EE101 Notes 2. December 27, The measurement devices to be studied are oscilloscope, function generator, dc power supply and spectrum analyzer.

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1 EE101 Notes 2 December 27, 2018 Measurement Devices The measurement devices to be studied are oscilloscope, function generator, dc power supply and spectrum analyzer. DC power supply It is a device used for generating DC voltages. It has three terminals as + terminal, - terminal and ground. Current limiter puts a limit on the current Figure 1: DC Power Signal supplied by the generator. If the LED for the current lights, this indicates a 1

2 current limit. That is more current is demanded than the device can supply. Usually, a short circuit can cause such a case. The maximum voltage that can be generated is 30V. Oscilloscope It is a measurement device that displays voltage vs. time graph. Figure 2: Oscilloscope display 2

3 Operation Steps: 1. Turn on the device. Figure 3: Oscilloscope display 2. Connect the probe to channel Connect the probe to the probe compensation contacts. Figure 4: Oscilloscope display 3

4 4. Push the autoset to observe the square wave. Figure 5: Oscilloscope display 5. If the square wave is not observed properly, this indicates a failure in the probe. 4

5 Measurement Steps for Tektronix TDS Series Oscilloscopes The following steps are applied for manual measurements (no autoset): 1. Click CH1, menu. Adjust the ground level from the position dial to the center. Figure 6: Oscilloscope display 2. Click the menu button until the coupling setting appears to be DC. DC shows DC + AC signals whereas AC shows only AC signals. 3. Rotate the vertical Volts/Div dial to adjust the Volts/Div. 4. Rotate the horizontal Sec/Dic dial to adjust the Sec/Div. 5

6 Function generator It is a electronic device that generates alternating (time varying) voltage signals. Time varying nature of the voltage signal is sinusoidal, i.e. where A(V ) is the voltage amplitude, ω(rad/sec) is the radian frequency, φ(rad) is the phase angle, V DC (V ) is the dc voltage. f(t) = Asin(ωt + φ) + V DC (1) We have the following relations: Figure 7: Sine wave ω = 2πf (2) where f is the Hertz frequency (number of cycles per second), and T = 1 f (3) is the period (time for 1 cycle). Also, V pp = 2A (4) 6

7 Proper selection of values for Volts/Div and Sec/Div in an oscilloscope are determined based on the available settings. These settings are given in the following figure: Figure 8: Oscilloscope display Also, note that in an oscilloscope display, there are total of 10 divisions horizontally, and 8 divisions vertically. The best oscilloscope setting can be considered to be one division greater than the minimum required setting for full coverage. The minimum require full coverage is: Horizontally, t=t (Period), and Vertically, v=vpp (Peak to peak Amplitude). 7

8 Examples Example 1: Oscilloscope displays which of the following graphs? a. Power vs. frequency b. Voltage vs. frequency c. Voltage vs. time d. Current vs. frequency The correct answer is C. Example 2: In oscilloscope measurement steps, which of the followings is the correct coupling setting? a. Ground b. Volts/div. c. DC d. Sec./div. The correct answer is C. Example 3: Find the best oscilloscope settings (Volts/div, sec./div.) for the following signal? V (t) = sin(2πt), a. 500mV/div, 250ms/div b. 2V/div, 100ms/div c. 200mV/div, 50ms/div d. 1V/div, 100ms/div T = 1/f = 1s. Since there are 10 horizontal divisions, The min. sec/div is: sec/div = T = 1 sec/div = 100ms/div The best sec/div is: sec/div > T = 100ms/div 10 So, from the table (fig.7), sec/div = 250ms/div 8

9 Also, amplitude is V = 1V, V pp = 2V Then, the min. volts/div is: volts/div > Vpp = 2 volts/div = 250mV/div. 8 8 Then, from the table (fig.7),volts/div = 500mvolts/div Then, the correct answer is A. Example 4: Find the best oscilloscope settings (volts/div, sec./div.) signal? V (t) = sin(2000πt), a. 100mV/div, 500us/div b. 500mV/div, 250us/div c. 200mV/div, 50us/div d. 100mV/div, 250us/div for the following T = 1/f = 1ms. Since there are 10 horizontal divisions, The min. sec/div is: sec/div = T = 1ms = 100us/div 10 10div The best sec/div is: sec/div > 100us/div So, from the table (fig.7), sec/div = 250us/div Also, amplitude is V = 1V, V pp = 2V Then, the min. volts/div is: volts/div > Vpp = 2 volts/div = 250mV/div. 8 8 Then, from the table (fig.7),volts/div = 500mvolts/div Then, the correct answer is B. Example 5: Find the best oscilloscope settings (volts/div, sec./div.) signal? V (t) = 0.2sin(40000πt), a. 50mV/div, 250us/div b. 100mV/div, 10us/div for the following 9

10 c. 200mV/div, 10us/div d. 100mV/div, 5us/div T = 1/f = 50us. Since there are 10 horizontal divisions, The min. sec/div is: sec/div = T = 50u = 5us/div 10 10div The best sec/div is: sec/div > 5us/div So, from the table (fig.7), sec/div = 10us/div Also, amplitude is V = 0.2V, V pp = 0.4V Then, the min. volts/div is: volts/div > Vpp = 0.4 volts/div = 50mV/div. 8 8 Then, from the table (fig.7),volts/div = 100mvolts/div Then, the correct answer is B. 10

11 FFT Analysis FFT analysis gives us amplitude vs. frequency graph. Power Analysis: Power is the time ratio of energy, i.e. Figure 9: Spectrum Analyzer display P = W t (Joules/sec) (5) In electrical engineering, the power consumed by a resistor R is: P = V I = V 2 /R = I 2 R(W ) (6) Many times, we end up having very large or very small values for the power. Thus, we often use decibel as the unit for power. Decibel (db): It is a logarithmic unit to measure power ratios: By definition: is called the decibel. The term is called the bell. P P ref P db = 10log10 (7) P P ref P db = log10 (8) 11

12 If not specified, we take P ref = 1W. Millidecibel (dbm): It is defined as P dbm = 10log P 1mW 10 (9) Then, or or AC Power Consider the voltage signal P dbm = 10log P (10) P dbm = 10log P log (11) P dbm = P db + 30 (12) V = Asin(ωt + φ) (V ) (13) applied across a resistor R as shown in the following figure. The power Figure 10: AC power consumed by the resistor R is given by or P = V 2 R = A2 sin 2 (ωt + φ) (W ) (14) R P = A2 R 1 [1 cos(2ωt + 2φ)] (W ) (15) 2 12

13 If we plot this power expression in (15), we obtain the following graph Figure 11: AC power Since the power is not constant, we define a concept of average power defined as P avg = A2 (W ) (16) 2R where A is the amplitude of the voltage. If A is the amplitude of the current, then we have the average power as P avg = A2 R (W ) (17) 2 13

14 Examples Example 1: Find the power in db for P=1000W? (Pref=1W) a. 0dB b. 10dB c. 20dB d. 30dB P P ref 10 P db = 10log or P db = 10log = 30dB. The correct answer is D. Example 2: Find the power in db for P=1W? (Pref=1W) a. 0dB b. 10dB c. 20dB d. 30dB P db = 10log 1 10 = 0dB. The correct answer is A. Example 3: Find the power in db for P=1nW? (Pref=1W) a. 0dB b. -60dB c. -90dB d. -120dB P db = 10log = 90dB. The correct answer is C. 14

15 Example 4: Find the power in dbm for P=1nW? (Pref=1W) a. 0dB b. -60dB c. -90dB d. -120dB P dbm = P db + 30 = = 60dB. The correct answer is B. Example 5: If the amplitude of the AC voltage across a 10Ω resistor is 10V, what is the average power? a. 1W b. 2W c. 4W d. 5W P R = V amp 2. P 2R R = 102 = 5W. The correct answer is D. 20 Example 6: FFT analysis gives us...? a. Amplitude vs. time graph b. Current vs. time graph c. Amplitude vs. frequency graph d. Resistance vs. time graph The correct answer is C. Example 7: What is the reason for using db and/or dbm for units of power? a. db or dbm are more convenient because the numbers are smaller. b. db or dbm are the only units for power. c. db or dbm can be used for many applications d. db or dbm are the only units accepted by many people. The correct answer is A. 15

16 Electromagnetic Waves and Radiation -Electromagnetic wave radiation is a propagation of energy due to time varying (sinusoidal) electric and magnetic fields. Electric and Magnetic Fields -Electric and/or magnetic field: The force field acting on charges. Electromagnetic Waves (Radio Waves) -Electromagnetic Waves (Radio Waves): Motion of electric and magnetic field as a wave (changing with sin or cos function). Capacitors and Inductors In addition to resistors, capacitors and inductors are the two major electric elements. A capacitor can be made of two conducting plates separated by an insulator. This is shown in Fig.12. Figure 12: Capacitor An inductor can be made by circularly winding a conducting wire. This is shown in Fig.13. Figure 13: Inductor 16

17 - Both capacitor and inductors can be used to store energy. - Modern batteries utilizes the concept of supercapacitance where large energies can be stored inside a capacitors usually through modification of its dimensions and the properties of the insulating material. - Capacitors have a resistance that is inversely proportional to the applied frequency. - Inductors have a resistance that is directly proportional to the applied frequency. - capacitors and inductors can be combined to obtain electric resonators. - Resonance is a concept where the resistance of an object becomes zero at a specific frequency. 17