuit c = RC = 10

11 V p C 0 R V o Fig. 2-4 RC Differentiator Circuit 11

12 Using a 5 V square wae input signal (use TTL output from the function generator on the design station), sketch the wae forms for the input and output signals for the following cases. Connect your input to CH 1 and output to CH 2 of the oscilloscope. i) RC << f = Hz, Period (T) = sec CH 1 CH 2 Time Base s/di ii). f = 100 Hz, Period (T) = sec CH 1 CH 2 Time Base s/di 12

13 Calculate the time constant of the circuit from the figure in (ii). Recall that the time constant of the RC circuit is the time it takes for the resistor to charge to 0.368Vp during the charging process. 13