Physics'241'Signal'Processing:'Lab'3' Sinusoidal esponse of, L ircuits In the previous lab, we studied the behavior of series combinations of and L circuits with input square and triangular waveforms. In this lab, we will study the same circuits, but with input sine waves. This will prove easier to calculate, since the complex math presented in class can be used for these circuits, and no integration or derivatives are needed. Background' v L v i ircuit Figure 1: and L circuits with input sine waves. L ircuit Kirchhoff's laws still apply in this case. With a single loop, the current throughout the loop will be sinusoidal, of the form i = Isinωt (1) If this current passes through an, L, or, the voltage drop across the element will be v = i = I sinωt (for a resistance) (2) v L = L di dt = ωli cosωt = ωli sin(ωt + 90 o ) (for an inductance) (3) v = 1 idt = I cosωt = I ω ω sin(wt 90o ) (for a capacitance) (4) The ratio of voltage amplitude/current amplitude = V/I is therefore given by, ωl, and 1/ω, for, L and respectively. These quantities all have units of ohms, and are called the magnitudes of the impedances Z, Z L and Z. Note, however, that the phases of voltage (referred to current) are different for the three elements: for a resistance the voltage and current phases are the same; in an L voltage LEADS current by 90, and in a, voltage LAGS current by 90. Therefore two numbers characterize an impedance: its magnitude Z and its phase angle φ. onsider the L circuit in Figure 1. The current throughout the circuit can be represented by i=i sin ωt. The voltage across the will be in phase with i, but the voltage across L will lead the current phase by 90. The two instantaneous voltages must add to the generator voltage V i, according to Kirchhoff's law, so the sine waves should look like Figure 2. EVISED 2016-02 M. DOBBS FOM OIGINAL BY J. AWFOD 1
Figure 2: Addition of V and V L in an L circuit. learly, we do not want to draw graphs of sine waves for every such addition. For single loops, a simple method of adding sinusoids is to add the vectors that represent their amplitudes, with the correct phase relationship. For the L circuit, the diagram would be: V L =ωli V i = V +V L φ = arctan ωl/ V =I Figure 3: Vector addition of the amplitudes V and V L. Phase of urrent We can see that the vector diagram tells us everything we might want to know about the voltage-current relationships in the loop. For example, V i = 2 + (ωl) 2 I (5) and φ = arctan ωl (6) We can see from Figure 3 that this angle φ represents the phase by which the generator voltage leads the current in the circuit. We can represent such relationships in a sort of vector shorthand by regarding the impedances themselves as vector quantities. For any element,, L or, v = Zi, where Z = 0 Z L = ωl 90 Z = 1/ω 90 This means that to find the voltage in an L, we multiply the current amplitude by Z L = ωl and change the phase by +90. Adding Impedances Note that in Figure 3, all the voltages were equal to (current amplitudes x impedances). If we divide all the vectors by current, we obtain a diagram with the same relative vector lengths and angles. We call this the impedance diagram. EVISED 2016-02 M. DOBBS FOM OIGINAL BY J. AWFOD 2
Z L =ωl Z i = Z +Z L φ = arctan ωl/ Z = Figure 4: Impedance diagram displaying the vector Addition of Impedances for the L ircuit. This shows us a simple way to do common calculations. The total impedance for the loop is Z i = Z i φ, where Z i = Z 2 + Z L 2 = 2 + (ωl) 2 (7) and φ = arctan ( Z L / Z ) (8) Then V i = Z i I, and the applied voltage leads the current in the loop by φ. For the circuit the impedance diagram will be Z = φ = arctan (œ1/ω ) Z =1/ω Z i = Z + Z Figure 5: Impedance diagram displaying vector Addition of Impedances for the ircuit. with Z i = Z 2 + Z 2 = 2 " 1 + $ # ω% and φ = arctan ( 1/ω) (10) (Here the applied voltage lags the current in the circuit by φ.) Frequency esponse of and L ircuits In our experiment, we will apply a constant amplitude sinusoidal voltage to the pair of impedances and vary its frequency. We can see from Figures 5 how the total impedance Z i should vary for an circuit: Z should decrease with ω while will stay constant. This affects the phase φ as well. We can use the circuit on the left of Figure 1 as an example. The two elements form a voltage divider; the voltage amplitude V is given by 2 (9) EVISED 2016-02 M. DOBBS FOM OIGINAL BY J. AWFOD 3
V V i = Z Z i = " 1 $ # ω% 2 1 + (ω) 2 = 1 1+ (ω) 2. (11) We know that has the units of time, so 1/ has the units of frequency. We define ω H = 1 and write eq. 11 as V 1 = V = 1 i " 1 + $ ω % 2 " ' 1 + $ f % ' # ω H & # f H & (12) 2 (13) where ω=2πf. The ratio of (output voltage/input voltage) is often called the transfer function H(f) of the circuit. Eq. 13 shows that at very low frequencies, with f<<f H, V V i. However, at very high frequencies, with f>>f H, V falls off inversely with frequency, with V /V i f H /f; if the frequency increases by a factor of 10, the amplitude V decreases by a factor of 10. This sort of behavior is very simply represented by plotting log H(f) vs log f. It is common in physics and engineering to use a decibel scale for this sort of log-log plot. For voltage ratios, H(f) in decibels is simply H(f) db = 20 log 10 [H(f)] In our example, H(f) db at very low frequencies will be 20 log [1] = 0 db; when f=f H, eq. 13 gives H(f) db = -3 db. At high frequencies H(f) drops a factor of ten for each frequency decade; this translates to a change of H(f) db = 20 db per decade. A graph of this behavior is shown in Figure 6. 20 0-3 H db -20-40 -60-80 -100-5 -4-3 - 2-1 0 1 2 3 4 5 log f/f H Figure 6: Bode plot for the circuit. EVISED 2016-02 M. DOBBS FOM OIGINAL BY J. AWFOD 4
This graph, of H(f) db vs. log f/f h in Figure 6 is called a Bode Plot. Note that at low frequencies the circuit 'passes' all the amplitude of the input sine wave to the output (capacitance) terminals. At f=f H (called the high frequency cutoff ) the output has dropped to 1/ 2 of the low frequency amplitude it is "down by 3 db". At high frequencies the slope of the falling amplitude is 20 db per frequency decade. Since the 3 db frequency is simply f H = 1/2π, one can reproduce the Bode plot by simply drawing two lines that intersect at f H as shown, but remembering that at f H, the graph actually curls down by 3 db. This circuit is a simple example of a low-pass filter. It allows low frequency signals to pass from the input terminals to the output, but it attenuates high frequencies (i.e., decreases their amplitude). The'Experiment' 1. An ircuit onnect the circuit of Figure 7 as shown, with =10k and =.01 µf. onnect the function generator s sine wave output to h1 and v to h2 of the scope. Trigger the scope externally with the generator sync signal. to h1 v to h2 Figure 7: ircuit for Experiment 1. ircuit Set the scope so that you can see the signals from both channels. Use a reasonably large amplitude sine wave (~ 10 V if your generator can provide it). Sweep the generator from very low frequencies to very high frequencies, keeping its amplitude constant, and observe the behavior of both the amplitude and phase of the v waveform. The amplitude should follow the prediction of the Bode plot of Figure 6; Figure 5 says that v should lag the generator voltage by φ = arctan (ω). Be sure to check the Bode plot prediction the V amplitude is essentially constant at low frequency, and at high frequency it falls off by a factor of 10 for every factor of 10 in frequency increase. Hint: it is important to check and understand this behavior before proceeding to the next step. If you messed up your circuit, its better to find out now rather than when you re preparing your lab report. Measure the frequency f H, for which v has dropped to 1/ 2 (~70%) of its low frequency amplitude. Measure the phase angle φ as accurately as possible at this frequency. You can do this by measuring the time displacement of the 2 sine waves. (Is it easier to do this by measuring the peak separations or the zero crossings?) EVISED 2016-02 M. DOBBS FOM OIGINAL BY J. AWFOD 5
2. A ircuit Interchange the and to make the circuit of Figure 8. to h1 v to h2 Figure 8: circuit for Experiment 2. ircuit Begin by sweeping the generator from very low to very high frequencies to get a qualitative idea of the behavior of the output v as a function of frequency. Now, make enough measurements of amplitude with frequency to enable you to make a Bode plot for this circuit. Don't be too painstaking: in the flat portion 2 or 3 points are enough; make more measurements in the region where the graph 'bends'. In this case, H(f) db will be 20 log 10 (V / V i ). The plot will be different from Figure 5, but should have a flat region, and one exhibiting a change with frequency. Measure the frequency f L at which V / V i = 1/ 2. Again, measure the phase angle between V and V i. Does V lead or lag V i? 3. A Blocking apacitor Leave all the components connected as in Experiment 2, but use the function generator to add a D voltage offset to the sine wave. This does the following to your circuit: to h1 v to h2 D offset + œ Figure 9: ircuit for Experiment 3. ircuit Both h1 and h2 inputs should be set to D so that they read instantaneous voltages correctly. Modify the value of the D offset in the generator menu. Do you observe the change on h1? On h2? After you have completed this experiment, be sure to turn off the D offset in the function generator menu. EVISED 2016-02 M. DOBBS FOM OIGINAL BY J. AWFOD 6
4. An L ircuit eplace the in the previous experiment with the inductor coil provided, leaving = 10k. to h1 L v to h2 Figure 10: ircuit for Experiment 4. L ircuit Make enough measurements so that you can make a Bode plot for this circuit. Measure the f H frequency as you did in Experiment 1, and measure the phase angle φ at this frequency. Questions' (Suggestion: complete the relevant questions for each experiment before you leave the lab room, so that you can consult your peers and get help from the TAs.) 1. (Suggestion: do this question BEFOE coming to the lab it will cement your understanding.) Using a 1 kω resistor and either an inductor or capacitor, draw a low-pass filter circuit that has a 3dB point of 100 khz. Specify the inductance or capacitance of the component you choose. Draw Bode plot for this circuit. 2. In Experiment 1, was your measured f H close to the predicted value? Did your measured φ agree with the predicted value? 3. Make a Bode plot for the measurements of Experiment 2. This can be done with a computer if you wish, or with a plot of calculated db values. emember that the Bode plot should show the frequencies on a log scale. ompare your Bode plot with the graph shown in Figure 6. 4. In Expt. 2, what was the phase angle between V and V i? Draw a vector diagram similar to Figure 5 (but scaled correctly) to show the relationship between Z, Z, and Z T for the frequency f L. What is the calculated phase angle between V and V i according to this plot? 5. In Expt. 3, you saw (we hope) a difference in the behavior of the h1 and h2 waveforms. Explain in words why the circuit behaved in this way. 6. In Expt. 4, make a Bode plot for the L circuit. What was the measured φ at the frequency f H? From your measurements, find the value of L. 7. For each of experiments 1,2,3 & 4, write down whether the output of hannel 2 is a low pass filter or a high pass filter. EVISED 2016-02 M. DOBBS FOM OIGINAL BY J. AWFOD 7