electronics - justanswer 22, 2015 · the aim of this electronics unit is to develop that basic...

________________________________________________________________________________________

INTRODUCTION________________________________________________________________________________________

The aim of this Electronics unit is to develop that basic understanding of

analogue electronics that has been gained through previous study. Particular

emphasis is placed on the use of manufacturers' data and modern circuit

analysis techniques.

This first lesson investigates the concept of 'noise' in electronic systems: how it

is generated and how the effects of noise can be minimised.

When a sub-circuit or circuit element is introduced into the transmission path

of a system it will usually cause a gain or loss in the signal level. The circuit

element will cause a noise gain or attenuation. The ratio of the signal level to

the noise level will thus be affected. Both signal and noise gain and

attenuation are commonly measured in logarithmic units, namely the decibel

and the neper.

In this lesson we investigate some of the properties and applications of

logarithmic units in electronics.

________________________________________________________________________________________

YOUR AIMS________________________________________________________________________________________

On completion of this lesson you should be able to:

• describe some advantages of using logarithmic units of measurement

• define the decibel and the neper

• perform gain and attenuation calculations using the decibel and neper

• perform calculations using a reference power

• distinguish between power and field gains

• perform calculations in relation to sound levels.

1

Teesside University Open Learning(Engineering)

© Teesside University 2012

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________________________________________________________________________________________

STUDY ADVICE________________________________________________________________________________________

For those students who have previously studied the unit Electrical and

Electronic Principles, they will find much of the work on logarithmic units in

this lesson revision.

Voltage and Current Sources

The symbol on the left is used to

represent a voltage source and that on

the right to represent a current source.

I

V

+

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________________________________________________________________________________________

LOGARITHMIC, FIELD AND POWER QUANTITIES________________________________________________________________________________________

We begin by redefining a few terms:

• Field quantity: quantities such as current (I), voltage (V), magnetic field

(H) that when squared are proportional to power (P ∝ V2, etc.).

Field quantities will be represented by the symbol F.

• Power quantity: power itself or a quantity directly proportional to power.

Examples include luminous intensity (I), sound intensity (J), apparent

electrical power (S) as well as, of course, power (P).

Power quantities will be represented by the symbol P.

• Logarithmic quantity: a quantity that is expressed as the logarithm of

the ratio of two quantities of the same kind. The ratio may be formed by

field or power quantities.

Examples of field ratios include

Examples of power ratios are P

P

Q

Q

S

S2

1

2

1

2

1

, . and

V

V

I

I

Z

Z2

1

2

1

2

1

, . and

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• Attenuation (A) is the property of a transmission path to decrease the

strength of a signal. Causes of signal attenuation include transmission

lines, pads, filters, connectors, etc.

When expressed as a logarithmic quantity,

Attenuation is also referred to as loss.

• Gain (G) is the property of a transmission path to increase the strength of

a signal. Gain is caused by amplifiers and amplifying circuits.

When expressed as a logarithmic quantity,

Gain is also referred to as amplification.

Both gain and attenuation may be expressed as the ratio of field quantities

(e.g. voltage gain) or as the ratio of power quantities (e.g. power gain).

________________________________________________________________________________________

1We are using the notation 'lg' for logarithms in general, 'logb' where b is any base, 'log' for common logs

(log10) and 'ln' for natural logs (loge).

G =⎛⎝⎜

⎞⎠⎟

log .outputinput

A =⎛⎝⎜

⎞⎠⎟

log .inputoutput

1

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________________________________________________________________________________________

LOGARITHMIC UNITS________________________________________________________________________________________

THE DECIBEL

When expressed as a logarithmic ratio in decibels (unit symbol: dB), power

gain is given by:

where the notation ‘log’ indicates that the common logarithm (to the base of

ten) is to be used.

If the power output is less than the power input, we say the signal has been

attenuated. Attenuation (symbol A) is also measured in decibels and is defined

as:

There are several good reasons for using the decibel, some of which we will

now explore.

power attenuation, power in

power outA = 10 log

⎛⎛⎝⎜

⎞⎠⎟

dB

power gainpower outpower in

dB=⎛⎝⎜

⎞⎠⎟

10 log

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Cascaded Systems

FIGURE 1 shows three cascaded elements in a system. The first and last

elements have respective power gain ratios of GP1 = 1000 and GP3 = 50.

The middle stage has a power attenuation of AP2 = 2000.

FIG. 1

The overall power gain ratio is given by the product of the individual gains and

the reciprocals of the attenuations, in this case:

For this example,

If the gains/attenuations were expressed logarithmically (+30, –33 and +17 dB

respectively) then the overall gain is given by:

(This result should be confirmed by taking the logarithm of 25 and multiplying

by 10).

Thus, in a cascaded system if the gains/attenuations are expressed in decibels,

then the overall gain/attenuation is given simply by their algebraic sum.

GPtotal dB= + = +30 33 17 14–

GPtotal = × × =10001

200050 25

G GA

GP PP

Ptotal = × ×12

31

Powerin

Powerout

GP1 AP2 GP3

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Very Large Power Gains

Frequently the power gains in electronics are very large. For example, the

signal power picked up by the aerial of a radio set might be 1 µW and the

output at the speaker 10 W. This gives a gain of 10 × 106. This is concisely

expressed as 70 dB!

Variation of Gain with Frequency

The variation of power gain with frequency is an important consideration in

amplifiers and filters. A gain vs. frequency graph is a convenient way of

showing a system's behaviour. It is common practice to plot the power gain in

decibels. The next example shows one reason why.

FIGURE 2 shows an electronic system consisting of two stages. Stage 1 has a

gain/frequency response of FIGURE 2(a) and stage 2 that of FIGURE 2(b). In

both cases the power gain is expressed as a straight ratio. Our problem is to

determine the overall frequency response of the stages cascaded.

To do this we must find the gain of each stage at some particular frequency and

multiply them together to get the overall gain at that frequency. The result can

be plotted on a new graph, FIGURE 2(c). Note the change of scale of the

vertical axis.

FIGURE 3 shows the same system but when the gain is measured in decibels.

One other subtle change has been made: the frequency axis is now plotted on a

log scale. This is done to maintain linearity: if gain is proportional to

frequency then log (gain) is proportional to log (frequency).

To get the overall response we simply have to add the two graphs together.

This is much easier than attempting to multiply the graphs together and has the

additional advantage that all three responses can be shown on the same set of

axes.

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FIG. 2

Pow

er g

ain

Frequency/Hz

200

400

600

800

1000

1000

800

600

400

200

Pow

er g

ain

Frequency/Hz

200

400

600

800

1000

30

20

10

Stage 2Stage 1

(a) Stage 1 (b) Stage 2

(c) Cascaded stages 1 & 2

Pow

er g

ain

Frequency/Hz

200

400

600

800

1000

10 000

8000

6000

4000

2000

12 000

14 000

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FIG. 3

(c) Cascaded stages 1 & 2

Pow

er g

ain/

dB

Frequency/Hz

(a) Stage 1 (b) Stage 2

Pow

er g

ain/

dB

Frequency/Hz

100010 100 100010 100

Pow

er g

ain/

dB

Frequency/Hz

100010 100

10

20

30

10

20

30

10

20

30

40

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POWER REFERENCE LEVELS

The decibel (dB) as such is simply the logarithm of a ratio of two powers. It

just gives a relative measurement of two powers. It is often convenient to

make the measurement with respect to some fixed power reference level.

The dB(mW)

In electronics a power level of one milliwatt is often used. The dB(mW) is a

logarithmic power level measured relative to 1 mW.

The dB(mW) is defined as:

Thus, 1 watt expressed in dB(mW) is

5 watts expressed in dB(mW) is

0.4 watt expressed in dB(mW) is

0.0001 watt expressed in dB(mW) is

Note the negative result when the power is below the reference level.

10 0 1 10log . –( ) = ( ) dB mW

10 400 26log ( ) = ( ) dB mW

10 5000 37log ( ) = ( ) dB mW

10 1000 30log ( ) = ( ) dB mW

power in dB mWpower1 mW

( ) = ⎛⎝⎜

⎞⎠⎟10 log

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To see the need for the dB(mW) try answering the following question.

FIGURE 4 shows a three stage system having gains of 4, 3 and 6 dB. Calculate the

power out of the final stage if 3 mW is fed into the first stage.

FIG. 4

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________________________________________________________________________________________

+ 4 dB + 3 dB + 6 dB3 mW watt ?

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First we will express the input power (3 mW) in dB(mW):

To find the power at any stage in the system, all that has to be done is to add the input

power, in dB(mW), to the gain of that stage (FIGURE 5):

FIG. 5

To convert the power back into watts, all we need do is divide the dB(mW) by 10 and find

the antilog (inverse log):

output power inv log dB(mW)

10

inv lo

= ⎛⎝⎜

⎞⎠⎟

=

P

gg10

mW

17 8

60 3

.

.

⎛⎝⎜

⎞⎠⎟

=

+ 4 dB + 3 dB + 6 dB4.8 dB(mW) 17.8 dB(mW)

8.8 dB(mW) 11.8 dB(mW)

input power mW mW

dB(mW)= ⎛⎝⎜

⎞⎠⎟ =10

31

4 8log .

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The dB(W)

If in some applications the reference level of one milliwatt is too small then the

reference of one watt can be used.

The dB(W) is defined as:

Thus, 1 watt expressed in dB(W) is

5 watts expressed in dB(W) is

0.4 watt expressed in dB(W) is

0.0001 watt expressed in dB(W) is

10 0 0001 40log . –( ) = ( ) dB W

10 0 4 4log . –( ) = ( ) dB W

10 5 7log ( ) = ( ) dB W

10 1 0log ( ) = ( ) dB W

power in db W logpower1 W

( ) = ⎛⎝⎜

⎞⎠⎟10

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A note on notation . . .

The symbols 'dBm' and 'dBW' are widely used instead of dB(mW) and dB(W)

but this is not to British Standards. There are occasions where 'dBm' and

'dBW' could be ambiguous and we will therefore only use the notation

'dB(mW)' and 'dB(W)'

Equipment in high quality audio systems might have signal level meters

measuring 'VUs', standing for Volume Units. Actually the meter is calibrated

in dB(mW) where one ‘Volume Unit’ is the same as 1 dB(mW). Thus an

output of 20 VU is the same as 20 dB(mW) or 100 mW.

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________________________________________________________________________________________

DECIBELS AND VOLTAGE GAIN________________________________________________________________________________________

With reference to the equivalent circuit of FIGURE 6, we can express the

power input and load power in terms of the input and load voltages:

FIG. 6

Taking the power gain as the ratio of the power in the load to the input power:

GP

P

GV

R

R

V

GV

V

R

R

P

P

P

=

∴ = ×

∴ = ×

L

I

L2

L

I

I2

L2

I2

I

L

VI RI RL

RO

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Expressing the gain in decibels:

For the special case of when the source and load resistances are matched

(RI = RL):

More generally, for a circuit element

matched input and output and having a

voltage gain , then its power

gain is given by:

This is quite a useful result, but is only valid if the two resistances in question

are of the same value. However, the result is so useful that it has become

common practice to express voltage ratios in this way, even when the

resistances concerned are not equal. But remember that when this is done the

result is not a power level (see the discussion on power and field quantities

which follows shortly).

GV

VGP V=

⎛⎝⎜

⎞⎠⎟

= ( )20 20log logO

I

dB .......................... 2( )

GV

VV = O

I

VOVI GV

GV

VP =⎛⎝⎜

⎞⎠⎟

20 log L

I

dB

GV

V

R

R

V

V

P = ×⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

+

10

102

log

log

L2

I2

I

L

L

I

110

20

log

log

R

R

GV

VP

I

L

L

I

dB

⎛⎝⎜

⎞⎠⎟

∴ =⎛⎝⎜

⎞⎠⎟

+ 110 logR

RI

L

dB .....................⎛⎝⎜

⎞⎠⎟

.... 1( )

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Example 1

For the circuit of FIGURE 7 determine the power gain in decibels and the ratio

for when:

(a) RL = 1200 Ω

(b) RL = 600 Ω

FIG. 7

VI 600 Ω RLVO

RO600 Ω

10 3VI

Amplifier

+

20 logV

VO

I

dB⎛⎝⎜

⎞⎠⎟

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Solution

(a) When RL = 1200 Ω,

Applying equation (1):

(i)

(ii)

(b) When RL = 600 Ω,

(i) GV

V

R

RP =⎛⎝⎜

⎞⎠⎟

+⎛⎝⎜

⎞⎠⎟

= ×

20 10

2012

log log

log

L

I

I

L

1101

10600600

3VV

GP

II

dB

×⎛⎝⎜

⎞⎠⎟

+ ⎛⎝⎜

⎞⎠⎟

∴

log

== 54 dB

V VO I= ×12

103

20 2023

1013log log

V

VV

VL

II

I

⎛⎝⎜

⎞⎠⎟

= × ×⎛⎝⎜

⎞⎠⎟

∴ 220 56 5log .V

VL

I

dB⎛⎝⎜

⎞⎠⎟

=

GV

V

R

RP =⎛⎝⎜

⎞⎠⎟

+⎛⎝⎜

⎞⎠⎟

= ×

20 10

2023

log log

log

L

I

I

L

1101

10600

12003V

V

G

II

dB

×⎛⎝⎜

⎞⎠⎟

+ ⎛⎝⎜

⎞⎠⎟

∴

log

PP = 53 5. dB

V VO I= ×23

103

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(ii)

In the latter case, as the source and load resistances are the same, the voltage

ratio and the power gain have the same numerical value.

20 2012

1013log log

V

VV

VL

II

I

⎛⎝⎜

⎞⎠⎟

= × ×⎛⎝⎜

⎞⎠⎟

∴ 220 54logV

VL

I

dB⎛⎝⎜

⎞⎠⎟

=

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________________________________________________________________________________________

HALF POWER________________________________________________________________________________________

The gain of an amplifier (and most other electronic systems) is frequency

dependent. A typical response might be that shown in FIGURE 8. The

response is substantially flat over much of the frequency range but falls off

sharply at either end.

FIG. 8

The frequencies f1 and f2 are the half-power frequencies.

At these frequencies the gain, , falls to half its maximum value.P

P2

1

Frequency

Gai

n 10

dBP

2

P1

⎛ ⎜ ⎝⎞ ⎟ ⎠

f2f1

Bandwidth

3 dB

P1 P2Amplifier

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This represents a fall of:

which is the same as

which simplifies to 10 log 2 dB = 3 dB.

Thus half power represents a fall of 3 dB.

This important observation is used extensively in electronics.

The bandwidth of the amplifier is defined as the frequency range between the

lower and upper half-power frequencies.

Bandwidth = f2 – f1

The half power frequencies are also called the 3 dB points or the corner

frequencies.

10 log dBP

P

PP

2

1

1

2

2

×

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

10 log 10 log dBP

P

P

P2

1

2

1

2⎛⎝⎜

⎞⎠⎟

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

–

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THE NEPER

The neper (unit symbol Np) is an alternative logarithmic unit used to express

the ratio of field quantities (e.g. current and voltage ratios). As the neper is a

ratio, it is dimensionless quantity.

For the circuit element represented by the

box opposite, the voltage gain (or loss) in

nepers is given by

Similarly the current gain (or loss) is given by

If the system is correctly matched so that R1 = R2, then

V

V

I

I

G GI

2

1

2

1

=

=and V

GI

II =⎛⎝⎜

⎞⎠⎟

ln 2

1

Np

GV

VV =⎛⎝⎜

⎞⎠⎟

ln 2

1

Np

V2V1

I1 I2

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The Relationship Between the Neper and the Decibel

The power gain (or loss)

of a system expressed in

decibels is:

However

So that

(We have used the notation GNp for the current and voltage gains to

emphasize that here they are being measured in nepers. Similarly GdB is

used to indicate a gain measured in decibels.)

∴ = ×( ) log dBdBNp NpG e eG GV I10

V

Ve

I

IeG GV I2

1

2

1

= = Np Np and

GV

VG

IV I =

⎛⎝⎜

⎞⎠⎟

=Np Np

Np and ln ln2

1

22

1I

⎛⎝⎜

⎞⎠⎟

Np

GP

P

V I

V I

P =⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

10

10

2

1

2 2

1 1

log

log dBB

log dB∴ = ( )G G GP V I10

V2V1

I1 I2

P1 P2

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For a correctly matched system GV = GI and so

Now log(e) = 0.4343

where GV)Np is the voltage gain expressed in nepers.

Similarly

where GI)Np is the current gain expressed in nepers.

For a correctly matched system:

A matched amplifier has a power gain of 100. Express this gain in:

(a) decibels

(b) nepers.

________________________________________________________________________________________

To convert from nepers to decibels, multiplyy by

To convert from decibe

20

ln 10 8.686

lls to nepers, multiply by ln 10

20 0.1151

G GP I = dB Np dB ....................8 686. ........................... (2b)

∴ = ×

∴ =

dB

dB Np

dB

G G

G

P V

P

20 0 4343

8 6

.

. 886 GV Np dB .............................................. (2a)

G e G ePG

VV = ( ) = ( )

dB Np log logNp10 202

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________________________________________________________________________________________

DECIBELS AND SOUND LEVELS________________________________________________________________________________________

One very widely used application of decibels is in the measurement of sound

levels. You will most likely have seen quoted, for example, that the noise of a

motor bike is 95 dB or that of a jet engine 100 dB or that of a pop concert

amplifier as being 105 dB, etc. What are we to make of these figures?

The use of the decibel in sound level measurements is exactly the same as that

in the measurement of power ratios in electronics. If, for example, we were to

compare the sound output of two loudspeakers, then

Thus if speaker 2 had twice the sound output of speaker 1, then it would have

3 dB more output than speaker 1.

In practice, the level of sound output of a loudspeaker would be measured with

a microphone. A microphone responds to sound pressure, p, a field quantity

and so sound power P is proportional to the square of sound pressure, P ∝ p2.

power ratio = 10 log dBP

P2

1

⎛⎝⎜

⎞⎠⎟

Power gain in decibels, log 100 dPG = ( ) =10 20 BB

Power gain in nepers, PG = × =20 0 1151 2 302. . NNp

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Sound Reference Level

All of the above examples for the noise levels of machines are quoted with

respect to a reference pressure level. This reference is taken as the change in

sound pressure that is on the threshold of human hearing. This level

corresponds to a sound pressure of 20 micropascals (20 µPa). This is an

extraordinary low pressure and shows how sensitive (and therefore delicate!)

an instrument the ear is; as atmospheric pressure is 105 Pa, the threshold of

hearing is at a sound pressure of about a million millionth of atmospheric

pressure. Thus, on this reference scale, a sound level of 0 dB is one that

corresponds to a pressure change of 20 µPa. The notation we use for

expressing decibels to this reference is dB(20 µPa). Thus, a sound of

100 dB(20 µPa) has a power of ten thousand million (1010) of that of the

reference level. With such a large span of sound levels over the audible range,

it can be seen why a logarithmic scale is used!

Sound loudness

When comparing the sound output of two loudspeakers it was tempting to use

the word 'loudness', one speaker is perhaps twice as loud as the other. We have

to be careful here though because a doubling in the sound power output of a

loudspeaker does not produce a doubling in loudness to our ear. The response

of the ear is very non linear and its 'output' approaches saturation at high sound

levels. 'Loudness' is a subjective measure of the perceived sound level of a

listener, to whom a sound power might have to be increased fourfold or more

to make it 'sound' twice as loud.

The situation is further complicated because the ear's response also depends

upon the frequency of the sound. Although humans can hear sounds over the

frequency range of about 20 Hz to 18 kHz, the ear is most sensitive to sound in

the range of 1 kHz to 4 kHz. The graph of FIGURE 9 shows the relationship

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between sound power and perceived loudness for different frequencies, the

lines being of equal loudness. The lowest line shows the threshold of hearing

and from this line it can be seen that at about 4 kHz a power level of

0 dB(20 µPa) can just be heard, whereas the equivalent power at 100 Hz is

about 30 dB(20 µPa).

FIG. 9

20 40 60 100 200 500 1000 5000 10k 20k2000

1301201101009080706050403020100

Frequency (Hz)

Soun

d le

vel/

dB (

rela

tive

to 2

0 µP

a)

Loudness

Thresholdof hearing

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SOUND METERS

A sound meter is an electro-acoustic device for the measuring of acoustic noise

levels. Hand-held sound meters are widely used for measuring noise pollution in

the work place. In such applications the meter is really a 'loudness meter' as it is

designed to have a gain-frequency response similar to that of the human ear.

This is achieved by incorporating a bandpass filter with a suitably shaped pass

band into the meter. FIGURE 10 shows a typical hand-held instrument and

FIGURE 11 gives a block diagram of the essential features of a sound meter.

FIG. 10

FIG. 11

The variations in sound pressure are picked up by a microphone which

converts them into an electrical signal. An input bandpass filter is used to

restrict the frequencies to the audible range. The signal is then amplified and

then passed to a second filter. It is this filter that has a gain-frequency response

that is intended to mimic that of the human ear. After passing through the

second filter, the signal then undergoes further signal processing, which will

Microphone

Bandpassfilter

AmplifierSignal

processingDisplay

Bandpassfilter

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likely include analogue-to-digital conversion (ADC), before being presented as

a numerical reading on the output display. The display will show the measured

audio signal level in decibels.

A-Weighting Filter

The common standard for the filter used to mimic the response of the ear is the

'A-weighted' filter. This filter has the gain-frequency response of that shown in

FIGURE 12. It can be seen that the filter has a maximum response in the

range of 1000 to 6000 Hz, which corresponds to the peak sensitivity of our

ears.

The response is modelled on the rather fearsome equation:

Despite the complexity of the equation, it is still a fairly crude model of the

ear's behaviour. It makes, for example, no attempt to represent the 'wobbles'

present in the curves of equal loudness given in FIGURE 9. Nevertheless, the

A-weighting has been adopted in most countries as the legal standard for noise

measurements made in both public and work places in the protection of

citizens and workers from exposure to damaging levels of noise and premature

deafness. The A-weighted model is the one that has stipulated by the

International Standards Organization (ISO) and therefore adopted by national

standard bodies such as British Standards (e.g. in BS EN 61672-1:20030).

Gf

f fA-scale( ) = ×

+( ) +20

1 872 10

20 6 12 2

8 4

2 2 2log

.

. 000 107 7 737 92 2 2 2 2( ) +( ) +( )⎛

⎝⎜⎜

⎞

⎠⎟⎟f f. .

dB

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FIG. 12 'A-weighted' filter response

When sound level measurements are A-weighted in this way, the decibel unit

often has an 'A' appended, e.g. '27 dBA' or '27 db(A)'. It can be seen that one

of the readings on the instrument in FIGURE 10 is given as 50.0 dBA. (The

other reading is given as 62.4 dBC. The 'C-weighting' is another, less

commonly used, weighted scale.)

20

0

–20

–40

–60

–80

–100

–120

–140

–1601 10 100 1000 10000 100000

f/Hz

Gai

n dB

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________________________________________________________________________________________

NOTES________________________________________________________________________________________

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________________________________________________________________________________________

SELF-ASSESSMENT QUESTIONS________________________________________________________________________________________

1. An amplifier increases the power of a signal from 100 µW at the input to

32 W at the output. Calculate the gain of the amplifier in decibels.

2. A cable attenuates a signal passing down it by 6 dB. If the input power is

50 mW, calculate the output power of the cable.

3. Express a power of 800 mW in dB(mW).

4. Express a signal of power 70 µW in in dB(mW).

5. A signal of level 20 dB(mW) is amplified by a factor one hundred.

Express the amplified signal in db(mW).

6. A signal of 3 dB(mW) is passed through an amplifier with a gain of

28 dB, along a cable with a loss of 6 dB, through a connector with a loss

of 1 dB, along a second cable with a loss of 14 dB and then through a

second amplifier with a gain of 20 dB. Find the final level of the signal in

dB(mW).

7. An amplifier has a voltage gain of 1000 and a current gain of 5.

Determine its power gain in decibels.

8. Express a power of 0 dB(W) in dB(mW).

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9. For the circuit of FIGURE 13, determine

(a) the voltage gain in nepers

(b) the current gain in nepers

(c) the power gain in decibels

(d) the power gain in nepers.

FIG. 13

VI 1 kΩ RL100 Ω

VO

50 Ω

10 3VI

Amplifier

+

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10. FIGURE 14(a) shows the 'front-end' of a sound level meter. Given the

information below calculate the required value of the power gain in

decibels of G2 to give an output power of 20 mW when the sound input

power is 80 dB(20 µPa) at a frequency of 100 Hz.

• The microphone has the response shown in FIGURE 13(b).

• It can be assumed that the microphone's output is frequency

independent over the measured frequency span.

• The bandpass filter has the response of that shown in FIGURE 13(c).

• The amplifier G1 has a gain of 27 dB that is constant over the

operational frequency range.

• The A-weighted filter has the response of that shown in FIGURE 11.

• All system components are matched to a resistance of 50 Ω.

FIG. 14(a)

Microphone

Bandpassfilter

A-weightedfilter

PI POG1 G2

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FIG. 14(b) FIG. 14(c)(

)

Pressure p/mPa

200 400

8

4

Volt

age

V/m

V

Frequency f/kHz

0.1 100

0

–8

20 lo

g

d

B

–12

–4

1 10

VO

VI

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________________________________________________________________________________________

ANSWERS TO SELF-ASSESSMENT QUESTIONS________________________________________________________________________________________

1.

2.

3.

4.

5.

6. Final signal level = 28 – 6 – 1 – 14 + 20

= dB mW

3

30

+

( )

Amplified signal level = 10 log 100

= dB m

20

40

+ ( )

WW( )

PP

P ⎛

⎝⎜⎞⎠⎟

×⎛⎝⎜

⎞⎠⎟dB

ref

= 10 log = 10 log

=

70 101

3–

–– .11 5 dB mW( )

PP

P ⎛

⎝⎜⎞⎠⎟

⎛⎝⎜

⎞⎠⎟dB

ref

= 10 log = 10 log

= d

8001

29 BB mW( )

Attenuation in decibels, = 10 logdBO

I

AP

P ⎛

⎝⎜⎞⎞⎠⎟ ×

⎛⎝⎜

⎞⎠⎟

∴ ×

= 10 log

=

O

O IdB

P

P PA

50 10

10

3–

1106

1050 10 10

0 2

3=

= W

× ×–

.

Gain in decibels = 10 log = 10 logO

I

P

P

⎛⎝⎜

⎞⎠⎟

32100 ××

⎛⎝⎜

⎞⎠⎟10

55

6–

= dB

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7.

8. 0 dB(W) is the same as one watt. Thus

9. (a)

(b)

(c) G G G

G G G

P V I

P V I

=

= ( ) = × ×dB

10 10 667 1000 667log log (( )

= 86 5. dB

IV

IV

V

GI

II

OI

I1000 and = =

+=

10

100 50667

3

=⎛⎝⎜

⎞⎠⎟

=

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

=

NpO

I

I

Iln ln

l

I

I

VV

667

1000

nn .667 1000 13 4×( ) = Np

V V V

GV

VV

O I I

NpO

I

= ×+

=

=⎛⎝⎜

⎞⎠⎟

10100

100 506673

ln ==⎛⎝⎜

⎞⎠⎟

=ln .667

6 5V

VI

I

Np

0 dB W = 10 log 1000

= dB mW

( ) ( )

( )30

GP dB= 10 log

= dB

×( )1000 5

37

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(d)

Note that the relationship cannot be used

in this example as the system is not matched.

10. The input sound level must be converted to its equivalent pressure level.

From FIGURE 13(b), 200 mPa gives a microphone output of 4 mV.

80 20

4

dB 20 Pa20 Pa

20 Pa

µµ

µ

( ) =⎛⎝⎜

⎞⎠⎟

=⎛⎝

log

log

p

p⎜⎜

⎞⎠⎟

=

= ( ) ×

= =

p

p

20 Pa

20 Pa

Pa

µ

µ

µ

10

10

200 000 200

4

4

mmPa

G GP V = dB Np8 686.

Alternatively: Np

Np

G G G

G

P V I

V

= ( )

=

ln

ln exp ×× ( )( )( )∴ = +

=

exp G

G G G

I

P V I

Np

Np Np Np

66 5 13 4 19 9. . .+ = Np

GP = × ×( )

=

Np

Np

ln

.

667 1000 667

19 9

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• In a resitance of 50 Ω, this represents a power of

• At 100 Hz the bandpass filter has a loss of 6 dB.

• At 100 Hz the A-weighted filter has a loss of 20 dB.

• A power output of 20 mW is equivalent to 10 log(20) = 13 dB(mW).

Thus – –

–

35 6 27 13

34 13

47

2

2

2

+ + =

=

=

G

G

G dB

4 10

500 32

10

3 2

3

×( )=

×( ) =

–

–

.

–

µW

or 10 log 0.32 335 dB mW( )

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________________________________________________________________________________________

SUMMARY________________________________________________________________________________________

• In general we can classify quantities such as voltage, current and sound

pressure as:

Field quantities (general symbol F). The square of a field quantity is

proportional to power (P ∝ V2, etc.).

Power quantities (general symbol P). A quantity directly proportional to

power (e.g. apparent electrical power (S), sound intensity (I) and of

course power (P) itself).

Logarithmic quantities. The logarithm of the ratio of two field or two

power quantities.

Power attenuation or loss is given by

• Power gain (or attenuation) can be expressed in terms of voltage gain:

For matched source and load (RI = RL):

G GP V= ( )20 log dB

G GR

RP V= ( ) +⎛⎝⎜

⎞⎠⎟

20 10log log I

L

dB

A =⎛⎝⎜

⎞⎠⎟

10 loginputoutput

dB

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• Logarithmic power levels can be expressed with respect to some reference

level:

• Voltage and current gains can be expressed in nepers:

For a correctly matched system:

To convert from nepers to decibels, multiply by ≈ 8.686

To convert from decibels to nepers, multiply by ≈ 0.1151

• Sound level is measured with respect to a hearing threshold pressure level

of 20 µPa.

Thus

• A-weighting filters are incorporated into sound meters in an attempt to

make them measure 'loudness' as perceived by the human ear, which is

highly non-linear.

Pp =

⎛⎝⎜

⎞⎠⎟

( )dB Pa dB Pa20

2020log

µµ

GV

V

GI

I

V

I

=⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

ln

ln

2

1

2

1

Np

Np

PP

PP

=⎛⎝⎜

⎞⎠⎟

( )

=⎛

dB

dB

mW dB mW

W

101

101

log

log⎝⎝⎜

⎞⎠⎟

( ) dB W

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