eee117l network analysis laboratory 2 passive low-pass filter

Dr. Milica Markovic Network Analysis Laboratory page 1

EEE117L Network Analysis

Laboratory 2

Passive Low-Pass Filter

Instructor: Dr. Milica MarkovicOffice: Riverside Hall 3028

Email: [email protected]:http://gaia.ecs.csus.edu/˜milica

Figure 1: http://www.home.agilent.com/

California State University Sacramento EEE117L revised: 6. September, 2012

Contents

1 Step I - Calculations 4

2 Step II - Simulations 62.1 Single Frequency AC Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

3 Step III - Measurements 10

4 Discussion 11

A Background Concepts Review 12

Background Concepts Review 12A.1 Sinusoidal Signals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12A.2 Decibels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15A.3 Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16A.4 Phasors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17A.5 Using Vectors to Represent Phasors in an example . . . . . . . . . . . . . . . . . . . . 21

B Equipment Operation 23B.1 Using DMM to measure AC voltages and currents . . . . . . . . . . . . . . . . . . . . 23B.2 Function Generator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23B.3 Oscilloscope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

B.3.1 Checking the probe settings . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25B.3.2 Vertical Controls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25B.3.3 Estimating Measurements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

B.4 Adding measurements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27B.5 Horizontal Controls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32B.6 Trigger . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

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Purpose and Learning Objectives

Monitor your learning in the laboratory. Periodically and throughout the lab, go back to this page,and check the learning objectives you have achieved. Write or ask questions and answers. Be anengineer. Look at the handouts on class web page. Search the Internet. Talk to your group. Askthe instructor.

1. Review sinusoidal signals, complex numbers and phasors.

(a) Describe sinusoidal signal

(b) Summarize how and why is the phasor theory applied to solution of circuits.

(c) Solve linear circuits using phasors

2. Use Agilent’s Digital Multimeter for AC voltage measurements

3. Use Tektronix Oscilloscope for AC voltage measurements

4. Use Agilent’s Signal Generator

5. Perform Single Freuency AC Analysis using National Instrument’s Multisim.

6. Perform segment of engineering design process: analysis, simulation, measurements, compari-son.

7. Use protoboard to make a simple RC circuit.

8. Practice the common laboratory procedures to document experimental and simulation results.

The purpose of this experiment is to provide the students with an introduction to the oscillo-scope, function generator, DMM and a low-pass filter.

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Chapter 1

Step I - Calculations

The first important task in this lab is to understand the background theory and operation of an RCcircuit, so that we know what to expect during simulations and measurements. Consider circuitsin Figure 1.1. The following steps are suggested to evaluate circuit performance for at least thefollowing frequencies: 1Hz, 10Hz, 100Hz, 500Hz, 1kHz, 10kHz, 1MHz.

1. Derive the equations for RC circuit current and voltages in terms of resistance R, capacitanceC, frequency f and input voltage Vin. See appendix.

2. Assume that the input signal to the circuit is Vp = 2V V, with zero phase.

3. Calculate the magnitude and phase expressions for all currents and voltages in the circuit,if R=1kΩ and C=1µF, at (at least) the 7 frequencies above. In order to do that, you willneed to put your data in an excel file, matlab, or some other software you know how to use.Alternatively you can do calculations with your calculator, but it’s going to take a bit longer.

4. Sketch all signals as a function of time (label the graphs, see appendix), and then sketch vectors(try to keep proportion approximately valid). See appendix.

5. Discuss how KCL and KVL differ for different frequencies? Why and how are the vectordiagrams different? What happens at lower frequencies? What happens at higher frequencies?

Measure your resistor and capacitor using RCL meter. How much are measured and nominalvalues different? Are the values within tolerance?

In the next step, we will simulate this circuit.

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(a) Voltage Divider.

Figure 1.1: Multisim Single Frequency AC Simulation.


Chapter 2

Step II - Simulations

2.1 Single Frequency AC Analysis

Build the circuit as shown in Figure 2.1. Place two two instruments on the scematic diagram: signalgenerator (red arrow) and osciloscope (blue dotted arrow). Right click on each instrument and selectproperties to display the diagrams. Then build the circuit, and connect it as shown. Select SingleFrequency AC Analysis by going to : Simulate, Analyses, Single Frequency AC Analysis, as shownin Figure 2.3. In oder to display two channels on the scope with different colors, change the color ofChannel B+s trace to green:

1. Right click on the wire that connects to Channel B+ (do not click on the scope icon itself)

2. Select ”Color Segment”

3. Choose green color

Perform the simulation. Set the voltage scale and time scale to an appropriate setting, so thatyou see several sinusoids. First estimate, then use markers to measure the signal magnitudes andtime-delay on the scopes.

To estimate the measurement, read the voltage scale V/div and time scale s/div. Count how manydivisions in x (to estimate amplitude - voltage peak value) and y direction (to estimate period), thenmultiply with the scale factor. Write the estimated values, then use markers to measure preciselyamplitude and period of the signal.

To use the markers, locate two triangles in the upper left corner of Multisim scope screen asshown in Figure 2.3 (a), then move the markers on one waveform. Look at the the three columns.The first column represents the time on the x-axis where the marker is positioned. The second andthird column represent the Y position of the marker of Channel A and Channel B. For example,in Figure 2.3 (b), I see that the position of the first marker is located on Channel A at (126.86 µs,1.037V), and on Channel B at (126.86 µs, 2.000V) the second marker is positioned at (623.34 µs,-1.019V) and on Channel B at (623.134 µs, -1.997V). You can also read the difference bewteen themarkers ∆T as (∆x,∆y)=(X2-X1, Y2-Y1)=(496.269 µs, -2.057V). From the time delay, calculatephases. If you don’t know how to calculate phase from time-delay, consult appendix. Repeat the

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Figure 2.1: Voltage and Current Divider.





simulation of the circuit at the same set of frequencies as in calculations. Record allamplitudes and time-delays and calculate phases.



(a) Locate markers (b) Move markers.



Chapter 3

Step III - Measurements

Review equipment operation appendix before attempting this section.Make sure the output impedance of the function generator at SYS menu is set to

High ZConnect the multimeter to the function generator. Remember that Function Generator displays

RMS values. Compare reading of Function Generator and DMM. They should show different num-bers, but describe the same quantity. Explain.

Build the R-C circuit that you previously calculated and simulated. Measure the RMS voltagesat the input of the generator and the capacitor using DMM. Estimate, then measure amplitude,period and phase between two voltages. Compare calculations, simulations and measurements.

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Chapter 4

Discussion

1. Was KVL verified?

2. How is KVL in this situation different than the one in the previous lab with resistive voltagedivider?

3. How does change of frequency affect the output voltage magnitude and phase? Relate theimpedance of capacitor for different frequency points you measured with your measurements.

4. Compare computed, simulated and measured values. How do they agree? Compute errors.Now measure the values for resistor and capacitor you used, re-simulate your circuit and see ifthe agreement between measurements and simulations improves?

5. Explain why measurement displays on DMM and scope show different value (ex. 0.7 and 1V)when measuring the same waveform.

1

1This laboratory was written by the following faculty at California State University Sacramento: Jim Simes, StevedeHaas, JP Bayard, Tom Matthews, Russ Tatro and Milica Markovic.

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Appendix A

Background Concepts Review

A.1 Sinusoidal Signals

Sinusoidal signals are important because all periodic signals can be represented with sinusoidal signalsof different amplitudes and phases using Fourier series.

Typical sinusoidal signal is shown in Figure A.1. On the y-axis is the instantaneous value of thesinusoidal voltage and on the x-axis is time. Instantaneous values of voltage change from -1V to1V with time. Sinusoidal signals can be characterized by the following parameters: peak amplitude,peak-to-peak, average, rms, period, time-delay and phase. Peak amplitude, peak-to-peak, averageand rms values, are read on the y-axis in Figure A.1, whereas period, time delay and phase are readon the x-axis.

1. Peak amplitude is measured on the y-axis as the length from the average value of the signal (inthis case zero) to the maximum value of the signal (in this case 1). For signal shown in FigureA.1, peak amplitude has a constant value of Vp = 1. Peak amplitude is NOT an instantaneousvalue of the signal, and it does not vary with time. Sometimes amplitude and peak-amplitudeare used interchangeably. Other times, amplitude is used to denote an instantaneous value ofthe signal, and peak-amplitude is meant to denote the maximum value of amplitude. We willuse amplitude to mean peak-amplitude. When we want to emphasize an instantaneous valueof the signal, we will call it an instantaneous value.

2. Peak-to-peak is measured from the minimum value of the function (in this case -1) to themaximum value of the function (in this case 1). For signal shown in Figure A.1, peak-to-peakvoltage has a constant value of Vpp = 2.

3. RMS or root-mean-square is defined as vrms = 1T

√∫ T0v(t)2dt. For signal shown in Figure A.1,

and other sinusoidal signals of this form, vrms = Vp√2

= 1√2

= 0.707. Root mean square value isimportant because it represents the equivalent amount of DC power.

4. Average value vave1 = 1T

∫ T0v(t)dt. For signal shown in Figure A.1, average value is Vave1 = 0

because the function has the same area under the function in the positive and negative cycle.

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5. Period is measured on the x-axis as the length of one full cycle of the sinusoidal signal. Forsignal shown in Figure A.1, this value is period = T

6. Time delay represents the lag (or lead) of one function with respect to another. For example,in Figure A.1, function cos(ωt − 90o) is time-delayed for τ = T

4with respect to cos(ωt). To

find the time delay for a sinusoidal signal from its phase, we look at the way to represent thephase 90o in terms of product of frequency and time. Since in the sinusoidal signal expressioncos(ωt + Θ) phase Θ is added to ωt term, the phase has the same units as ωt, and can berepresented as the product of ωτ = θ, τ = θ

ω, where τ represents the time delay.

7. Phase of the signal is shown in Figure A.2(d)-(e).

Review signals shown in Figure A.2. Signals in Figure A.2(a)-(b) are shown as a function of time,whereas signals in Figure A.2(c)-(d) are shown as a function of angle. See how are the graphs thesame and how are they different.

Figure A.1: Vocabulary used in describing sinusoidal signals.



(a) sin(ωt) (b) Sinusoidal signal shifted for time delay −π/4ω

(c) Sinusoidal signal as a function of angle ωt (d) Sinusoidal signal as a function of angle ωt with aphase shift of −π/4

(e) Sinusoidal signal as a function of angle ωt with aphase shift of +π/4

Figure A.2: Sinusoidal signal as a function of time (a)-(b) and angle (c)-(e).



(a) Sinusoidal signals of different frequencies sin(ωt) (b) Sinusoidal signals of different amplitudes sin(ωt)

Figure A.3: Comparison of sinusoidal signals.

A.2 Decibels

We will often represent the ratio off the to voltages in decibels. Decibel is a unit that is derived fromthe unit Bell. Bells are describing an order of magnitude difference between two values. For example,one Bell represents the ratio of 10. 1B = log P1

P2. This ratio is too large to represent quantities in

microwave engineering. Decibel is defined as the ratio of two powers 10 log P1

P2. For example if we

want to say that the output power is twice the input power we say that the power gain is 3dB.

G = 10logPoutPin

G = 10 log 2

G = 3dB (A.1)

The ratio of voltages is derived from the definition above, because voltages are proportional topower.

G = 10 logV 2out

V 2in

G = 20 logVoutVin

(A.2)

if the output voltage is 1.4 times the input voltage the voltage gain is equal to 3dB, whichrepresents the power ratio of two.



A.3 Complex Numbers

1. A complex number z can be represented in Cartesian coordinate system, Equation A.3 or Polarcoordinate system, Equation A.4 form.

Figure A.4: Complex number z in rectangular and polar coordinates.

z = x+ jy (A.3)

z = |z|ejΘ (A.4)

x is the real part, y is the imaginary part, |z| is the magnitude and Θ is the angle of the complexnumber. Note that we can represent this complex number as a vector.

2. Euler’s Identity is used to convert time domain signal to phasors.

ejΘ = cosΘ + jsinΘ (A.5)

3. Cartesian and polar form representation are used in phasors.

|z| =√x2 + y2 (A.6)

Θ = arctgy

x(A.7)

4. Complex Conjugate is often seen when finding the conditions for maximum power transfer.

z∗ = (x+ jy)∗ = x− jy = |z|e−jΘ (A.8)



5. Complex number addition and subtraction is often seen went to complex impedances are placedin series and the equivalent complex impedance has to be found. the easiest way to add twocomplex numbers is to find Cartesian representation of both and then add the real partsseparately and the imaginary part separately.

z1 = x1 + jy1 (A.9)

z2 = x2 + jy2 (A.10)

z1 + z2 = x1 + x2 + j(y1 + y2) (A.11)

6. Multiplication and Division are often seen the in calculation of the transfer function of a circuit.The easiest way to divide two complex numbers is to find the polar representation of both andthen divide the amplitudes and subtract the phases.

z1 = |z1|ejΘ1 (A.12)

z2 = |z2|ejΘ2 (A.13)

z1

z2

=|z1||z2|

ejΘ1−Θ2 (A.14)

7. Some examples. Find the magnitude and phase of a complex number

−1 = ej180o (A.15)

j = ej90o (A.16)

(A.17)

A.4 Phasors

The learning objective of this section is to show how to solve a circuit in frequency domain usingphasors. we will review of fat phasers with an RC circuit example shown in Figure A.6. To solvethis circuit in the time domain we apply Kirchoff’s voltage law as shown in Equation A.18 -A.19.

The circuit in Figure A.6 is a simple RC circuit. KVL equation in the time domain is given inEquation A.18.

vs(t) = vR(t) + vC(t) (A.18)

vs(t) = Ri+1

C

∫i(t)dt (A.19)

Where

vs(t) = Acos(ωt+ Θ) (A.20)




In order to solve this circuit we have to solve a differential equation. Use of phasors simplify theequations significantly. Differential equations become a set of linear equations. Phasor is anothername for a complex number in polar coordinate system.

In order to use phasors, the circuit has to be linear. Circuits that have only capacitors, inductorsand resistors are linear circuits. In a linear circuit, all currents and voltages are at the frequencyof the generator. That means that we don’t have to keep track of the frequency of voltages andcurrents when we are solving the circuit. We know the frequency once we know the frequency of thegenerator. The quantities that will differ for different currents and voltages is the amplitude andphase of the signal. Phasors allow us to drop the information about the frequency of the signal, andonly keep track of the magnitude and phase of the signal. In order to remove cos(ωt) term fromthe equations we have to use complex numbers. To write cos(ωt) 1 in a concise form as a complexnumber, we add to the forcing function a sinusoidal imaginary term.

A cos(ωt+ Θ) + jAsin(ωt+ Θ) (A.21)

We have to make sure later when we are done with our calculations with complex number, thatwe only take the real part of the final expression. It seems that we made the above expression morecomplicated, however, if we remember Euler’s identity, the expression becomes

vs(t) = V cos(ωt+ ΘV ) = ReAcos(ωt+ Θ) + jAsin(ωt+ Θ) = ReAej(ωt+Θ) = ReAejΘejωt(A.22)

In Equation A.22 we extracted the phase and amplitude information and separated it from thefrequency. The amplitude and phase information is called phasor VS(jω) = AejΘ. Why is thisexpression better then the one with a cos(ωt) and how can we remove t from Equation A.19? Toanswer this question, we first have to write general expressions for voltage v(t) and current i(t) inthe Equation A.19:

v(t) = ReV cos(ωt+ ΘV ) + jV sin(ωt+ ΘV ) = ReV ej(ωt+ΘV ) = ReV ejΘV ejωt (A.23)

1It is customary to use cos(ωt) for our time-domain signal. If signals in a circuit are given in terms of sin(ωt),the sin function has to be converted to a cosine. To do that, subtract 90o from the phase of the sinusoid, becausesin(ωt) = cos(ωt− 90o)



If we look at the first and last expression in Equation A.23 we see that the time domain signal isthe real part of the product of phasor and the ejωt term.

v(t) = ReV ejΘV ejωt (A.24)

Similarly for current

i(t) = I cos(ωt+ ΘI) = ReIcos(ωt+ ΘI) + jI sin(ωt+ ΘI) = ReIej(ωt+ΘI) = ReIejΘIejωt(A.25)

If we look at the first and last expression in Equation A.25 we get a similar expression for current.

i(t) = ReIejΘIejωt (A.26)

In Equations A.23-A.25 V and I are voltage and current amplitudes, and ΘV and ΘI are voltageand current phases. The voltage on the resistor is then given in Equation A.27

vR(t) = R× i(t) = R×ReIejΘIejωt = ReR× IejΘIejωt (A.27)

The voltage on the capacitor is a bit more complicated. We know that i(t) = ReIejΘIejωt, butwhat is the integral of i(t)?

vC(t) =1

C

∫ReIejΘIejωtdt (A.28)

Now if integral and Re exchange places, and if we take all time-independent quantities in frontof the integral,

vC(t) =1

C

∫i(t)dt = Re 1

C

∫IejΘIejωtdt = Re 1

CIejΘI

∫ejωtdt (A.29)

vC(t) = Re 1

CIejΘI

1

jωejωt = Re 1

jωCIejΘIejωt (A.30)

We now replace the time-domain quantities in equation A.19 with these newly developed expres-sions.

vs(t) = vR(t) + vC(t) (A.31)

ReAejΘejωt = ReR× IejΘIejωt+Re 1

jωCIejΘIejωt (A.32)

A common term in the previous equation is ejωt, and we can now drop Re, as long as we laterremember to take only the real part of the expresion for the phasor of voltage and current to get thetime domain expression. We can now write the equation as



VSejΘVS = R× IejΘI +

1

jωCIejΘI (A.33)

VS(jω) = RI(jω) +I(jω)

jωC(A.34)

Since this is a linear equation, we can easily solve it:

I(jω) =VS(jω)

R + 1jωC

(A.35)

Let’s say that the values are given for R, C, ω and VS such that the phasor of the current isI = 3ej45o . To obtain the signal in the time domain, we multiply the phasor I with the ejωt term,and then we find the real part of the expression to obtain its current in the time domain, as shownin Figure A.36

i(t) = Re3ej45oejωt = Re3ejωt+45o = Re3 cos(ωt+ 45o) + j3 sin(ωt+ 45o = 3 cos(ωt+ 45o)(A.36)

In case we have an inductor in the circuit, the voltage on an inductor can be derived as shown inFigure A.37.

vL(t) = L∂i(t)

∂t= L

ReIejΘIejωt∂t

dt = ReLIejΘI∂ejωt

∂t = ReLIejΘI jωejωt = RejωLIejΘIejωt(A.37)


Step-by-step instructions on how to solve circuits using phasors is given as follows:

1. Adopt cosine reference for generator voltage or current.

2. replace all impedances with their phasor expressions,

3. write KVL and KCL, or use other Network Analysis techniques.

4. find the phasor expression for the required current or voltage.

5. multiply the phasor with ejωt

6. find the real part of the above expression to get the current in the time domain.



cicruit element impedance low frequencies f → 0 high frequencies f → infcapacitor 1

jωC∞ 0

inductor jωL 0 ∞

Table A.1: Impedance of the capacitor and inductors and their equivalent impedances at high andlow frequencies.

A.5 Using Vectors to Represent Phasors in an example

1. Calculate on paper the magnitude and phase of the current and voltages in a series RC circuitshown in Figure A.7 (a) if the circuit is driven with a frequency of 1 GHz (phase is zero) and

R=1kΩ, C= 12π

10−12

F. The vector plot in Figure A.7 (c)-(d) shows how the three vectors Vs,VC and VR add if you want to show their absolute phases as well as magnitudes. When yousketch your phasor diagram, you should start from the voltage on the resistor on the horizontalaxis, then add the magnitude of the voltage on the capacitor at the end of the first vector asin Figure A.8 to represent R− j 1

ωCI as shown in Figure ??. Then the source voltage will be

the sum of the two vectors.

(a) RC circuit in ADS (b) Sinusoidal signal as a function of angle ωt

(c) Points represent polar plot of com-plex voltages in RC circuit.

(d) Vectors are drawn in the polar plotof complex voltages in RC circuit. Notehow they add up to 1V.

Figure A.7: Adding phasors using vectors in Agilent’s ADS.

(a) What are the magnitude, phase and time delay of the source voltage?

(b) What are the magnitude, phase and time delay of the voltage across the resistor?



Figure A.8: Skeching phasors using vector magnitudes on paper.

(c) What are the magnitude, phase and time delay of the voltage across the capacitor?

(d) The simulated voltage across the resistor is about 0.5V and the simulated voltage acrossthe capacitor is about 0.85V. If we use KVL 0.85V + 0.5V 6= 1V . Why? Look at FigureA.7 to help you with answer this question.

(e) If you know the current through circuit, capacitance, resistance and the frequency ofoperation, could you plot the phasor diagram to find the input voltage? How would youdifferentiate between the voltage on an inductor and a capacitor?


Appendix B

Equipment Operation

B.1 Using DMM to measure AC voltages and currents

DMM display RMS not amplitude. See appendix for definition of RMS, amplitude etc. Explorethe front pannel of DMM and set it to measure AC voltage. You will still attach connectors to theterminal for voltage as in the last lab.

B.2 Function Generator

Signal generators are devices designed to produce signals with controlled parameters (e.g., frequency,amplitude, waveform, etc.). Types of the signal generators are: function generators, arbitrary wave-forem generators, vector signal generators and sweepers. Front pannel of the function generator inour lab is shown in Figure B.1 (a). Explore the front pannel and set a sinusoidal voltge frequency to1Khz and amplitude to 2V.

The function generator has a fixed output impedance of 50 ohms on the OUTPUTterminal. You can specify whether you are terminating the output into a 50Ω load oran open circuit. Incorrect impedance matching between the source and load will resultin an output amplitude or dc offset which does not match the specified value. FollowFigure B.1 (b) to change the impedance to high-z, turn on the menu, move across tothe D:SYS MENU, move down a level to the OUT TERM, move down a level and thenacross to the HIGH Z. Save the change and turn-off the menu. We will discuss thisin greater detail later. For now, just note that the function generator displays the value of theoutput voltage Vout based on calculation of the OUT TERM and Vp setting. If you set the OUTTERM to HIGH Z and connect resistance of over 100Ω or so to the output of the generator, thefunction generator display will be accurate. If you set the OUT TERM to 50Ω and attach a 50Ωresistor at the output, the function generator display will be accurate as well. The problems appearwhen you set OUT TERM to 50Ω and attach a resistor of over 100Ω to the output, and vice versa.

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(a) Signal Generator.

(b) After moving the Vertical Position button

Figure B.1: Agilent Signal Generator.



B.3 Oscilloscope

The oscilloscope is the most versatile measurement instrument used in the electrical engineeringlaboratory. It allows the user to not only view waveshapes but also shows the relative shape andpositions of several waveshapes. There are four main groups of controls on the oscilloscope, shownin Figure B.2. First is the input (vertical) control section (red rectangle) that controls the V perdivision. Second is the time base (horizontal) that controls the time displayed on the horizontal axis(green rectangle). Third is the trigger section that synchronizes the horizontal sweep time with theinput signals (yellow rectangle). The final section is the measurement section that provides voltageand time measurements (purple square). Use Autoset to quickly display the signal on your screen.

B.3.1 Checking the probe settings

To make sure your probe is set to the appropriate attenuation factor, press the channel you wantto check, for example blue button 2. Then press soft-key More on the bottom of the screen, thenkeep clicking on the More softkey until Probe Setup is highlighted. You should be able to setup theprobe attenuation to x1, x10 etc if your probe is not Tek6139A. Tek6139A is automatically set toattenuation x10.

B.3.2 Vertical Controls

Attach the probe 1 from the signal generator to Channel 1 of the scope. Notice that the trace onthe scope screen is color coded to the channel where the scope probe is connected. In this case, thetrace on the screen should be yellow. If no trace is observed push Autoset.

From the vertical control section select Channel 1 by pushing the yellow button. You will see aselection of menu choices at the bottom of the screen. From the on-screen menu labeled Couplingselect the ground symbol (it has a symbol and not a name). There should be a trace on the screenwith a caret on the left vertical axis. This caret shows where 0 volts is located. Observe the trace andcaret as the vertical position knob (in the vertical control section) for channel 1 is rotated. Adjustthe vertical position so that ground is in the center of the screen.

Now set the function generator for a 1 kHz sinusoid with a (function generator front panel displaysetting of) 1 volt peak-to-peak amplitude, zero volts DC offset. Connect the probe from channel 1 tothe function generator. Set Channel 1 coupling to DC. The probe will auto-sense to X10 setting. Setthe trigger source to Channel 1. Hint: remember Menu of the Trigger section. Adjust the oscilloscopetime base control for about 4 cycles of the sine wave. Add a DC offset of 1 volt on the functiongenerator. Now you see both DC and AC signals together. Check where zero is by setting couplingto ground. Observe 0V. Then change the channel coupling (from the on-screen menu) from DC toAC. Now you see both AC and DC voltages.

B.3.3 Estimating Measurements

To estimate measurements, use the Volts/Division and/or Seconds/division (division is one largesquares on the scope screen). For example on the screen above, the second (bottom, blue) trace is set



Figure B.2: To make sure your probe is set to the appropriate attenuation factor, press the channelyou want to check, for example blue button.



Figure B.3: To make sure your probe is set to the appropriate attenuation factor, press the channelyou want to check, for example blue button.

to 1.00V/division, so that means that the amplitude of the signal is about 0.5V (about 0.5 of a square),and the Seconds/division is set to 400micro-seconds, so the period is about 2.5 squares, so the periodof the signal is about 2.5*400 microseconds=1000 microseconds=1ms, and the frequency is 1/period,so 1kHz. This is a manual way to estimate the frequency and the period of the signal. You can alsoestimate what is the time delay of one signal with respect to the other. It is about 0.6 of a horizontalsquare, which is 0.6*400microseconds=240microseconds. Since the frequency is 1kHz, the angular fre-quency is 2*pi*f=6.28kradsec, and the phase is omega*t=6.28kradsec*240microseconds=360*240*10(−3) 86degrees. What should be the exact value of phase shift between the voltage on the resistor andcapacitor in an RC circuit?

B.4 Adding measurements

Follow the visual guide on Figure B.10 to add measurements to your graphs:

1. Press Measure in Wave Inspector Manu section

2. First remove all previous measurements, if any as shown in Figure B.6. To do that click onbottom softkey Remove Measurements, then on right softkey Remove All Measurements.

3. To add measurement, click on bottom softkey Add Measurement as shown in Figure B.7

4. Use Multipupose button a to select the measurement you want to display. For example letsmove it until you find Peak-to-peak measurement. Click from the right softkey menu OK AddMeasurement

5. The screen should show peak-to-peak measurements.

6. You can also add cursors to measure specific feature of the signal you are interested in. Followvisual directions in Figure B.8.



(a) To change the vertical position of the signal on the screen press Positionbutton under vertical menu..

(b) After moving the Vertical Position button

Figure B.4: Vertical Control Before and After.



Figure B.5: To change the V/div of the signal on the screen press move Scale button under Verticalmenu.

(a) . (b)

Figure B.6: Removing measurements from a previous experiment.



(a) To use cursors for more accuratemeasurements, select the trace youwant to measure

(b) Then click on cursors button,next to multipurpose button a

(c) Then use Multipurpose buttonsa and b to move cursors to specificpoints on the waveform, and see thecursor points and the delta betweenthem on the window shown

(d) Then use Multipurpose buttonsa and b to move cursors to specificpoints on the waveform, and see thecursor points and the delta betweenthem on the window shown

Figure B.7: Adding Measurements to the Screen.



(a) To use cursors for more accuratemeasurements, select the trace youwant to measure

(b) Then click on cursors button, next to multipurpose button a

(c) Then use Multipurpose buttons a and b to movecursors to specific points on the waveform, and seethe cursor points and the delta between them onthe window shown

(d) Then use Multipurpose buttons a and b to move cursors to specificpoints on the waveform, and see the cursor points and the delta betweenthem on the window shown

Figure B.8: Adding Cursors.



B.5 Horizontal Controls

To change s/div use horizontal controls. Change the scale button until you can see couple of periods.You can also move the waveform using Position button.

(a) To change the horizontal scale

(b) After changing the horizontal scale

Figure B.9: Displaying .

B.6 Trigger

If your measurement is moving accross the screen as shown in Figure B.11 it is very likely that youhave triggering problem. The possible solution to triggering problem is visually shown in Figure B.11(b) and (c).

We will not explore this function in more detail. Set the function generator to 1 Vp-p, 1 kHz, 0volts DC offset and connect it to the Channel 1 of the scope. Set the trigger SOURCE to 1. Next



(a) Use cursors for more accuratemeasurements, select the trace youwant to measure

(b) Click on cursors button, next to multipurpose button a

(c) Use Multipurpose buttons a and b to move cur-sors to specific points on the waveform, and see thecursor points and the delta between them on thewindow shown

(d) Use Multipurpose buttons a and b to move cursors to specific pointson the waveform, and see the cursor points and the delta between themon the window shown

Figure B.10: Adding Measurements.



select trigger MODE and from the on-screen menu select Auto (Untriggered Roll). If no trace is seenon the oscilloscope, turn the level control in the triggering control section. You should momentarilysee a straight line across the screen. The oscilloscope will trigger at the point where this line intersectsthe signal voltage. Continue to turn the level control until a trace is observed.

Now observe how the triggering point changes as the trigger level changes. From the triggeringcontrol section select Slope and from the on-screen menu select the symbol for rising and falling slope(arrow up or down).

Explain what happens to the triggering point as the slope is toggled from rising to falling slope.The slope of the trigger determines which part of the input signal is used to trigger the start of thetime sweep of the oscilloscope. T Move the trigger level with both positive slope and negative slopeselected.

Explain what happens when the triggering mode is toggled from Auto (Untriggered Roll) toNormal. This observation should be made for triggering levels within the signal voltage and withtriggering levels outside of the signal voltage range.



(a) Triggering problem

(b) Press Menu Key

(c) Move Level until the waveform stops sweeping

Figure B.11: Triggering problem and solution.


eee117l network analysis laboratory 2 passive low-pass filter

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