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Communication Systems
Prof. Kuo, Chungming
Chapter 1
Decibel Computations (cont.)
Decibel Computations



Widely employed in the communications
industry.
Decibel forms are vital to understanding
the many system specifications and
performance standards.
Arguably not as essential today, but the
practice of utilizing decibel forms is so
widespread that the tradition will likely
continue.
Decibel Computations (cont.)

This module covers basic decibel
definitions and how they are applied in
systems analysis.
Important Logarithmic Identities
1
log   log x
x
log x k  k log x
log xy  log x  log y
x
log  log x  log y
y
Block Diagram for Defining Gain
I1
I2

G

V1 P1
or
P2 V2

GdB

R
Power Gain Definitions
• Absolute Power Gain:
P2
G
P1
• Decibel Power Gain:
P2
GdB  10 logG  10 log

P1
Power Loss Definitions
• Absolute Power Loss:
P1 1
L 
P2 G
• Decibel Power Loss:
P1
LdB  10 log L  10 log
P
2

• Note: LdB  GdB
Useful Decibel Patterns
Absolute Absolute
Gain
Loss
Decibel
Gain
Decibel
Loss
>1
<1
+
-
<1
>1
-
+
1
1
0
0
Conversion from Decibel Gain
to Absolute Gain
• Start with:
• Divide both
sidesby 10
GdB  10 log G
GdB
 log G
10
Conversion from Decibel Gain
to Absolute Gain (cont.)
• Raise both
sides to power
of 10
G 10
• By a similar

approach
L 10 LdB 10
G dB 10
Decibel Voltage and Current Forms
• Assume a resistance R at both input and output
2
2
2
1
V
GdB  10 log
V
2
V2 
R
V2
 10 log   20 log
R
V1
V1 
Let V2 V1 voltage gain  Av
Decibel Voltage and Current Forms (cont.)
• Assume a resistance R at both input and output
GdB  20log Av
Av 10
G dB 20
Let I 2 I1  current gain  Ai


GdB  20log Ai
Ai 10
G dB 20
Some Common Decibel Values
Power Ratio
2
4
8
10
100
10n
102n
1/2 = 0.5
1/4 = 0.25
1/8 = 0.125
1/10 = 0.1
1/100 = 0.01
1/10n = 10-n
1/102n = 10-2n
Voltage or
Current Ratio
2 = 1.414
2
22 = 2.828
10 = 3.162
10
10n/2
10n
1/2 = 0.7071
1/2 = 0.5
1/(22) = 0.3536
1/10 = 0.3162
1/10 = 0.1
1/10n/2 = 10-n/2
1/10n = 10-n
Decibel Values
3 dB
6 dB
9 dB
10 dB
20 dB
10n dB
20n dB
-3 dB
-6 dB
-9 dB
-10 dB
-20 dB
-10n dB
-20n dB
Some Common Decibel Values
(cont.)
• Example 1:
An amplifier has an absolute power gain of 175.
Determine the decibel gain.
GdB  10 log G  10 log175  10  2.243  22.43 dB
Some Common Decibel Values (cont.)
• Example 2:
An amplifier gain is given as 28 dB. Determine the
absolute power gain.
GdB  10 log G
28  10 logG
2.8  logG
G  10
2.8
 631.0
Some Common Decibel Values
(cont.)
• Example 3:
Assuming equal input and output
determine the voltage gain in
resistances,
Av  G  631.0  25.12
Some Common Decibel Values (cont.)
• Example 4:
In a lossy line, only 28% of the input power
reaches the load. Determine the decibel
gain and
loss.
P2
GdB  10 log  10 log 0.28  10  0.5528
P1
LdB  GdB  5.528  5.528 dB  5.528 dB
Decibel Reference Levels



In its basic form, the decibel involves a
logarithmic ratio and is dimensionless.
However, there are various portions of the
industry that have adopted decibel measures
relative to some standard reference level.
All of these forms have some modifier
attached to the unit; e.g., dBm, dBf, etc.
Typical Decibel Reference Levels
power level (W)
power level (dBW)  10 log
1W
power level (mW)
power level (dBm)  10 log
1 mW
power level (fW)
power level (dBf)  10 log
1 fW

Conversion Between Decibel Signal
Levels
Level in dBm = Level in dBW + 30
Level in dBf = Level in dBW + 150
Level in dBf = Level in dBm + 120
Example 5
• A signal has a power level of 100 mW.
• Express the level in dBm, dBW, and dBf.
P(mW)
100 mW
P(dBm)  10log
 10log
 20 dBm
1 mW
1 mW
P(W)
0.1 W
P(dBW)  10log
 10log
 10 dBW
1W
1W
P(fW)
1014 fW
P(dBf)  10log
 10log
 140 dBf
1 fW
1 fW
Decibel Gain Combined with Decibel
Signal Levels
PO  GPS
• Divide both sides by the same reference level, take
logs of both sides, multiply by 10, and expand. The
quantity 
x below represents any reference
standard; e.g., m.
PO dBx   PS dBx   GdB
Cascade System
G1
G2
G3
Gn
or
or
or
or
G1dB
G2dB
G3dB
GndB
• It is assumed that impedance (resistance)
levels are matched at all junctions.
Cascade Decibel Gain Analysis
G  G1G2G3
Gn
• Take logs of both sides, expand, and multiply
by 10. Apply dB forms to all terms.

GdB  G1dB  G2dB  G3dB 
 GndB
Example 6
• For system below, determine (a) system absolute
gain, (b) system decibel gain from (a), and (c)
system decibel gain from individual stages.
L
G1
G1  5000
G1dB  37 dB
G2
L  2000
G2  400
LdB  33 dB
G2dB  26 dB
Example 6 (cont.)
(a)
1 
1
G  G1  G2  5000 
 400  1000
L 
2000
(b)
G dB  10 log1000  10  3  30 dB
Example 6 (cont.)
(c)
G1dB  10 logG1  10 log 5000  10  3.7  37 dB
LdB  10 log L  10 log 2000  10  3.3  33 dB
G2dB  10 logG2  10 log 400  10  2.6  26 dB
GdB  G1dB  LdB  G2dB  37  33  26  30 dB
Example 7
• The source below is connected to the system of Example
6. Find (a) power levels in watts at all junctions and (b)
corresponding dBm levels.
Ps  0.1 mW
Ps (dBm) 
10 dBm
G1
P1
G1  5000
G1dB  37 dB
L
P2
G2
L  2000
G2  400
LdB  33 dB
G2dB  26 dB
Po
Example 7 (cont.)
(a)
P1  G1 PS  5000  0.1 mW  500 mW  0.5 W
P1 500 mW
P1  
 0.25 mW
L
2000

PO  G2 P2  400  0.25 mW  100 mW
Example 7 (cont.)
(b)
PS (mW) 
PS (dBm)  10 log
 10 dBm
 1 mW 
P1 (dBm)  PS (dBm)  G1dB  10  37  27 dBm
P2 (dBm)  P1 (dBm)  LdB  27  33  6 dBm
PO (dBm)  P2 (dBm)  G2dB  6  26  20 dBm
Example 8
• For the system below, determine power and voltage levels
at each point and the required gain of the output receiving
amplifier based on 75- matching at all junctions.
CABLE
SIGNAL
LINE
SOURCE AMPLIFIER SECTION
A
10 dBm
OUTPUT
13 dB
GAIN
26 dB
LOSS
CABLE
BOOSTER
AMPLIFIER SECTION
B
20 dB
GAIN
29 dB
LOSS
OUTPUT
AMPLIFIER
6V
GrdB  ?
75 W
Data for Example 8
Gain
Signal
Source
Line
Amplifier
Cable
Section A
Booster
Amplifier
Cable
Section B
Output
Level
10 dBm
Voltage
0.8660 V
13 dB
23 dBm
3.868 V
-26 dB
-3 dBm
0.1939 V
20 dB
17 dBm
1.939 V
-29 dB
-12 dBm
68.79 mV
Example 8: Computations
P(mW) 
 P(W) 
P(dBm)  10 log
 10 log

3
 1 mW 
110 W 
P(W)  1103 10 P (dBm) 10
V2 V2
P

or V  75P
R 75
2
6

P0 
 480 mW
75
Example 8: Computations (cont.)
480 mW 
P0 (dBm)  10 log
 26.81 dBm
 1 mW 
GrdB  26.81 dBm  12 dBm  38.81 dB
Alternately,
6V
Av 
 87.22
3
68.79 10 V
GrdB  20 log 87.22  38.81 dB
Decibel Signal-to-Noise Ratios
• P = average signal power in watts
• N = average noise power in watts
P
S N
N
S N dB  10 logS N 
S N dB  P(dBx)  N (dBx)
Example 9
• At a given point, signal power is 5 mW and noise
power is 100 nW. Determine absolute and dB S/N
ratios.
4
S
N

10
log
S
N

10
log
5
10
 47 dB
 dB
 
5 mW
4
 5 10
S N   4
10 mW
Example 9 (cont.)
• At a given point, signal power is 5 mW and noise
power is 100 nW. Determine absolute and dB S/N
ratios.
• Alternately,
P(dBm)  7 dBm
N(dBm)  40 dBm
S N dB  P(dBm)  N(dBm)  7  40  47 dB
Summary


A decibel is not an absolute unit, but is based
on a logarithmic power ratio.
Decibel measures are widely employed
throughout the electronics industry, but
especially in the communications area.
Summary (cont.)



Decibel level units are based on a standard
reference and dB is always accompanied by a
modifier in that case, namely, dBm.
The decibel gain of a complete system is the
sum of the individual dB gains.
A decibel level output is the sum of the decibel
level input and the decibel gain.