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.