communication systems prof. kuo, chungming. chapter 1 decibel computations (cont.)
DESCRIPTION
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.TRANSCRIPT
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
log 1x log x
log x k k log xlog xylog x log y
log xylog x log y
Block Diagram for Defining Gain
1V
2V
2P1P
1I 2I
dB
orG
GR
Power Gain Definitions• Absolute Power Gain:
• Decibel Power Gain:
G P2
P1
GdB 10 logG 10 logP2
P1
Power Loss Definitions
• Absolute Power Loss:
• Decibel Power Loss:
• Note:
L P1
P2
1G
LdB 10 logL 10 log P1
P2
LdB GdB
Useful Decibel PatternsAbsolute
Gain Absolute
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 logG
GdB
10logG
Conversion from Decibel Gain to Absolute Gain (cont.)• Raise both
sides to power of 10
• By a similar approach
G 10GdB 10
L 10LdB 10
Decibel Voltage and Current Forms
• Assume a resistance R at both input and output
GdB 10 logV22 R
V12 R
10 log V2
V1
2
20 logV2
V1
Let V2 V1 voltage gain Av
Decibel Voltage and Current Forms (cont.)
• Assume a resistance R at both input and output
GdB 20 log Av Av 10GdB 20
Let I2 I1 current gain Ai
GdB 20 log Ai Ai 10GdB 20
Some Common Decibel ValuesPower Ratio Voltage or
Current Ratio Decibel Values
2 2 = 1.414 3 dB 4 2 6 dB 8 22 = 2.828 9 dB
10 10 = 3.162 10 dB 100 10 20 dB 10n 10n/2 10n dB 102n 10n 20n dB
1/2 = 0.5 1/2 = 0.7071 -3 dB 1/4 = 0.25 1/2 = 0.5 -6 dB 1/8 = 0.125 1/(22) = 0.3536 -9 dB 1/10 = 0.1 1/10 = 0.3162 -10 dB
1/100 = 0.01 1/10 = 0.1 -20 dB 1/10n = 10-n 1/10n/2 = 10-n/2 -10n dB
1/102n = 10-2n 1/10n = 10-n -20n dB
• Example 1:An amplifier has an absolute power
gain of 175. Determine the decibel gain.
Some Common Decibel Values (cont.)
GdB 10 logG 10 log175 102.243 22.43 dB
• Example 2:An amplifier gain is given as 28 dB.
Determine the absolute power gain.
Some Common Decibel Values (cont.)
GdB 10 logG28 10 logG2.8 logG
G 102.8 631.0
• Example 3: Assuming equal input and output
resistances, determine the voltage gain in
Some Common Decibel Values(cont.)
Av G 631.0 25.12
• Example 4: In a lossy line, only 28% of the input power reaches the load. Determine the decibel gain and loss.
Some Common Decibel Values (cont.)
GdB 10 log P2
P1
10 log0.28 10 0.5528
LdB GdB 5.528 5.528 dB 5.528 dB
Decibel Reference Levels In its basic form, the decibel involves a logarit
hmic ratio and is dimensionless. However, there are various portions of the ind
ustry 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 (dBW) 10 log power level (W)1 W
power level (dBm) 10 log power level (mW)1 mW
power level (dBf) 10 log power level (fW)1 fW
Conversion Between Decibel Signal Levels
Level in dBm = Level in dBW + 30Level in dBf = Level in dBW + 150Level 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.
dBf 140fW 1
fW 1010logfW 1(fW)10log(dBf)
dBW 10 W 1W 0.110log
W 1(W)10log(dBW)
dBm 20mW 1
mW 10010logmW 1(mW)10log(dBm)
14
PP
PP
PP
Decibel Gain Combined with Decibel Signal Levels
• 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 GPS
PO dBx PS dBx GdB
Cascade System
1
1dB
orG
G
3
3dB
orG
G dB
orn
n
G
G
2
2dB
orG
G
• It is assumed that impedance (resistance) levels are matched at all junctions.
Cascade Decibel Gain Analysis
• Take logs of both sides, expand, and multiply by 10. Apply dB forms to all terms.
G G1G2G3Gn
GdB G1dB G2dB G3dB GndB
Example 6
1
1dB
500037 dB
GG
2
2dB
40026 dB
GG
dB
200033 dB
LL
1G 2GL
• For system below, determine (a) system absolute gain, (b) system decibel gain from (a), and (c) system decibel gain from individual stages.
Example 6 (cont.)
(a)
G G11L
G2 5000 1
2000400 1000
(b)GdB 10 log1000 10 3 30 dB
Example 6 (cont.)
(c)G1dB 10 logG1 10 log5000 10 3.7 37 dBLdB 10 logL 10 log2000 10 3.3 33 dBG2dB 10 logG2 10 log 400 10 2.6 26 dBGdB G1dB LdB G2dB 37 3326 30 dB
Example 7
1
1dB
500037 dB
GG
2
2dB
40026 dB
GG
dB
200033 dB
LL
1G 2GL
0.1 mW(dBm)
10 dBm
s
s
PP
1P 2P oP
• 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.
Example 7 (cont.)
(a)P1 G1PS 50000.1 mW 500 mW 0.5 W
P1 P1
L500 mW
20000.25 mW
PO G2P2 4000.25 mW 100 mW
Example 7 (cont.)
(b)
PS (dBm) 10 log PS (mW)1 mW
10 dBm
P1(dBm) PS (dBm)G1dB 10 37 27 dBmP2 (dBm) P1(dBm) LdB 27 33 6 dBmPO (dBm) P2 (dBm)G2dB 6 26 20 dBm
Example 8
75
SIGNALSOURCE
10 dBmOUTPUT
LINEAMPLIFIER
CABLESECTION
A
CABLESECTION
B
BOOSTERAMPLIFIER
OUTPUTAMPLIFIER
13 dBGAIN
26 dBLOSS
20 dBGAIN
29 dBLOSS
6 V
dBrG ?
• 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.
Data for Example 8
Gain Output Level
Voltage
Signal Source
10 dBm 0.8660 V
Line Amplifier
13 dB 23 dBm 3.868 V
Cable Section A
-26 dB -3 dBm 0.1939 V
Booster Amplifier
20 dB 17 dBm 1.939 V
Cable Section B
-29 dB -12 dBm 68.79 mV
Example 8: Computations
P(dBm) 10log P(mW)1 mW
10log P(W)
110 3 W
P(W) 110 3 10P (dBm) 10
P V2
RV
2
75 or V 75P
P0 6 2
75480 mW
Example 8: Computations (cont.)
Alternately,
P0 (dBm) 10 log 480 mW1 mW
26.81 dBm
GrdB 26.81 dBm 12 dBm 38.81 dB
Av 6 V68.79 10 3 V
87.22
GrdB 20 log87.22 38.81 dB
Decibel Signal-to-Noise Ratios
• P = average signal power in watts
• N = average noise power in watts
S N PN
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.
S N dB 10 log S N 10 log5 104 47 dB
S N 5 mW10 4 mW
5 104
Example 9 (cont.)
• Alternately,
•At a given point, signal power is 5 mW and noise power is 100 nW. Determine absolute and dB S/N ratios.
P(dBm) 7 dBm N(dBm) 40 dBmS 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 ref
erence 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.