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
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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
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Operation Steps:1. Turn on the device.
Figure 3: Oscilloscope display
2. Connect the probe to channel 1.3. Connect the probe to the ”probe compensation” contacts.
Figure 4: Oscilloscope display
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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 theprobe.
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Measurement Steps for Tektronix TDS 210-220 SeriesOscilloscopes
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”.
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Function generator
It is a electronic device that generates alternating (time varying) voltage sig-nals.
Time varying nature of the voltage signal is sinusoidal, i.e.
f(t) = Asin(ωt+ φ) + VDC (1)
whereA(V ) is the voltage amplitude,ω(rad/sec) is the radian frequency,φ(rad) is the phase angle,VDC(V ) is the dc voltage.
Figure 7: Sine wave
We have the following relations:
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,
Vpp = 2A (4)
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Proper selection of values for Volts/Div and Sec/Div in an oscilloscopeare determined based on the available settings. These settings are given inthe following figure:
Figure 8: Oscilloscope display
Also, note that in an oscilloscope display, there are total of 10 divisionshorizontally, and 8 divisions vertically.
The best oscilloscope setting can be considered to be one division greaterthan the minimum required setting for full coverage.The minimum require full coverage is:Horizontally, t=T (Period), andVertically, v=Vpp (Peak to peak Amplitude).
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Examples
Example 1:Oscilloscope displays which of the following graphs ?a. Power vs. frequencyb. Voltage vs. frequencyc. Voltage vs. timed. Current vs. frequency
The correct answer is C.
Example 2:In oscilloscope measurement steps, which of the followings is the correct cou-pling setting ?a. Groundb. Volts/div.c. DCd. Sec./div.
The correct answer is C.
Example 3:Find the best oscilloscope settings (Volts/div, sec./div.) for the followingsignal ?V (t) = sin(2πt),a. 500mV/div, 250ms/divb. 2V/div, 100ms/divc. 200mV/div, 50ms/divd. 1V/div, 100ms/div
Answer:T = 1/f = 1s.Since there are 10 horizontal divisions, The min. sec/div is:sec/div = T
10= 1
10sec/div = 100ms/div
The best sec/div is:sec/div > T
10= 100ms/div
So, from the table (fig.7),sec/div = 250ms/div
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Also, amplitude isV = 1V ,Vpp = 2V Then, the min. volts/div is:
volts/div > Vpp
8= 2
8volts/div = 250mV/div.
Then, from the table (fig.7),volts/div = 500mvolts/divThen, the correct answer is A.
Example 4:Find the best oscilloscope settings (volts/div, sec./div.) for the followingsignal ?V (t) = sin(2000πt),a. 100mV/div, 500us/divb. 500mV/div, 250us/divc. 200mV/div, 50us/divd. 100mV/div, 250us/div
Answer:T = 1/f = 1ms.Since there are 10 horizontal divisions, The min. sec/div is:sec/div = T
10= 1ms
10div= 100us/div
The best sec/div is:sec/div > 100us/divSo, from the table (fig.7),sec/div = 250us/divAlso, amplitude isV = 1V ,Vpp = 2V Then, the min. volts/div is:
volts/div > Vpp
8= 2
8volts/div = 250mV/div.
Then, from the table (fig.7),volts/div = 500mvolts/divThen, the correct answer is B.
Example 5:Find the best oscilloscope settings (volts/div, sec./div.) for the followingsignal ?V (t) = 0.2sin(40000πt),a. 50mV/div, 250us/divb. 100mV/div, 10us/div
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c. 200mV/div, 10us/divd. 100mV/div, 5us/div
Answer:T = 1/f = 50us.Since there are 10 horizontal divisions, The min. sec/div is:sec/div = T
10= 50u
10div= 5us/div
The best sec/div is:sec/div > 5us/divSo, from the table (fig.7),sec/div = 10us/divAlso, amplitude isV = 0.2V ,Vpp = 0.4V Then, the min. volts/div is:
volts/div > Vpp
8= 0.4
8volts/div = 50mV/div.
Then, from the table (fig.7),volts/div = 100mvolts/divThen, the correct answer is B.
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FFT Analysis
FFT analysis gives us amplitude vs. frequency graph.
Figure 9: Spectrum Analyzer display
Power Analysis:Power is the time ratio of energy, i.e.
P =W
t(Joules/sec) (5)
In electrical engineering, the power consumed by a resistor R is:
P = V I = V 2/R = I2R(W ) (6)
Many times, we end up having very large or very small values for thepower. Thus, we often use decibel as the unit for power.
Decibel (dB):It is a logarithmic unit to measure power ratios:By definition:
PdB = 10logP
Pref
10 (7)
is called the ”decibel”. The term
PdB = logP
Pref
10 (8)
is called the ”bell”.
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If not specified, we take Pref = 1W .Millidecibel (dBm):
It is defined as
PdBm = 10logP
1mW10 (9)
Then,PdBm = 10logP.10
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10 (10)
orPdBm = 10logP10 + 10log10
3
10 (11)
orPdBm = PdB + 30 (12)
AC Power
Consider the voltage signal
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
P =V 2
R=A2sin2(ωt+ φ)
R(W ) (14)
or
P =A2
R
1
2[1 − cos(2ωt+ 2φ)] (W ) (15)
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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 definedas
Pavg =A2
2R(W ) (16)
where A is the amplitude of the voltage. If A is the amplitude of the current,then we have the average power as
Pavg =A2
2R (W ) (17)
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Examples
Example 1:Find the power in dB for P=1000W ? (Pref=1W)a. 0dBb. 10dBc. 20dBd. 30dB
Answer:
PdB = 10logP
Pref
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orPdB = 10log100010 = 30dB. The correct answer is D.
Example 2:Find the power in dB for P=1W ? (Pref=1W)a. 0dBb. 10dBc. 20dBd. 30dB
Answer:PdB = 10log110 = 0dB. The correct answer is A.
Example 3:Find the power in dB for P=1nW ? (Pref=1W)a. 0dBb. -60dBc. -90dBd. -120dB
Answer:PdB = 10log10
−9
10 = −90dB. The correct answer is C.
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Example 4:Find the power in dBm for P=1nW ? (Pref=1W)a. 0dBb. -60dBc. -90dBd. -120dB
Answer:PdBm = PdB + 30 = −90 + 30 = −60dB. The correct answer is B.
Example 5:If the amplitude of the AC voltage across a 10Ω resistor is 10V, what is theaverage power ?a. 1Wb. 2Wc. 4Wd. 5WAnswer:PR =
V 2amp
2R. PR = 102
20= 5W. The correct answer is D.
Example 6:FFT analysis gives us ... ?a. Amplitude vs. time graphb. Current vs. time graphc. Amplitude vs. frequency graphd. Resistance vs. time graphAnswer: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 applicationsd. dB or dBm are the only units accepted by many people.Answer:The correct answer is A.
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Electromagnetic Waves and Radiation
-Electromagnetic wave radiation is a propagation of energy due to time vary-ing (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 magneticfield as a wave (changing with sin or cos function).
Capacitors and Inductors
In addition to resistors, capacitors and inductors are the two major electricelements.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 isshown in Fig.13.
Figure 13: Inductor
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- Both capacitor and inductors can be used to store energy.
- Modern batteries utilizes the concept of supercapacitance where large en-ergies can be stored inside a capacitors usually through modification of itsdimensions and the properties of the insulating material.
- Capacitors have a resistance that is inversely proportional to the appliedfrequency.
- Inductors have a resistance that is directly proportional to the appliedfrequency.
- capacitors and inductors can be combined to obtain electric resonators.
- Resonance is a concept where the resistance of an object becomes zeroat a specific frequency.
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