Lesson12 Using dB and dBmV

INTRODUCTION/WORKING WITH dBmV Page 1

COPYRIGHT - NCTI, INC. USING dB AND dBmV

INTRODUCTIONThe previous lesson on decibels (dBs) presented the fundamentalconcepts necessary for understanding how dBs apply to broadbandcable communications. Power ratios are the basis for the dBmeasurement system and prepare the technician to understand dBsand decibel millivolts (dBmVs). The changes in signal power due toattenuation as the signal passes through the cable were thoroughlycovered in the previous lesson. The next step is to examine thechanges in the signal voltage and current as the signal passesthrough the cable. Broadband cable communications signals aremeasured in dBmVs. Therefore, you must clearly understand thevoltage ratios. Once the power and voltage ratios and dBmVs arepresented, some signal level calculations are shown.

WORKING WITH dBmVAll the ratios, percentages, and fractions discussed thus far havenot indicated any specific level values. When the term dB ordecibel is used, it only indicates a ratio and does not indicate anyspecific amount of voltage or power. By combining millivolt withthe decibel unit to form a decibel millivolt (dBmV), it can noweffectively indicate specific amounts of voltage and power.

Lesson Objectives:

Upon successful completion of thislesson, the student will be able to:

1. Describe a TV sets minimumacceptable input voltage leveland its equivalent dBmV ref-erence level;

2. Explain the difference betweenpositive and negative decibelmillivolt (dBmV) values;

3. Identify the equivalent powerpercentage increase for anattenuation increase of 1 dB;

4. Explain why dBmV values donot progress in a linear fashion;

5. Describe the significant differ-ence in dB between 1 millivolt(mV) decreases in signal level at40 dBmV compared to 6 dBmV;

6. Identify and convert valuesfrom a mV, dBmV, andnanowatt (nW) equivalencychart;

7. Calculate the minimum signallevel required at the tap portto provide the minimum sig-nal level for one TV set;

8. Calculate the minimum signallevel required at the tap portto provide a minimum signallevel for two TV sets;

9. Calculate the maximum tapvalue for a customer drop; and

10. Calculate changes to the TVset input caused by changes inthe line extender output.

tconnorsHighlightWhen the term dB ordecibel is used, it only indicates a ratio and does not indicate anyspecific amount of voltage or power. By combining millivolt withthe decibel unit to form a decibel millivolt (dBmV), it can noweffectively indicate specific amounts of voltage and power.

Establishing aReference Level (0 dBmV)All that is necessary to indicate specific values of voltage and power is toestablish a reference to which other amounts can be compared. In theearly days of broadband cable, it was determined that TV sets requiredat least 0.001 volt that is, 1 mV of signal from the coaxial cableto produce a good quality picture. This amount of signal was adopted asthe minimum acceptable level at each customers TV set (Figures 1Aand 1B). A level of 1 mV across 75 impedance is the broadband cablereference for the dB measurement system. Therefore, we can say that0 dBmV equals 1 mV across 75 and has a power of 13.33 nanowatts(nW). Using Ohms law and checking the figures we can see that:

Defining Positive and NegativedBmV ValuesWhen comparing mV and dBmV val-ues, pay particular attention towhether the value is positive or nega-tive (Figure 2). The dBmV values thatrepresent voltages greater than 1 mVare positive dBmV values (e.g.,+1 dBmV, +3 dBmV, etc.). Often, theplus sign is omitted. If no sign is indi-cated, the value is understood to bepositive (e.g., 1 dBmV, 3 dBmV, etc.).

P = E2 R = 0.0012 75 = 13.33 nW (or 1.333 108 watts)

Page 2 WORKING WITH dBmV

USING dB AND dBmV COPYRIGHT - NCTI, INC.

NOTES

1 mV (0 dBmV)required for agood picture

Broadbanddrop cable(75 Wimpedance)

A B

1 mVBroadband drop cable(75 W impedance)

Measuring 0 dBmV (1 mV across 75 W )

Figure 1. A level of 1 mV across 75 at the customers TV set equals 0 dBmV.

Measuring TV set input signal level Ensure minimum of 1 mV (0 dBmV) at input to customer TV set

+4

+3

+2

+1

0

1

3

2

4

1.413

1.122

0.891

0.708

1.585

1.259

1.0

0.794

0.631

dBmV

3 dB greater than 1 mv (0 dBmV) = +3 dBmV

1 dB greater than 1 mV (0 dBmV) = +1 dBmV

1 dB less than 1 mV = 1 dBmV

3 dB less than 1 mV = 3 dBmV

Reference level

mV

Figure 2. Signal voltages that are 1 dB greater, 3 dB greater, 1 dB less, and 3 dB lessthan 1 mV (0 dBmV).

NTI

NTI

tconnorsHighlightIn theearly days of broadband cable, it was determined that TV sets requiredat least 0.001 volt that is, 1 mV of signal from the coaxial cableto produce a good quality picture. This amount of signal was adopted asthe minimum acceptable level at each customers TV set (Figures 1Aand 1B). A level of 1 mV across 75 impedance is the broadband cablereference for the dB measurement system. Therefore, we can say that0 dBmV equals 1 mV across 75 and has a power of 13.33 nanowatts

tconnorsHighlight(nW). Using Ohms law and checking the figures we can see that:Defining Positiveand NegativedBmV ValuesWhen comparing mV and dBmV values,pay particular attention towhether the value is positive or negative(Figure 2). The dBmV values thatrepresent voltages greater than 1 mVare positive dBmV values (e.g.,+1 dBmV, +3 dBmV, etc.). Often, theplus sign is omitted. If no sign is indicated,the value is understood to bepositive (e.g., 1 dBmV, 3 dBmV, etc.).P

tconnorsHighlightE2 R = 0.0012 75 = 13.33 nW (or 1.333 108 watts)

The dBmV values that represent voltages less than 1 mV arealways shown as negative dBmV values (e.g. , 1 dBmV,3 dBmV, etc.). Remember that all dBmV values represent avoltage level that is measured across a 75 impedance. RFsignal voltages measured across other impedance valuescannot be directly compared to voltages measured across75 and cannot be represented by dBmVs.

It was established earlier that the power ratio for 1 dB is 1.2589:1. InFigure 3, the equivalent power for 0 dBmV is 13.333 nW. To find theresulting power after an increase of 1 dB, the amount of power ismultiplied by the power ratio 1.2589:1. We previously learned thatratios are actually percentages. When 13.333 nW is multiplied by1.2589 (1 dB power ratio), it is increased by a certain percentage.

The percentage of increase is found by dividing the increase in powerinto the original power level: 3.452 13.333 = 0.2589 = 25.89%.

A 1 dB increase can be significant because it represents slightly morethan a one-fourth increase in power. However, an accuracy of 1 dB isnormal and considered sufficient for many measurements made in abroadband cable system.

13.333 1.2589 = 16.785 nW

16.785 13.333 = 3.452 nW increase

10 mV across 75 = +20 dBmV10 mV across 300 (is not equal to) +20 dBmV

WORKING WITH dBmV Page 3


1.122

1.000

0.8910.875

1.12516.785

13.333

10.59610.208

16.875

1 dB increase results in a25.89% increase in power

13.333 1.2589 = 16.785 nW

16.785 13.333 = 3.452 nW increase

3.452 13.333 = 0.2589

= 25.89% increase in power

0

1

dBmVmV nW

+1

Figure 3. A 1 dB increase results in a 25.89% increase in power.

NOTES

NTI

tconnorsHighlightvoltages less than 1 mV arealways shown as negative dBmV values (e.g., 1 dBmV,3 dBmV, etc.). Remember that all dBmV values represent avoltage level that is measured across a 75 impedance.

tconnorsHighlightA 1 dB increase can be significant because it represents slightly morethan a one-fourth increase in power.

tconnorsHighlight10 mV across 75 = +20 dBmV

tconnorsHighlightFigure 3. A 1 dB increase results in a 25.89% increase in power.NOTESN

WORKING WITH dBmV


Understanding the Nonlinearity of dBmVsIt was stated earlier that measurements of signals in a broadbandcable system can be expressed using standard units of voltage or power.In fact, a signal level meter does exactly that; it measures the signalsvoltage. Lets consider what difficulties would emerge without using thedBmV measurement system. If we use only voltages, and the outputsignal level of a line extender that should be 100 mV is measured at99 mV, the output signal is 1 mV low. Another measurement is made atthe ground block of a customer premises drop. To adequately serve twoTV sets, a measurement of 2 mV is required at the input to the two-way splitter. The actual voltage measurement of the splitter inputsignal indicates only 1 mV 1 mV low. The signal is low by the sameamount in both the line extender output and the ground block.

To know how significant the missing mV is in each case, compare it tothe desired amount of signal. In Figure 4A, the signal should be 100 mV.Clearly, it is 1100 or 1% low (1 100 = 0.01 or 1%). If the meter wascapable of 1% accuracy, it would be a very accurate meter, but it stillcould not indicate with certainty that the signal is 1 mV low. A voltagevariation of 1% would not cause any noticeable effects. Using the dBmVsystem, we find that in the first case 100 mV is +40 dBmV (Figure 4B)and +39 dBmV is 89.125 mV, almost 11 mV less. We are considering asignal that is 1 mV low, less than 110 of 1 dB (0.1 dB) low (39.9 dBmV).

Less than0.1 dB

B

Line extender

100.0

99.0

98.0

97.0

96.0

95.0

94.0

93.0

92.0

91.0

90.0

89.125

88.0

Desired signal

Measured signal1 mV

1 dB10.875

mV

Designed output signal level = 40.0 dBmVActual output signal level = 39.9 dBmV

39.9 dBmV

Line extender #1

100 mV 40.0 dBmV 99 mV 39.9 dBmV

1 mV 0.1 dBdifference

A

+39

dBmVmV

+40

NOTES

1 mV low at line extender output 1 mV difference < 0.1 dB

Figure 4.The dB equivalent of a 1 mV difference at output of line extender.

NTI

tconnorsHighlightLess than0.1 dBBLine extender100.099.098.097.096.095.094.093.092.091.090.089.12588.0Desired

In Figure 5A the signal should be 2 mV. The measured signal is only 1 mVor 1 mV too low, which means it is 12 or 50% low (1 2 = 0.50 = 50%).Certainly, a 50% loss of signal is more significant than a 1% loss of signal.Notice especially that to make this determination, it is necessary to knownot only the amount by which the signal is low, but the amount that itshould be. In the second case, 2 mV is +6 dBmV and 1 mV is 0 dBmV(Figure 5B). This signal is 6 dB low.

Earlier, it was stated that 0 dBmV is the minimum amount of signalthat still consistently produces a good picture on a TV set. The groundblock where we measured 1 mV rather than the desired 2 mV isfollowed by a two-way splitter. It is easy to see that when the signal is1 mV low, there is not enough signal for the two TV sets. In the case ofthe line extender output level, the 1 mV shortfall equals less than110 dB and is not very significant. However, in the case of the splitterinput level, the 1 mV shortfall equals 6 dB and is quite significant.

By using dBmVs, the need to make a comparison to the desired signallevel is eliminated. The fact that in the first case the signal is 110 dBlow, and in the second case it is 6 dB low, clearly shows the instanceof greater significance. It also shows that dBmV values do notprogress in a linear fashion.

If the signal in the first measurement is +39 dBmV (89.125 mV)rather than the desired +40 dBmV (100 mV), it is 1 dB low. This 1 dB



NOTES

1 mV low at two-way splitter input 1 mV difference = 6 dB

Figure 5.The dB equivalent of a 1 mV difference at the input to the two-way splitter.

B

6 dB

Designed two-way splitter input signal level = 6.0 dBmV (2.0 mV)Actual two-way splitter input signal level = 0.0 dBmV (1.0 mV)

Line extender #2

A

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6-O

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In

3.5

dB3.

5 dB

dBmVmV

To ground

Groundblock

Should be 2 mV,but is 1 mV

Two-waysplitter

2.0 mV 6.0 dBmV1.0 mV 0.0 dBmV 1.0 mV 6.0 dB

difference

Desiredsignal

Measuredsignal

1 mV

Ground block

1.0

+6

+5

+4

2.0

+3

+2

+1+1

0

1.778

1.61.585

1.4131.3

1.259

1.122NTI

tconnorsHighlightThe measured signal is only1mVor 1 mV too low, which means it is 12 or 50% low

tconnorsHighlightCertainly, a 50% loss of signal is more significant than a 1% loss of signal.

tconnorsHighlightEarlier, it was stated that 0 dBmV is the minimum amount of signalthat still consistently produces a good picture on a TV set. The groundblock where we measured 1 mV rather than the desired 2 mV isfollowed by a two-way splitter. It is easy to see that when the signal is1 mV low, there is not enough signal for the two TV sets.

WORKING WITH dBmV


equals 10.875 mV (100 89.125). If the second measurement is+5 dBmV (1.778 mV) rather than the desired +6 dBmV (2 mV), it isalso 1 dB low. However, this 1 dB equals only 0.222 mV. A 1 dBdifference may be either a large or a small voltage difference, but thetwo voltage values making up that difference will always form a1.122018:1 ratio.

Figure 6 shows equivalent signal level values in three different units mV, dBmV, and microwatt (W) or nanowatt (nW). Along the left side ofeach of the three columns is a linear mV scale, and along the right sideis a nonlinear power scale (in W or nW). The power scale is nonlinearbecause power represents a mathematical squaring of the voltage. Thecorresponding amounts of voltage and power shown are correct for a75 impedance. For instance, the left-hand column shows that 10 mVhas the power of 1.333 W. Remember, this is true only when theimpedance is 75 (P = E2 R for 75 ). The numbers in the center ofthe columns are the equivalent dBmV values. At the 10 mV level thedBmV value is +20. It is apparent that the dBmV values also are notspaced linearly, but are progressively farther apart as they becomelarger. This is because dBmV represents the logarithm of a powerratio.

A thorough study of the columns in Figure 6 and the graph in Figure 7(on page 8) provides an understanding of the nonlinearity of dBmVscompared to the linearity of mVs. The resulting appreciation of thesignificance of the various signal level measuring units will help guideyour analysis of problems and offer insight on how best to solve them.

NOTES

NOTE

The text on this page refers to Figures 6and 7 on pages 7 and 8 respectively.

Review Questions, Part I

1. What is the broadband signal reference level for thedB measurement system?

2. What is the reference level in dBmV and power levelin nW that is represented by 1 mV?

3. All dBmV values represent a voltage level that mustbe measured across what value impedance?

4. A 1 dB increase results in a _______________________ %increase in power.

5. What three different units can be used to representsignal level values?

6. Why are the dBmV values in Figure 6 not spacedlinearly the way the mV values are?

tconnorsHighlightThis is because dBmV represents the logarithm of a powerratio.



4 213.33Standard referencepoint other dBmVvalues are based upon

Center and right-hand columnshave expanded scales

To convert any power value in watts to anequivalent voltage value in microvolts useOhms law.

P (watts) = E2 (voltage)

R (75 W resistance)

mV dBmV m W mV dBmV nW mV dBmV nW

nW

+5

+24

+30

+28

33

3231.62

31

30

29

28

27

26

25.12

24

23

22

21

19.95

19

18

17

15.85

15

14

1312.59

12

11

10.00

9

7.94

76.31

6

5.014

3.981

+26

+22

+20

+18

+16

+14

3.981

3.548

3.162

3

2.818

2.512

2.239

21.995

1.778

1.585

1.413

1.259

1.122

1.000

211.33

167.86

133.33

120.00

105.91

84.13

66.82

53.3353.08

42.16

33.49

26.60

21.13

16.79

13.33

+12

+11

+10

+9

+8

+7

+6

+4

+3

+2

0

+1

01.000

0.891

0.794

.0708

0.631

0.562

0.501

0.446

0.398

0.316

0.251

0.1995

0.158

0.1250.100

13.33

10.596

8.413

6.682

5.308

4.216

3.349

2.660

2.113

1.333

0.8413

0.5308

0.3349

0.21130.1333

1

3

4

5

6

7

2

8

10

12

14

16

1820

+12

14.520

13.65313.333

12.000

10.453

9.013

8.413

7.680

6.453

5.308

4.320

3.349

2.613

2.1131.920

1.333

841.3

530.8480.0

334.9213.33

211.3

Figure 6. Voltage and corresponding power values shown on a linear scale, with dBmV values indicated. The nonlinearity of the dBmVvalues is due to the logarithmic relationship with the respective voltage and power ratios.

NTI

tconnorsHighlightQuestion #6 Chart

WORKING WITH dBmV


Am

plifi

er #

1

% P

ow

er10

0%50

.12%

25.1

2%10

%5%

2.5%

1.0%

0.63

1%%

Vo

ltage

100%

70.7

9%50

.12%

31.6

2%

2

2.39

%

1

5.85

%

10

.0%

7.94

%

01

23

45

67

89

1011

1213

1415

1617

1819

2021

22 dB

2.82

mV

1.12

2:1

ratio

(1 dB

)1.

259:

1ra

tio(2

dB)

1.41

3:1

ratio

(3 dB

)1.

122:

1ra

tio

1.12

2:1

ratio

35.4

8 m

V

31.6

2 m

V

28.1

8 m

V

25.1

2 m

V

dBm

V m

V

+31

+30

+29

+28

+27

+26

+25

+24

+23

+22

+21

+20

+19

+18

+17

+16

+15

+14

+13

+12

+11

+10

+ 9

In th

e fir

st 1

dB

sect

ion

of c

able

the

sign

al vo

ltage

dec

reas

es fr

om

35.4

8 m

V to

31.

62 m

V.

35.4

8 1.

122

= 31

.62

mV

In th

e se

cond

1 d

B se

ctio

n of

cab

le:

31.6

2 1.

122

= 28

.18

mV

In th

e th

ird 1

dB

sect

ion

of c

able

:

28.1

8 1.

122

= 25

.12

mV

Beca

use

the

redu

ctio

n is

bas

ed o

n lo

garit

hm o

f a ra

tio, t

he s

igna

llo

ss is

not

line

ar. In

the

first

6 d

B lo

ss o

f sig

nal o

n th

e ca

ble

half

the

sign

al v

olta

ge is

lost

. Hal

f of t

he r

emai

ning

sig

nal v

olta

geis

lost

in th

e ne

xt 6

dB

of c

able

, et

c.

Not

e th

at w

e st

arte

d ou

t w

ith 3

1 dB

mV

and

at t

he e

nd o

fth

e ca

ble

span

we

have

9 dB

mV.

31 d

BmV

9

dBm

V =

22 d

B

This

22

dB is

the

tota

l los

s fro

m th

e ca

ble.

36 34 32 30 28 26 24 22 20 18 16 14 12 10 8 6 4 2 0

Figu

re 7

.Am

plitu

de o

f sig

nal v

olta

ge as

the

sign

al p

asse

s th

rou

gh 2

2 dB

of c

able

fro

m o

ne

ampl

ifier

to a

noth

er.

NT

I

PERFORMING SIGNAL LEVEL CALCULATIONS Page 9


PERFORMING SIGNAL LEVEL CALCULATIONS

Now that the basis for and the meaning of dBs and dBmVs isestablished, the practical use of this measurement system can berealized. It is no longer necessary to convert from mV or nW to dBmVnor to think in terms of specific voltage and power ratios. Because0 dBmV already refers to a specific amount of voltage and power, it isonly necessary to add or subtract dBs and dBmVs to determine theamount of signal that is present anywhere in the system.

As mentioned earlier, the 1 mV reference level for 0 dBmV existsbecause it is the minimum signal level that consistently provides agood quality picture on a TV set.

Estimating Tap Port Signal Level for One TV SetIf a system requires a minimum signal level at the TV set of+3 dBmV, it is easy to determine the necessary minimum signal levelat the tap port (Figure 8). A drop that consists of 100 feet of Series 6drop cable (from the tap to the TV set) has a loss of 5.65 dB in thedrop cable at 750 MHz (Series 6 attenuation rate = 5.65 dB/100 feet@ 750 MHz). Thus, the minimum amount of signal from the tap port@ 750 MHz must be:

3.00 dBmV @ 745.25 MHz (system minimum level at the TV set)+ 5.65 dB/100 feet of Series 6 @ 750 MHz (cable loss)

8.65 dBmV (minimum tap port signal level at 745.25 MHz)

NOTES

Minimum of+8.65 dBmV at745.25 MHz

+3 dBmV@ 745.25 MHzrequired atthe TV set

Tap

To ground

5.65 dB loss/100 feet ofSeries 6 @ 750 MHz

Figure 8. Minimum tap port signal level required for one TV set.

NTI

NOTE

Remember that a given attenuation rate isfor 100 feet. For drop lengths greater than100 feet, you must multiply the rate by thedesired footage and divide by 100 to obtainthe total loss of the cable.

tconnorsHighlightEstimating Tap PortSignal Level for One TV Set

Estimating Tap Port Signal Level for Two TV SetsIf a two-way splitter is needed to provide service to an additional TVset (Figure 9), and the longest amount of cable from the tap to eitherTV set is 100 feet, the minimum signal needed from the tap is:

Estimating the RequiredTap Value for Two TV SetsIf this drop is connected to a tap on the output of a line extender,which has a flat output signal level of 45 dBmV (as shown inFigure 10), the highest value tap that can be used and still providethe required minimum level is:

45.0 dBmV @ 745.25 MHz (signal into the tap from the line extender) 12.7 dBmV (minimum signal required from the tap port)

32.3 dB (maximum tap value)

3.00 dBmV @ 745.25 MHz (system minimum level at the TV set)5.65 dB/100 feet @ 750 MHz (loss from Series 6 drop cable)

+ 4.00 dB (approximate loss from the two-way splitter)

12.65 dBmV (12.7 rounded off) (minimum signal level from the tap port @ 745.25 MHz)

Page 10 PERFORMING SIGNAL LEVEL CALCULATIONS


NOTES

5-6

00 M

Hz

2-W

ay

Sp

litte

r

S2D

GH

6-O

P

In

3.5

dB3.

5 dB

Tap

+12.7dBmV

To ground

Two-way splitter

4.0 dB@ 750 MHz

100 feet Series 6cable 5.65 dB loss

+3 dBmVrequired

+3 dBmVrequired

Set-topterminal #2

Set-topterminal #1

Figure 9. Minimum tap port signal level required for two TV sets.

NTI

NOTE


PERFORMING SIGNAL LEVEL CALCULATIONS Page 11


In practice, the nearest tap value lower than 32.3 dB is used. If this isa 32 dB tap value, the amount of signal arriving at the TV set is:

This is 0.35 dB higher than the required minimum.

Learning the Effects ofReduced Line Extender Output In Figure 11 on page 12 the input to the line extender is shown as+20 dBmV. The line extender gain of 25 dB produces an output of+45 dBmV, and the other signal levels along the drop are indicated.If the gain of the line extender decreased to 22 dB, the output wouldbe +42 dBmV, 3 dB lower than normal. All levels from that point on

45.00 dBmV @ 745.25 MHz (tap input level) 32.00 dB (tap value)

13.00 dBmV @ 745.25 MHz (tap port output) 5.65 dB (cable loss)

7.35 dBmV @ 745.25 MHz (splitter input level) 4.00 dB (approximate splitter loss)

3.35 dBmV @ 745.25 MHz (actual TV set input level)

NOTES

32.3 dB maximum tap value

Line extender

5-6

00

MH

Z

2-W

ay

Sp

litt

er

S2D

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6-O

P

In

3.5

dB3.

5 dB

JerroldStarlineJerroldStarline GENERALINSTRUMENT

32 dB actual tap value32 four-waytap

TV set #1

TV set #2

+13.0 dBmV@ 745.25 MHz (actual)

To ground

+45 dBmV@ 745.25 MHz

100 feet Series 65.65 dB

+3 dBmV@ 745.25 MHz(desired)

+12.7 dBmV@ 745.25 MHz (desired)

Two-waysplitter

+3.35 dBmV@ 745.25 MHz(actual)

4.0 dBsplitter loss

Figure 10. Required theoretical and actual video carrier signal levels.

NTI

NOTE


PERFORMING SIGNAL LEVEL CALCULATIONS


will be reduced by the same amount (3 dB). Therefore, instead ofhaving +3.35 dBmV at the TV set, the signal is +0.35 dBmV(2.65 dB lower than system-required minimum).

NOTES

Line extender

5-6

00

MH

Z

2-W

ay

Sp

litt

er

S2D

GH

6-O

P

In

3.5

dB3.

5 dB

32 four-waytap

TV set #1

TV set #2

+42 dBmV @ 745.25 MHz (ch. 119)3 dB less+22 dB gain3 dB less

10 dBmV @ 745.25 MHz (ch. 119)3 dB less

7.35 dBmV @ 745.25 MHz (ground block)4.35 dBmV @ 745.25 MHz (ground block)3 dB less

To ground

100-foot Series 65.65 dB/100 feet @ 750 MHz

+45 dBmV

Two-waysplitter

4.0 dBsplitter loss

+20dBmV@ 745.25MHz

+25 dB gain

13 dBmV @ 745.25 MHz

3.35 dBmV @ 745.25 MHz0.35 dBmV @ 745.25 MHz (ch. 119)3 dB less

Figure 11. Signal level to the TV set with reduced line extender output.

Review Questions, Part II

7. What is the minimum amount of signal needed@ 750 MHz from the tap port through a 100-foot Series 6drop if the minimum signal level required at a single TVset is +2 dBmV?

8. Referring to Figure 11, how much signal would arriveat TV set #2 if a 29 dB tap value is used instead of a32 dB tap when the amplifier gain is 25 dB?

9. Again, referring to Figure 11, if the line extender gainis a normal 25 dB, but the input signal level is instead+15 dBmV, how much signal from the 32 dB tap portwill be present at TV set #2?

NTI

SUMMARIZING dB AND dBmV Page 13


SUMMARIZING dB and dBmVSummarizing, dB (decibel) and dBmV (decibel millivolt) are termsused daily by the broadband cable technician. Just remember thatfractions, percentages, ratios, and dBs are all simply methods ofcomparing values.

The items listed below are derived from this basic foundation and aretherefore the most important things to understand and remember.

Finally, there is nothing mysterious or unexplainable about dBsand dBmVs. Some mathematics is required to understand andconvert dBs to their equivalent ratios and dBmVs to their

The foundational starting point of the entire decibelmeasurement system is that 1 bel equals a 10:1 power ratio.

1/10 bel (1 dB) is a power ratio of 1.2589:1.

1/10 bel (1 dB) is a voltage ratio of 1.122:1.

The voltage ratio is always the square root ofthe power ratio.

A 1 dB increase is equal to a 25.89% powerincrease.

A 1 dB increase is equal to a 12.2% voltageincrease.

When dBs and dBmVs are added, theirrespective power and voltage ratios aremultiplied. (1.258910 = 10 dB)

When dBs are added to dBmVs, the voltage ismultiplied by the corresponding dB ratio.

When dBs are subtracted from dBmVs, thevoltage is divided by the corresponding dBratio.

Voltage and dBmV values can be compared ina meaningful way only if measured across thesame impedance value.

A level of 0 dBmV equals 1 mV across 75 .

Signal values greater than 1 mV across 75 are indicated by positive dBmV values (e.g.,+3 dBmV or 3 dBmV).

Signal values smaller than 1 mV across 75 are indicated by negative dBmV values (e.g.,3 dBmV).

dBmV values do not progress in a linearfashion because they are based on thelogarithms of power ratios.

When the signal from an amplifier changes bya dB amount, 2 dB for instance, the signal ischanged by that same dB amount from thatpoint on.

3 dB is essentially a 2:1 power ratio (twice asmuch or one-half as much).

6 dB is essentially a 4:1 power ratio.

6 dB is a 2:1 voltage ratio (twice as much orone-half as much).

NOTES

SUMMARIZING dB AND dBmV


equivalent voltage and power values. However, with a goodunderstanding of ratios, the groundwork for understanding highermathematical techniques is established. Figures 12 and 13 provideconvenient reference material for finding equivalent dBmV,voltage, power, and ratio values.

NOTES

UNIT CONVERSION CHART

MICROVOLTS

PICOWATTS

dBmVRMS(mV)

Ratioof Voltageto 0 dBmV

RMSPower(nW)

Ratioof Power

to 0 dBmV

+20.00+19.00+18.00+17.00+16.00+15.00+14.00+13.00+12.00+11.00+10.00+9.00+8.00+7.00+6.00+5.00+4.00+3.00+2.00+1.00

0.001.002.003.004.005.006.007.008.009.00

10.0011.0012.0013.0014.0015.0016.0017.0018.0019.0020.00

10.00008.91257.94337.07956.30965.62345.01194.46683.98113.54813.16232.81842.51192.23871.99531.77831.58491.41251.25891.12201.0000

891.25794.33707.95630.96562.34501.19446.68398.11354.81316.23281.84251.19223.87199.53177.83158.49141.25125.89112.20100.00

10.00008.91257.94337.07956.30965.62345.01194.46683.98113.54813.16232.81842.51192.23871.99531.77831.58491.41251.25891.12201.00000.89130.79430.70790.63100.56230.50120.44670.39810.35480.31620.28180.25120.22390.19950.17780.15850.14130.12590.11220.1000

1333.331059.10841.28668.25530.81421.64334.92266.03211.32167.86133.33105.9184.12866.82553.08142.16433.49226.60321.13216.78613.33310.5918.4136.6825.3084.216 3.3492.6602.1131.6791.3331.059

841.28668.25530.81421.64334.92266.03211.32167.86133.333

100.0079.43363.09650.11939.81131.62325.11919.95315.84912.58910.0007.94336.30965.01193.98113.16232.51191.99531.58491.25891.00000.794330.630960.501190.398110.316230.251190.199530.158490.125890.100000.079430.063100.050120.039810.031620.025120.019950.015850.012590.01000

NTI

Figure 12. Converting dBmV to voltage and power ratios.

To find the voltage ratio for 16 dB multiply the voltage ratio for 10 dB by the voltage ratio for 6 dB.

To find the voltage ratio for 4 dB divide the voltage ratio for 10 dB by the voltage ratio for 6 dB.

16 dB = 10 dB + 6 dBVoltage ratio for 16 dB = (3.162:1)(2:1) = 6.324:116 dB = 6.324:1 voltage ratio

For 36 dB:36 dB = 20 dB + 10 dB + 6 dBVoltage ratio for 36 dB = (10:1)(3.162:1)(2:1) = 63.24:136 dB = 63.24:1 voltage ratio

SUMMARIZING dB AND dBmV Page 15


3 dB = *2:1 power ratio and *1.414:1 voltage ratio6 dB = *4:1 power ratio and *2.000:1 voltage ratio

10 dB = 10:1 power ratio and 3.162:1 voltage ratio20 dB = 100:1 power ratio and 10.000:1 voltage ratio

*These values are rounded off. The accuracy is within 0.5% (one-half of 1%).

To multiply ratios, simply add decibels.To divide ratios, simply subtract decibels.

To find the power ratio for 16 dB multiply the power ratio for 10 dB by the power ratio for 6 dB.

To find the power ratio for 4 dB divide the power ratio for 10 dB by the power ratio for 6 dB.

16 dB = 10 dB + 6 dBPower ratio for 16 dB = (10:1)(4:1) = 40:116 dB = 40:1 power ratio

For 36 dB:36 dB = 20 dB + 10 dB + 6 dBPower ratio for 36 dB = (100:1)(10:1)(4:1) = 4000:136 dB = 4000:1 power ratio

4 dB = 10 dB 6 dBPower ratio for 4 dB = (10:1) (4:1) = 2.5:14 dB = 2.5:1 power ratio

4 dB = 10 dB 6 dBVoltage ratio for 4 dB = (3.162:1) (2:1) = 1.581:14 dB = 1.581:1 voltage ratio

Figure 13. Adding and subtracting dB values using power and voltage ratios.

NTI

ANSWERS TO REVIEW QUESTIONS/GLOSSARY


Part II

7. 2.00 dBmV (required level at the TV set)+ 5.35 dB (cable loss through 100 feet of Series 6

cable @ 750 MHz)

7.35 dBmV (minimum signal required from the tap port @ 750 MHz)

8. A signal level of 6.35 dBmV would arrive at TV set #2if a 29 dB tap value were used instead of a 32 dB tap.

9. The signal level at TV set #2 would be +1.35 dBmV.

ANSWERS TO REVIEW QUESTIONS

Part I

1. The broadband signal reference level for the dBmeasurement system is 1 mV across 75 of imped-ance or 0 dBmV.

2. A level of 1 mV across 75 represents the 0 dBmVreference level and a power level of 13.33 nW.

3. All dBmV values represent a voltage level that ismeasured across a 75 impedance.

4. A 1 dB increase results in a 25.89% increase inpower.

5. The three different units that can be used to repre-sent signal level values are mV, dBmV, and W or nW.

6. The dBmV values in Figure 6 are not spaced linearlybecause the dB measurement system is based on log-arithms of power ratios.

Impedance The combined effect of induc-tance, capacitance, and resistance in anelectronic circuit or transmission line thatopposes current flow. The value of impedanceis stated in ohms () and is the ratio of volt-age in volts to current in amperes.

Logarithm A type of shorthand to repre-sent a number expressed as the exponent ofa base number (usually 10) that equals theoriginal number.

Negative dBmV values Negative ()dBmV values represent signal level volt-ages of less than 1 mV. The minus sign isnever omitted for negative values. A dBmVvalue without a polarity sign is always con-sidered positive.

Positive dBmV values Positive (+) dBmVvalues represent signal level values of morethan 1 mV. The plus sign is frequently omit-ted; however, the value remains positive(more than 1 mV) when no polarity sign isused (e.g., 1 dBmV, 16 dBmV, 30 dBmV, etc.).

0 dBmV A signal voltage value of 1 mVacross a 75 impedance. Other amounts ofsignal voltage (both smaller and greater)are compared to this reference by using thedB measurement system.

GLOSSARYBelow are the definitions of the key terms occurring in bold italic and regular type in the text.

USING dB AND dBmVEXAMINATION QUESTIONS FOR LESSON 217-23-1

1. A minimum acceptable signal level of 0 dBmV at a TV set inputA. Only indicates a ratio and does not indicate any specific

amount of voltage or power.B. Equals 1 mV across 75 .C. Equals 0 mV across 300 .D. Equals 0 mV across 75 .E. Is 1 dB less than 1 dBmV.

2. Values expressed in dBmV that are preceded by a minus sign ()representA. Values of voltage less than 1 mV across 75 .B. Values of voltage subtracted from one another.C. No voltage.D. Signals that are too low by the amount indicated.E. Values of power more than 1 W across 75 .

3. A signal increase of 1 dB represents a _______ increase in power.A. 10%B. 50%C. 25.89%D. 12.5%E. 90%

4. Values expressed in dBmV progress in a nonlinear fashion whencompared to a linear mV scale becauseA. They are based on a 300 impedance.B. They are based on the logarithms of power ratios.C. 1 bel equals a 1.122:1 voltage ratio.D. They are based on a sine wave ratio.E. Both C and D.

5. A 1 mV decrease in a signal level of 40 dBmV is approximatelyequivalent to a decrease in level ofA. 10 dB.B. 20 dB.C. 3 dB.D. 0.1 dB.E. 6 dB.

6. What is the equivalent voltage in mV for a +20 dBmV signal usingthe chart in Figure 6?A. 10 mV.B. 1.333 mV.C. 3.162 mV.D. 0.100 mV.E. 12.59 mV.

7. How much signal is necessary from the subscribers tap port@ 750 MHz to provide a minimum of 3.0 dBmV to the TV set whenthe drop is 150 feet of Series 6 cable and just one outlet (use5.65 dB per 100 feet for the attenuation rate and round off tonearest tenth of a dBmV)?A. 15.0 dBmV.B. 7.7 dBmV.C. 11.5 dBmV.D. 4.5 dBmV.E. 8.7 dBmV.

8. How much signal is needed from the subscribers tap port@ 750 MHz to provide a minimum of +3.0 dBmV to each of the TVsets when there is 132 feet of Series 6 cable and a two-waysplitter? (Use 5.65 dB per 100 feet for the Series 6 cableattenuation rate and 4.0 dB for the splitter attenuation.)A. 14.5 dBmV.B. 10.3 dBmV.C. 8.8 dBmV.D. 12.7 dBmV.E. 17.2 dBmV.

9. What is the maximum tap value that can be installed when the tapis connected directly to a line extender amplifier output and: (1) theline extender output is 45 dBmV @ 745.25 MHz; (2) the longestamount of Series 6 coaxial cable from the tap to either of the twoTV sets is 150 feet; (3) the Series 6 coaxial cable attenuation rate is5.65 dB/100 feet @ 750 MHz; (4) the drop includes a splitter with a4.0 dB loss; and (5) the system minimum signal level required ateach of the two TV sets is 3.0 dBmV?A. 32 dB.B. 20 dB.C. 23 dB.D. 29 dB. E. 35 dB.

10. When the output of a line extender is 3 dB lower than it normallyshould be,A. Only the subscriber drops that have a two-way splitter are

affected.B. The signal power is 25.89% low from that point on.C. The signal is 3 dB lower than normal from that point on.D. The signal voltage is 1 millivolt lower than normal from that

point on.E. The signal at the TV set is always too low to provide a good

picture.

USING dB AND dBmV EXAM 217-23-1

Table of Contents

INTRODUCTION................................................................................. 1

WORKING WITH dBmV .................................................................... 1 Establishing a Reference Level (0 dBmV)Defining Positive and Negative dBmV ValuesUnderstanding the Nonlinearity of dBmVs

PERFORMING SIGNAL LEVEL CALCULATIONS ..................... 9 Estimating Tap Port Signal Level for One TV SetEstimating Tap Port Signal Level for Two TV SetsEstimating the Required Tap Value for Two TV SetsLearning the Effects of Reduced Line Extender Output

SUMMARIZING dB AND dBmV..................................................... 13 Converting dBmV to Voltage and Power RatiosAdding and Subtracting dB Values Using Power andVoltage Ratios

ANSWERS TO REVIEW QUESTIONS .......................................... 16

GLOSSARY ......................................................................................... 16 Key terms occurring in bold italic and regular type in the text are listed with definitions specific to this lesson.

217-23 USING dB AND dBmV

Copyright 2000 by Versacom, Inc., dba NCTI, Inc., 801 W. Mineral Ave.,Littleton, CO 80120-4501. All rights reserved. No part of this lesson may bereproduced in any form without permission in writing from the publisher.

Printed in U.S.A.

Your Broadband Communications Training Partner

217-23

Title PageTable of ContentsCredits/CopyrightLesson ObjectivesIntroductionWorking With dBmVEstablishing a Reference Level (0 dBmV)Defining Positive and Negative dBmV ValuesUnderstanding the Nonlinearity of dBmVsReview Questions, Part I

Performing Signal Level CalculationsEstimating Tap Port Signal Level for One TV SetEstimating Tap Port Signal Level for Two TV SetsEstimating the Required Tap Value for Two TV SetsLearning the Effects of Reduced Line Extender OutputReview Questions, Part II

Summarizing dB and dBmVConverting dBmV to Voltage and Power RatiosAdding and Subtracting dB Values Using Power and Voltage Ratios

Answers to Review QuestionsGlossaryExam 217-23-1

Lesson12 Using dB and dBmV

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