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Just Before Start
Analog CommunicationSem V EXTC
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Evaluation System
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Chapters
1. Basics of Communication System
2. Amplitude Modulation & Demodulation
3. Angle Modulation & Demodulation
4. Radio Receivers
5. Sampling Techniques
6. Pulse Modulation and Demodulation
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Analog CommunicationBy
Vaqar Ansari
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Electronics and
Telecommunication
Engineering
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What is a circuit?
Combination of electronic parts, wires connectedbetween power sources.
It's like a physical program.
It's also like setting up dominoes in sequence.
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What is a breadboard?
What are they good for?
Creatings, organizing, and prototyping a circuit.
Literally started out as a bread board with nails.
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What are LEDs?
Light Emitting DiodesDiode Symbol + Arrows for lightPoints to ground
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Diodes A semiconductor diode isformed with pieces of N and
P-type material are joined. The P material is called the
anode.
The N material is called thecathode.
The resulting structure is called
a PN junction.
A PN junction (or diode) is aswitch or component throughwhich electrons will flow
easily in one direction but notin the opposite direction.
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0.3(Ge) 0.7(Si)
(Ge)(Si)
VD(V)
ID(mA)
Comparison of Si and Ge semiconductor diodes
Is(Si)=10nA
Is=reverse saturation current
Is(Ge)
1
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Transistor Operation The basic operation will be described using the pnp transistor.
The operation of the pnp transistor is exactly the same if the
roles played by the electron and hole are interchanged.
One p-n junction of a transistor is reverse-biased, whereas theother is forward-biased.
Forward-biased junctionof a pnp transistor
Reverse-biased junctionof a pnp transistor
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Junction FETs
(JFETs)
JFETs consists of a piece ofhigh-resistivity semiconductormaterial (usually Si) whichconstitutes a channelfor the
majority carrier flow.
Conducting semiconductorchannel between two ohmiccontacts source & drain
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Classification scheme for
Field Effect Transistors.
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General R-L-C AC circuit
AC
R
AC
LC
Volt and current relations
tVV o cos
tVtVtVtV cRLo cos
c
tq
Vc
dt
tdq
RVR
dt
tdI
LVL
2
2
dt
tqd
L
tV
c
tq
dt
tdqR
dt
tqdL o cos2
2
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t
Z
V
dt
tdqtI o cos,
tZVtq o sin
22 Lc XXRZ 2
2 1
L
C
R
R
XX LctanR
LC
1
tIocos
XC- X
L
R
Solution of the differential equation
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-20
-10
0
10
20
30
40
50
60
0 5000 10000 15000 20000
(rad/sec)
OhmorOhm^
Xc=1/wC
XL=wL
Xc-XL
(Xc-XL)^2
R^2+(Xc-XL)^2
Z=sqrt(R^2+(Xc-XL)^2)
C =10 FL = 1mH
R = 5 ohm
LC
1Capacitive circuit Inductive circuitLC
1
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-90
-75
-60
-45
-30
-15
0
15
30
45
60
75
90
0 5000 10000 15000 20000
(rad/sec)
degr
ees
C =10 FL = 1mH
R = 5 ohm
Capacitive circuit
Inductive circuit
R
XXLc
tanR
LC
1
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0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 5000 10000 15000 20000
(rad/sec)
I
o
Amp C =10 F
L = 1mHR = 5 ohm
Vo = 5 Volt
Capacitive circuit
Inductive circuit
Z
VI oo
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-1
0
1
2
3
4
5
0 0.0005 0.001 0.0015 0.002
Time (sec)
P(t)(W
att)
w=5000 rad/sec w=10000 rad/sec w=15000 rad/sec
C =10 FL = 1mH
R = 5 ohm
Vo = 5 Volt
Capacitive case, 0 Inductive case, 0Resistive case, 0
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0
0.25
0.5
0.75
1
1.25
1.5
1.75
2
2.25
2.5
0 5000 10000 15000 20000
(rad/sec)
PavW
C =10 F
L = 1mHR = 5 ohm
Vo = 5 Volt
Capacitive circuit
Inductive circuit
RI
P oav2
2
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Root MeanSquare
2
orms
II
rmsR RIVrms
2rmsav RIP
rmsCX
IXV Crms
rmsLX IXV L
rms
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Resonance
At resonance, the capacitive reactance is equalto the inductive reactance.
LC XX LC
1
LCres
1
22 Lc XXRZ RZ res
R
XX Lctan 0res
The impedanceis minimum, and it is totally resistive.
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Resonance
-20
-10
0
10
20
30
40
50
60
0 5000 10000 15000 20000
(rad/sec)
Ohmor
Ohm^
Xc=1/wC
XL=wL
Xc-XL
(Xc-XL)^2
R^2+(Xc-XL)^2
Z=sqrt(R^2+(Xc-XL)^2)
C =10 FL = 1mH
R = 5 ohm
Z=R=5 ohm
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Resonance
-90
-75
-60
-45
-30
-15
0
15
30
45
60
75
90
0 5000 10000 15000 20000
(rad/sec)
degr
ees
C =10 FL = 1mH
R = 5 ohm
Capacitive circuit
Inductive circuit
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Resonance
At resonance, the capacitive reactance is equalto the inductive reactance, thus
The impedance is minimum.
The current is maximum.
tR
VtI res
ores cos,
t
Z
VtI o cos,
tVV resores cos
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Resonance
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 5000 10000 15000 20000
(rad/sec)
I
o
Amp C =10 F
L = 1mH
R = 5 ohm
Vo = 5 Volt
Capacitive circuit
Inductive circuit
Io=Vo/R =5/5 = 1 Amp.
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Resonance
0
0.25
0.5
0.75
1
1.25
1.5
1.75
2
2.25
2.5
0 5000 10000 15000 20000
(rad/sec)
PavW
C =10 FL = 1mH
R = 5 ohm
Vo = 5 Volt
Capacitive circuit
Inductive circuit
Pav=V2
o/2R =25/(2*5) = 2.5 W
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Ideal Op-Amp Characteristics
Open Loop Voltage Gain (Avol) is
infinite
Input offset Voltage is zero
Input bias current is zero
No power supply limits Input impedance is infinite
Simple model is a voltage
controlled voltage source with
high gain.
Avol= -
+-
+Vosi = 0V
Vout
IB-= 0
IB+= 0
Avol= ZIN=
-
+-
+
Vout
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Close Loop GainVpos Vneg Avol Vout= (1) Input x gain = output
(2) Voltage divider equation
(remember Ib =0)Vneg Vout
R1
R1 Rf
=
Vpos Vout
R1
R1 Rf
Avol Vout= (3) Substitute 2 into 1
Vout
AvolVpos
AvolR1
R1 Rf 1
=
(4) Rearange 3
Vout
Vpos
1
R1
R1 Rf
1
Avol
=(5) Multiply num and
denom with 1/Avol
(6) Take the limit
for high gainAvol
1
R1
R1 Rf
1
Avol
lim
1
R1
R1 Rf
0
=
Rf
R1
1=
Vpos
Vneg
Avol-
+-
+
Vout
Rf 1kR1 1k
IB = 0
Vpos-
+
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Rin 1k
RF 2k
Vin
2V
Vout
0VVirtual
Short to
GND
+ 2V -
+
-
Iin= 2V/1kIin= 2mA
Simple Analysis for Inverting Amp
Rin 1k
RF 2k
Vin
2V
Vout
+
-
Iin= 2mA
2mA
VRF= 2mA x 2k = 4V
+ 4V -
Vout = -4V
With respect to GND
Gain = Vout / Vin
Gain = (-4V) / (2V) = -2
IB= 0A
Vin across Rin
Iin flows through Rf
Vout is the voltage across Rf
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R1 1k
Rf 99k
+ Vout-
+
Vin
+
Vout-
+
Vin
R1 1k
Rf 99k
+
Vout-
+
VinBuffer
Vout = Vin
Common Amplifiers
Noninverting Amp
Vout
Vin
Rf
R1
1=
Inverting Amp
Vout
Vin
Rf
R1
=
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Superposition principle
Used for circuits with multiple input sources.
Analyze the output response for one sourceat a time.
Short unused voltage sources
Open unused current sources
Repeat the analysis for each source
Add all the response from each analysis toget the overall system respones
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VrefVref
+5VR1 1k
R2 1k
+
VM1+
VG1
+
-
U1 OPA333
V
Vref
Vref
+5VR1 1k
R2 1k
+
VM1
+
-
U1 OPA333
V
Vref
VrefVref
+5VR1 1k
R2 1k
+
VM1+
VG1
0.1V
+
-
U1 OPA333
V
-1 x Vref = -Vref
W.R.T.G
2(Vg1 + Vref)
W.R.T.G
(2Vg1 + 2Vref) + (-Vref)=2Vg1 +Vref W.R.T. GND
(2Vg1 +Vref)Vref =2Vg1 W.R.T. Ref
2Vg1
w.r.t Ref
2Vg1 +Vref
w.r.t.g
Superposition Example: Single Supply Amp
(Noninverting)
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R3 40k
R4 40k
R5 40k
R6 40k
Vout
Va1
Va2
-
+
A3
Output Stage
Dif Amp
R3 40k R5 40k
Vout
Va1
-
+
A3
R3 40k
R4 40k
R5 40k
R6 40k
Vout
Va2
-
+
A3
Va1
Va2
Va2+ Vref
Find Vout Through Superposition
Vout = Va2Va1+ Vref
Inverting Amp
Gain = -1
Non-inverting Amp
Gain = 2
Voltage Divider
Gain = 1/2
Va1
Vin_dif
Va2
-Va1
Vref
Va22
+Vref2
Vref
Superposition Example : Diff Amp
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Cf Filter (High Gain)
At high frequency CF will short
RF
High Freq Gain = 0/1k+1 = 1
At low frequency CF acts like
an open
DC Gain = (99k/1k)+1 = 100
-15V
+15V
R1 1k
Rf 99k
+
Vout
Cf 10n
-
+
U1 OPA827
Vin
fp1
2 99 k10 nF160.8Hz
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Gain(dB)
-10.00
0.00
10.00
20.00
30.00
40.00
Frequency (Hz)
1.00 10.00 100.00 1.00k 10.00k 100.00k 1.00M
Phase[deg]
-80.00
-60.00
-40.00
-20.00
0.00
fp1
2 99 k10 nF160.8Hz
High Frequency
Gain = 1 (0dB)
Low Frequency
Gain = 100 (40dB)
-45o/ decade
+45o/ decade
Zero at f= 16.08kHz
40dB / 20dB/dec = 2 decades.
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Cf Filter (Low Gain)
At high frequency CF will
short RF
High Freq Gain = 0/1k+1 = 1
At low frequency CF acts like
an open
DC Gain = (2k/1k)+1 = 3
-15V
+15VR1 1k
Rf 2k
+
Vin
Vout
Cf 10n
-
+
fp1
2 2 k10 nF8 10
3 Hz
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Gain(dB)
-2.00
0.00
2.00
4.00
6.00
8.00
10.00
Frequency (Hz)
1.00 10.00 100.00 1.00k 10.00k 100.00k 1.00M
Phase
[deg]
-40.00
-30.00
-20.00
-10.00
0.00
High Frequency
Gain = 1 (0dB)
Low Frequency
Gain = 3
(9.54dB)
-45o/decade +45o/
decade
Zero
fp1
2 2 k10 nF8 10
3 Hz
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Analog
Discrete
Digital
Signals
Periodicity
Periodic
Non periodic
Nature ofsample
Random
Deterministic
Odd/Even
EnergySignal
Power Signal
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System
Linearity
Causality
StabilityTime
Invariant/Variant
Static/Dynamic
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FT
Spectrum
ParsivalsRelationship
DFT & FFT
LT
Initial Value Theorem
Final Value Theorem
ZT Pole & Zeros on ROC Residue Theorem
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Just Start
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