Lab 6 Black Box. Lab Performed on November 19, 2008 by Nicole Kato, Ryan Carmichael, and Ti Wu Report by Ryan Carmichael and Nicole Kato

Transcription

1 Lab 6 Black Box Lab Performed on November 19, 2008 by Nicole Kato, Ryan Carmichael, and Ti Wu Report by Ryan Carmichael and Nicole Kato E11 Laboratory Report Submitted December 16, 2008 Department of Engineering, Swarthmore College 1

2 General Procedure: In Lab: 1. Determined resistances between node A and ground, node B and ground, as well as between nodes A and B, as seen in Figure Attached node A to a function generator and node B to an oscilloscope, recorded oscilloscope output. This will be referred to as normal setup. Figure 1: General Black Box Setup Outside of Lab: 1. Solved for the resistances of Z 1, Z 2, and Z 3 by solving the system: Z 1 + Z 2 = resistance between A and ground Z 2 + Z 3 = resistance between B and ground Z 1 + Z 3 = resistance between A and B 2. Used the resistance values along with the shape of the oscilloscope output graph to determine what elements were present in the circuit 3. Curve fit the oscilloscope data to determine a s value for the circuit 4. Used the s value to determine values for any capacitors or inductors present in the circuit 5. Created a Multisim simulation of the circuit to verify results 2

3 Box 2: Lab Data: Measured Resistances Between Circuit Terminals A - B A - Ground B - Ground 994 Ω Table 1: Measured Resistances for Box 2 Oscilloscope Output For Box 2 (Normal Setup) Input - dark line Output -light line Figure 2: Box 2 Oscilloscope Output (Normal Setup) 3

4 Analysis: Z 1 + Z 2 = Z 2 + Z 3 = Z 1 + Z 3 = 994 Ω From these equations and the oscilloscope output graph, we determine that the circuit is first order and that Z 2 contains a capacitor in series with a resistor. This is true because Z 2 has an infinite resistance and the graph stays near the origin until it reaches.5v. Additionally, Z 1 contains one resistor because the circuit is first order and the Z 1 must contain at least one element. Z 3 contains a wire because the circuit cannot have more than three elements. We then curve fit our oscilloscope output shown below in Figure 3. From this fit we determined a value for s, from which we solved for the values of the circuit elements using the math shown below the curve fit. Curve Fit for Curve 2 Voltage (v) Box 2 Normal General model: f(x) = a*(1-exp(b*x)) Coefficients (with 95% confidence bounds): a = (0.4227, ) b = (-8759, -8638) Goodness of fit: SSE: R-square: Adjusted R-square: RMSE: Time(s ) Figure 3 4

5 From the above formulas we see that Figure 4 below represents the circuit in Box 2. We then simulate this proposed circuit with Multisim (Figure 5). The results of the simulation match our lab data, which leads us to believe that our proposed circuit is indeed the circuit in Box 2. Figure 4 5

6 Figure 5 6

7 Box 3: Lab Data: Measured Resistances Between Circuit Terminals A - B A - Ground B - Ground 8.11 kω 8.14 kω 25 Ω Table 2: Measured Resistances for Box 3 Oscilloscope Output For Box 3 (Normal Setup) Input - dark line Output -light line Figure 6: Box 3 Oscilloscope Output (Normal Setup) 7

8 Analysis: Z 1 + Z 2 = 8.14 kω Z 2 + Z 3 = 25 Ω Z 1 + Z 3 = 8.11 kω From these equations and the oscilloscope output graph we determine that the circuit is second order. Because there are no infinite resistances and because the graph begins at the origin, the capacitor must be in parallel with the inductor. From this, we know that Z 3 contains a wire because we are limited to three total elements. Then looking at our resistances, we can clearly see that Z 1 must contain an 8.1 kω resistor and Z 2 contains the L / C combination. We then curve fit our oscilloscope output shown below in Figure 7. From this fit we determined a value for s, from which we solved for the values of the circuit elements in the math shown below the curve fit. Curve Fit for Circuit 3 Voltage (V) Circuit 3 normal General model: f(x) = a*exp(b*x)*cos(c*x+offset)+d Coefficients (with 95% confidence bounds): a = ( , ) b = (-645.6, ) c = 8932 (8917, 8948) d = ( , ) offset = (-1.613, -1.58) Goodness of fit: SSE: R-square: Adjusted R-square: RMSE: Time (s) Figure 7 8

9 From the above math we see that Figure 8 represents the circuit in Box 3. We then simulate this proposed circuit with Multisim (Figure 9). The results of the simulation match our lab data, which leads us to believe that our proposed circuit is the circuit in Box 3. Figure 8 9

10 Figure 9 10

11 Box 4: Lab Data: Measured Resistances Between Circuit Terminals A - B A - Ground B - Ground 3.16 kω 5.54 kω 4.26 kω Table 3: Measured Resistances for Box 4 Oscilloscope Output For Box 4 (Normal Setup) Input - dark line Output -light line Figure 10: Box 4 Oscilloscope Output (Normal Setup) 11

12 Analysis: Z 1 + Z 2 = 5.54 kω Z 2 + Z 3 = 4.26 kω Z 1 + Z 3 = 3.16 kω From the oscilloscope output graph we determine that the circuit is made of all resistors. Then from the equations above we solve for Z 1-3, finding that Z 1 contains a 2.18 kω resistor, Z 2 contains a 3.28 kω resistor, and Z 3 contains a 0.98 kω resistor. From this knowledge, we see that Figure 11 can represent the circuit in Box 4. We then simulate this proposed circuit with Multisim (Figure 12). The results of the simulation match our lab data, which leads us to believe that our proposed circuit is indeed the circuit in Box 4. Figure 11 12

13 Figure 12 13

14 Box 5: Lab Data: Measured Resistances Between Circuit Terminals A - B A - Ground B - Ground 25 Ω 2.69 kω 2.67 kω Table 4: Measured Resistances for Box 5 Oscilloscope Output For Box 5 (Normal Setup) Input - dark line Output -light line Figure 13: Box 5 Oscilloscope Output (Normal Setup) 14

15 Analysis: Z 1 + Z 2 = 2.69 kω Z 2 + Z 3 = 2.67 kω Z 1 + Z 3 = 25 Ω From the oscilloscope output graph we determine that the circuit is first order. Because Z 1 + Z 3 = 25 Ω, we deduce that Z 3 must contain a wire and because the graph stays near the origin until.5v, Z 1 must contain an inductor in series with a resistor. Due to our element rules, Z 2 must then contain a single resistor. We then curve fit our oscilloscope output shown below in Figure 14. From this fit we determined a value for s, from which we solved for the values of the elements in the math shown below the curve fit. Curve Fit for Circuit 5 Voltage (V) Box 5 normal General model: f(x) = a*(1-exp(b*x)) Coefficients (with 95% confidence bounds): a = (0.4122, ) b = e+004 (-1.399e+004, e+004) Goodness of fit: SSE: R-square: Adjusted R-square: RMSE: Time (S) Figure 14 15

16 From the above math we see that Figure 15 can represent the circuit in Box 5. We then simulate this proposed circuit with Multisim (Figure 16). The results of the simulation match our lab data, which leads us to believe that our proposed circuit is indeed the circuit in Box 5. Figure 15 16

17 Figure 16 17