chapter 1-part 2.pdf

Principles of CommunicartionSystem by N Hasan
Principles of Communications
System
Chapter 1 (Part 2): Gain, Loss And
Noise
by
Nurulfadzilah Hasan
Faculty of Electrical & Electronics Engineering
nurulfadzilah@ump.edu.my

Gain, Attenuation, and Decibels
• Amplifier is an electronic device that increase the power of a
signal fed into it.
• Gain represent this amplification process. It gives the ratio of
output of a circuit to its input.
AV = gain = =
output
input
Vout
Vin
Amplifier
Vin
Output signal
Input signal
Vout
• Power gain (Ap) = Pout / Pin
where Pin is the power input and Pout is the power output.

Example 1
The power output of an amplifier is 10 watts (W). The power
gain is 80. What is the input power?
SOLUTION:
Ap = Pout / Pin
therefore Pin = Pout / Ap
Pin = 10 / 80 = 0.125 W = 125 mW

• When two or more amplifiers are connected
together (cascaded), total gain is the product of
each individual gain at every stage
A1=2
Vin=2mv
Vout
A2=3 A3=4
Total gain, 1 2 3 2 3 4 24 /
Thus 2 24 48

Example 2
A circuit has three cascaded amplifiers, with power gains of 3, 2,
and 7 respectively. If the input power is given as 50 mW, find the
total gain and output power
Ap = A1 × A2 × A3 = 3 × 2 × 7 = 42
Ap = Pout / Pin
therefore Pout = ApPin Pout = 42 (50 × 10-3) = 2.1 W

Attenuation (Loss)
• The opposite of gain is Attenuation
• It refers to refers to reduction (loss ) of output power
at the of a circuit or component.
• As with gain previously, attenuation can be found by:
A = output/input = Vout/Vin
• If A is less than 1, then the circuit causes attenuation

As with gain, for circuit with several loss stages, total
attenuation is the product of individual attenuations of
each cascaded circuit.
A1=0.2
Vin=2v
Vout
Loss
stage 1
Loss
stage 2
Loss
stage 3
A2=0.3 A3=0.5
Total loss, 1 2 3 0.2 0.3 0.4 0.024 /
Thus 2 0.024 0.048 48

Total gain & attenuation
• A circuit can also consists of combination of
amplifiers and attenuators.
• Thus, total gain is the product of all stages of gains
and attenuations
A1=0.2
Vin=2v
Vout
Loss
stage 1
Loss
stage 3
A2=3 A3=0.5
Total gain, 1 2 3 0.2 3 0.4 0.24 /
Thus 2 0.24 0.48

Decibel (dB) – ration of two powers
• If 2 powers are expressed in the same units (e.g. watt, miliwatt), their
ratio is a dimensionless quantity that can be expressed in decibel form as
follow
Where P1 : power level 1 (watts)
P2 : power level 2 (watts)
• the dB value is for the power of P1 with respect to the reference power P2
• the dB value shows the difference in dB between power P1 and P2
÷
ø
ö
ç
è
æ
=
2
1
10
log
10
P
P
dB (1)

Gain & Loss (dB)
• We can measure the ratio between the power at the output
and input of any electronic circuit or device by using eq (1) as
below:
where Ap(dB) : power gain (unit in dB) of Pout with respect to
Pin
Pout : output power level (watts)
Pin : input power level (watts)
Pout/Pin : absolute power gain (unitless)
÷
ø
ö
ç
è
æ
=
in
out
dB
p
P
P
A 10
)
( log
10
(2)

Gain & Loss (dB)
• Positive (+) dB value: output power is greater than the
input power à power gain or amplification
• Negative (-) dB value : output power is less that the
input power àpower loss or attenuation
• If Pout = Pin, (power gain is 1), dB power gain is 0 (unity
power gain)

Gain, Loss, and Decibels
Decibels: Decibel Calculations
– Voltage Gain or Attenuation
dB = 20 log Vout/ Vin
– Current Gain or Attenuation
dB = 20 log Iout/ Iin
– Power Gain or Attenuation
dB = 10 log Pout/ Pin

dBm and dBc
dBM is a decibel value when the base value is referenced to
1 mW:
P(dBm) = 10 ⋅ log10( P(mW) / 1mW)
dBc is a decibel gain attenuation value where the reference is
the carrier.

Example
• To calculate the ratio of 1 kW (one kilowatt, or 1000 watts)
to 1 W in decibels, use the formula
• To calculate the ratio of 1 mW (one milliwatt) to 10 W in
decibels, use the formula
• To find the power ratio corresponding to a 3 dB change in
level, use the formula

Example 3
If an amplifier has input of 3 mW and an output of 6 W.
What is the gain in decibels?
GdB = 10 log 6/0.003
= 10 log 2000
= 10 (3.301)
= 33.01

Example 4
Let’s say a power amplifier with a 30 dB gain has output power 100 W.
Calculate the input power.
Solution:
dB = 10 log (Pout/ Pin )
30 =10 log (Pout/ Pin )
3 = log (Pout/ Pin )
Antilog 3 = antilog (log (Pout/ Pin ) )
(Pout/ Pin ) =10^3= 1000
Pin =Pout/1000
= 100/1000
= 0.1 W
= 100 mW
antilog = log^(-1)

NOISE, INTERFERENCE AND
DISTORTION

Noise, interference and distortion
• Noise
– unwanted signals that coincide with the desired signals.
– Two type of noise: internal and external noise.
• Internal noise
– Caused by internal devices/components in the circuits.
• External noise
– noise that is generated outside the circuit.
– E.g. atmospheric noise, solar noise, cosmic noise, man made
noise.

Noise, interference and distortion
(Cont’d)
• Interference
– Contamination by extraneous signals from human sources.
– E.g. from other transmitters, power lines and machineries.
– Occurs most often in radio systems whose receiving antennas
usually intercept several signals at the same time
– One type of noise.
• Distortion
– Signals or waves perturbation caused by imperfect response of
the system to the desired signal itself.
– May be corrected or reduced with the help of equalizers.

Noise Temperature
• Thermal noise, Pn can be found using:
Where
Pn @ N = noise power (Watt)
k = Boltzman constant (1.38 x 10-23 J/K)
T = T(K) = T(°C) + 273
environmental temperature (K)
B = Bandwidth of system (Hz)

Example 5
In a circuit, a resistor with 25 °C temperature is connected in across the input of
an amplifier. The bandwidth of the system is 50 kHz. Calculate the amount of
noise caused by the resistor to the input of the amplifier.
SOLUTION
T(K) = T(°C) + 273
= 25 + 273
= 298 K
PN = kTB
= 1.38 × 10−23 × 298 × 50 × 103
= 2.06 × 10−16 W
= 0.206 fW

Signal-to-Noise Ratio
• Signal-to-noise ratio is one of the most important characteristics
of a communication system
• It represents the ratio of a signal strength to interference.
• Generally it is expressed in decibels.
• Signal-to-noise ratio is found by:
• In dB:
N
S
P
P
N
S log
10
)
dB
(
/ =
N
S
P
P
N
S =
/

Example 6
For an amplifier with output signal of 10 W and an output noise
power of 0.01 W, determine the signal-to-noise power ratio in dB:
dB
P
P
N
S
N
S
30
01
.
0
10
log
10
log
10
)
dB
(
/ =
=
=

Noise Figure
• Since thermal noise is produced by all conductors and active
devices, it follows that any stage in a communication system
will add noise
• Noise Factor (in ratio) and Noise Figure (in dB) shows how
much SNR degrades as a signal pass through a circuit
• Ratio of SNR at the input to SNR at the output of a circuit

Signal to Noise Ratio
— SNR is ratio of signal power, S to noise power, N.
— Noise Factor, F
— Noise Figure, NF
dB
N
S
SNR log
10
=
o
o
i
i
N
S
N
S
F =
)
(
log
10
log
10
dB
N
S
N
S
F
NF
o
o
i
i
=
=

Equivalent noise Temperature
• Equivalent noise temperature, (Te) is a
hypothetical value, used to indicate thermal
noise
Te = T(F-1)
Where T = environmental temperature = 290 (kelvin)
F = Noise factor
• Te is often used in low noise, sophisticated radio
receivers rather than noise figure.

Example 3
The signal power at a receiver is 100 μW and the noise power is
1 μW. The components in the receiver has an additional 80 μW
of noise, Nd, and has a power gain of 100 W. Find the receiver’s:
(i) input SNR
(ii) output SNR
(iii) noise figure.

Solution
100
10
1
10
100
/ 6
6
=
´
´
=
= -
-
N
S
i
P
P
N
S
dB
N
S i 20
100
log
10
)
dB
(
/ =
=
W
S
A
S i
p
o
2
6
10
1
)
10
100
)(
100
( -
-
´
=
´
=
´
=
W
N
N
A
N d
i
p
o
4
6
10
8
.
1
80
)
10
1
)(
100
( -
-
´
=
+
´
=
+
=
(i) Input SNR in dB
(ii) Output SNR in dB

Solution (cont.)
dB
F
dB
dB
N
S
N
S
F
db
o
o
i
i
db
55
.
2
45
.
17
20
55
.
2
)
(
56
.
55
100
log
10
)
(
log
10
=
-
=
=
=
=
(iii) The receiver’s noise figure in dB
56
.
55
10
8
.
1
10
1
/ 4
2
=
´
´
=
= -
-
N
S
i
P
P
N
S
dB
N
S O 45
.
17
56
.
55
log
10
)
dB
(
/ =
=
OR:

Collaborative authors:
Nurulfadzilah Binti Hasan
Noor Zirwatul Ahlam Binti Naharuddin
Norhadzfizah Binti Mohd Radi
Mohd Hisyam Bin Mohd Ariff
Faculty of Electrical & Electronics Engineering,
UMP

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