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Signal Processing - E051
Professor Dr Ir Mostafa Afifi
mailto:[email protected]:[email protected]8/7/2019 Signal Processing E051
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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
8R is replaced by 47K
16R is replaced by 100K
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
8/7/2019 Signal Processing E051
60/61
Lab Verification for the Inversion-Non-Inversion Op Amp Operations
8/7/2019 Signal Processing E051
61/61
Bi-stable Op Amp Multi-vibrator
f= 1 / R1 C