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

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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

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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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