PEEII Lab Manual

332:224 Principles of Electrical Engineering II Laboratory

Instructions for Lab Reports

1. General

• Each student must submit his/her own report despite the fact that a group of students may

collaborate in the conduction of the experiment. It is understandable that lab partners will

have similarities in their data but the presentations cannot be identical and certainly editorial

items such as the Summary are expected to be completely individualized1.

• The report is due within two weeks from the day the experiment was conducted2 unless prior

permission is given otherwise3.

1 Plagiarism, i.e. copying others’ work and presenting as one’s own, is a form of cheating and it

is treated as such

2 Except of the last lab that is due within 2 weeks or by 4:30pm on the last day of classes,

whichever is earlier.

3 One missed lab or missed report deadline results in an automatic maximum of “C” for the

course, regardless of the

grade in the remaining experiments; two such omissions result in an automatic “F”.

2. Content

All reports must contain the following basic aspects in addition to any other specifically required

for the particular experiment:

1. Cover sheet with the following:

a. title of the experiment,

b. student name,

c. name(s) of lab partner(s) who participated,

d. lab section number and instructor,

e. date when the experiment was carried out.

2. Title and aim of the experiment in student’s own words.

3. The procedure followed along with the circuit diagrams.

4. Tables of the experimental data.

5. Determination of any parameters deduced from the experimental data.

6. Explicit answers to the questions posed in section 5, titled “Report”.

7. Conclusions written in student’s own words.

8. References, if any were used in theoretical considerations.

9. Software simulations appended to the report.

3. Style

• All reports must be written neatly.

• Separate report in sections and label with numbers corresponding to those of the instructions.

• Always use passive voice: “The voltage was measured” not “We measured the voltage”.

Department of Electrical & Computer Engineering 94 Brett Rd. Piscataway NJ 08854-2820

© Rutgers UniversityAuthored by P. Sannuti

Latest revision: December 16, 2005, 2004by P. Panayotatos

School of EngineeringDepartment of Electrical and Computer Engineering


Experiment 1

Series and Parallel Resonance

1 Introduction

Objectives • To introduce frequency response by studying thecharacteristics of two resonant circuits on either side ofresonance

Overview

In this experiment, the general topic of frequency response is introduced by studying thefrequency-selectivity characteristics of two specific circuit structures. The first is referred toas the series-resonant circuit and the second as the parallel-resonant circuit.The relevant equations and characteristic bell-shaped curves of the frequency responsearound resonance are given in section 2.Prelab exercises are designed to enhance understanding of the concepts and calculateanticipated values subsequently measured in the lab.Current and voltage are then measured in the two resonant circuits as functions of frequencyand characteristic frequencies (resonance and 3-dB points) are experimentally determined.

PEEII-III-2/12

2 Theory

2.1 Frequency Domain AnalysisIn electrical engineering and elsewhere, frequency domain analysis or otherwise known asFourier analysis has been predominantly used ever since the work of French physicist JeanBaptiste Joseph Fourier in the early 19th century. The pioneering work of Fourier led towhat are now known as Fourier Series representations of periodic signals and FourierTransform representations of periodic signals. A periodic signal of interest in engineeringcan be represented in terms of a Fourier Series1 which is a weighted linear combination ofsinusoids of harmonically related frequencies. Each frequency among harmonically relatedfrequencies is an integer multiple of a particular frequency known as the fundamentalfrequency. The number of harmonically related sinusoids present in the Fourier Seriesrepresentation of a periodic signal could be finite or countably infinite. Since a periodicsignal can be viewed as being composed of a number of sinusoids, in order to specify aperiodic signal, one could equivalently specify the amplitude and phase of each sinusoidpresent in the signal. Such a specification constitutes the frequency domain description of aperiodic signal. Similarly, in Fourier Transform representation, under some naturalconditions, an aperiodic signal or a signal which is not necessarily periodic can be viewed asbeing composed of uncountably infinite number of sinusoids or a continuum of sinusoids. Inthis case, instead of being a weighted sum of harmonically related sinusoids, an aperiodicsignal is a weighted integral of sinusoids of frequencies which are all not harmonicallyrelated. Again, instead of specifying an aperiodic signal in terms of the time variable t, onecan equivalently specify the amplitude and phase density of each sinusoid of frequency !contained in the signal. Such a description obviously uses the frequency variable ! as anindependent variable, and thus it is said to be the frequency domain or !"domain descriptionof the given time domain signal. In this way, a time domain signal is transformed to afrequency domain signal. Of course, once the frequency domain description of a signal isknown, one can compose all the sinusoids present in the signal to form its time domaindescription. However, it is important to recognize that the frequency domain description issimply a mathematical tool. In engineering, signals exist in a physically meaningful domainsuch as time domain. The frequency domain description only serves to help for the betterunderstanding of certain signal characteristics.

1 A more detailed treatment can be found in the text starting with section 16.1.

PEEII-III-3/12

2.2 Series ResonanceThe basic series-resonant circuit is shown in fig. 1. Of interest here in how the steady stateamplitude and the phase angle of the current vary with the frequency of the sinusoidalvoltage source. As the frequency of the source changes, the maximum amplitude of thesource voltage (Vm) is held constant.

Vs++- VR

VL

+

++

-

--VC

R

CLi

Vs = Vmcos(!t) i = Imcos(!t + #)

The frequency at which the reactances of the inductance and the capacitance cancel eachother is the resonant frequency (or the unity power factor frequency) of this circuit. Thisoccurs at

!o =1LC (1)

Since i = VR /R, then the current i can be studied by studying the voltage across the resistor.The current i has the expression

i = Imcos(!t + #)where

Im =

Vm

R2 + !L " 1!C

#$%

&'(2 (2A)

and

! = " tan"1

#L " 1#C

R

$

%

&&&

'

(

)))

(2B)

The bandwidth of the series circuit is defined as the range of frequencies in which theamplitude of the current is equal to or greater than 1 / 2 = 2 / 2( ) times its maximum

amplitude, as shown in fig. 2. This yields the bandwidth B = !2-!1= R/L

Fig. 1 The SeriesResonant Circuit

PEEII-III-4/12

Where !2,1 =R2L

"#$

%&'2

+1LC

±R2L (3)

!2,1 are called the half power frequencies or the 3 dB frequencies, i.e the frequencies at whichthe value of Im equals the maximum possible value divided by 2 = 1.414 .

The quality factor Q =!o

B=1R

LC (4)

Then the maximum value of :1- VR occurs at ! = !o (5A)

2- VL occurs at

!o

1" R2C2L

(5B)

3- VC occurs at !o 1" R2C2L (5C)

Im

!!01! 2!

Imax =VmR

m( 2 )1/2 B = R

L 2! 1!-= I m a x/1 .414

Fig. 2 Frequency Response of a Series - Resonant Circuit

PEEII-III-5/12

2.3 Parallel ResonanceThe basic parallel-resonant circuit is shown in fig. 3. Of interest here in how the steady stateamplitude and the phase angle of the output voltage V0 vary with the frequency of thesinusoidal voltage source.

! R V0

+

-

I s C L

Is = Imcos(!t) Vo = Vmcos(!t+#)

If Is = Imcos(!t), then Vo = Vmcos(!t+#) where

Vm =Im

1R2

+ !C " 1!L

#$%

&'(2 (6A)

and

! = " tan"1 R #C "1#L

$%&

'()

$%&

'() (6B)

The resonant frequency is !o =1LC

The 3 dB frequencies are: !2,1 =12RC

"#$

%&'2

+1LC

±12RC (7)

The bandwidth B = !2 - !1 = 1/RC.

The quality factor Q =!o

B= R C

L (8)

Fig. 3 The ParallelResonant circuit

PEEII-III-6/12

!!01! 2!

Vm

B = RL 2! 1!-=

RIm

Im(2) 1/2

R

Fig. 4 Frequency Response of the Parallel - Resonant Circuit

2.4 A More Realistic Parallel Resonance CircuitA more realistic parallel-resonant circuit is shown in fig. 5. It is a more realistic modelbecause it accounts for the losses in the inductor through its d.c. resistance RL.

! R V0

+

-

I s L

LR

C

In this case : !o =1LC

"RL

L#$%

&'(2

(9)

Z(!o ) =RL

RLRC + L (10)

and

Fig. 5 A More RealisticParallel - Resonant Circuit

PEEII-III-7/12

Vo(!o ) = Is (!o )RL

RLRC + L (11)

An analysis of the amplitude of the output voltage as a function of frequency reveals that theamplitude is not maximum at !0. It can be derived that |V0| is maximum when

! = !m = (x - y)1/2 (12)

where x = (a + b)1/2

a =1LC( )2

1+ 2RL

R!"#

$%& b =

RL

L!"#

$%&2 2LC and y =

RL

L!"#

$%&2

This analysis can be followed by first expressing Vo as a function of !, differentiating thisexpression with respect to ! and then finding the value of ! that makes the derivative zero.

3 Prelab Exercises3.1 Derive equations 1, 2, 3, and 4 for the series-resonant circuit in fig. 1.

3.2 Derive equations: 5A, 5B, and 5C for the series-resonant circuit in fig. 1.HINT: | VR |=| I | R where I = Im is given by equation 2A. So VR is maximum whenIm R is maximum i.e., Im is maximum (since R is constant). Similarly solve forVL = | I | ZL and VC = | I | ZC .

3.3 For the series-resonant circuit shown in fig. 6, use equations: 5A, 5B, and 5C to determine

the frequencies at which VR, VC, and VL+RL are maximum.

PEEII-III-8/12

4 Experiments

Suggested Equipment:Tektronix FG 501A 2MHz Function Generator2

Tektronix 504A Counter - TimerHP 54600A or Agilent 54622A OscilloscopeProtek Model B-845 Digital MultimeterLS-400A Inductance Substituter Box620 Ω Resistor0.1 µF CapacitorBreadboardOther circuit elements to be determined by the students.

4.1 Series Resonance

Any function generator used has internal resistance. Also, the inductor has internal resistance.Both need to be determined since all resistances affect the behavior of the circuit.

Function generator resistance

The internal resistance of the function generator will affect the damping of an RLC circuit towhich it is connected. Check the resistance in the following way:

a- With a sine wave output, set the open circuit voltage to some convenient value, say 1V.

b- Connect a pure variable resistance load (potentiometer) thus forming a voltage divider.Adjust R until the terminal voltage falls to one-half the open circuit value. At this point thetwo resistances of the voltage divider have to be equal. Therefore, the resistance of thepotentiometer should now be equal to the internal resistance of the function generator.Disconnect the potentiometer from the circuit and measure its resistance.

Inductor internal resistance

Use the digital ohmmeter to measure the internal resistance of the inductor used.

Measure Rs and RL.

Rs = Ω. RL= Ω.

2 NOTE: The oscillator is designed to work for a very wide range of frequencies but may not be stable at very lowfrequencies, say in the order of 100 Hz or 200Hz. To start with it is a good idea to have the circuit working at somemid-range frequency, say in the order of 1K Hz or 2K Hz, and then change the frequency slowly as needed.

PEEII-III-9/12

Build the circuit shown in fig. 6 using R = 620 Ω, L = 100 mH, and C = 0.1 µF.Apply a sinusoidal input to the circuit and display both input and output on the screen of theoscilloscope.

LR L +

+

-

-VC

R

C

Vs++-

-

VL+ LR

Rs

+

V0

With the frequency varied from 600 Hz to 2,500 Hz in increments of 100 Hz (using thefrequency counter), measure the rms values of VR, VL+RL, and VC using the DVM and thephase angle from the scope (take the phase angle of Vs as the reference). Download thescope trace for your report.

The phase angle between two sinusoidal signals of the same frequency can be determined asfollows: Trace both signals on two different channels with the same horizontal sensitivities(the same horizontal scale). To calibrate the horizontal scale in terms of degrees, one canuse the fact that the angular difference between the two successive zero crossing points of asinusoidal signal is 180 degrees. Thus, by measuring the distance between the successivezero crossing points of either sinusoidal signal, one can calibrate the horizontal scale interms of degrees. To determine the phase difference between the two sinusoidal signals,determine the distance between the zero crossing point of one signal to a similar zerocrossing point of another signal and convert it into degrees.

Also, to save tedious calculations later, set the rms values of Vs to 1.00 volt before eachreading. Make sure that you use the frequency counter for all frequency measurements, andto note the exact frequencies at which VR, VC, and VL+RL are maximum.

Once the maximum output voltage (V0 = VR) is known, vary the frequency and find the 3 dB(the half power) frequencies, f1,2.

Before dismantling the equipment, check your results against those obtained from thetheoretical relationships in eqs 3 & 5. (Make sure to account for the internal resistance of thefunction generator and the d.c. resistance RL of the inductor L in all calculations.)

Fig. 6 A Series -Resonant Circuit

PEEII-III-10/12

f nominal f (Hz) VR VL+RL VC $600700800900

1,0001,1001,2001,3001,4001,5001,6001,7001,8001,9002,0002,1002,2002,3002,4002,500

4.2 Parallel Resonance

Using source transformation, the parallel - resonant circuit in fig. 5 can be represented asshown in fig. 7 where Rs is the internal resistance of the function generator.

R

V0

+

-

L

LR

CVs++-

Rs

Build the circuit of fig. 7 using R = 620 Ω, L = 100 mH, and C = 0.1 µF.

Apply a sinusoidal input to the circuit and display both input and output on the scope. Set therms value of Vs = 1.00 volts.

Fig. 7 A Parallel- Resonant Circuit

PEEII-III-11/12

With the frequency of the source varied from 600 Hz to 2,500 Hz in increments of 100 Hz(using the frequency counter), measure V0 using the DVM, and the phase angle using thescope. Download the scope trace for your report.

Make sure to note the exact frequency, fm, at which V0 is maximum.

Once the maximum output voltage is known,, increase the frequency from 200 Hz and findthe 3 dB frequencies, f1,2.

Before dismantling the equipment, check the measured fm against the theoretical oneobtained from eq. 12.

f nominal f (Hz) Vo $600700800900

1,0001,1001,2001,3001,4001,5001,6001,7001,8001,9002,0002,1002,2002,3002,4002,500

PEEII-III-12/12

5 Report

5.1 In pre-lab exercise 3.3, by using equations: 5A, 5B, and 5C, the frequencies weredetermined at which VR, VC, and VL+RL are maximum. Compare them with thoseexperimentally observed.

5.2 Tabulate the frequency f, VR, VC, and VL+RL and the phase angle measured in Section 4.1.Print out the scope trace and show how the phase angle was measured.

5.3 Plot VR, VC, VL+RL vs frequency on the same graph paper with rectangular coordinates. Circle, on the plot, the resonant frequency and the 3 dB frequencies.

5.4 Use eqs. 1 & 3 to determine the theoretical resonant frequency, the 3 dB frequencies, and the bandwidth. Compare with the experimental ones.

5.5 Tabulate f, V0 and the phase angle measure in Section 4.2. Print out the scope trace andshow how the phase angle was measured.

5.6 Using eqs. 9 & 12, determine the theoretical f0, and fm for the resonant circuit shown in fig. 7. Compare with the experimental ones.

5.7 Plot V0 vs f on a graph paper with rectangular coordinates. Circle, on the plot, f0, fm, and the 3 dB frequencies, f1,2.

5.8 Simulate the series-resonant circuit of fig. 6 in PSpice, and plot VR, VC, and VL+RL vs frequency. Vary the frequency from 600 Hz to 2,500 Hz in increments of 100 Hz. Compare with the experimental plot.

5.9 Simulate the parallel-resonant circuit of fig. 7 in PSpice, and plot V0 vs frequency. Varythe frequency from 600 Hz to 2,500 Hz in increments of 100 Hz. Compare with theexperimental plot.

5.10 Prepare a summary.





Experiment 2

Frequency Response of Filters

1 Introduction

Objectives • To introduce frequency response by studying thecharacteristics of two resonant circuits on either side ofresonance

Overview

This experiment treats the subject of filters both in theory as well as with realized circuits.The Low Pass, High Pass, Band Pass, Band Reject and All Pass filters are introduced. Thesefilters are characterized by their frequency response that indicates how near-ideal their filteroperation actually is. Filters of different specifications are realized as mostly 2nd order activefilters utilizing op-amps. Their frequency response (mostly the magnitude part but the phasepart as well) is measured and the cut-off frequencies are determined.

PEEII-IV-2/15

2 Theory

2.1 Filters1

An ideal filter is a network that allows signals of only certain frequencies to pass whileblocking all others. Depending on the regime of frequencies that are allowed through or notthey are characterized as low-pass, high-pass, band-pass, band-reject and all-pass. There aremany needs for electric filters, some of the more common being those used in radio andtelevision sets, which allow tuning in a certain channel by passing its band of frequencieswhile filtering out those of other channels. The frequency response is divided into magnitude(amplitude) and phase parts. The amplitude curve of a filter will indicate how closely thepractical circuit imitates the ideal filter characteristics that are as follows:

Notice that, depending on the relative magnitude of their corner frequencies, a low pass inseries with a high pass can either completely block all signals or give a band reject. By thesame token, a low pass in parallel with a high pass can give either a bandpass or an all pass.

These ideal characteristics will at best be approximated by real circuits. How closely thiswill be achieved will depend on the frequency response of the particular circuit. The designof electric filters is based on a compromise between deviation from ideality vs. complexity(and cost). The order of the polynomial being used (thus the order of the filter) is the major

1 The subject is treated in detail beginning with section 14.1 of the text.

!!c

!!c2!c1

Low pass

!!c

High pass

!!c2!c1

Bandpass

!!c

All passBandreject

|V(j!)|

|V(j!)|

|V(j!)|

|V(j!)||V(j!)|

PEEII-IV-3/15

indicator: the higher the order2, the more complex the circuit and the closer the frequencyresponse to the ideal curve. Filter operation is considered when the amplitude of the outputsignal of the circuit is relatively large for some frequencies and small for others. Then onecan say that for the former the signal passes and for the latter it does not.

2.2 Practical Filters

2.2.1 Low Pass Filter

A low pass filter is a circuit whose amplitude (magnitude) function decreases as !increases, that is, the circuit passes low frequencies (relatively large amplitudes at theoutput) and rejects high frequencies (relatively small amplitudes at the output) asshown in fig. 1.

H j( )!

H j( )! max

H j( )! max(2)1/2

!c !0

In fig. 1, !c is defined as the (3 dB) frequency, that is the frequency at which theamplitude is (1/2)1/2 = 0.707 times the maximum amplitude. It is traditional to considerthe 3 dB frequency as the cutoff frequency. That is, a low pass filter is said to passfrequencies lower than !c and reject those that are higher than !c. In other words, thepass(ing) band is ! < !c.

2.2.2 High Pass Filter

A high pass filter is a circuit whose amplitude response increases with ! as shown infig. 2.

2 The subject is treated in detail beginning with section 15.4 of the text.

Fig. 1 Low Pass FilterMagnitude Portion ofFrequency Response

PEEII-IV-4/15

H j( )!

H j( )! max

H j( )! max(2)1/2

!c !0

This filter passes frequencies that are higher than the cutoff frequency !c and rejectsthose that are lower than !c. That is, the pass band is ! > !c.

2.2.3 Band Pass Filter

A band pass filter is a circuit which passes the band of frequencies centered around !0

as shown in fig. 3.

c1! c2!!0 !

H j( )!

H j( )! max

H j( )! max(2)1/2

0

Fig. 2 High Pass FilterMagnitude Portion ofFrequency Response

Fig. 3 Band Pass FilterMagnitude Portion ofFrequency Response

PEEII-IV-5/15

!0 is the frequency at which the maximum amplitude occurs, and is called the centerfrequency. The band of frequencies that passes, or the pass band, is defined to be !c1

≤ ! ≤ !c2, where !c1 and !c2 are the cutoff (3 dB) frequencies.

The width of the pass band, given by B = !c2 - !c1 is called the bandwidth.

2.2.4 All Pass Filter

The all pass filter is one whose amplitude is constant, thus, it passes all frequenciesequally well as shown in fig. 4. It is usually put in a cascade when it is desired to keepthe amplitude part of the frequency response unaltered but shift the phase as desired.

H j( )!

H j( )! max

!0

2.2.5 Band Reject Filter

The band reject filter is one that passes all frequencies except a single band. A typicalamplitude response of a band reject filter is shown in fig. 5.

H j( )! max(2)1/2

H j( )!

H j( )! max

c1! c2! !0

Fig. 4 All Pass FilterMagnitude Portion ofFrequency Response

Fig. 5 Band Reject FilterMagnitude Portion ofFrequency Response

PEEII-IV-6/15

As shown in fig. 5, the band reject filter passes all frequencies that are greater than !c2

and smaller than !c1. Its stop band is !c1 < ! < !c2.

2.3 Review of Op Amps3

Filters realized with RLC circuits have limited performance. Filters incorporating suchpassive elements in addition to active elements such as op amps exhibit bettercharacteristics and are called active filters4.

-

+Av=-∞Vi

+

-

V-

V+R i

+

-V0

R0

Inverting Input

Non-Inverting Input

Fig. 1 Operational Amplifier

Characteristics of an Ideal Op Amp1. Input Resistance Ri = ": An infinite input resistance means that no current flows into

or out of either of the input terminals.2. Output Resistance Ro = 0: In this case the output voltage Vo is independent of the

output current3. Open Loop Voltage Gain µ =AV = ±"5:

In order to predict the behavior of an operational amplifier when circuit elements areexternally connected to its terminals, one must understand the constraints imposed on theterminal voltages and currents by the amplifier itself. Those imposed on the terminalvoltages are as follows :

V0 = AV (Vp – Vn) (1)and

3 A somewhat more detailed description can be found in part 2 of Principles of EE I lab IV and a full description insections 5.1-5.2 of the text.4 The subject is treated in detail in chapter 15 of the text.5 The sign of Av (whether positive or negative) depends on the definition of Vi i.e. which of the two input terminalsis defined as the reference. If Vi is defined as (Vn –Vp) then Av is <0; if defined as (Vp-Vn) then it is >0.

Vn

Vp

PEEII-IV-7/15

V- =-VCC ≤ V0 ≤ +VCC = V+ (2)

Eq. 1 states that the output voltage is proportional to the difference between Vp and Vn.If the output voltage Vo is to be finite6 it follows from the definition of voltage gain, thatVi = Vo / Av will go to zero when Av is infinite. This, however, assumes that there is someway for the input to be affected by the output. Indeed this will only happen if there isnegative feedback in the form of a connection between the output and the invertingterminal (closed loop operation). For closed loop operation, it is said that a virtual shortexists between the inverting and noninverting input terminals. This means that if an OpAmp is operating in its linear region (if it is unsaturated) then Vi = 0, or equivalently Vn =Vp.

Eq. 2 states that the output voltage is bounded. In particular, V0 must lie between ±VCC, thepower supply voltages. Else V0 will be at either limiting value, and the Op-Amp is thensaturated. The amplifier is operating in its linear range so long as |V0| < |VCC|.

-

+741

+15V

- 15V

2

3

7

4

6

1

2

34

876

5

Top View

Fig. 2 741 OP-AMP

The chip layout is shown in Fig. 2. The standard procedure on a DIP is to identify pin 1 withthe notch in the end of the chip package. The notch always separates pin 1 from the last pinon the chip. In the case of 741, the notch is between pins 1 and 8. Pins 2, 3, and 6 are theinverting input Vn, the non-inverting input Vp, and the amplifier output Vo respectively.These three pins are the three terminals that normally appear in an op-amp circuit schematicdiagram. The null offset pins (1 and 5) provide a way to eliminate any offset in the outputvoltage of the amplifier. The offset voltage is an artifact of the integrated circuit. The offsetvoltage is additive with pin Vo (pin 6 in this case), can be either positive or negative and isnormally less than 10 mV. Because of its small magnitude, in most cases, one can ignore thecontribution of the offset voltage to Vo and leave the null offset pins open.

6 or rather unsaturated since when Vo tries to become larger than V+ or smaller than V- it gets clamped to V+ or V-

respectively (or to a constant voltage somewhat less).

PEEII-IV-8/15

3 Prelab Exercises3.1 Show why the cutoff frequency (or the frequency at which the output amplitude is (1/2)1/2

times the maximum output amplitude) is called the 3 dB frequency.

3.2 (a) Determine the amplitude response and the phase response functions for the low pass filter

shown in fig. 7.(b) Determine the maximum amplitude of the output.(c) Determine the 3 dB frequency and its corresponding amplitude.

3.3 Repeat 3.2 for the high pass filter shown in fig. 8.

3.4 Repeat 3.2 for the band pass filter shown in fig. 9.

PEEII-IV-9/15

4 Experiments


Tektronix PS 503 Power SupplyTektronix DC 504A Counter-TimerHP 54600A or Agilent 54622A Oscilloscope2: 620 Ω, 2: 1 KΩ, 2: 2KΩ, 2: 2.2KΩ, 20 KΩ Resistors0.05 µF, 3: 0.1 µF, 0.01 µF Capacitors741 Op-AmpProtoboard

4.1 Low Pass Filter

Build the low pass filter circuit shown in fig. 7. Apply a sinusoidal input of amplitude 1Vrms and display both input and output on the scope.

Vs++-

-

+

V0+

-

2.2 2.2KΩKΩ

0.1µF

0.05 µF

Fig. 7 A Second Order Low Pass Filter

Vary the frequency of the input voltage from 100 Hz to 3000 Hz, and measure the rmsoutput voltage and the phase angle8. Take as many readings as are necessary to develop awell-defined plot. Make sure to set Vs = 1 V rms before each reading.Find the exact cutoff (3 dB) frequency and the exact corresponding output voltage V0.

7 NOTE: The oscillator is designed to work for a very wide range of frequencies but may not be stable at very lowfrequencies, say in the order of 100 Hz or 200Hz. To start with it is a good idea to have the circuit working at somemid-range frequency, say in the order of 1K Hz or 2K Hz, and then change the frequency slowly as needed.8 The phase angle between two sinusoidal signals of the same frequency can be determined as follows: Trace bothsignals on two different channels with the same horizontal sensitivities (the same horizontal scale). To calibrate thehorizontal scale in terms of degrees, one can use the fact that the angular difference between the two successive zerocrossing points of a sinusoidal signal is 180 degrees. Thus, by measuring the distance between the successive zerocrossing points of either sinusoidal signal, one can calibrate the horizontal scale in terms of degrees. To determine thephase difference between the two sinusoidal signals, determine the distance between the zero crossing point of onesignal to a similar zero crossing point of another signal and convert it into degrees.

PEEII-IV-10/15

Frequency (Hz) Vo (rms) V Phase Angle º

PEEII-IV-11/15

4.2 High Pass Filter

Build the high pass filter circuit shown in fig. 8, and repeat the same procedure followed inSection 4.1.

KΩ

Vs++-

-

+0.1 0.1µFµF

1

1

KΩV0+

-

Fig. 8 A Second Order High Pass Filter

Take readings with the frequency varied from 100 Hz to 4000 Hz. Take also a few readingsbetween f = 5 KHz and f = 20 KHz. Make sure to take as many readings as are necessary todevelop a well - defined plot. Also, make sure to set Vs = 1 V rms before each reading.Find the exact 3 dB frequency and the exact corresponding output voltage.


PEEII-IV-12/15

4.3 Band Pass Filter

Build the band pass filter circuit shown in fig. 9, and repeat the same procedure followed inSection 4.1.

Vs++-

-

+620

620 Ω

Ω

0.1

0.01

µF

µF

20KΩ

-

+V0

Fig. 9 A Second Order Band Pass Filter

Take readings for the output voltage and the phase angle with the frequency varied from1100 Hz to 3000 Hz. Take also a few readings between f = 100 Hz and f = 800 Hz, and

PEEII-IV-13/15

between f = 4000 Hz, and f = 15 KHz. Make sure to take as many readings as are necessaryto develop a well-defined plot. Also, make sure to set Vs = 1 V rms before each reading.Find the center frequency, the 3 dB frequencies, and the corresponding output voltages.


PEEII-IV-14/15

4.4 Band Reject Filter (Optional)

Vs++-

-

+

1

1 2

0.1

0.1

0.1

KΩ KΩ

KΩ

µF

µF µF

V0+

-

Fig. 11 A Third Order Band Reject Filter

Build the band reject filter circuit shown in fig. 11.

(a) Show that the circuit is indeed a band reject filter.(b) Determine the maximum amplitude of the output.(c) Determine the minimum amplitude of the output and its corresponding frequency.(d) Determine the 3 dB frequencies and their corresponding output amplitudes.(e) Determine the stopband.(f) Read some phase angles at some particular frequencies.

4.5 All Pass Filter (optional)

Build the all pass filter circuit shown in fig. 10. Take readings of the output voltage and thephase with the frequency varied from 100 Hz to 10 KHz.

-

+Vs++-

V0+

-

1

1

2

2

0.1

0.1

KΩKΩ

KΩ KΩµF

µF

Fig. 10 A Second Order All Pass Filter

PEEII-IV-15/15

5 Report

5.1 For the low-pass filter, using the functions determined in pre-lab exercise 3.2 (a), determinethe theoretical values of the amplitude and the phase at the frequencies that were used to takereadings during the experiment. Tabulate the theoretical and the experimental values of bothamplitude and phase. On graph paper with rectangular coordinates, plot the theoretical and theexperimental data. Compare the theoretical plot with the experimental one.

5.2 For the high-pass filter, repeat 5.1 using the functions determined in pre-lab exercise 3.3(a).

5.3 For the band-pass filter, repeat 5.1 using the functions determined in pre-lab exercise 3.4(a).

5.4 Simulate, in PSpice, the low pass filter circuit in fig. 7, and plot the amplitude and thephase response in the interval 0.001 Hz ≤ f ≤ 2000 Hz.

5.5 Simulate in PSpice, the high-pass and band pass filter circuits of figures 8 and 9 inappropriate frequency ranges.






Experiment 3

Steady State Frequency Response Using Bode Plots

1 Introduction

Objectives • To study the steady state frequency response of stabletransfer functions of certain simple circuits using Bodeplots

Overview

This experiment treats the subject of frequency response by the use of Bode plots. The basicequation and logic of Bode plots are introduced in section 2 for transfer functions with realand complex-conjugate poles and zeros. The asymptotes for the magnitude and phase plotsare introduced, as is the method for correcting around the corner frequency to obtain realisticplots.Four circuits with transfer characteristics that exhibit illustrative examples of the pole andzero combinations covered in section 2 are realized in section 4. Their theoretical Bode plotsare derived in section 3 and data for plotting their actual ones are extracted in section 4 byvarying the frequency and measuring output/input (magnitude of gain) and phase.

PEEII-V-2/16

2 Theory

2.1 Introduction to Bode plots1

Bode plots are commonly used to display the steady state frequency response of a stablesystem. Let the transfer function of a stable system be H(s). Also, let M(!) and "(!) berespectively the magnitude and the phase angle of H(j!). In Bode plots, the magnitudecharacteristic M(!) and the phase angle characteristic "(!) of the frequency response areplotted vs. log10(!). Also, in the magnitude plot, instead of plotting M(!), vs. ! one plots 20log10M(!) which is called the decibel2 (abbreviated as dB) value of M(!). By utilizing thelogarithm, one accommodates a wide range of data on a compact graph. Moreover, as will beseen shortly from equations (3) and (4), the logarithmic operation permits the decompositionof the transfer function into simpler parts in a natural way. By analyzing each of the simplerparts individually, and then combining them appropriately, one can plot relatively easily theoverall frequency response of a composite transfer function. Note here that the DC frequency!=0 cannot be represented in the logarithmic scale since log10(0) =-#.

2.2 Transfer function standard formConsider a rational transfer function H(s) in which all the poles and zeros are shown in afactored form as

H (s) =Ksr s + z1( ) s + z2( )..... s + zm( )s + p1( ) s + p2( )..... s + pn( ) (1)

where m, n, and r are some integers. It is convenient to rewrite H(j!) in the form known as astandard form for Bode plots,

H ( j! ) =Ko j!( )r 1+ j! / z1( ) 1+ j! / z2( )..... 1+ j! / zm( )

1+ j! / p1( ) 1+ j! / p2( )..... 1+ j! / pn( ) (2A)

where

Ko =Kz1z2 ....zmp1p2 ....pn

(2B)

If any of the poles or zeros are complex, each of the complex conjugate pairs of such poles or zerosof H(s) can be combined to get a quadratic form of the type s2 + (2$/!0)s + !0

2, and then, byfactoring out !0

2 and letting s =j!, it can be reduced to a standard form as

1 The subject is treated in detail in Appendix E of the text.2 The decibel is treated in detail in Appendix D of the text.

PEEII-V-3/16

1 + (2$/!0)j! - !2/ !02

2.3 Real Poles and Zeros

First assume that all the poles and zeros are real. Then, one can easily determine the magnitudeand phase functions 20 log10M(!) and "(!) as

20 log10M(!)= 20 log10(|K0|)+ 20 r log10(j!) + 20 log10(|1+j!/z1|) + 20 log10(|1+j!/z2|) + ...... - 20 log10(|1+j!/p1|) - 20 log10(|1+j!/p2|)-.... (3)

"(!) = %K0 + r 90o +tan-1(!/z1) + tan-1(!/z2) + ...... - tan-1(!/p1) - tan-1(!/p2) - ..... (4)

The phase of K0 is zero if K0>0 and 180o if K0<0.

The above equations reveal that the magnitude and phase functions 20log10M(!) and "(!) areobtained by simply adding the contributions due to several individual factors. In what follows,we examine individually each of such factors. Once this is done, Bode plots of compositetransfer functions can easily be determined. For clarity, we denote 20 log10M(!) by MdB(!).

Note that 20 log10 1+j!p1

= 20 log10 1+ ! / p1( )2

If ! << p1 then !/p1<<1 and 20 log10 1+j!p1

= 20 log10 1+ ! / p1( )2 " 1

If ! >> p1 then !/p1>>1 and

20 log10 1+j!p1

= 20 log10 1+ ! / p1( )2 " 20 log10 !p1

#$%

&'(

The latter represents a line with slope of 20dB per decade or 6db per octave3 since:

3 A decade is a ten-fold increase in frequency, while an octave is a two-fold increase in frequency.

PEEII-V-4/16

If !=10 p1 then 20 log10!p1

"#$

%&'= 20 log10 10 = 20dB

If !=2 p1 then 20 log10!p1

"#$

%&'= 20 log10 2 ( 6dB

Exactly the same is true for zeros. The only difference is that the contribution of poles issubtracted (negative slope –20dB/dec) and that of zeros is added (+20dB/dec).

These asymptotes of MdB(!) meet at != p1. Thus, the frequency p1 is called the cornerfrequency. At this corner frequency, the actual value of MdB(!) can be computed easily as 3dB rather than the 0 dB predicted by either asymptote. At half of the corner frequency, i.e., at! = 0.5p1, one can compute MdB(!) as 1 dB rather than the 0 dB given by the low frequencyasymptote. On the other hand, at twice the corner frequency, i.e., at ! = 2 p1, one cancompute MdB(!) as 7 dB rather than the 6 dB given by the high frequency asymptote. Usingthese actual values as a guide, an exact plot of MdB(!) can easily be drawn4.

In terms of the angle, one can choose three cases;

If ! << p1 then !/p1<<1 and tan-1 !

p1

"#$

%&'( 0

If ! = p1 then !/p1=1 and tan-1 !

p1

"#$

%&'= tan-11 = 45 o

If ! >> p1 then !/p1>>1 and tan-1 !

p1

"#$

%&'( tan-1) = 90 o

This means that the maximum contribution of any one zero or pole is 90º (+ for zeros, - forpoles). If one approximated “much smaller than p1” by p1/10 5and as “much larger than p1”by 10 p1,6 then the angle plot has a slope of -45º/dec7 and is flat elsewhere.

4 Figure E-6 in the text shows both the low and high frequency asymptotes of MdB(!) as well as the actual characteristic.5 tan-1(0.1) = 5.3o≈ 06 tan-1(10)= 84.7o≈ 90º7 +45º/dec for a zero

PEEII-V-5/16

2.3.1 Constant gain K0:

If H(s)=K0, then MdB(!) = 20 log10|K0|

which is a constant with respect to log10 !. If K0 is positive, then the phase angle ofH(j!) is zero for all !; if K0 is negative, then the phase angle of H(j!) is 180o for all !.

2.3.2 Pole or zero at the origin:

For an ideal differentiator, the transfer function is s, while for an ideal integrator, thetransfer function is 1/s. Thus, the ideal differentiator has a zero at s=0, and an idealintegrator has a pole at s=0.For an ideal differentiator, H(j!)= j!, and MdB(!) = 20 log10 !.

Thus, the plot of MdB(!) as a function of log10 ! is a straight line with a slope of 20 dBper decade or approximately 6 dB per octave. The straight line passes through 0 dB at!=1 since 20 log10 ! equals 0 at !=1. The phase angle of H(j!)= j! is 90o for all !.

For an ideal integrator, the transfer function is 1/s which is the reciprocal of that of thedifferentiator. The Bode plots of 1/s are the negative replicas of those of s. That is, themagnitude plot is a straight line with a slope of -20 dB per decade and passes through 0dB at !=1, while the phase angle is -90o for all !.

2.3.3 Simple Real Pole or Zero:

Then the discussion of section 2.3 above is relevant and the asymptotes are at zero dband ±20dB/decade for zero and pole respectively.

2.3.4 Complex Conjugate Pairs of Poles or Zeros:We consider next a transfer function having a pair of complex conjugate poles whichtogether give rise to a quadratic factor, i.e.,

H (s) = 1+ 2!"o

s + s2

"o2

#

$%

&

'(

)1

where $ and !0 are respectively called as the damping coefficient and the naturalfrequency of the two complex conjugate poles. For this case,

PEEII-V-6/16

MdB (! ) = "20 log10 1"! 2

!o2 + 2#

j!!o

and !(" ) = # tan#1

2$ ""o

1# " 2

"o2

%

&

''''

(

)

****

.

Two asymptotes of MdB(!), one for low frequencies and another for high frequencies,can easily be ascertained. For !/! 0 << 1, we have

MdB(!) ≈ -20 log10 1 = 0.

Similarly, for !/! 0 >> 1, we have MdB(!) ≈ -40 log10 (!/! 0).

That is, for high enough frequencies such that ! >> ! 0, the magnitude characteristicMdB(!) is a straight line with a slope of -40 dB per decade and intersects the frequencyaxis at != ! 0.. Both the low and high frequency asymptotes of MdB(!) meet at thecorner frequency ! = ! 0. The exact magnitude characteristic of MdB(!) depends on thedamping coefficient $. As shown in Figure 1, for small values of $, there is significantpeeking in the neighborhood of the corner (or natural) frequency ! 0. For ! < 1 / 2 andfor certain representative frequencies, one can compute exactly MdB(!) as outlinedbelow:1. At the corner frequency !0, MdB(!) has a value of -20 log10[2 $].2. At ! = 0.5 ! 0, MdB(!) has a value of -10 log10[$2+0.5625].3. It can be shown that the peak of MdB(!) occurs at ! =!o 1" 2# 2 and that the peakvalue is -10 log10[4 $2(1-$2)].

4. MdB(!) crosses the 0 dB axis at the frequency ! =!o 2 1" 2# 2( )

For ! > 1 / 2 the exact characteristic of MdB(!) lies entirely below its asymptoticapproximation.

For the angle, as shown in Figure 1, the exact plot of "(!) depends on the dampingfactor $ . Without taking into account the value of $ , one can coarsely8 approximate

8 There exist in the literature certain better mid frequency approximations which take into account the value of $ .

PEEII-V-7/16

the phase angle plot by three asymptotes, similar to the ones derived for the simple realpole, only now the total contribution of the pair will be 180º rather than 90º of thesimple pole or zero:Low frequency approximation: For ! < 0.1 ! 0, "(!) can be approximated by 0.High frequency approximation: For ! > 10 ! 0, "(!) can be approximated by -180o .Mid frequency approximation: For 0.1 ! 0 < ! <10 ! 0, "(!) can be approximated by astraight line having a slope of -90o per decade and passing through -90o at ! = ! 0.

Consider next a transfer function with a pair of complex conjugate zeros,

H(s)= 1 + (2$/!0)s+ s2/ !02 .

For this case, the magnitude and phase plots are negative replicas (inverted versions) ofthe corresponding plots derived for the case of complex conjugate poles. In particular,the high frequency asymptote of the magnitude characteristic has a slope of +40 dB /decade, and the mid frequency asymptote of the phase characteristic has a slope of+90o/decade.

0.1

0.2

0.3

0.50.71.0

=

=

=

=

=

=

=

=

= = =

= 0.10.2

0.3

0.5

0.7

1.0

20

15

10

5

0

!5

!10

!15

"

0

!45

!90

!135

!180

dB

#

"

"

"

"

0.1 0.2 0.4 0.6 0.8 1 2 4 6 8 10

$$k

% % %

%

%

%

%%

%

%

%

%

"

Fig. 1 Log-Magnitudeand Phase-Angle curves

for a pair of complexconjugate poles

!/!&

PEEII-V-8/16

3 Prelab Exercises3.1 Derive the transfer functions H(s) for the five circuits in Section 4. Note that the transfer

function of circuit #2 is not the product of two simple transfer functions of type #1circuits.This is the result of the so called loading effect. Circuits #4 and #5 differ in the sense that thetransfer function of circuit #4 has a quadratic factor with complex roots in its denominator,whereas the transfer function of circuit #5 has a quadratic factor with real roots in itsdenominator.

3.2 Put each of the transfer functions derived in 3.1 in the appropriate standard form of thetransfer functions in section 2. Derive or state any necessary arguments that are essential ornecessary for plotting the theoretical Bode diagrams for each transfer function. Also, plot thetheoretical Bode diagrams for each transfer function. Make sure that you use a semi-log graphpaper for all your plots. Make sure that both plots, the straight line approximation and thecorrected plot, are shown on the same graph paper for every circuit.

PEEII-V-9/16

4 Experiments


Tektronix PS 503 Power SupplyTektronix DC 504A Counter-TimerHP 54600A or Agilent 54622A Oscilloscope741 Op-AmpProtoboard

9 NOTE: The oscillator is designed to work for a very wide range of frequencies but may not be stable at very lowfrequencies, say in the order of 100 Hz or 200Hz. To start with it is a good idea to have the circuit working at somemid-range frequency, say in the order of 1K Hz or 2K Hz, and then change the frequency slowly as needed.

PEEII-V-10/16

4.1 Circuit #1

Construct the circuit shown in fig. 2 using R = 2 KΩ and C = 0.1 µF. Apply a sinusoidal inputof amplitude 1 V to the circuit (recheck that it is 1 V whenever changing frequency). Usingthe oscilloscope, measure the ratio of the output voltage to the input voltage, and the phaseangle10 over the frequency range from 20 Hz to 20 KHz. Use frequencies of 1, 2, 4, 6 and 8 ineach decade.

V0Vs

R

C

Frequency (Hz) output voltage (V) input voltage (V) Phase angle º

10 The phase angle between two sinusoidal signals of the same frequency can be determined as follows: Trace bothsignals on two different channels with the same horizontal sensitivities (the same horizontal scale). To calibrate thehorizontal scale in terms of degrees, one can use the fact that the angular difference between the two successive zerocrossing points of a sinusoidal signal is 180 degrees. Thus, by measuring the distance between the successive zerocrossing points of either sinusoidal signal, one can calibrate the horizontal scale in terms of degrees. To determinethe phase difference between the two sinusoidal signals, determine the distance between the zero crossing point ofone signal to a similar zero crossing point of another signal and convert it into degrees.

Fig. 2 Circuit # 1

PEEII-V-11/16

4.2 Circuit #2Repeat Section 4.1 for the circuit shown in fig. 3, except that the frequency range should befrom 20 Hz to 100 KHz. Use R1 = 1 KΩ, R2 = 2 KΩ, C1 = 0.1 µF, and C2 = 0.1 µF. Usefrequencies of 1, 2.7, 4.6, 6, and 8 in each appropriate decade.

Vs 1CR 1 2R

2C V0


Fig. 3Circuit # 2

PEEII-V-12/16

4.3 Circuit #3

Repeat Section 4.1 for the circuit shown in fig. 4, except that the frequency range should befrom 10Hz to 20 KHz. Use R1 = 1 KΩ, R2 = 10 KΩ, R3 = 1 KΩ, and C = 0.1 µF. Usefrequencies of 1, 1.6, 4, 6, 8, and 9.5 in each appropriate decade.

R 1 2RR3

C

V0Vs


Fig. 4Circuit # 3

PEEII-V-13/16

4.4 Circuit #4

Repeat Section 4.1 for the circuit shown in fig. 5, except that the frequency range should befrom 100 Hz to 100 KHz. Use an R so that Rg + R + RL is around 65 Ω (Rg is the internalimpedance of the function generator, and RL is the d.c. resistance of the inductor), L = 10mH, and C = 0.1 µF. Use frequencies of 1, 3, 4, 5, 6, and 8 in each appropriate decade.

Vs V0

LR LR g R

C


Fig. 5Circuit # 4

PEEII-V-14/16

4.5 Circuit #5

Repeat 4.4 for the circuit shown in fig. 5, except that the frequency range should be from 100Hz to 20 KHz. Use Rg + R + RL = 200 Ω, L = 10 mH, and C = 1 µF. Use frequencies of 1,1.6, 3, 6, and 8 in each appropriate decade.


4.6 A Band Reject Filter (optional)

This is a Band-Reject Filter Circuit. You can attempt this only after completing all theexperiments in Section 4.

Vs++-

-

+

1

1 2

0.1

0.1

0.1

KΩ KΩ

KΩ

µF

µF µF

V0+

-

Apply a sinusoidal input of rms value of 1 V and display both input and output on the screenof the oscilloscope.

Fig. 6 A BandReject Filter

PEEII-V-15/16

Vary the frequency of the input, read the output on the DVM, and show your instructor thatthe circuit actually behaves like a band reject filter.Find the following:

(1) The maximum value of the output.(2) The 3 dB frequencies and the corresponding value of the output.(3) The stop band.(4) The phase angle at some particular frequencies.

PEEII-V-16/16

5 Report

5.1 Tabulate all experimental data obtained in Section 4.

5.2 Plot the Bode diagrams (magnitude and phase) for the five circuits in Section 4 using theexperimental data. Make sure that you use a semi-log graph paper for all your plots.

For each circuit, compare the experimental Bode diagrams (plots) with the theoretical Bodediagrams (plots) previously completed in pre-lab exercise 3.2.






Experiment 4

The R-C series circuit

1 Introduction

Objectives • To study the behavior of the R-C Series Circuit underdifferent conditions

• To use different methods for the determination of the RCtime constant from experimental results

Overview

The aim of this experiment is to study the R-C Series Circuit under different conditions byobserving input and output waveforms and studying their interrelation.In particular the following are explored:(a) Natural response of an R-C Circuit: The capacitor is charged to a certain value and itsdecay is observed as a function of time. Such a decay is known as the natural response ofthe R-C Circuit as there are no forcing inputs applied to the circuit.(b) Response of an R-C Circuit to a periodic square wave input: When the input to a circuitis a periodic signal (wave), the output voltage is a periodic wave as well but not necessarilyof the same waveform as that at the input. The output voltage across the capacitance isstudied for a square wave input.(c) Differentiation And Integration Properties: Under certain appropriate conditions, an R-CCircuit can function approximately as an integrating circuit or as a differentiating circuit.These properties are also observed.

PEEII-I-2/11

2 Theory

2.1 The Natural Response of an R-C Circuit 1

Consider the R-C circuit of fig. 1 where the source voltage Vs is a DC voltage source.Assuming that the switch has been closed for a long period of time, the circuit has reached asteady state condition. In steady state, the capacitor has been charged to

Vg = Vs R/(R+Ri)

Therefore, when the switch opens, at t = 0, the initial voltage on the capacitor is Vg volts.With the capacitor so charged, it would be desirable to compute the natural response of theR-C Series Circuit shown in fig. 2.

Vs R

iR

++-

t=0+

-

CR!

iVg

-

+

-

C

Fig. 1 Fig. 2

Analysis of the circuit in fig. 2 yields the general solution

vc(t) = vc(0) e-t/RC , t > 0

With initial condition: vc(0-) = vc(0+) = Vg = Vs R/(R+Ri) !from which

vc(t) = Vg e-t/!, (1)

where ! = RC is the time constant. This means that the natural response of the R-C circuit isan exponential decay of the initial voltage. The rate of this decay is governed by the timeconstant RC. The graphical plot of Eq. 1 is given in fig. 3 where the graphical interpretationof the time constant is also shown.

1 A more detailed description can be found in section 7.2 of the text.

PEEII-I-3/11

The time constant ! = RC can be measured in several ways. Assuming that R and C are notknown, ! can be measured from the discharge data of the capacitor as will be seen in section5 below.

!

0 t

Vg

V(t)

-t/"eVgV(t) =

"

Fig. 3 Exponential Decay Of An R-C Circuit

2.2 Square Wave Response 2

Let the source voltage Vs be a square wave of frequency f and amplitude A, applied to the R-C circuit as shown in fig. 4.

-

+

C++-

R

A

A- V(t)

Since the input is a periodic wave, the output voltage across the capacitor is also a periodicwave, albeit not a square wave. In each period, the output voltage across the capacitorconsists of two parts:

• During the half period when the input is a positive constant, the capacitor getscharged exponentially. Hence the output voltage v(t) during this half period is anexponentially increasing signal. At the end of this half period, v(t) has attained acertain positive peak value.

2 A detailed description of the step response can be found in section 7.3 of the text.

Fig. 4 A square waveinput applied across a series

R-C Circuit

PEEII-I-4/11

• During the half period when the input is a negative constant, the capacitor getsdischarged exponentially. Hence the output voltage v(t) during this half period is anexponentially decreasing signal. By the end of this half period, v(t) has attained acertain negative peak value.

The difference between the positive and negative peaks is called the peak-to-peak voltage ofthe capacitor. It can be shown that

VCPP = VPP (1-e-K)/(1+e-K) (2)

where VCPP is the peak-to-peak voltage of the capacitor, VPP is the peak-to-peak voltage of the input square wave, and K is a number such that K! is the input square wave half-period when ! = RC.

2.3 Differentiation and Integration PropertiesConsider the loop equation of the R-C series circuit,

vR(t) + vC(t) = v(t),

where v(t) is the source voltage vs which could be any time varying signal.

If vC(t) is kept small with respect to vR(t), then v(t) " vR(t) = R i(t), and!

vc (t) =1C

i(t)dt =!1RC

vR (t)dt "!1#

v(t)dt!i.e. the capacitor voltage is very closely proportional to the integral of the source voltage. Ifv(t) is a periodic function, vC(t) can be kept small by making the period T « !. In this way vC

never gets time to grow large.

On the other hand, when the period T » !, vC tends to follow v(t) almost exactly. In such acase,

vR (t) = Ri(t) = RCdvC (t)dt

! " dv(t)dt

i.e. the resistor voltage is very closely proportional to the derivative of the source voltage.

PEEII-I-5/11

3 Prelab Exercises

3.1 Following the discussion in section 7.2 of the textbook, write down the differentialequation of the series R-C circuit in the absence of any forcing input. Then, explain orderive equation (1) in your own way.

3.2 For R = 10MΩ and C = 15µF, determine the expected time constant ! = RC.

3.3 Equation (2) for VCPP is rather difficult to prove at this time. Take it as a challenge toderive it as you learn increasingly more on the topic of differential equations.

3.4 Explain in your own words why an R-C series circuit can act approximately as anintegrator as well as a differentiator and under what conditions.

PEEII-I-6/11

4 Experiments

Suggested Equipment:TEKTRONIX FG 501A 2MHz Function GeneratorHP 54600A or Agilent 54622A OscilloscopeProtek Model B-845 Digital MultimeterTEKTRONIX DC504A Counter-TimerTEKTRONIX P503A Dual Power Supply100Ω, 10KΩ, 100KΩ, Resistors0.001µF, 0.01µF, 1µF, 15µF Capacitors

BreadboardOne 3.5” diskette

NOTE: The oscillator is designed to work for a very wide range of frequencies but may notbe stable at very low frequencies, say in the order of 100 Hz or 200Hz. To start with it is agood idea to have the circuit working at some mid-range frequency, say in the order of 1KHz or 2K Hz, and then change the frequency slowly as needed.

4.1 Time Constant of an R-C Circuit

Construct the R-C circuit shown in fig. 5.

Vs++-

!t=0

+

-

CVo VR

Let the source voltage Vs be a DC voltage of 10 V, C = 15µF, and RV = 10MΩ is the internalresistance of the DVM. Neglect the internal resistance Ri of the source. Then, when theswitch is closed, the capacitor charges quickly to the source voltage Vs.

At t = 0, the switch opens, and the voltage source gets disconnected from the R-C circuit.The capacitor will now discharge through the internal resistance of the DVM. Using a timeror a stopwatch, record the DVM readings for an interval of 5 minutes taking data in 15second intervals. Repeat the same procedure until you are assured that you have arepresentative set of data. Fill Table 1 with the data and use the set of data that, in youropinion, corresponds to the run that is mostly representative of the capacitor discharge.

Fig. 5 Natural Responseof an R-C Circuit

PEEII-I-7/11

T (min) v(t) V v(t) V v(t) V v(t) V v(t) V f(t)=v(t)/v(0) ln(f(t))0.0

0.15

0.30

1.00

1.15

1.30

1.45

2.00

2.15

2.30

2.45

3.00

3.15

3.30

3.45

4.00

4.15

4.30

4.45

5.00

Table 1 Capacitor Discharge Data

PEEII-I-8/11

4.2 Square Wave Response

a- Take C = 0.01µF and use a resistor R so that one-half period of a 1.00KHz square wavewill be 5! where the time constant ! =RC. Arrange such an R-C circuit (fig. 4) with thefunction generator as the source. Observe the function generator output on Channel 1 and vCon Channel 2. R = Ω.

b- Set the generator to square wave and set f = 500Hz3. Set the scope sensitivities to 1volt/div (CAL), AC input. Set sweep to 0.5ms/div (CAL). Adjust the generator amplitude for6 V peak-to-peak (VPP = 6 V, i.e. the square wave amplitude A=3 V) and center both imagesvertically on the screen.

c- Since f = 500Hz, a half period is 10!. Notice that the capacitor has time to fully charge oneach half-cycle (see eq. (2) and note that e-10 is negligible.)Change the sweep to 0.1ms/div. Now a half period is 10 divisions wide.Let us write a theoretical expression for the capacitor voltage vC. Taking the origin of time, t=0, when the square wave jumps from -A to A, we can determine the capacitor voltage vCduring the half cycle that follows t =0 as

vC(t) = A-2A e-t/!

Verify that the plot on the screen follows the above equation. This can be achieved bychecking whether four or so representative points of vC(t) on the screen are as predicted bythe equation. Download the waveform for your report.

d- Increase R and decrease C by a factor of 10, and verify that the circuit still behaves as inpart c. Download the waveform for your report.

e- From their original values, decrease R and increase C by a factor of 100 and again verifythat the circuit behavior is substantially the same as in part c. However you may notice thatthe 50-ohm internal impedance of the function generator causes some distortion at thebeginning of each half-cycle when the current is large. Download the waveform for yourreport.

f- Return to the original values of R and C and adjust the amplitude for VPP = 6.0 ifnecessary. Change the frequency to 2.00KHz (check frequency again). The half period isnow 2.5!. Measure the peak to peak capacitor voltage VCPP and keep this value for theReport.

VCPP = V. 3 Because of the inaccuracy of the oscillator frequency knobs of the function generator, it is essential to actuallymeasure the exact frequency or period of the signal generated. Both the counter and the digital oscilloscope can beused for this purpose. Either should be used in all experiments where a frequency or a period reading is required, inorder to obtain a correct and accurate result. In this case, use the counter.

PEEII-I-9/11

4.3 Integration and Differentiation

4.3.1 Integration of a square wave:Choose a frequency at which you are satisfied with the performance of the circuit asan integrator (adjust sensitivity and sweep rate as needed). Measure the peak to peakcapacitor voltage VCPP and the frequency of the waveform using the frequencycounter. Study the relation of the function and its integral and download the scopeimage for your report.

4.3.2 Integration of a triangular wave:With the circuit functioning well as an integrator, switch to triangular input. Adjust thegenerator output to 6 V peak-to peak. Study the relation of the function and its integraland copy the scope image for your report.The image of the integral on the scope may look like a sine wave but in fact it isparabolic. To prove this, make adjustments with the sweep, sensitivities, and variablesuntil the image spans 8 divisions peak-to-peak and the half period is 4 divisions wide.Notice that the amplitude at 1/8 period is 3/4 of peak instead of 0.707 of peak as itwould be in a sine wave. Download the waveform.

4.3.3 Integration of a sine wave:Change to sine wave input. Set the sensitivity variables and the sweep variable toCAL. Study the image and download as before.

4.3.4 Differentiation of a triangular wave:Return to triangular wave. Interchange the capacitor and the resistor in the circuit.Channel 2 should now be receiving vR. Adjust the sensitivity of the scope so vRbecomes visible. Reduce frequency gradually, making changes in scope sensitivity andsweep rate as you go, until you are satisfied with the performance of the circuit as adifferentiator.Measure the peak to peak resistance voltage VRPP and the frequency of the wave usingthe frequency counter. Study the images and download them for your report.

4.3.5 Differentiation of a square wave:Change to a square wave and decrease Channel 2 sensitivity until the derivativebecomes visible. Study and copy the images. With 1 volt/div on Channel 1 and 2volts/div on Channel 2, notice that the peak VR is twice the peak input.

4.3.6 Differentiation of a sine wave:Change to sine wave. Adjust input to VPP = 6.0 and make the usual positioningadjustments. Tweak the frequency to make the phase difference stand out. Study theimages and download them.

PEEII-I-10/11

5 Report

5.1 Determine the time constant ! in the following two ways:(a). Plot f(t) data from table 1 on a graph paper with rectangular coordinates.

The value of ! is the time at which f(t) = f(0)e-1 = 0.368f(0)

(b). f(t) = A e-t/! u(t) !o => ln(f(t)) = (-1/!)t + ln(A) for t > 0 which is in the form:

Y = A1 t + A0 and ln(f(t)) vs t should plot as a straight line.o Plot ln(f(t)) vs t on a graph paper with rectangular coordinates and find the

best straight line fit to the data.o The slope of the straight line must be (-1/!), hence the value of ! can be

computed from the slope.

5.2 Determine the time constant by integration using the following method:(a) From the equation f(t) = Ae-t/!, it is easy to show that the area under the

complete f(t) curve is equal to A!.(b) The trapezoidal rule is used to find the area. For a set of observations f1....fN

spaced at a common interval #t, the area ! in the interval t1 < ti < tN is givenapproximately by:

Area = ! = "t f1 + fN2

+ fii=2

N#1

$%&'

()*

5.3 Determine the time constant by differentiation using the following method:(a) At any point on the graph of f(t) vs t, the slope line (tangent) will always reach

zero in a time t = !. Given

f (t) = f (to )e!t!to"

it can be shown that (do not prove)

! = f (to )"t

f (to )# f (to + "t) (b) Apply the above equation for ! to each of the data points to the first 3 minutes and

average the values of ! to obtain the time constant.

5.4 From the value of VCPP measured in Section 4.2f, determine VPP/VCPP. Compare with the theoretical value obtained from Eq. 2.

PEEII-I-11/11

5.5 Submit all images copied from the scope, using the appropriate scales and labelwith the appropriate descriptive labels.

5.6 From Section 4.3.1, determine the minimum !/T for good integration of a squarewave. Determine the minimum VPP/VCPP. Compare with the theoretical valueobtained from Eq. 2.

5.7 From Section 4.3.2, Show that the integration of the triangular wave is a parabolicwave, i.e, show that the amplitude of the wave at 1/8 period is 3/4 of peak.

5.8 From Section 4.3.4, determine the minimum T/! for good differentiation of thetriangular wave. Determine the minimum VPP/VRPP. What does the outputwaveform look like?

5.9 Simulate in PSpice all parts of section 4.3. Use the frequencies obtained in theLab. Plot the output waveforms. Compare with the experimental ones.

5.10 Design a first order RC circuit that produces the following response:

vc(t) = 10 - 5 e-3000t V for t ≥0.



Latest revision:December 16, 2005, 2004by P. Panayotatos



Experiment 5

Step response of an RLC series circuit

1 Introduction

Objectives • To study the behavior of an underdamped RLC SeriesCircuit for different damping coefficients

Overview

This experiment is a study of the step response of an underdamped RLC series circuit. Thenatural frequency is chosen and that determines the values of L and C. Then the dampingcoefficient is varied by changing the value of R.A short theory is presented. The prelab exercise is rudimentary but crucial for theperformance of the experiment which includes comparison of experimentally determined andtheoretically derived waveform maxima and minima.

PEEII-II-2/6

2 Theory1

Consider an RLC series circuit subject to a unit step voltage as shown in fig. 1.

++-1 u(t)

L R

C

+

-

VC(t)

For a second order linear differential equation with step function input

a2d 2y(t)dt 2

+ a1dy(t)dt

+ a0y = Au(t)

the step response is the general solution for t > 0. This step response can be partitioned intoforced and natural components. Natural response is the general solution of the homogeneousequation (with u(t) = 0), while the forced response is a particular response of the aboveequation.

Following the notation of the textbook, the following notation is used for a second ordercircuit: ! = The damping factor

" = The damping coefficient#n = The damped frequency#o = The natural frequency

For the RLC series circuit of fig. 1, the expression for the capacitance voltage in theoscillatory (underdamped) case is

vC(t) = 1 + Ke-atcos(#d t + $) (1)

where ! = R/2L = "#o and

!d = !o2 "# 2 = ( 1

LC"R2

4L2)1/2 =!o(1"$

2 )1/2

1 A more detailed treatment can be found in the text starting with section 8.4.

Fig. 1A Series RLC Circuitwith a Unit Step Input

PEEII-II-3/6

Also, !o =1LC and ! =

R2

CL

In equation (1) for vC(t), the term Ke-atcos(#d t + $) represents the natural response and theunity represents the particular solution.

Consider the initially quiescent state (both the initial inductance current and the initialcapacitor voltage are zero). In this case,

! = " tan"1 #$d

%&'

()*= " tan"1 +

1++ 2%

&'

(

)* (2)

K = ! 1cos"

= ! 11!# 2 (3)

The maxima and minima occur alternately when

tan(#d +$) = -!/#d, i.e. #d t = k !; k = 0, 1, 2, ....... (4)

From the above, the maximum and the minimum values of vC are given by

vc k! /"d( ) = 1# (#1)k exp #k!$1#$ 2

%

&''

(

)** k = 0, 1, 2, ....... (5)

(odd values of k give the maxima)

3 Prelab Exercises

3.1 For L = 100mH, compute the value of C needed for the natural frequency #o = 2!fo = 104!.To vary the damping coefficient " , vary the value of R . Compute the values of R for eachof the six different values of the damping coefficient " = 0.1, " = 0.2, "=0.4, " = 0.6, " = 0.8, and "=1.0, (see Table 1 under section 4.4).

3.2 From Eq. 5, calculate and fill the theoretical values of the maxima and minima in Table 1 under section 4.4.

PEEII-II-4/6

4 Experiments

Suggested Equipment:TEKTRONIX FG 501A 2MHz Function GeneratorHP 54600A or Agilent 54622A OscilloscopeDC 504A Counter-TimerProtek Model B-845 Digital MultimeterResistance/Capacitance BoxInductance Box

NOTE: The oscillator is designed to work for a very wide range of frequencies but may notbe stable at very low frequencies, say in the order of 100 Hz or 200Hz. To start with it is agood idea to have the circuit working at some mid-range frequency, say in the order of 1KHz or 2K Hz, and then change the frequency slowly as needed.

Any function generator used has internal resistance. Also, the inductor has internal resistance.Both need to be determined.

4.1 Function generator resistance

The internal resistance of the function generator will affect the damping of an RLC circuit towhich it is connected. Check the resistance in the following way:

a- With a sine wave output, set the open circuit voltage to some convenient value, say 1V.

b- Connect a pure variable resistance load (potentiometer) thus forming a voltage divider.Adjust R until the terminal voltage falls to one-half the open circuit value. At this point thetwo resistances of the voltage divider have to be equal. Therefore, the resistance of thepotentiometer should now be equal to the internal resistance of the function generator.Disconnect the potentiometer from the circuit and measure its resistance.

4.2 Inductor internal resistance

Use the digital ohmmeter to measure the internal resistance of the inductor used. The valuesof R that should be used in the series RLC circuit are those computed in Section 3 (Pre-labexercise) minus the internal resistance of the function generator and minus the internalresistance of the inductor.

4.3 Step response of RLC branch

Arrange a series RLC series circuit using values calculated for " = 0.1 in Section 3 (Pre-labexercise). Set the scope sweep rate to 0.5ms/div (CAL), sensitivity to 500mV/div. With a

PEEII-II-5/6

square wave input to the circuit, set f = 250Hz and increase the amplitude of the input untilyou see the typical oscillatory response to a step input. (Instead of a step input, a squarewave of low enough frequency is used so that the repetitive wave form of the capacitorvoltage can be easily plotted on the scope). Adjust the oscillator voltage so that the steadystate value of the capacitor voltage is 1V corresponding to the particular solution for a unitstep input. Download the waveform.

4.4 Measurement of peaks

Read the values of the maxima and the minima based on a unit step input. (Use the cursor).Repeat 4.3 for different values of " , fill all the values in Table 1 and compare with thetheoretical values that were calculated in pre-lab exercise 3.2. In each case download thewaveform.

Table 1 Unit step response of 2nd order system

4.5 Natural frequency

Change the sweep to 0.1ms/div and reduce damping as much as possible by reducing thevalue of R. Notice that the period of the oscillation is close to 0.2ms which corresponds tothe natural frequency #o used to determine L and C values.

! 1st Max 1st Min 2nd Max

0.1

0.2

0.4

0.6

0.8

1.0Critically Damped

Theoretical

Measured

Measured

Measured

Measured

Measured

Measured

Theoretical

Theoretical

Theoretical

Theoretical

Theoretical

"

PEEII-II-6/6

5 Report

5.1 Derive Equation 1 in Section 2 for the underdamped case of a series RLC circuit.

5.2 Derive and prove Equation 4 in Section 2.

5.3 In pre-lab exercise 3.2 you have already filled the theoretical values of the maxima andminima in Table 1. Compare them now with the experimental values. Print out thedownloaded waveforms and label them appropriately.


PEEII Lab Manual

Documents

Transcript of PEEII Lab Manual