POLYTECHNIC UNIVERSITY Electrical Engineering Department. EE SOPHOMORE LABORATORY Experiment 5 RC Circuits Frequency Response

Transcription

1 POLYTECHNIC UNIVERSITY Electrical Engineering Department EE SOPHOMORE LORTORY Eperiment 5 RC Circuits Frequency Response Modified for Physics 18, rooklyn College I. Overview of Eperiment In this eperiment you will measure the frequency response of an RC circuit by driving the circuit with a variable frequency sine wave and measuring the change in amplitude and phase as the signal passes through the circuit from input to output. II. Equipment Required 1 Oscilloscope - Dual trace capability and with -y mode feature 1 Function Generator 1 Digital Multimeter 1 Impedance meter for capacitance measurement 1 Protoboard 1 10kΩ Resistor μF Capacitor 1 Decade resistance o III. Measurement of phase differences using the oscilloscope There are two standard ways to measure the phase difference between two sinusoids of the same frequency by using an oscilloscope. The first method, which requires a dual trace scope, eploits the fact that a phase difference is equivalent to a time shift, as we now describe. Consider the two sine waves given below with angular frequency ω and period T = π/ω. ( t) = sinωt y( t) = sin( ωt+ φ) (5-1) in which the angle φ represents the phase difference between the two signals. We may also refer to φ as the phase angle of y(t) relative to the reference signal (t). We can rewrite y(t) as φ T y (t) = sin ω t + or, since ω = π/t, y (t) = sin ω t + φ. If we now compare the two ω π signals again 9

2 (t) = sin ωt T y(t) = sin ω t + φ π (5-) we can see that, if φ is in the range 0 < φ < π, y(t) is a time advanced version of (t). The terminology we use is that y(t) leads (t) by T φ seconds. You should convince yourself of π this by verifying that if y(t) takes a certain value at time t, (t) takes that same value at time t + T T φ, i.e. t + φ = ) π π y(t. In a similar way, if φ is in the range -π < φ < 0, then y(t) lags (t). The figure below depicts a situation in which y(t) leads (t) by T/4 seconds or, equivalently, π/ radians of phase. y(t) (t) t In general, if we let τ denote the time lead of y(t) with respect to (t), and let φ denote the corresponding phase lead, we have the simple relationships τ = T π φ (5-3) and, correspondingly φ = π τ T. (5-4) It should be clear that the discussion above applies equally well to the case in which the amplitudes of the two sinusoids are different. In this case the eample above would look like the following 30

3 y(t) (t) t and y(t) still leads (t) by T/4 seconds or π/ radians of phase. The second method for determining the phase difference between two sinusoids relies on the well known Lissajous pattern and can be applied even with a single trace oscilloscope, provided the scope has the -y mode feature. Suppose again that we are given two sine wave signals of the same frequency, but perhaps with different amplitudes: ( t) = cosωt y( t) = cos( ωt+ φ) (5-5) If we apply (t) to the horizontal input of an oscilloscope in the -y mode, and simultaneously apply y(t) to the vertical input, then the horizontal deflection of the electron beam at any time instant, t, will be = cosω t and the vertical deflection at the same time will be y = cos( ωt+ φ ). s a function of time both and y will be continuously varying, and the electron beam will trace out some continuous path on the screen of the oscilloscope. In order to see what this path will look like, let us rewrite y as y = cosφcosωt sin φsinω t. (5-6) Now, since = cosω t, we can write cos ωt = / (5-7) and, using the trigonometric identity sin ωt + cos ωt = 1, we also have sinωt = 1 cos ωt = 1 /. (5-8) Now, by substituting (5-7) and (5-8) into (5-6) and performing some algebraic manipulations, we can eliminate the time variable, and arrive at the following equation between and y directly y y + cosφ = sin φ. (5-9) 31

4 s the electron beam traces out a path on the oscilloscope screen, it will pass through precisely all those (,y) points for which (5-9) is satisfied. Therefore, we see that the path will have an elliptical shape which depends on the amplitudes of the and y signals, as well as on the phase angle between the two signals. This elliptical path is called a Lissajous pattern. Notice that if the phase difference is φ = 0o then (5-9) reduces to y y + = 0 or y = 0 y = and the oscilloscope screen will appear as follows. - - If the phase difference is φ = 90o then (5-9) becomes y + = 1 (5-10) and the oscilloscope screen looks like 3

5 - - t φ = 180o, (5-9) becomes + y = 0 y = nd the Lissajous pattern becomes - - Finally, at some intermediate phase difference, say between 0o and 90o, the pattern would look like

6 Notice from (5-9) that, since both cosφ and sinφ are even functions of φ, the Lissajous pattern will not change if φ is changed to -φ, so it is impossible to determine the sign of φ from the Lissajous pattern. However, we can determine the magnitude of φ as follows. If we set y equal to zero in (5-9) we find that the -ais zero crossings are at =± sinφ =± sin φ. lso, since the maimum ecursion of the pattern in the direction is ± (see equation (5-5)), we can see that length between ais crossin gs total length of Lissajous pattern = sin φ. In the figure below therefore, we can find φ as sin φ= L L1 ` (5-11) L 1 L IV. Preparation for the Eperiment a) Using the phasor technique, analyze the circuit of figure 5-1 to show that the transfer V 1 function V/Vs is given by =, in which V is the phasor voltage across the Vs 1 + jωrc capacitor and Vs is the phasor input voltage. 34

7 R I V s C V Figure 5-1 b) Find the magnitude and phase of the transfer function V 1 = Vs 1 + jωrc 1 as a function of f. 1+ jπfrc = c) On a sheet of graph paper, plot Log10 V V s versus Log10f for the frequency range 100 Hz < f < 105 Hz using appropriate values of R and C. d) On a separate sheet of graph paper, plot the phase φ of the transfer function versus Log10f using the same element values and frequency range as in part (c). V. Eperimental Procedure a) Measure the internal impedance of your function generator using the procedure followed in eperiment 1, and record the value. b) On your protoboard, wire the circuit shown in Figure 5-. In this figure, Rs represents the internal resistance of the function generator, which you have measured in part (a). Measure the precise value of the 10kΩ resistor with your multimeter before you place it on the board. Measure the value of your capacitor, and record it also. 35

8 R s i(t) 10k v (t) s v i (t) 0.01uF v(t) Function Generator Figure 5- c) Turn on the function generator and observe its output, vi(t), with your oscilloscope. Set the function generator to the sine wave setting and adjust the peak-to-peak amplitude to 5V and the frequency to khz. In order to set these parameters accurately, use your oscilloscope to measure the amplitude and frequency as you adjust them. Now, observe the capacitor voltage, v(t), on the second channel of your scope and make an accurate measurement of its peak-to-peak amplitude. Take the ratio of the capacitor voltage peakto-peak amplitude and the input peak-to-peak amplitude, and record the value in the appropriate bo in table 5-1 below. Net, observe the capacitor voltage and the input voltage simultaneously, using the dual trace feature of your scope and measure the phase lead of the capacitor voltage with respect to the input voltage. Do this by measuring the time lead and multiplying by π/t as discussed in section III above. Make sure, when doing this measurement, that both sine waves are centered vertically. If you find that the capacitor voltage actually lags the input voltage, then enter the number as a negative phase lead in the third row of table 5-1. Now repeat these amplitude and phase measurements for the rest of the frequencies in table 5-1 to complete the second and third rows of the table. For each measurement, be sure to measure the input voltage amplitude again, and accurately measure the frequency by using the oscilloscope. f(hz) V/V i phi (rad) phi (rad) Table

9 You will now fill in the last row of the table by measuring the phase difference again, this time using the Lissajous pattern method. To do this, set the input frequency to khz again. Net, put the oscilloscope in the -y mode and check that both the vertical and horizontal inputs are set to DC coupling. Now, apply the input sine wave to the - channel and the capacitor voltage to the y-channel and observe the Lissajous pattern on the screen. You may have to increase the intensity to produce the pattern on the screen. Make sure the pattern is centered in the y-ais direction (it is not necessary to center the pattern in the -ais direction) and measure the values of L1 and L in equation (5-11) and calculate φ. Enter the result in table 5-1, but enter it as a negative number since we know from the third row of table 5-1 that φ is, in fact, negative. Repeat this measurement for the rest of the frequencies indicated, and complete the table. VI. Report a) On a sheet of graph paper, make a plot of Log10 V V i versus Log10f for the measured data from table 5-1. How well does this graph compare with the theoretical plot from the lab preparation, part (c)? If the graphs are different, eplain why you think this is so. b) On separate sheets of graph paper, plot φ versus Log10f for the third and fourth rows of table 5-1 and compare the results to the plot from the lab preparation, part(d). Which of the two phase measurements is more accurate? Can you eplain why? 37