Signal Systems Lab Manual Fall 2013

5/20/2018 Signal Systems Lab Manual Fall 2013

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Jacobs University Bremen

Natural Science LaboratoryAdv. EE Lab - Signals and Systems

Fall Semester 2013

Course 300221

Instructors - Prof. Dietmar Knipp and- Uwe Pagel

e-mail - [email protected] tel.: +49 421 200 3570- [email protected] tel.: +49 421 200 3114

Website - http://www.faculty.jacobs-university.de/upagel/AdvEELab

September 4, 2013
http://www.faculty.jacobs-university.de/upagel/AdvEELabhttp://www.faculty.jacobs-university.de/upagel/AdvEELab


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Contents

I General remarks on the course 3

1 Lab Guidelines 41.1 Safety . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2 Attendance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 Preparations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.4 Supplies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.5 Prelab and Lab Report . . . . . . . . . . . . . . . . . . . . . . . . . . 4

II Experiments 5

2 Lab Experiment 1 : RLC-Circuits - Transient Response 62.1 Introduction to the experiment . . . . . . . . . . . . . . . . . . . . . 62.2 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.3 Prelab Transient response of RLC-Circuits . . . . . . . . . . . . . . . 162.4 Execution Transient response of RLC-Circuits . . . . . . . . . . . . . 18

3 Lab Experiment 2 : RLC-Circuits - Frequency Response 20

3.1 Introduction to the experiment . . . . . . . . . . . . . . . . . . . . . 203.2 Handling of the function generator and the oscilloscope . . . . . . . . 283.3 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293.4 Prelab RLC Circuits - Frequency response . . . . . . . . . . . . . . . 303.5 Execution RLC Circuits - Frequency response . . . . . . . . . . . . . 31

4 Lab Experiment 3 : Fourier Series and Fourier Transform 344.1 Introduction to the experiment . . . . . . . . . . . . . . . . . . . . . 344.2 Prelab Fourier Series and fourier Transform . . . . . . . . . . . . . . 484.3 Execution Fourier Series and fourier Transform . . . . . . . . . . . . 50

5 Lab Experiment 4 : Sampling 535.1 Introduction to the experiment . . . . . . . . . . . . . . . . . . . . . 535.2 Prelab Sampling. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 635.3 Execution Sampling. . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

6 Lab Experiment 5 : Modulation 676.1 Introduction to the experiment . . . . . . . . . . . . . . . . . . . . . 676.2 Prelab AM Modulation . . . . . . . . . . . . . . . . . . . . . . . . . . 826.3 Prelab FM Modulation . . . . . . . . . . . . . . . . . . . . . . . . . . 836.4 Execution AM Modulation . . . . . . . . . . . . . . . . . . . . . . . . 84

6.5 Execution FM Modulation . . . . . . . . . . . . . . . . . . . . . . . . 88

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III Additional Information 90

A Appendix 91A.1 Hardcopy from oscilloscope screen . . . . . . . . . . . . . . . . . . . . 91

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Part I

General remarks on the course

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1. Lab Guidelines

1.1 SafetyRead the safety instructions before attending the lab.

1.2 AttendanceAbsence of a group member will be accepted only after providing a medical certifi-cate.

1.3 PreparationsEach student has to be prepared before attending the lab. All students have to befamiliar with the subject, the objectives of the experiment and have to be able toanswer questions related to the experiment handout and prelab. If a member of thegroup or the whole group is not prepared, the student or the group will be excludedfrom the lab for this specific experiment. The prelab has to be prepared in a writtenform by the group and has to be presented before conducting the experiment.

1.4 Supplies

All equipment, cabling and component you need should be in your work area. If youcant find it, ask your lab instructor or teaching assistant, dont take it from anothergroup. Before leaving the lab, put everything back, where you found it! Please bringyour notebook so that you can readout the oscilloscope.

1.5 Prelab and Lab ReportA lab report has to be prepared after each lab. The lab report has to be written insuch a form that the instructors and the teaching assistants can follow it. The labreport includes the experimental data taken during the lab, the analysis of the data,a discussion of the results and answers of all questions. You should also include

your Matlab codes and all the required plots, sketches and hardcopies. All groupmembers are in charge of preparing and finalizing the lab report. Please divide theworkload amongst the group members. The lab report and the prelab have to besubmitted one week after the experiment. This is a hard deadline. If you miss thedeadline, the experiment will be downgraded by 1 point each day (excluding theweekends).

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Part II

Experiments

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2. Lab Experiment 1 : RLC-Circuits - Tran-

sient Response2.1 Introduction to the experiment

2.1.1 Objectives of the experiment

The goal of the first experiment of the Signals and Systems Lab, is to study thetransient response of second-order systems. Particularly, we will in study the be-havior of second-order electrical systems. A typical second-order system in electricalengineering is an RLC circuit. As part of the prelab the transient response of suchsystems will be studied using Matlab. As part of the experiment, different RLCcircuit configurations will be implemented and tested. The experimental and thesimulation results will be compared and the differences will be discussed.

2.1.2 Introduction

Second-order systems are very common in nature. They are named second-ordersystems, as the highest power of derivative in the differential equation describingthe system is two.In electrical engineering, circuits consisting of two energy storage elements, capac-itors and inductors, for example RLC circuits, can be described as second-order

electrical system. These circuits are frequently used to select or attenuate partic-ular frequency ranges, as in tuning a radio or rejecting noise from the AC powerlines.The handout is divided into two parts. Throughout the first part of the handoutthe following topics will be discussed.

1. The equation describing a second-order system in its general form.

2. The complete solution for a second-order differential equation (D.E.) repre-senting the complete response of the system. The complete solution consistsof two responses, that is

(a) Transient responseIn transient response, depending on the circuit parameters the circuitoperates under,

i. Over-damped condition

ii. Critically damped condition

iii. Under-damped condition

(b) Steady-state responseThe steady state response due to a constant input signal (DC source).

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The second part of the handout describes the practical part of the experiment. Itwill explain how to develop a second-order differential equation describing a seriesRLC circuit configuration and how the differential equation can be solved in orderto have a complete solution consisting of the transient response and steady-stateresponse.

2.1.3 Differential equations describing second-order systems

Circuits with two energy storage elements as the RLC circuits are described by asecond-order ordinary D.E.

a2d2y(t)

dt2 +a1

dy(t)

dt +a0y(t) =x(t) (2.1)

where,

y(t) is the response of the system to an applied input x(t).a0, a1 and a2 are the system parameters.

In the context of the response of second-order systems, it is more useful to rewriteEqu.(2.1) in the form of a linear constant coefficient non-homogeneous differentialequation

d2y(t)

dt2 + 2n

dy(t)

dt +2ny(t) =K

2nx(t) (2.2)

Thus, the system parameters become

n=

a0a2

(2.3)

= a1

2

a0a2(2.4)

K= 1

a0(2.5)

where,

n is the natural frequency. is the damping ratio.K is the gain of the system.

2.1.4 The complete solution for a second-order D.E.

The complete solution of the second-order non-homogeneous differential equation,Equ.(2.2), is given by the following sum

y= yh+yf (2.6)

The solution consists of two terms, where yh

is the homogeneous solution of thesecond-order non-homogeneous differential equation, Equ.(2.2), where the input

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x(t) = 0 . This solution satisfies the initial condition of the system. Therefore,the homogenous solution of the second-order non-homogeneous differential equationdescribes the transient response of the system.The termyfis the forced solution of the second-order non-homogeneous differentialequation, Equ.(2.2), where an external input signal x(t)

= 0 is applied to the sys-

tem. Therefore, the forced solution of the second-order non-homogeneous differentialequation describes the steady-state response of the system.In the following we will derive in details the homogenous solution and the forcedsolution for a constant input signal (DC source).

a. The homogeneous solution

The homogeneous solution of the second-order non-homogeneous differential equa-tion can be found by rewriting Equ.(2.2), where the applied input x(t) is equal to0, so that the non-homogeneous D.E is reduced to a homogeneous equation with

constant coefficientsd2y(t)

dt2 + 2n

dy(t)

dt +2ny(t) = 0 (2.7)

This equation has a solution of the form

y(t) =C et (2.8)

By substitutingy(t) from Equ.(2.8) in Equ.(2.7), we get

Cet(2 + 2n +2n) = 0 (2.9)

From Equ.(2.9), we get

2 + 2n +2n = 0 (2.10)

The above equation is called the characteristic equation. To find the homogeneoussolution, we need to solve the characteristic equation. The characteristic equationhas two roots (solutions)

1 = n+n

2 1 (2.11a)2 =

n

n2 1 (2.11b)

By substituting in Equ.(2.8), we get

y1(t) =C1exp ((+

2 1)nt) (2.12a)y2(t) =C2exp ((

2 1)nt) (2.12b)

Each root 1 and2 contributes a term to the homogeneous solution. The homoge-neous solution is

yh=C1e1t +C2e

2t (2.13)

where,

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C1 and C2 are unknown coefficients determined by the initial conditions.1 and 2 are unknown constants determined by the coefficients of the

D.E. (depends on the circuit parameters R, L and C).

The time-dependent response of the circuit depends upon the relative values of the

damping ratio and the undamped natural frequency n (radians/sec). Accordingto the relative values of and n, we can classify the transient response into threecases:

1. Under-damped case : 0< 1, 1 and 2 are real and unequal

Now, let us describe each case separately.

1. Under-damped Case: 0<


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2. Critically damped Case: = 1When the damping ratio is equal to one, the general solution is of the form

y(t) =C1exp (nt) +C2t exp(nt) (2.17)

where, C1 andC2 are unknown coefficients derived from the initial conditions.Thus for a critically damped system, the response is not oscillatory. It ap-proaches equilibrium as quickly as possible.

3. Over-damped Case: >1When the damping ratio is greater than one, the general solution to the ho-mogenous equation is

y(t) =C1exp ((+

2 1)nt) +C2exp ((

2 1)nt) (2.18)

where, C1 andC2 are unknown coefficients derived from the initial conditions.

This indicated that the response is the sum of two decaying exponentials.The total solution of a second-order system for the previously discussed casescan be summarized as shown in Fig.(2.2) for = 0.1, 0.5, 1, 2 and 3.

Figure 2.2: Solution of a second-order homogenous system

Fig.(2.2) shows the normalized step response, where A is the amplitude of theconstant input signal. Generally, the aim of normalizing (scaling) is to comparethe output (the response) to a reference value. In our case, the reference valueis the step function and A represents the input voltage (DC source).

Note: When= 0, the response becomes undamped and oscillations continueindefinitely at frequency n.

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b. The forced solution

The response to a forcing function will be of the same form as the forcing function.The forced solution of the non-homogeneous second-order differential equation isusually given by a weighted sum of the input signal x(t) and its first and second

derivatives. If the input x(t) is constant, then the forced response yf is constant aswell. Ifx(t) is sinusoidal, then yF is sinusoidal.However, we will not discuss the forced solutions of second-order differential equa-tions as part of this lab. Here we will deal only with constant input signals (DCsources). The necessary steps for determining the steady-state response of RLCcircuits with DC sources will be described later in the handout.

2.1.5 Solving a second-order differential system

The behavior of a series RLC circuit shown in Fig.(2.3) can be determined from asimple circuit analysis.

Figure 2.3: Second order system based on a serial RLC circuit

The input voltage is the sum of the output voltage and voltage drops across theinductor and resistor

Vin=VR+VL+Vout (2.19)

The current i is related to the current flowing through the capacitor

i= iC=CdVout

dt (2.20)

The voltage drop across the resistor is given by

VR=iR = RCdVout

dt (2.21)

the voltage drop across the inductor is

VL=Ldi

dt=L

d

dt(C

dVoutdt

) =LCd2Vout

dt2 (2.22)

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Substituting Equ.(2.21) and (2.22) into (2.19) yields

LCd2Vout

dt2 +RC

dVoutdt

+Vout = Vin (2.23)

Thus, from Equ.(2.3), (2.4) and (2.5), the undamped natural frequency (radians/sec),the damping ratio and the gain of the circuit are:

n= 1

LC(2.24)

=R

2

C

L (2.25)

K= 1 (2.26)

Example:

Lets assume that the components in Fig.(2.3) have the values R= 50, C= 1Fand L= 50mH. By substituting in Equ.(2.24) and (2.25)

n= 1

LC= 4472rad/sec

=R

2

C

L = 0.1 (under-damped case)

2.1.6 Definitions and Practical Hints

Step response

The step response is the response of a system upon applying an input signal in theform of a step function.

The steady-state value

The steady-state value is the magnitude of the voltage, or current, after the systemhas reached stability.

RingingRinging is the oscillation phenomenon that occurs if the system is under-damped.

Overshoot

An overshoot is observed if the transient signal exceeds the final steady state value.The overshoot is often represented by a percentage of the final value of the stepresponse. The percentage overshoot is:

Percentage overshoot =Vmax VSteadyState

VSteadyState 100%

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The undamped natural frequency, n

It is the frequency of oscillation of the system without damping. For the stepresponse of an under-damped system shown in Fig.(2.4), the transient responsespecifications are:

Peak time, TpIt is the time required for the response to reach the peak of the overshoot.

Rise time, TrIt is the time required for the step response to rise from 10% to 90% of itsfinal value for critical and over-damped cases, and from 0% to 100% for under-damped cases.

Settling time, TsIt is the time required for the step response to settle within a certain percentage

of its final value. The percentage can be chosen to be 2% or 5%.Note: Not all these specifications apply to all cases of system response. For example,for an over-damped system, the terms ringing, peak time and maximum overshootdo not apply.

Figure 2.4: Under-damped second-order system

2.1.7 Initial conditions of switched circuits

The switched circuit is a circuit with one or more switches that open or close at acertain point in time. We are interested in the change of the current and voltageof energy storage elements (L and C) after the switch changed from open to closeor vice versa. So if the time of the switch is t = 0, we want to determine thecurrent through the inductor and voltage across the capacitor at t = 0 and t = 0+

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immediately before and after the switching. These values together with the sourcesdetermine the behavior of the circuit for t >0.Before continuing with a solved example two important notes should be mentionedwhich are:

a. The current through an inductor cannot change instantaneously, whereas thevoltage drop across an inductor can change instantaneously.

b. The voltage drop across a capacitor cannot change instantaneously, whereasthe current flow through a capacitor can change instantaneously.

Example:Consider the circuit shown in Fig.(2.5). The switch has been closed and steady

state conditions were reached. In order to findvC(0) and iL(0

) the capacitor isreplaced by an open circuit and the inductor by a short circuit as shown in Fig.(2.6).

Figure 2.5: Switched circuit

Figure 2.6: Steady-state conditions

It can be easily calculated that

iL(0) =

VinR1+R2

(2.27)

vC(0) =Vin

R2R1+R2

(2.28)

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Since the current through an inductor cannot change instantaneously and the voltageacross a capacitor cannot change instantaneously, therefore,

iL(0) =iL(0

+) = Vin

R1+R2(2.29)

vC(0) =vC(0

+) =Vin R2

R1+R2(2.30)

2.1.8 The Complete Response

The necessary steps to determine the complete response of a second-order systembased on RLC network with DC sources are:

For transient response1. Using Ohms law, KVL, and/or KCL, obtain a second order differential

nonhomogeneous equation. Another way is to obtain two first-order dif-ferential equations, and then combine them to a second order differentialnon-homogeneous equation.

2. Solve the homogeneous equation corresponding to the obtained secondorder differential non-homogeneous equation.

3. Obtain the complete solution by adding the forced solution to the ho-mogeneous solution. The complete solution still contains unknown coef-ficientsC1 and C2.

4. Use the initial conditions to determine the value ofC1 and C2.

For DC steady state response1. Replace all capacitances with open circuits.

2. Replace all inductances with short circuits.

Solve the remaining circuit.

2.2 References1. A. V. Oppenheim, A. S. Willsky, S. H. Nawab, Signals and Systems, Prentice

Hall, Second Edition (1997)

2. Sarma, M.S., Introduction to Electrical Engineering, Oxford University Press,2001.

3. R.A. DeCarlo, P-M. Lin, Linear Circuit Analysis, Oxford press, 2nd edition.

4. Allan R. Hambley ,Electrical Engineering: Principles and Applications,Prentice Hall,Second Edition.

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2.3 Prelab Transient response of RLC-Circuits

2.3.1 Problem 1: Switch-On of a RLC Circuit

A supply voltage of 1V is switched on at t= 0 (Fig. 2.7).

1V

100Ohm 100mH

1n5F

Figure 2.7: Series RLC circuit

1. Obtain the differential equation for the voltage vc(t) across the capacitor,identify the damping nature of the circuit and determine the values for thecoefficients C1 and C2.

2. Plot the voltagevc(t) using Matlab.

3. Calculate the resistor value to obtain a critically damped case and obtain thecorresponding equation describing the voltagevc(t) including the values forC1and C2.

2.3.2 Problem 2: Switch-Off of a second order RLC circuit

The switch in circuit (2.8) is opened at t= 0.

3K

100nF12V

1K

100mHt=0

i c

Figure 2.8: Second order RLC circuit

1. Obtain the differential equation for the current ic(t) through the capacitor,identify the damping nature of the circuit and determine the values for thecoefficients C1 and C2.

2. Plot the current ic(t) using Matlab.

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2.3.3 Problem 3 : Oscilloscope Probe

1. Draw an equivalent circuit describing the oscilloscope voltage probe togetherwith the oscilloscope input impedance. The circuit should include the probecable and the probe compensation. Explain the purpose of each component in

the circuit.

2. Use the oscilloscopes manual to obtain available values for your model and therange of the probe compensation.

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2.4 Execution Transient response of RLC-Circuits

2.4.1 Problem 1 : Influence of the oscilloscope probes ontransient measurements.

The oscilloscope probes and impedances from the surrounding circuit have an in-fluence on the transient measurement of a signal. The influence of the oscilloscopeprobe on the measurement should be investigated in the following problem. Mainlythe oversized influence of the ground strap. The problem is a continuation ofproblem 3 in the Prelab.

1. Use the auxiliary signal generator from the experiment box.

+10V

Gnd

sine

square

Auxiliary FunctionGenerator

10V

L

Connect oscilloscopechannel 1

Use L = 10mH. Connect the oscilloscope probe and the inductor in theshortest possible way! Use 1x attenuation.

2. Switch on and observe the shape of the displayed waveform. Measure thenatural frequency n and take a hardcopy. Take a second hardcopy fromwhich you can determine and the damped natural frequency.

3. Replace the 10mH inductor with a 100mH inductor and repeat the measure-

mentes from before.

In the lab report:

1. Determineand the damped natural frequency for both cases from the hard-copies.

2. The capacitance representing the probe cable is 72pF. Its resistance is about500. The capacitance at the oscilloscope input is about 40pF. You can ig-nore the oscilloscope input resistance! The internal resistance of the auxiliaryfunction generator is about 1.5K. The ground strap is represented by the

inductor 10mH.Obtain the differential equation for the voltage vscope(t) seen by the oscillo-scope, identify the damping nature of the circuit and determine the values forC1 and C2.

3. Plot the voltagevscope(t) using Matlab.

4. Compare the experimental results obtained in the lab with the calculatedresults and Matlab plots. Please provide a detailed explanation if the exper-imental results and the Matlab simulations deviate. Discuss the origin of thedeviation.

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2.4.2 Problem 2 : Design of an RLC circuit

Implement the RLC circuit shown in Fig.(2.7) of the Prelab: C= 1.5nF and L =100mH. Instead of using a fixed resistance, use the R-decade.

1. Set the function generator to produce a 100Hzsquare wave with an amplitudeof 0.5V and an offset of 0.5V. Check that he signal modulates between 0Vand 1V. Set the R-decade to 100.

2. Measure the damped radian frequencyd. The frequencyfdcan be determinedby measuring the time period of the exponentially damped sinusoid. Take ahardcopy focusing on the ringing phenomenon.

3. Calculate the damped radian frequencyd. In your calculations, consider theinternal resistance of the function generator to be 50.Compare the calculated value with the measured value in step (2). If they are

consistent, proceed with the next steps.4. Calculate the value of the R-decade, so that the circuit is critically damped.

5. Adjust the R-decade, so that the circuit is critically damped. Display thetransient voltage across the capacitor and take a hardcopy.

6. Adjust the R-decade to 30k, so that the circuit is over-damped. Display thetransient voltage across the capacitor and take a hardcopy.

7. Replace the 1.5nFcapacitor with a 22nFcapacitor.Repeat the steps startingfrom (1) till (6). In step (6), set the R-decade to 40k.

In the lab report:

1. Compare the experimental results obtained in the lab with the Matlab simula-tions in the prelab. Please provide a detailed explanation if the experimentalresults and the Matlab simulations deviate. Discuss the origin of the deviation.

2. Discuss the influence of the capacitance on the transient response.

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3. Lab Experiment 2 : RLC-Circuits - Fre-

quency Response3.1 Introduction to the experiment

3.1.1 Objective of the experiment

The goal of the experiment is to study the frequency response of RLC circuitsand their application as analog filters and resonators. As part of the prelab, thefrequency response of different RLC circuit configurations will be studied usingMatlab. As part of the experimental procedure, different RLC circuit configurationswill be implement and tested. The experimental and the simulation results will becompared and the differences will be discussed.

Introduction

In the experiment RLC Transient Response, we studied the time dependent re-sponse of RLC circuits including the transient response and the steady-state responsefor a constant dc input signal. In the second experiment, we will only explore thesteady-state response of an RLC circuit for a periodic sinusoidal input signal. Asthe frequency of the periodic sinusoidal input signal changes the circuit responsechanges, that is why the second experiment is called RLC frequency response.

In electronics, resonating circuits are often used to select or attenuate particularfrequency ranges, as in tuning a radio. In its easiest form, a resonator can berealized using a resistor, inductor and a capacitor. Therefore, the circuit consists ofat least two different energy storage devices.

Series and parallel RLC configurations

In this experiment, we will in particular study the series and the parallel configura-tion of the RLC resonating circuits shown in Fig.(3.1) and (3.2).First, we will write down the impedance of the series RLC circuit in Fig. (3.1).

ZS=RS+ jL+ 1jC

(3.1)

The admittance of the parallel RLC circuit in Fig.(3.2) can be expressed in an analogway

YP =GP+jC+ 1

jL (3.2)

The absolute value of the impedance and the admittance is given by

|ZS| = R2S+ L 1C2

(3.3a)

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Vc

VL

I

V

C

L

RS VR

Figure 3.1: Series resonator basedon a serial RLC circuit.

C L RP

Ic IL IR

I

U

Figure 3.2: Parallel resonator based ona parallel RLC circuit.

|YP| =

G2P+

C 1

L

2(3.3b)

The phase of the complex impedance and admittance can be expressed by

= arctan

L 1/C

RS

(3.4a)

= arctan

C 1/L

GP

(3.4b)

Based on the phasor plot of the series and parallel resonators in Fig. (3.3) and(3.4), the amplitude and the phase of the complex impedance and admittance canbe determined for a given frequency.In general, the impedance and the admittance can be written as

Z() =R+jX() (3.5a)

Y() =G+jB () (3.5b)

In the phasor plot, the tip of the vector corresponds to the impedance Z or theadmittance Y of the circuit. The following information can be extracted from thephasor plot.

The circuit is in resonance if the oscillation parameter is maximized. In thecase of a series resonator the oscillation parameter is the current, whereas fora parallel resonator the oscillation parameter is the voltage.

The oscillation parameter is maximized and the resonance point is reached,when the phasor intersects with the real axis of the graph. In this case thefrequency is getting equal to the resonance frequency 0.

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Figure 3.3: Phasor diagram of a se-ries resonator based on a RLC cir-cuit. In this case, the impedance ofa series resonator is plotted [3].

Figure 3.4: Phasor diagram of a par-allel resonator based on a RLC cir-cuit. In this case, the admittance ofa parallel resonator is plotted [3].

The frequency0is the frequency associated with the resonance point of the circuit.Consequently, the impedance results in

Im(Z(= 0)) = 0 (3.6)

and the admittance results in

Im(Y(= 0)) = 0 (3.7)

In both cases, the phase gets zero

= 0 (3.8)

Furthermore, the following circumstances apply for the resonance points of a seriesresonator:

ZS(=0)) =Re(ZS()) =RS (3.9)If= 0, the impedance is minimized and|ZS(0)| =min.The parallel resonator exhibits an analog behavior.

YP(=0)) =Re(YP()) =GP = 1/RP (3.10)

Due to the opposite characteristics of an inductive reactance (the inductive reac-tance increases as the frequency is increased) and a capacitive reactance (capacitivereactance decreases with higher frequencies) the reactance XL equals XC for theresonance frequency. Consequently, the inductive and the capacitive impedancescompensate each other, so that the reactive power is getting zero, which means thatonly effective (real) power is consumed by the circuit.

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3.1.2 Application of series and parallel RLC circuits

Reactive power compensation

The reactive power is minimized under the following conditions:

The reactive power gets minimized for a series resonator if the circuit isdriven by a current source. In such a case the input current is constant (|I| =const.) and the impedance determines the voltage drop across the circuit. Asthe impedance is minimized Z( = 0) min. it follows that the voltagedrop is minimized as well|V(0)| = |I| |Z(o)| min.!

The reactive power gets minimized for a parallel resonator if the circuitis driven by a voltage source. In such a case the input voltage is constant(|V| =const.). As the admittance is minimized Y( =0) min. it followsthat the current is minimized as well|I(0)| = |V| |Y(o)| min.!

The reactive power is maximized under the following conditions:

The reactive power gets maximized for a series resonator if the circuit isdriven by a voltage source. In such a case the input voltage is constant ( |V| =const.) and the impedance determines the current flow. As the impedance isminimized Z( = 0) min. it follows that the current flow through thecircuit is maximized|I(0)| = |V|/|Z(o)| max.!

The reactive power gets maximized for a parallel resonator if the circuit isdriven by a current source. In such a case the input current is constant (|I| =const.). As the admittance is minimizedY(=

0)

min.it follows that thevoltage drop across the circuit is maximized as well|V(0)| =|I|/|Y(o)| max.!

Therefore, resonators can be applied for reactive power compensation. Very oftenelectrical consumers have an inductive character. This is the case if several in-ductive consumers like motors, pumps or heaters are in operation. The reactivepower consumption of a load can be reduced or compensated by using a reactiveload, which has the opposite reactive impedance. For example: A capacitive loadcan be used to compensate an inductive load. As a consequence the impedance ofthe whole system is getting reduced, which means the impedance is getting nearly

real.

Filters

Analog Filters can be constructed based on RLC circuits, which transmit certainfrequencies (resonance frequency) in an optimized fashion. Other frequencies (higherand lower frequencies) can be attenuated or blocked. By combining RLC circuitswith slightly different resonance frequencies band-pass filters and band-stop filterscan be design.

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RLC filter design The frequency response of an RLC circuit can be representedby a magnitude and a phase diagram. Two magnitude and phase plots for a seriesand a parallel resonator are shown in Fig.(3.5) and (3.6). The upper graphs showthe magnitude diagram and the lower graphs the phase diagram of the series andthe parallel resonator.

The magnitude of the signal is normalized to simplify visualization and facilitate acomparison of different magnitude plots. Furthermore, the frequency is normalizedin all graphs by the resonance frequency 0. We know from our previous discus-sion that the current is maximized for a series resonator in resonance, whereas fora parallel resonator the voltage is maximized. Consequently, Fig.(3.5) exhibits thenormalized current for the series resonator and Fig.(3.6) shows the normalized volt-age for a parallel resonator.The magnitude and the phase for the two circuits in Fig.(3.5) and (3.6) were cal-culated for two different resistors. The dashed lines correspond to the resistors Rs1and Rp1, whereas the solid lines correspond to the resistors Rs2 and Rp2. With

increasing resistance of the resistor Rs the width of the magnitude for the seriesresonator is enhanced. The opposite behavior is observed for a parallel resonator.With increasing resistance of the parallel resistor Rp the width of the magnitude isreduced. It follows that, Rs1< Rs2 and Rp1> Rp2.The phase diagram exhibits a corresponding behavior. With increasing series re-sistance the transition region from/2 to /2 is widened, whereas for increasingparallel resistance the transition region is getting narrower. Therefore, the series andthe parallel resistance have a distinct influence on the bandwidth of the resonators.

Bandwidth and quality factor The bandwidth is a measure of the frequencyselectivity of a resonating circuit. The bandwidth B of the resonators can directlybe extracted either from the magnitude or the phase plot in Fig.(3.5) and (3.6). Inthe magnitude plot, the bandwidth corresponds to the full width of the curve athalf maximum, which means that the magnitude is dropped by a factor sqrt2. Fromthe phase plot, the bandwidth B can be extracted by taking the difference of thephase between +45 +/4 and45 /4). Furthermore, the bandwidth canbe determined by mathematical means.

In the following, we will focus on a series resonator In the case of a phase

difference of 45

the real part and the imaginary part of the impedance are equal.It follows,

Im(ZS(1)) =I m(ZS(2)) =RS (3.11)

The equation can be expressed in different terms for negative values of the imaginarypart,

|Im(Z(1))| = 11C

1L= RS (3.12)So the frequency 1 can be determined by:

1= R

S2L +

RS

2L2

+ 1

LC (3.13)

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10-2

10-1

100

101

102

0

0.5

0.707

1

Voltage const.

/ 0

I / I0

10-2

10-1

100

101

102

-1.57

-0.79

0

0.79

1.57

/ 0

V

- I

RS1

RS2

B1/

B2/

RS1

RS2

Figure 3.5: Magnitude and phase plot ofa series resonator (RLC circuit) drivenby a voltage source (|V| =const.). Thenormalized magnitude and the phaseare shown as a function of the nor-malized frequency. Rs is the series re-sistor of the RLC circuit. Rs1 is thelarger and Rs2 the smaller series resis-tor: Rs1 > Rs2 [3].

10-2

10-1

100

101

102

0

0.5

0.707

1

Current const.

/ 0

U / U0

10-2

10-1

100

101

102

-1.57

-0.79

0

0.79

1.57

/ 0

V

- I

RS1

RS2

B1/

B2/

RS1

RS2

Figure 3.6: Magnitude and phase plot ofa parallel resonator (RLC circuit) drivenby a current source (|I| = const.). Thenormalized magnitude and the phaseare shown as a function of the normal-ized frequency. Rp is the parallel re-sistor of the RLC circuit. Rp1 is thelarger and Rp2 the smaller series resis-tor: Rp1 > Rp2 [3].

For the positive imaginary part the following equation applies:

|Im(Z(2))| =2L 12C

=RS (3.14)

So the frequency 2 can be determined by:

2=RS2L

+

RS2L

2+

1

LC (3.15)

Based on the two resonance frequencies the bandwidth B can be determined as:

B =2 1= RSL

(3.16)

Besides the bandwidth, the quality-factor (Q-factor) is also an important measureof the frequency selectivity. For example, filters with high Q-factors are important

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and necessary for applications in wireless communications to separate or filter outclosely spaced channels/bands. The quality factor is a representation of the widthof the resonance peak (the larger the Q value, the narrower the peak). Hence, thereis a relation between the Q-factor and the Bandwidth, where high-Q circuit has asmall bandwidth and low-Q circuit has a large bandwidth.

In terms of energy consumption, the Q-factor is a measure for the ratio between thereactive power (inductive or capacitive) and the real power of a resonator. In anotherword, it is a measure of the energy-storage in relation to the energy dissipation ofthe circuit. The Q-factor can be determined by:

Q=QCPR

=QLPR

(3.17)

where,

Q is the Q-factor of the resonator.

QC and QL are the reactive power of the capacitor or inductor.PR is the real (effective) power.

The quality factor of a series resonator can be expressed by other terms:

QS=X0RS

(3.18)

X0 is the reactive resistance of series RLC circuit under resonance conditions.

As the circuit is driven in resonance frequency, the following equation applies:

X0 = 0L= 10C

(3.19)

or it can be expressed by the following terms,

0= 1

LC(3.20)

and

X0 =

L

C (3.21)

It can be seen from Equ.(3.17)-(3.21) that the resonance frequency is defined by theinductance and capacitance of the circuit, whereas the series resistance of the circuitdetermines the Q-factor.

3.1.3 Practical hints

Phasor diagram

Generally, the phasor diagram represents a complex variable G(j) by a rotatingphasor as varies from zero to infinity. where,

G(j) =Re[G(j)] +Im[G(j)] =R() +jX() = |G(j)|G(j)

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|G(j)|2 = [R()]2 + [jX()]2

G(j) = = arctan

X()

R()

Each point on the curve represents the complex value of the variable G(j) for a

given frequency. In the phasor diagram provides information about the locus of thecomplex variable and the circulation of the phasor as the frequency increases. Theprojections of G(j) on the axis are the real and imaginary components at thatfrequency.

Magnitude and phase plot of frequency response

Magnitude and phase plots are used to represent the frequency response of any givencircuit. Both plots are usually shown together and called Bode diagram.

a. Plot of the magnitude of|G(j)|, indicating the amplitude change imposed ona sinusoidal input signal as a function of frequency.

b. Plot of the phase angle G(j) , indicating the phase change imposed on asinusoidal input signal as a function of frequency.

Both graphs are plotted against the frequency rad/sec.

Lissajou figure

An oscilloscope can be used to determine the phase difference between two periodicsignals. However, an oscilloscope is only able to measure a voltage signal. A currentcan be measured by creating a voltage drop cross a resistor. The phase differencecan be extracted by applying both of these signals to the input channels of theoscilloscope.However, for small phase differences it is difficult to extract the phase differencedirectly from the screen of the oscilloscope. As an alternative, a Lissajou figure canbe used to determine the phase difference of two signals. A Lissajou figure is moreaccurate if it comes to smaller phase differences. The concept of a Lissajou figuregoes back to the old days when people were using cathode ray tubes (CRT) basedoscilloscopes. The position of a spot on the screen of a CRT oscilloscope is controlledby the voltages applied to the x and y deflection capacitors of the cathode ray tube.In normal operation (a voltage is shown as a function of time), a triangular voltage

is applied to the x-deflection capacitor so that the spot moves from the left to theright side of the screen. In the case of a Lissajou figure we directly apply the secondsignal to the x-deflection capacitor.The x-deflection of the signal can be described by:

Vx(t) =Vpp sin(t) (3.22)whereas the y-deflection can be described by:

Vy(t) =c Ipp sin(t+) (3.23)Vpp

is the peak amplitude of the voltage, c is a constant factor (resistance) and Ippis the peak current.

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Figure 3.7: Extraction of the phase dif-ference between two sinusoidal signals.

Figure 3.8: Extraction of the phase dif-ference by using a Lissajou figure.

For t = 0, we can extract: 2 c Ipp sin= aFurthermore, we can determine 2 ymax= 2 c Ipp=b.As a consequence we can deduce the phase difference from the parameters a and b

by the following simple expression

= arcsina

b

(3.24)

The accuracy of this method is distinctly higher than a direct comparison of thesignals. This is in particular true for small phase differences.

3.2 Handling of the function generator and theoscilloscope

Throughout the procedure of the lab the frequency response of several circuits hasto be taken. In order to measure the frequency response of a circuit the sweep modeof the function generator can be used. In this case the function generator changesits frequency over time.Using and setting up the sweep mode:

a. Settings of the Function generator:Select the Sweep Menu from the function generator and set:

Sweep frequency from 100Hz (1: START F)Sweep frequency to 100KHz (2: STOP F )Sweep time to 500msec (3: SWP TIME )Sweep mode to logarithmic (4: SWP MODE )

b. Settings of the Oscilloscope:To see the full sweep at the oscilloscope you have to set the time base to500ms/10div = 50ms/div).

c. Synchronization of the function generator and the oscilloscope:In order to measure a frequency response with the oscilloscope the functiongenerator and the oscilloscope have to be synchronized. To do so the synchro-nization signal of the signal generator has to be connected with the triggerinput of the oscilloscope and the oscilloscope has to be set to external trig-ger.

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d. Grounding of the function generator and the oscilloscopeThe grounds of the function generator and the oscilloscope have to be con-nected together throughout all measurements in order to have the correctoutput on the oscilloscope screen. You have to construct the circuit in thisway that one of the terminals of the component under test is always on ground

while measuring the voltage across the component.

3.3 References1. M.S. Sarma, Introduction to Electrical Engineering, Oxford Series in Electrical

and Computer Engineering, 2000.

2. J. Keown, ORCAD PSpice and Circuit Analysis, Prentice Hall Press (2001).

3. F. Hohls, Lab experiments, Resonators, University Hannover, Spring 2003.

4. R.A. DeCarlo, P-M. Lin, Linear Circuit Analysis, Oxford press, 2nd edition.

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3.4 Prelab RLC Circuits - Frequency response

3.4.1 Problem 1: Steady state response of an RLC circuit

Given is the following circuit:

1Ohm

0.25H 0.8F

~ Vou tVin

The input voltage is

Vin= 0.375cos(t) + 0.5sin(t) = 2 rad/sec

Find the steady state voltage Vout using two methods:

1. The differential equation describingVout.

2. Phasors.

Note: Vout should be in the form a cos(t) +b sin(t).

3.4.2 Problem 2: Bandwidth of an RLC circuit

Derive a mathematical expression for the bandwidth of a RLC parallel resonatorcircuit.

3.4.3 Problem 3: RLC resonator

For a series RLC resonant circuit based onR = 560, C= 100nF andL = 100mH.

1. Use Matlab to plot the voltage drop across the resistor vR, the capacitor vC,the inductor vL, and across both the capacitor and the inductor together vCLvs. frequency. Use 5V peak and vary the frequency starting 100Hz to 100KHz.

2. Taking the voltage across the resistance represents a band-pass filter. Measurethe bandwidth and calculate the Q-factor.

3. Replace the 560 resistance with a 220 resistance. Measure the bandwidthand calculate the Q-factor.

4. Connect the 560 resistance back to the circuit. Replace the 100mH inductorwith a 10mH inductor. Measure the bandwidth and calculate the Q-factor.

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3.5 Execution RLC Circuits - Frequency response

3.5.1 Problem 1 : Bandwidth of an oscilloscope probe

Determine the frequency response of an oscilloscope probe set to 1x attenuation.

1. Connect the reference signal from the frequency generator to channel 1 of theoscilloscope. Use a BNC cable with a proper 50 termination resistor.

2. Determine the type of probe you got together with your oscilloscope. It iseither a Testec TT-LF 312 or a Tektronix P2100 probe.

3. Connect the probe under test to channel 2. Do not forget to switch the probeand the corresponding setting in the channel menu of the oscilloscope to 1x.

4. Measure reference and probe channel voltage at 100Hzand maximum gener-ator frequency. If possible try to find the frequency where the probe signaldrops to3dB. Take a hardcopy for every setting.Note : The last point may not be possible depending on the probe or thefunction generator!

In the lab report:

1. What is the magnitude (indB!) at minimum/maximum frequency? Did youreach/exceed the3dB frequency of the probe?

2. Explain the strategy you have used to find the3dB frequency.

3. What statement can you give about the the bandwidth of the 1x probe?

4. Compare your findings to the manual of the used probe.

3.5.2 Problem 2 : Characterization of an RLC resonator

Implement a series RLC resonant circuit based on R = 560, C = 100nF, L =100mH. Use the function generator as source.

1. Set the function generator to sine, 5Vpp, and no offset. Use the sweep mode.Vary the frequency in 500msfrom 100Hzto 100kH z. Use log sweep mode.

Note: Setting of the sweep mode is described in 3.2!

2. Obtain hardcopies of the voltage across

(a) the resistorVR.

(b) the capacitor VC.

(c) the inductor VL.

(d) both the capacitor and the inductor togetherVCL.

Note: Special settings and grounding of the oscilloscope is described in 3.2!

3. Change the circuit in a way that you get a band pass!

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4. Measure the resonance frequency of the circuit.

5. Determine the phase of the output signal below resonance (500Hz), at reso-nance and above resonance (2kH z) using two methods:

(a) Use the cursors of the oscilloscope.(b) Use Lissajous figure.

Take hardcopies for both methods.

In the lab report:

1. Determine the phase shift for both methods. (Take care of the sign of thephase.)

2. Compare the obtained curves with the theoretical curves obtained using Mat-lab.

3. Compare the measured resonance frequency with the calculated resonancefrequency using the provided values for R, L and C.

3.5.3 Problem 3 : Bandwidth and quality factor of RLCBand Pass

A series RLC resonance circuit represents a band-pass filter. Use the circuit imple-mented in the problem before.

1. Measure the resonance frequency. Find the upper and lower

3dB frequencies

to determine the bandwidth of the band-pass filter. Take hardcopies!

2. Replace the 560 resistor by a 220 and repeat the measurements. Takehardcopies!

3. Replace the 220 resistor by a 560 resistor. Furthermore, replace the 100mHinductor by a 10mH inductor. Measure the resonance frequency, the upperand lower -3 dB frequency. Take hardcopies.

In the lab report:

1. Calculate the bandwidth for all three cases.

2. Calculate the Q-factor for all three cases by using only the resonance frequency,the upper and the lower3dB frequencies.

3. Compare the experimental results obtained in the lab with the Matlab simu-lations in the prelab.

4. Describe the influence of the resistor and the inductor to the properties of abandpass?

5. Provide an explanation if the measured parameters deviate from the parame-ters in the prelab. If necessary provide a calculation that explains the differ-ence.

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3.5.4 Problem 4 : The Black Box Resonator

Characterize the unknown circuit provided by the instructor or the teaching assis-tant. The circuit consists only of a single resistor, a single capacitor and a singleinductor connected together to form a series resonant circuit. The values for the

resistor, the inductor and the capacitor are unknown. Determine the resonancefrequency, R, L and C by carrying out measurements.In the lab report:

1. Explain the strategy you have used in the lab.

2. Show detailed calculations how to determine the values of R, C, and L.

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4. Lab Experiment 3 : Fourier Series and

Fourier Transform4.1 Introduction to the experiment


The goal of the second experiment of the Signals and Systems Lab is to study differ-ent signals in terms of their Fourier coefficients and to get a deeper understandingof the Fourier transform. The handout will provide the basic theory and describesthe various variables and concepts involved. A more detailed description of the the-ory can be found in reference [1]. The prelab and the experimental procedure willconcentrate on the simulation and implementation of the Fast Fourier Transform(FFT) rather than the detailed mathematical description. The experimental andthe simulation results will be compared and differences will be discussed.

4.1.2 Introduction

A signal can be represented in the time domain or in the frequency domain. The fre-quency domain representation is also called the spectrum of the signal. The Fourieranalysis is the technique that is used to decompose the signal into its constituentsinusoidal waves, i.e. any time-varying signal can be constructed by superimposing

sinusoidal waves of appropriate frequency, amplitude, and phase. The knowledge ofthe frequency content of a signal can be very useful. For example, the frequency con-tent of human speech can be filter, the quality of transmitted signals can be improvedand noise can be removed. The Fourier transform is used to transform a signal fromthe time domain to the frequency domain. For certain signals, Fourier transformcan be performed analytically with calculus. For arbitrary signals, the signal mustfirst be digitized, and a Discrete Fourier Transform (DFT) is performed. On theother hand, the inverse Fourier transform is used to transform a signal from thefrequency domain to the time domain. The handout is divided into two parts. Thefirst part introduces Fourier series representation and Fourier transform for periodic

continuous-time signals. The second part describes Fourier series representation andFourier transform for periodic discrete-time signals.

4.1.3 Part I: Continuous time signals

A. Fourier series representation

A continuous-time periodic signal can be described by the sum of basic signals, i.e.the sum of sine or cosine waves.

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Periodic signals

A signal is defined as periodic, if for some positive value of T, the signal can bedescribed by Equ.(4.1),

x(t) =x(t+T) (4.1)

This must hold for all t. The fundamental period is the minimum positive, nonzerovalue of T for which the above equation is satisfied. The value0 = 2/Tis referredto as the fundamental frequency.

Determination of Fourier series coefficients

Given a functionx(t) , its Fourier series coefficients ak can be obtained by using thefollowing equation.

ak = 1TT

x(t)ejk0tdt (4.2)

Thus,x(t) can be expressed in terms ofak as follows,

x(t) =+

k=

akejk0t (4.3)

On substitutingk = 0 into Equ.(4.2), the dc or the constant component of the signalx(t) is obtained,

a0 = 1T

T

x(t)dt (4.4)

So far we expressed the signal x(t) as a sum of superimposed complex exponentialfunctions. However, a number of other ways can be used to represent x(t). Asexample, Equ.(4.3) can be rewritten as,

x(t) =a0

2 +

k=1

[Akcos (k0t) +Bksin (k0t)] (4.5)

In such a case, each continuous time periodic function can be described by the sumof superimposing sine and cosine functions. The required Fourier coefficients are a0(dc component), Ak and Bk.In the following section, the Fourier series coefficients and the Fourier series for acontinuous time periodic square wave will be obtained analytically.

Continuous time periodic square wave

A continuous time periodic square wave is given in Fig.(4.1) with a period T and apulse width of 2T1.

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y = x(t)

tTT/2T1-T1-T/2

Figure 4.1: Continuous periodic square wave

The signal is described as follows:

x(t) =

1, |t| T10, T1< |t| < T /1

(4.6)

The Fourier series coefficients are obtained using,

ak = 1T

T

x(t)ek0tdt (4.7)

To calculate the constanta0, usek = 0 and perform the integration over any intervalof length T,

a0 = 1

T

T/2T/2

1 dt= 1

T

T/4T/4

1 dt= 1

T

T

2

=

1

2 (4.8)

Due of the symmetry ofx(t) about t = 0, it is suitable to integrate fromT/2 t < T/2. The signal is only equal to 1 in the range of

T /4 t < T/4, so that

Equ.(4.7) results to,

ak = 1

T

T

x(t)ek0tdt= 1

T

TT1

x(t)ek0tdt

= 1jk0T

ejk0tT1

T1

= 2

k0T

ejk0T1 ejk0T1

2j

(4.9)

Making use of the trigonometric identity,

sin(x) = e

jx

ejx

2j (4.10)

Equ.(4.9) becomes

ak = 2

k0T sin(k0T1) =

1

ksin (k0T1) (4.11)

where, T is the period of the signal and T1 is the width of the periodic pulses.For more demonstration, Matlab was used to plot the scaled Fourier coefficient T akgiven by Equ.(4.11) for k =50 to k = 50 . The width of the pulses were keptconstant atT1, and the period of the signal was varied fromT = 4T1,T = 8T1, andT = 20T1 as shown in Fig.(4.2), Fig. (4.3) and Fig.(4.4), respectively.

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-50 0 50-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Scaled Fourier Series Coeffic ients Takfor x(t)

Harmonic Number

Amplitude

Figure 4.2: Scaled Fourier coefficients T ak for T1 fixed and T = 4T1

-50 0 50-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

Scaled Fourier Series Coeffic ients Takfor x(t)

Harmonic Number

Amplitude


Now, we will continue to describe x(t) in terms of sine and cosine functions asdescribed by Equ.(4.5).The coefficientsAk and Bk can be determined by

Ak = 2Re (ak) (4.12)

Bk = 2Im (ak) (4.13)

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-50 0 50-0.05

0

0.05

0.1

0.15

0.2

Scaled Fourier Series Coefficients Takfor x(t)

Harmonic Number

Amplitude


Using Equ.(4.11), for T = 4T1.

Ak = 2

ksin (k

2) =

2k

k = 1, 3, 5, ...(odd)

0 k = 2, 4, 6, ... (even)(4.14)

Bk = 0 (4.15)

Substituting into Equ.(4.5),

x(t) =1

2+

2

k=1,3,5,...

sin(k

2) cos(k0t)

k (4.16)

For more demonstration, Matlab was used to plot the first three harmonics as shownin Fig.(4.5).Fig.(4.6) shows the first 50 harmonics (terms in Equ.(4.16)) and the corresponding

square wave after the summation. Fig.(4.7) shows the first 1000 harmonics (termsin Equ.(4.16)) and the corresponding square wave after the summation.A comparison of Fig.(4.6) and Fig.(4.7) indicates that the shape of the reconstructedsquare wave can already be recognized after the summation of the first 50 harmonics.However, the reconstructed signal exhibits a lot of ringing at each step change inthe square wave, i.e. the Fourier series exhibits a peak followed by rapid oscillations.The phenomenon is called Gibbs effect. With increased number of terms, e.g.1000harmonics, reconstructed signal is getting closer to the original signal. Generally,this phenomenon is due to the discontinuities in the square wave and many highfrequency components are required to construct the signal accurately. More infor-

mation about Gibbs effect can be found in reference [1].

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0 0.2 0.4 0.6 0.8 1-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Cosine Oscillations

Time t/T

Amplitude

k =1

k=3

k=5

Figure 4.5: The first three harmonics of a square wave. The time axis is normalized.The time t = 1 corresponds to t= T

Figure 4.6: The first 50 harmonics of a square wave given in Equ.(4.16) and the sumof the harmonics. The time axis is normalized. Time t= 1 corresponds to t= T.

B. Continuous time Fourier transform

The continuous time Fourier transform is a generalization of the Fourier series. Itcan be also called Fourier series representation for continuous time aperiodic signals.It only applies to continuous time aperiodic signals. The Fourier transform of a given

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Figure 4.7: The first 500 harmonics of a square wave given in Equ.(4.16) and thesum of the harmonics. The time axis is normalized. Time t = 1 corresponds tot= T.

signal x(t) is defined as

X(j) =

x(t)ejtdt (4.17)

Using the inverse Fourier Transform the original signal can be obtained using thefollowing equation,

x(t) = 1

2

X(j)ejtd (4.18)

As an example, we will calculate the Fourier transform of the square pulse shown inFig.(4.8).

y = x(t)

t

Figure 4.8: Square pulse with a pulse width of

The signal in Fig.(4.8) is 0 everywhere except in the range /2 t < /2, wherethe signal is equal to 1. After calculating the Fourier transform we get:

X(j) = /2/2

e

jtdt= ejt

j/2

/2=ej /2 ej/2j (4.19)

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Again making use of the trigonometric identity (Equ.(4.10)),

X(j) =

sin(2

)2

= sinc

2

(4.20)

NOTE

Fourier declared that an aperiodic signal could be viewed as a periodic signal withan infinite period.As an example, we studied the Fourier series of a square wave and the Fouriertransform of a rectangular pulse. As the period becomes infinite, the periodic squarewave approaches the Fourier transform of the rectangular pulse.

4.1.4 Part II: Discrete time signals

A. Fourier series representation

In this part, we will discuss the Fourier series representation for discrete time signals.The discussion will closely follow the discussion in the first part.

Periodic signals

A signal is defined as periodic, with period N if,

x[n] =x[n+N] (4.21)

This must hold for all n. The fundamental period is the smallest positive integerN for which the above equation holds. The parameter 0 = 2N

is referred to as thefundamental frequency.

Determination of Fourier series coefficients

Given a function x[n], its Fourier series coefficients ak can be obtained using thefollowing equation.

ak = 1

N n=(N)a[n]ejk0n (4.22)

Thus x[n] can be expressed in terms ofak as follows,

x[n] = 1

N

n=(N)

akejk0n (4.23)

In the following section, the Fourier series coefficients for a discrete time periodicsquare wave will be obtained analytically.

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nNN1-N1-N 0

Figure 4.9: Discrete periodic square wave with a period of N and a pulse width of2N1

Discrete time periodic square wave

Given a discrete time periodic square wave as shown in Fig.(4.9) with a period Nand a pulse width of 2N1. The signal can be described as follows:

x[n] = 1 for N1 n N1 (4.24)

The Fourier series coefficients can be obtained using

ak = 1

N

2N1n=N1

ejk(2/N)n (4.25)

where, m= n+N1, so that the Fourier coefficients can be described as

ak = 1

N

2N1m=0

ejk(2/N)(mN1) = 1

Nejk(2/N)N1

2N1m=0

ejk(2/N)m (4.26)

The summation term is a geometric series. On using the equation for the sum of a

geometric series

ak =N1n=0

n =

N = 11n

1 = 1 (4.27)

The following equation is obtained

ak = 1

Nejk(2/N)N1

1 ejk2(2N1+1)/N

1 ejk(2/N)

=

1

Nejk(2/2N) ejk2(N1+1/2)/N

ejk2(N1+1/2)/Nejk(2/2N) (ejk(2/2N) ejk(2/2N)) (4.28)

The equation can be rewritten as

ak =

1N

sin2k(N1+1/2)

N

sin(/kN ) k= 0, N, 2N,...

2N1+1N

k= 0, N, 2N,...(4.29)

The scaled Fourier series coefficients Nak are plotted in Fig.(4.10) for 2N1+ 1 = 5and N= 40.

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-80 -60 -40 -20 0 20 40 60 80-0.04

-0.02

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

Fourier Series Coeffic ients Nakfor x(n)

Harmonic Number

Amplitude

Figure 4.10: Fourier series coefficients of the discrete periodic square wave

NOTE

There are some important differences between Fourier series representation for con-tinuous time periodic signals and Fourier series representation for discrete time peri-odic signals. The Fourier series representation for discrete time periodic signals is afinite series, while Fourier series representation for continuous time periodic signals

is infinite series. Also, an important property that must be noted is that discreteFourier series coefficients are periodic with period N, i.e.

ak=ak+N (4.30)

Compare Fig.(4.4) (T = 20T1) and Fig.(4.10) (N = 20N1) and you can notice thetwo differences clearly.

B. Discrete time Fourier transform

Computers and other digital electronic based systems cannot handle continuous time

signals. These systems can only process discrete data. Therefore, a discrete formof the Fourier Transform, i.e. a numerical computation of the Fourier transform,is needed to give us spectral analysis of discrete signals. This transform is calleddiscrete Fourier transform.The discrete Fourier transform converts a time domain sequence resulting fromsampling a continuous time signal into an equivalent frequency domain sequencerepresenting the frequency content of the given signal. Thus, the DFT is given bythe following equation.

X[k] =N1

n=0

x[n]ej2kn/N k= 0, 1, 2,...,N

1 (4.31)

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The Inverse Discrete Fourier Transform (IDFT) performs the reverse operation andconverts a frequency domain sequence into an equivalent time domain sequence

x[n] = 1

N

N1

k=0 X[k]ej2kn/N n= 0, 1, 2,...,N 1 (4.32)

The DFT plays an important role in many applications of digital signal processingincluding linear filtering, correlation analysis and spectrum analysis.The computation of the DFT is computationally expensive as the DFT computesthe sequenceX[k] ofNcomplex valued numbers given another sequence of data x[n]of length N. From the equations for the DFT, it can be seen that to compute allN values N2 complex multiplications and N2 Ncomplex additions are required.The direct computation using the DFT is inefficient because it does not exploit thesymmetry and periodicity properties. Thus a number of algorithms exist that makesthese computations more efficient. An important algorithm for computer the DFTis the Fast Fourier Transform (FFT).The FFT algorithms exploit these two basic properties and make the computationmore efficient.

NOTEIt should be clear that FFT is not an approximation of the DFT. It yield the sameresult as the DFT with fewer computations required.

In MATLAB, the FFT is performed using the function (fft). The output of the fft()function by itself is a vector of complex numbers. The following formula returnsa vector of the magnitudes of each of the frequencies contributions to the signalsamplitude.

y= 2 abs(fft data)length(data)

(4.33)

To achieve the absolute value of the complex magnitude we need the abs() function.Since the fft() returns the complex amplitudes scaled by the overall length of thedata, we need to divide by length of the data. Finally the equation has to bemultiplied by 2 (because of Eulers Relation!?). Only the first half of the vector ycontains relevant data, so

y= y

1 :

length(data)

2

(4.34)

The highest frequency that can be perceived in a signal is given by the NyquistFrequency:

fnyqu =Fs

2 where Fs is the sampling frequency (4.35)

The frequencies that correspond to the y vector range from 0 Hz to the NyquistFrequency can be generated by:

f = linspace (0, fnyqu, length(y)) (4.36)

Finally the plot command plot(f, y) shows the frequency components from the yvector. A detailed description of fft() function can be found along with examples inthe Matlab documentation [5].

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4.1.5 Definitions and practical hints

The sampling process Sampling is the transfer of a continuous time signal into adiscrete time signal or the transfer from the world of analog signal processing to theworld of digital signal processing. In practice, most signal processing is performed

on discrete time signals and not on continuous time signals. This applies despite thefact that most of the signals encountered in science and engineering are analog innature. In the next experiment, the sampling theory and sampling techniques willbe discussed in more details.

Continuous Fourier transform

The Continuous Fourier transform is used to transform a continuous time signal intothe frequency domain. It describes the continuous spectrum of a non-periodic timesignal.

Discrete Fourier transform

Is used in the case where both the time and the frequency variables are discrete.

Fast Fourier transform

Is a special algorithm which implements the discrete Fourier transform with con-siderable savings in computational time. It must be clear that the FFT is not adifferent transform from the DFT, but rather just a means of computing the DFTwith a considerable reduction in the number of calculations required.

FFT using the oscilloscope

As part of the experiment, the FFT has to be obtained using the oscilloscope. Inorder to obtain the FFT, the following steps should be carried out. First the timedomain signal should be set properly by the following steps:

a. Press AUTOSET to display the time domain waveform.

b. Position the time domain waveform vertically at center of the screen to getthe true DC value.

c. Position the time domain waveform horizontally so that the signal of interestis contained in the center eight divisions.

d. Set the YT waveform SEC/DIV to provide the resolution you want in the FFTwaveform. This decides how high the frequency content of the transform willbe. The smaller the time scale used the higher the frequency content.

Then the FFT can be performed as follows:

a. Locate the math button on the front panel of the oscilloscope.

b. On pressing it you will enter the Math menu.

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c. In function choose fft.

d. The following screen will appear. (Clearly the waveform in your case will bedifferent.)

Figure 4.11: Hardcopy from oscilloscope screen

1. Frequency value at center position of the screen (where the arrow is belowTrigd).

2. Vertical scale in dB/division

3. Horizontal scale in frequency/division

4. Sample rate in number of samples/sec

5. FFT window type.

The Fourier series represents a periodic waveform as an infinite series of harmonicallyrelated sinusoids. Since the Fourier series contains only discrete frequencies, eachsinusoidal component of the waveform is represented by a vertical line on a plotof the signal magnitude versus frequency. The height of the line represents themagnitude of the contribution from that particular frequency. The location of theline along the horizontal axis identifies its frequency.For all FFTs in this experiment always use the Hanning window, as it is best suitedamong the given options for a periodic signal. A windowing function is basically afunction used to cut out a part of the signal in time domain so that an FFT can becarried out on it. Thus basically it is multiplication by some kind of rectangular pulse

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in time domain, which implies convolution with some kind of a sinc in frequencydomain. Using a windowing function affects the transform but is the only practicalmethod for obtaining it.Once the time domain signal has been set up as discussed previously and you are inthe FFT screen. The following steps must be carried out:

a. Bring the region of the frequency domain that you are interested in towardsthe middle of the screen.

b. Adjust the FFT zoom button in order to zoom into the transform sufficientlytill you reach the magnitude of frequency that you desire. At this point youshould normally make your hardcopy, as it is where you would be able to seethe transform most clearly.

4.1.6 References

1. A. V. Oppenheim, A. S. Willsky, S. H. Nawab, Signals and Systems, PrenticeHall, Second Edition (1997).

2. Raymond A. DeCarlo, Pen-Min Lin, Linear Circuit Analysis, Oxford Uni-versity Press, Second Edition (2001).

3. J. G. Proakis, D. G. Manolakis, gDigital Signal Processingh, Prentice Hall,Third Edition (2002).

4. Sarma, M.S., Introduction to Electrical Engineering, Oxford University Press,2001.

5. MATLAB Documentation.

6. TDS220 manual.

7. TDS200-Series Extension Modules manual.

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4.2 Prelab Fourier Series and fourier Transform

4.2.1 Problem 1 : Decibels

In the lab, you must be able to express the signal amplitude in Vpp and Vrms, also

you have to know what dBVrms corresponds to.

1. Givenx(t) = 2 cos(21000t) ,

a. What is the signal amplitude inVpp?

b. What is the root-mean-square value of the provided signal inVrms?

c. What is the amplitude of the spectral peak indBVrms?

2. For a square wave of 4Vpp and the voltage level changes between 2Vand 2V,a. What is the signal amplitude inVrms?

b. What is the amplitude in dBVrms?

3. For a rectangular wave of 8Vpp with 20% duty cycle and the voltage levelchanges between4V and 4V,

a. What is the signal amplitude inVrms?

b. What is the amplitude in dBVrms?

4.2.2 Problem 2 : Fourier Transform of a Square/Rectangular

Wave1. Use Matlab to generate a square wave of 2Vpp amplitude, no offset. Use 1ms

for the period and 100kH zfor the sampling frequency.

2. Plot the square wave in time domain.

3. Obtain the FFT spectrum using Matlab FFT function. Use N, the FFT lengthto be the length of the square wave data vector.

4. Plot the spectrum magnitude for the first half of FFT showing the fundamentalfrequency and the first four harmonics.

Hint: Use the Matlab command xlim.

5. Plot the spectrum magnitude in dB for the first half of FFT showing thefundamental frequency and the first four harmonics.Hint: Use the Matlab command xlim.

6. Repeat the previous steps starting 2.1 till 2.5 using 20% and 33% duty cycles,respectively. The period and amplitude are kept constant at 1msand 1Vpp.

7. Discuss the changes for smaller pulse width.Hint: Use the subplot command to plot the spectrum magnitude for the three

cases 50%, 33% and 20% duty cycle to ease the comparison.

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4.2.3 Problem 3 : Calculate the Fourier Transform of asound sample

The Matlab command Y = WAVREAD(FILE) reads a wave file specified by thestring FILE, returning the sampled data in Y. To get the sample rate FS in

Hertz and the number of bits per sample NBITS used to encode the data in thefile use:[Y, FS, NBITS] = WAVREAD(FILE).

1. Download the sound file (Sound Sample.wav) from the Campusnet webpage.

2. Using Matlab, read the sound file and plot the first 10msof the signal in timedomain.

3. Use the Matlab FFT function to compute the spectrum and plot only the firsthalf of the FFT spectrum.

4. What are the tones forming this signal?

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4.3 Execution Fourier Series and fourier Trans-form

4.3.1 Problem 1 : FFT of Single Tone sinusoidal wave

1. Use the function generator to generate a sinusoidal wave having 500Hz fre-quency, 2Vpp amplitude and no offset.Take a hardcopy in time domain. Use the measure function to document allproperties.Obtain the FFT spectrum using the oscilloscope FFT MATH function andtake hardcopies of the complete spectra and the spectra peak.

2. Set the generator to sinusoidal wave with 2KH z frequency, 2Vpp amplitude,and +1V dc offset.Take a hardcopy in time domain with all properties shown on screen.

Obtain the FFT spectrum and take hardcopies of the complete spectra andthe spectra peak.Note : Do not forget to align the ground of the time domain wave exactly inthe vertical center of the screen (see4.1.5)!!!

3. Generate a sinusoidal wave having 0dBspectrum peak, 2KHzfrequency, with-out a dc offset. What is the amplitude value? Again take hardcopies of timeand frequency domain.

In the lab report:

1. What is the relation between the bandwidth, i.e. highest frequency that canbe measured accurately by a digitizing oscilloscope and the sampling rate?

2. Use Matlab to calculate the expected FFT spectra for the parameters given inpart4.3.1.1. Is the calculated spectra consistent with the measured spectra?

3. Use Matlab to calculate the expected FFT spectra for the parameters given inpart4.3.1.2. Is the calculated spectra consistent with the measured spectra?

4. Use Matlab to calculate the amplitude value of a spectrum with a 0dB peak.Is the calculated amplitude consistent with the measured amplitude?

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4.3.2 Problem 2 : FFT of square wave

1. Use the function generator to generate a square wave having 1msperiod, 2Vppamplitude, and no offset.

2. Obtain the FFT spectrum. Instead of using the time base (sec/div) controlto accurately measure the frequency components, use the FFT zoom controlthat provides a zoom factor up to 10 and use the cursors to determine theamplitudes of the fundamental frequency and the first four harmonics. Takehardcopies of the signal in time and frequency domains.

3. Obtain the FFT spectrum for 20% duty cycles. Determine the amplitudes ofthe fundamental frequency and the first four harmonics. Take hardcopies ofthe signal in time and frequency domains.

In the lab report:

1. For frequency domain measurements, the frequency scale needs to be expandedin order to accurately measure the frequency components. This could be donewith the time base (sec/div) control. What is the effect of doing this on themeasured bandwidth? Information can be found in reference [6] and [7].

2. Use the hardcopies taken to discuss the effect of changing the duty cycle onthe FFT results.

3. What is the effect of the DC offset in problem4.3.2.3if you look at the hard-copy?

4.3.3 Problem 3 : FFT of Multiple-Tone sinusoidal wave

1. Combine the signal from the sine output of the auxiliary signal generator anda 2Vpp, 10KHz sinusoidal wave from the Agilent signal generator. Use thefollowing circuit.

+10V

Gnd

sine

square

Auxiliary FunctionGenerator

10V

R110K0

R210K0

R3100K

Uout

~Sine 2Vpp10KHz

2. Take a hardcopy of the signal in time domain.

3. Take a hardcopy of the FFT spectrum of the signal.

In the lab report:

Use the hardcopy of the spectrum and discuss the linearity of the FFT.

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4.3.4 Problem 4 : Measure the Fourier Transform of a soundsample

1. Download the sound file (Sound Sample.wav) from the Campusnet web page.

2. Use any wave file player (e.g. media player) and set it to repeat playingthis sound file. Connect the audio-out from your laptop to the input of theoscilloscope. Adjust the amplitude in such a way that you get a signal withabout 1Vpp up to 2Vpp.

3. Take a hardcopy of the signal in time domain.

4. Obtain the FFT spectrum and take a hardcopy.

In the lab report:

1. What are the frequencies used to create this sound wave?

2. Compare the measured spectra with calculated spectra. Explain the differ-ences if necessary.

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5. Lab Experiment 4 : Sampling

5.1 Introduction to the experiment


The goal of this experiment is to introduce the basics of sampling and to implementdifferent sampling circuits.The foundations of sampling will be discussed on the signals and systems and onthe component level. Different sampling schemes like impulse train sampling, rect-angular pulses sampling and sample and hold will be introduced. In addition, theconsequence of the different sampling schemes on the reconstructed signals will be

described.Furthermore, two different types of sampling circuits will be discussed. The firstsampling circuit is the sampling bridge used for high-speed sampling applications,e.g. as part of a digital oscilloscope. The second sampling circuit is a samplingcircuit based on Metal Oxide Semiconductor Field Effect Transistors (MOSFETs).

5.1.2 Introduction

Most of the signals in nature exist in an analog form. Sampling is the transfer ofa continuous time signal into a discrete time signal. Sampling is the first and avery important step to provide signals, which can be digitally processed. Therefore,sampling is the connecting element between the world of analog and digital signalprocessing.The handout is divided into three parts, where the first part of the handout in-troduces the sampling theory and discusses the difference between sampling andquantization of signals. The second part will explain different ways of sampling likeimpulse train sampling, rectangular pulses sampling and sample and hold schemes.Finally, different implementations of sampling circuits will be discussed.

5.1.3 Sampling Theory

Sampling step versus quantization step

Sampling is the transfer of a continuous time signal into a discrete time signal.However, sampling should be clearly distinguished from the quantization of a signal.Both steps are necessary to carry out an analog-to-digital (A/D) conversion.Quantization is the transformation of a continuous signal into a discrete signal interms of discrete amplitudes or discrete signal levels. The quantization leads todiscrete values for the amplitude, whereas the sampling leads to values of discretetimes. The difference between sampling and quantization is illustrated in Fig. (5.1).Fig.(5.1) provides a schematic description of an A/D conversion process. In orderto transfer continuous time signals into a stream of bits, the signal can be first

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quantized and afterwards sampled or the signal can be first sampled and quantizedafterwards.In both cases, we get a signal that is discrete in terms of time and amplitude. Afterthe quantization and the sampling the signal is coded, meaning the signal is trans-formed in a stream of bits, which can be processed or transmitted. After carrying

Continuous-time signal(analog signal)

Discrete time andamplitude signal(digital signal)

t

x

[t]

t

x

[t]

t

x

[t]

t

x

[t]

t

x

[t]

Sampling

Sampling Quantization

Quantization

Coding

Data stream(digital signal)

Discrete time oramplitude signal(digital signal)

Figure 5.1: Schematic illustration of an A/D conversion process using a sampling,quantization and coding steps.

out certain processing steps the digital signal is very often converted back into ananalog signal. Under certain circumstances, it is possible to completely recover theinitial signal. This very important property follows the sampling theorem. Thistheorem is simple, but very important and useful, because based on the samplingtheorem we can decide whether a signal can be completely recovered or not.

The sampling theorem

We assume that x(t) is a band-pass limited signal which has a Fourier TransformX(j) = 0 for frequencies larger than a maximum cut-off frequency M. The initialsignal can be reconstructed from the sampled signal if the sampling frequencyS istwo times larger than the maximum cut-off frequency of the band pass filter.

S>2M (5.1)

The sampling frequencyScan be described by

S=2

T (5.2)

where T is the period of an impulse train sampling signal. If the sampling frequencyis smaller than 2 times the maximum frequency of the band-pass limited signal,

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the initial signal cannot be completely reconstructed afterwards. In this case, thesample rate is not high enough and the term under-sampling or aliasing is used. Ifthe sampling frequency is exactly equal to 2 times the maximum frequency of theband-pass limited signal, then we speak about the Nyquist frequency or the Nyquistrate. The sampling frequency has to be higher than the Nyquist rate. Otherwise,

the signal cannot be reconstructed.Remark: Sampling is not only important when we deal with voltages or currents.There are several other areas, where we have to deal with sampling. Take forexample the area of digital photography or digital image processing. The transferof an analog to a digital picture requires a sampling step. The original picture issampled for example by a digital camera and the sampling signal is in this casedefined by the size of the pixel of your camera in combination with the optics ofyour camera. So, if you complain about the resolution of your digital camera youalready made your experience with the sampling theorem.

5.1.4 Sampling Methodes

1. Impulse Train sampling (Ideal sampling)

We will first discuss the ideal sampling, where a periodic series of unit impulses isused as the sampling signal. The sampling scheme is therefore called Impulse Trainsampling. The schematic sampling procedure is shown in Fig.(5.2).

Figure 5.2: Schematic illustration of Impulse Train sampling [1]. Top: Continuoustime signal, Middle: Impulse train, Bottom: Sampled signal

The continuous signal x(t) is sampled by the signals p(t), which represents the

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impulse train signal. The periodic impulse train p(t) is described by

p(t) =

n=

(t nT) (5.3)

The sampling signal is multiplied with the input signal x(t). After multiplication,we get the signal xp(t). The index pindicates that an impulse samples the signal.

xp(t) =x(t)

n=

(t nT) (5.4)

We can rewrite Equ.(5.4), so that we get the following equation

xp(t) =

n=

x(nT) (t nT) (5.5)

The sampled signal xp(t) is illustrated in Fig.(5.2). Further information regardingImpulse Train sampling can be found in chapter 7 of reference [1].Now, let us discuss how the impulse train sampled signal can be reconstructed asshown in Fig.(5.3). We still assume that the initial band-pass limited signal wassampled by an impulse train and that the sampling frequency was higher than 2times the maximum frequency of the band-pass limited input signal. The outputsignal xp(t) in the frequency domain can be described by

XP

(f) =X(f)

n=

fn

T (5.6)

The initial signal is convolved with an impulse train in the frequency domain. It isassumed that the input signal in the frequency domain corresponds to a triangle.It can be seen that the signal is reproduced at integer multiples of the samplingfrequency. The output signalXp(j) of the sampling circuit is shown in Fig. (5.3c).The input signal can be recovered if the signal XP(j) is filtered by a low-pass filterwith a gain of T and a cut-off frequency greater than Mand less than s M tocut-off the redundant part of the signal as indicated in Fig.(5.3d). The final outputsignal, which is a perfect reconstruction of the input signal is shown in Fig.(5.3e).

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=

=n

nTttp )()(

)(txp)(tx

r

)(tx )( jH

( )a

)( jXr

)( jXp

MM

S SMM

( )e

Signal Systems Lab Manual Fall 2013

Documents

lab reporta lab report

lab guidelines1

lab instructor

lab report4ii experiments

ee lab signals

experiment handout

specific experiment

prelab fourier series