signal processing e051

8/7/2019 Signal Processing E051

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Signal Processing - E051

Professor Dr Ir Mostafa Afifi

[email protected]
mailto:[email protected]:[email protected]


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The Weekly Schedule

3 HrsLectures every week (one week , twoconsecutive periods, # 1 (at S-MEC) & # 2 (at L1)ofThursdays, with otherThursday , period # 2

2 Hours of exercises every week.4M2 on Wednesday , period 1 S154M1 on Wednesday , period 4 S144M3 on Sunday, period 3 S1

1 Hour of Laboratory work every week 4M1,2 on Wednesday , period 3

4M3 on Even Thursdays, period 2


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Grading

Final exam is 60% Exercise works 10%

Laboratory Works 20% Attendance 5% Midterm Exam 5%


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INTRODUCTION

The handling of operating electric and processing circuits bymechanical engineers is proper for the advance of civilprosperity. In all aspects of our life; as electromechanicalintegration is now very evident.

Mechanical engineers were the first to use electric motors tooperate and manipulate engine operations.

Also they were the first to apply digital handling of informationusing rotating shafts and timing activities.

This course ofSignal Processing introduces the practicalprocessing requirements for Mechanical Engineers, with toolsthat facilitate practical Supervisory Applications.


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Examples of many applications forProcessing and Engineering in Automobiles

For Safety: Antiskid brakes, inflatable restraints, Collision warning andavoidance, Blind-zone detections, infrared night vision systems, Heads-updisplays and accident notifications.

For Communications and entertainment: AM/FM Radios, Digital audioBroadcastings, CD / Tape players, Cellular phones, Computer /e-mail.

For convenience: Electronic navigation, Personalized seat/mirror/radiosettings, electronic door locks.

For Emission, performance & Fuel economy: Vehicle instrumentation,electronic ignition, Tire inflation sensors, computerized performance

evaluation and maintenance scheduling and adaptable suspension systems. For alternative propulsion systems: Electric vehicles, advanced batteries

and Hybrid vehicles.


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Output of a transducer for aKnocking Engine


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Other important applications are inHousehold Appliances (Mechatronics)

with intimate integration and harmoniousblending of many different technologies

Keypads for operator control

Sensors Electronic displays Microcomputer Chips

Electronic digital switches Heating Elements Motors


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Major application subdivisions Signal Processing:where transducers are used to convert physical

phenomenon to electrical signals that can be processed for desiredperformance by computers. Major example of this is the control of ignition incars to optimize performance and avoid damaging knock of engines.

Control Systems:gathering information with sensors and processing forphysical control, examples are the heating and air cooling in buildings,temperature and flow rates in chemical processes and control of motion intall buildings

Electronics:applied in every field of engineering and science, many usefulcircuits are planned for this course to introduce and build major engineeringcapabilities.

Computer Systems:Process and store of information in digital formats,Examples of wide scale applications is the control systems of modern cars,including more than 50 microcomputers in one car

Communication systems: with major modern Information technology usingInternet broadcastings in ground and Satellite systems (including roadsidesensors and GPS global positioning Systems)

Electromagnetism:using microwaves in ovens, manufacturing of plywood,cell phones and satellite applications.

Power systems:for interchanging conversion of mechanical and electricalenergies with applications in transportation and increase of efficiency.


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Processing Procedures of signals

the block diagram for the processing procedures ofsignals

DigitalSignal

Processing

Digital toAnalog

Converter

Analog toDigital

Converter(A/D)

Analog Input Digital Input Digital Output Analog Output


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Fourier Signal HandlingThe Fourier series expansion has three formats:The trigonometric series 1

f(x) = (ao / 2 ) + n = 1 n= [an cos (n x) + bn sin (n x) ]

With an = (1 / ) 02 f(x) cos(n x) dx and bn = (1 / ) 02 f(x) sin(n x)

Note that ao = (1 / )02 f(x) dx

With cn = ( an2 + bn2 )1/2 , co = ao and n = tan-1 ( - bn / an )

these can be transformed to the complex formatf(x) = (Co/2) + 1 Cn cos(n x + n ) = - dn e j n x

with dn = (an j bn) / 2 = (cn /2 ) ej n , do = Co /2 = ao /2, and d n = dn*

The average power= (1 / 2)02 f(x)2 dx =- dn2 = (Co /2)2 + n = 1 n= (cn2)/2


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Fourier Transform of Square Pulse,T= 0.02 Sec, = 0.01 Sec

in sec pi T Pulse Amplitude

Hz 0.01 3.1415927 0.02 1

n f n / T ( / T)sinc( / 2) |sinc( / 2)|

-8 -400 -12.566 0.000 0.000

-7 -350 -10.996 -0.045 -0.091

-6 -300 -9.425 0.000 0.000

-5 -250 -7.854 0.064 0.127

-4 -200 -6.283 0.000 0.000

-3 -150 -4.712 -0.106 -0.212

-2 -100 -3.142 0.000 0.000

-1 -50 -1.571 0.318 0.637


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Fourier Transform of Square Pulse,T= 0.02 Sec, = 0.01 Sec

Sinc Function & Fourier series (T/t=2)

-0.400

-0.200

0.000

0.200

0.400

0.600

0.800

1.000

1.200

-400

-300

-200

-100 0

100

200

300

400

Frequency in Hz

RelativeAm

plit

Fourier Series of repeated pulse Fourier Transform "Sinc"


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Rectangular Wavef(t) = 0.5 + (2/pi) sin (100pi t)+ (2/3pi) sin (300pi t) + (2/5pi)sin (500pi t) +

pi 3.141593

t f1(t) f3(t) f5(t) f7(t)

0 0.500 0.500 0.500 0.500

0.001 0.697 0.868 0.996 1.069

0.002 0.874 1.076 1.076 0.990

0.003 1.015 1.081 0.953 0.981

0.004 1.105 0.981 0.981 1.0340.005 1.137 0.924 1.052 0.961

0.006 1.105 0.981 0.981 1.034

0.007 1.015 1.081 0.953 0.981

0.008 0.874 1.076 1.076 0.990

0.009 0.697 0.868 0.996 1.069

0.01 0.500 0.500 0.500 0.500

0.011 0.303 0.132 0.004 -0.069

0.012 0.126 -0.076 -0.076 0.010


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The Fourier Transform

dn = (1/2 ) 02 f(x) cos(n x) dx - (j / 2 ) 02 f(x) sin(n x) dx

dn = (1/ 2 ) 02 f(x) e-jnx dx

for x = o t dn = (1/To) 02 f(ot) e-jnot dt = (1/To) F()

f(t) = (1/To) - F() e j n ot = (o /2 ) - F() e j n ot ,

with o = and = n o

f(t) = (1/2) - F() e jt d and F() = - f(t) e - jt dt

for a single square pulse of amplitude A we have F() = - /2 /2 A e - jt dt

or F() = A sinc (/2) = A [sin (/2)] / (/2)


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The Sinc Function


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Basics of the Numbering Systems

N = k=-r mAkbk ,

Where b is an integer ( > 1 ), called the base (or Radix),

Ak is an integer 0, 1, 2, ., b-1,

m and r are determined by the extent of the numberN,

r is not zero with existence of a decimal point(the decimal point exists between A0 and A-1).

The most common basis are:

Decimal (base 10)Binary (base 2)Octal (base 8)Hexadecimal (base 16)Duodecimal (base 12)


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The Decimal, the binary and theHex presentations of Numbers

123.4510 = 1 x 102 + 2 x 101 + 3 x 100 + 4 x 10-1 + 5 x 10-2 123.452 = 0x27 + 1x26 + 1x25 + 1x24 + 1x23 + 0x22 +

1x21 + 1x20 + 0x2-1 + 1x2-2 + 1x2-3 + 1x2-4 +

0x2-5

+ 0x2-6

+ 1x2-7

+ 1x2-8

+ a remainder of 0.0007813

= 01111011.011100112 + 0.0007813

(a 16 bit representation)

123.4516 = 7x161 + Bx160 + 7x16-1 + 3x16-2 + 3x16-3 + 0.0000488= 7B.73316 + 0.0000488.


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BCD Coded Numbers & Codes

Decimal Octal OctalBCD

Hexadecimal

Hex BCD Excess 3 Self Ninecomplement ofexcess 3

Gray Code

0 0 000 0 0000 0011 1100 0000

1 1 001 1 0001 0100 1011 0001

2 2 010 2 0010 0101 1010 0011

3 3 011 3 0011 0110 1001 0010

4 4 100 4 0100 0111 1000 0110

5 5 101 5 0101 1000 0111 0111

6 6 110 6 0110 1001 0110 0101

7 7 111 7 0111 1010 0101 0100

8 8 1000 1011 0100 1100

9 9 1001 1100 0011 1101

10 A 1010 1111

11 B 1011 1110

12 C 1100 1010


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Negative binary, Max & Min andRules of addition and subtractionN- = 2n N+

maximum positive number is (2n-1 1)and a minimum negative number is (-2n-1 )

For n = 8 bits 128 N 127

Addition 0 + 0 = 0 Subtraction 0 0 0

0 + 1 = 1 0 1 = 1 with borrow of 11 + 0 = 1 1 1 = 11 + 1 = 0 carry 1 1 1 = 0

1 + 1 + 1 = 1 carry 1


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USA Standard 7 Bit Code forInformation Interchange (ASCII)

# 0 1 2 3 4 5 6 7

3 MSBs 000 001 010 011 100 101 110 111 +4 LSB

0 NUL DLE SP 0 @ P ` p 0000

1 SOH DC1 ! 1 A Q a q 0001

2 STX DC2 " 2 B R b r 0010

3 ETX DC3 # 3 C S c s 0011

4 EOT DC4 $ 4 D T d t 0100

5 ENQ NAK % 5 E U e u 0101

6 ACK SYN & 6 F V f v 0110

7 BEL ETB ' 7 G W g w 0111

8 BS CAN ( 8 H X h x 1000

9 HT EM ) 9 I Y i y 1001

10 LF SUB * : J Z j z 1010

11 VT ESC + ; K [ k { 1011

12 FF FS , < L \ l | 1100

13 CR GS - = M ] m } 1101

14 SO RS . > N ^ n ~ 1110

15 SI US / ? O _ o DEL 1111


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Analog to Digital conversion Rules

The number of used bits determines the resolution of theA/D converter.The number of digitized levels N = 2n

The resolution as the minimum signal variation that can berecognized by the digital signal is determine by the analogrange divided by the number of digital recognizable levels.

i.e. Resolution = Vrange

/ 2n,

where n is the number of bits.If the voltage range is between 1 and 10 volts,

then the Resolution = (10-1)/ 212 = 2.2 mV


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Graphical Coding Verify Ease of MMI andSupervisory Control and Data Acquisition

Automation in present day technology made it difficult to havecompetitive production using the traditional Man Machine Interface(MMI) and Supervisory Control and Data Acquisition (SCADA).

These named systems include much complex functionality withinvolved textual programming. In most of the cases theprogrammers have limited knowledge about the physical process.

To overcome this complexity the graphical programming (G) was

lately invented to permit the development of complex applicationswithout the requirement of advanced computer programmingknowledge.

Graphical programming uses virtual instruments (VIs) resulting intointuitive MMI resembling a typical measurement and control blockdiagram.


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Use of LabVIEW to solve First OrderDifferential Equation; for step transition in a

Simple High Pass Digital Filter

Generate a G-code (for LabVIEW simulation) that implements theemulation of the output voltage Vout (n) for an input step voltage Vinof 10volts (starting at t= 0).

Vout

= Ri and Note that the current i = C d (Vin

- Vout

) / dtWhen time is divided into equal small intervals, with equal separations of

" t, starting at t=0, the voltages at any interval n, are related as follows:Vout (n) = (CR/ t) {[Vin (n) - Vin (n-1)] [Vout (n) Vout (n-1)]}

Separating the output voltage (at instant n) and denoting the time constant (CR = ) yieldsVout (n) = K [ Vin(n) Vin(n-1) ] + K Vout (n-1),

Where K = ( / t) / [1+( / t)] , with simulation shown asfollows:


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The Control and Indication Panel


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The ImplementationGraphical Code Connections


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Digital filters

The Low Pass R-C filter has the differential equation:Vi - Vo = RC dVo/dt

The Difference Equation at any two successive timeintervals separated by t is:Vo(n)=K Vo(n-1) + K1 Vi(n), Vi Vowith K = (RC/ t ) [1+ (RC/ t ) ]-1,and K1 = [1+ (RC/ t ) ]-1

This difference equation describes the equivalent digitalform describing the action in discreet time intervals

separated by the time interval (t ), or what is called thedigital filterof first order.The general equation for a digital filter is given by:Vo(n)=Ka a=1 N Vo(n-a) + Kb b=0 M Vi(n-b)

in this 1st

order filter N=1, M=0

C

R


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The Front Panelof the LPF simulation

Front Panel


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Superposition ApplicationVoltage AddersSuperposition can be applied to

generate the transfer function

to vfrom v1and v

2.

vo1=v1RR2 /(RR1+RR2+R1R2),vo2=v

2RR

1/(RR

1+RR

2+R

1R2)

Then

v = v1 RR2 / (RR1+RR2+R1R2)+ v

2RR

1/ (RR

1+RR

2+R

1R2)

=(v1R2+v

2R1)R/(RR

1+RR

2+R

1R2)

When R1=R

2=R

o

v= v +v R 2R+R


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Performance ofthe Differential Amplifier

differential input voltage vd = Vcc / A

= approximately (15 / 100000 = 0.15 mV)


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Non-Inverting OP-Amp

V(-) = Vin

Vin = Vout R1 /(R1+Ff)

(Vout / Vin) = (R1+Ff) / R1

The Unity Gain Op-Amp Circuit


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Negative Feed Backin Car Power Steering


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Inverting Op-Amp

Vout / V1 = - Rf/ R1

i1 = V1 / R1 = - if


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Difference Op-Amp Circuit

V+ = V2 [ Rf/ (R1+ Rf) ]

i1= [V1-V+] / R1if = - [ Vout - V+ ] / Rf

if

= - i1

Vout = (Rf / R1)(V2 - V1)


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Discharge of the Capacitor

C dvc /dt + vc /R = 0

vc= K es t

RCK s e s t + K e s t = 0RC s + 1 = 0 s = - 1/RC

K = Vi

vc = Vi e t / RC


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Charging of a Capacitor

RC d vc / d t + vc = Vs

vc= K1 + K2es t

)1+RC s K2es t + K1 = Vs

)1+RC s= 0 s = - 1/ RC K1 = Vs

Vc= Vs Vs e t / RC = Vs (1 e t / RC

)


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Integrator and differentiatorUsing the Capacitors

Vo

0

t

tVin

R C

d:=iin = vin / R

iin =C d vin / dt

vo = - RC d vin / dt


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Reactance of Capacitor andInductor with Alternating Voltage

For the Capacitor v(t) = Vm sin (t) and for the Inductor

i(t) = - Vm cos (t) / L

= -Vm sin (t+/2) / L

= - j Vm sin (t) / L

i(t) = Vm sin (t) / j L

i(t)= C dv / dt = C Vm cos (t)

= Vm cos (t) / (1 / C)

= - j Vm sin (t) /(1 / C).

i(t)= Vm sin (t) / (1 /jC)

Capacitor reactance = 1 /jC = XC

Inductor reactance = jL = XLi(t) = Vm sin (t) / Xi(t) = Vm sin (t) / X

v(t) = Vm sin (t)


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RLC Circuitsand its Mechanical Analogy

L d2 Q / d t2 + R dQ / d t + (1/C) Q = Vsm d2 x / d t2 + K d x / d t + Cm x = F

X = D + A e s1 t + B e s2 t Qc= D + A es1 t + B e s2 t

the mass (m) equivalent to the inductance (L),

spring action (Cm) equivalent to the inverted capacity (1/C)

viscous damping (K) equivalent to the Resistance (R)

Displacement xis equivalent to the

Charge Q


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The Digital form of the 2nd order DE

Vc(n) = K1 Vc(n-1) +K2 Vc(n-2) +Ko Vi (n),

with K1 = (2LC/t2 + RC/ t) (LC/t2 + RC/ t)-1,

K2 = - (LC/t2) (LC/t2 + RC/ t)-1 ,

andKo = 1 / (LC/t2 + RC/ t)

This can be G programmed with LabVIEW dual for loops.


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The Base Operating lines and the


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The Base Operating lines and theCollector Inversion Swing,

Av

= - 4/0.8 =-5


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AC Circuit of the BJT

The AC resistance at the input of the base ( for vb < vT) isgiven by IB+ ib = Is e(VB+ vb)/vT = Is eVB/vTevb/vT

= IB ( 1 + vb/vT) and VT=k T/q

kis Boltzman constant = 1.38x10-23 J/Ko, q = 1.6x10-19C, T in Ko

meaning ib = IBvb/vT i.e. r = vT/IB & RB=R1//R2


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AC Circuit of the MOSFET

ID + id = K (VGS + vgs Vto)2

= ID+ 2K (VGS Vto) vgs

or id

=2K (ID/K)1/2v

gs=2 )K I

D)1/2v

gs=g

mv

gs

gm = 2 (K ID)1/2


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Bode Patterns and Frequency Response ofLP, HP, BS & BP Filters

V1cij i R C

1 j i R C+: =

Vci1

1 j i C R+:=

0 1 2 3 4 5 6 7 8 9 100

0. 1

0. 2

0. 3

0. 40. 5

0. 6

0. 7

0. 8

0. 91

1

10 104

Voii

V1oii

9.9011 10 3 i

V o bij i C R

1 i( )2

L C j i C R+

: =

V o si1 i( ) 2L C

1 i( ) 2 L C j i C R+: =

0 1 2 3 4 5 6 7 8 9 100

0.1

0.2

0.3

0.40.50.6

0.7

0.8

0.91

1

1 103

Vobi

Vosi

9.9011 10 3 i


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The Frequency Bandwidth of Filters

Qii L

R:= Q503 3.16=

0 100 200 300 400 500 600 700 800 900100

80

60

40

20

0

20

40

60

80

100

1i

2i

0 100 200 300100

80

60

40

20

0

20

40

60

80

100

1i

2if2 f1 159.155=

o

Q503 2 1 5 9 . 2 4 7=

v03i 1

1 i( )2

L C j i C R+:=

v04ii( ) 2 C L

1 i( )2

L C j i C R+:=

0 1 0 0 2 0 0 3 0 0 4 0 0 5 0 0 6 0 0 7 0 0 8 0 0 9 0 0 1 0 0 00

0 . 4

0 . 8

1 . 2

1 . 6

2

2 . 4

2 . 8

3 . 2

3 . 6

43 . 2 0 3

3 . 948106

v 0 3i

v 0 5i

v 0 4i

31 i

R

2 L:=

o1

L C: =


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Combinational Logic Networks

A B A B (AB) (A+B) A B AA BB0 0 0 0 0 00 1 0 AND OR NOT (Invert) 0 1 11 0 0 1 0 11 1 1 1 1 1

001001

0 1 1 NAND NOR 0 1 01 0 1 1 0 01 1 0 1 1 0

Monitoring of closing car doors using a CLN.

An XOR Gate

Th H lf Add L i d It


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The Half Adder Logic and ItsCombinational Logic Network

The AND, the XOR and HA logics

A B A .B A B A B A + B C S0 0 0 0 0 0 0 + 0 = 0 00 1 0 0 1 1 0 + 1 = 0 11 0 0 1 0 1 1 + 0 = 0 11 1 1 1 1 0 1 + 1 = 1 0

S = AB + AB

C = A.B


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The Full Adder

S =A B C +A B C +A B C +A B C

C = AB + AC + BC


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Karnaugh map and the simplification ofcombinational logic expressions

The combinational logic expression F = A B D + A B D + B C D +A B CCan be simplified to:

F = F1 = A D + B C D + A B CWhich can also be simplified to:

F = F2 = A D +A B C

the four variable KARNAUGH mapeasily show the validity of the shortexpression

K h M Si lifi th


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Karnaugh Map Simplifies theCombinational Logic Expressions

Z = W X Y + W X Y + W X Y + W X YD = A B C + A B C + A B C + A BE = A B C D + A B C D + A B C D + A B C D

A G l P Fli Fl


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A General Purpose Flip-FlopRegister Memory Unit


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Laboratory Experiments

Diode Characteristics Light Emitting Diode (LED) Characteristics Bipolar Junction Transistor (BJT) Characteristics

BJT Collector-Base Current Relation First Order Digital Filter LabVIEW Experiment Inversion-Non Inversion Op Amp switching

4-bit Digital to Analog Op Amp conversion Bi-stable Op Amp Multi-vibrator


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Diode Experiment - 1

Vd Id R K-Ohm r K-Ohm

Volt mA Vd / Id dV/dI

0.5 0.1 5.000

0.53 0.2 2.650 0.3

0.547 0.3 1.823 0.17

0.56 0.4 1.400 0.13

0.57 0.5 1.140 0.1

0.585 0.6 0.975 0.15

0.587 0.7 0.839 0.02

0.59 0.8 0.738 0.03

0.596 0.9 0.662 0.06

0.6 1 0.600 0.04

0.62 1.5 0.413 0.04

0.63 2 0.315 0.02

0.65 3 0.217 0.02

0.7 16 0.044 0.004

0

2

46

8

10

12

14

16

18

0.5

0.54

70.57

0.58

7

0.59

60.62

0.65

Voltage (Volts

Currentin

mA/orRinOhm

Diode V oltage currents

rasistance

Dynamic rasistance K-O


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Light Emitting Diode Experiment

Vd Id R K-Ohm r K-Ohm

Volt mA Vd / Id dV/dI

1.76 0.22 8.000 8

1.8 1 1.800 0.0513

1.87 2 0.935 0.07

1.9 3 0.633 0.03

0

0.5

1

1.5

2

2.5

3

3.5

1.76 1.8 1.87 1 .9

Diode V ol ta

DildeCurrentin"mA"/RinOhm

Light Em i t t ing Diode Resis tance - K O

Dynami c Res i s tance


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Digital to Analog Experiment

b 1

1 6 R

V o

b 0

V i

b 3 2 R

8 R

O P A M P

+

-

O U T

Rb 2 4 R

R is 7.2K PredominaingResistance for

tolerance

2R is replaced by 10K

4R ie replaced by 22K



Computations are based on the replaceable values

Measurements indicate the tolerances of the resistor values( 10%)

ConclusionResistor values and its tolerances are important for building this kind of D/A converter


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Digital to Analog ExperimentVo/Vi = - (R/Ro)bo - (R/R1)b1 - (R/R2)b2 - (R/R3)b3

R in K Ohm 10 22 47 100 7.2 Vo Vo Ass Bin

Bin Value b3 b2 b1 bo Vo / Vi Measured Output

0 0 0 0 0 0.000 0.000 0.000 0.000

1 0 0 1 -0.072 -0.360 -0.416 -0.313

2 0 0 1 0 -0.153 -0.766 -0.875 -0.625

3 0 0 1 1 -0.225 -1.126 -1.291 -0.938

4 0 1 0 0 -0.327 -1.636 -1.857 -1.250

5 0 1 0 1 -0.399 -1.996 -2.273 -1.563

6 0 1 1 0 -0.480 -2.402 -2.732 -1.875

7 0 1 1 1 -0.552 -2.762 -3.148 -2.188

8 1 0 0 0 -0.720 -3.600 -4 -2.500

9 1 0 0 1 -0.792 -3.960 -4.416 -2.813

10 1 0 1 0 -0.873 -4.366 -4.875 -3.125

11 1 0 1 1 -0.945 -4.726 -5.291 -3.438

12 1 1 0 0 -1.047 -5.236 -5.857 -3.750

13 1 1 0 1 -1.119 -5.596 -6.273 -4.063

14 1 1 1 0 -1.200 -6.002 -6.732 -4.375

15 1 1 1 1 -1.272 -6.362 -7.148 -4.688

Lab Verification for the Inversion


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Lab Verification for the Inversion-Non-Inversion Op Amp Operations


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Bi-stable Op Amp Multi-vibrator

f= 1 / R1 C

signal processing e051

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