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ROYAL SCHOOL OF ARTILLERYBASIC SCIENCE & TECHNOLOGY SECTION

GUNNERY CAREERS COURSES

Properties of Electrical Signals

DWR Properties of Electrical Signals

E05-1 E05_Electrical_Signals.QXD

INTRODUCTION

An electrical signal could be defined as a voltage or

current that conveys some information. This distin-

guishes it from a voltage or current whose primary pur-

pose is to power some equipment and from voltages or

currents that have arisen from noise and interference.

The signal may be amplified to increase its power

(e.g. prior to being connected to a loudspeaker), manip-

ulated to change its balance of frequencies (e.g. tone

control), processed to present its information in another

form (e.g. when a monitor converts electrical signals toa picture) or transmitted to another place (e.g. by cable

between two equipment trailers. Its information content

remains essentially unchanged by these processes.

This handout describes the basic properties of infor-

mation signals in preparation for the work on systems

that use them.

POWER

One important property of a signal is its power. A

signal must have sufficient power for its purpose

and there is no simple answer to the question, “How

much power is required?” When the signal exists tocarry information, only, then the power might be much

less than a millionth of a Watt. When the signal has to

operate a device such as a radar transmitter then a

power of many kilo-Watts might be needed.

In practice, the power of a signal can usually be

increased using an amplifier of some sort. Electrical

power is the product of Volts and Amps - an amplifier

may increase either or both when it amplifies a signal.

The essential requirement is that the required signal is

detectable amongst any other signals that might be

present.

NOISE & INTERFERENCE

Any unwanted signal can be called noise although this

is usually reserved for signals that are generated by

natural processes inside and outside the equipment.

Interference is a signal that has been generated by other

pieces of equipment or electrical devices. Both types of

unwanted signal can obscure the required signal and,

therefore, prevent the use of the information in it.

Sounds of passing traffic can interfere with a con-

versation when the traffic sound becomes so loud that

speech becomes difficult to understand. Hiss and

crackle on old 78 rpm records is a noise that reduces

the effectiveness of the recording. Generally, noisearises from the equipment itself whereas interference

comes from other devices. For effective use of a signal

8 Jun 04

then the power in the signal must usually be much

greater than the combined power of the noise and inter-

ference.

Note that the term ‘noise’ is not limited to audible

signals: noise on a television picture makes it seem

‘speckly’ whilst noise on one of the old types of radar

display (A-Scope) resembles grass (the original display

was green).

SIGNAL TO NOISE RATIO

This is calculated by dividing the power in the signal bythe power in the noise. When the signal level is liable

to vary (e.g. voice and music) then the maximum signal

power is normally used as this gives the best figure.

S/N Ratio = 10 Log Signal Power dB

Noise Power

Analogue television reception (using an ordinary TV

aerial), for example, requires a signal to noise ratio (S/N

Ratio) at the input to the TV set of at least 50 dB. This

means that the signal power is 100,000 times greater

than the noise (and interference) power. Car radios(Stereo FM) require a similar S/N Ratio for good perfor-

mance whilst Compact Disks and other digital recordings

are capable of producing an output with S/N Ratios bet-

ter than 90 dB (signal 109

or 1 000 000 000 times greater

power than the noise). However, all the above examples

are intended for leisure activities where pictures and

music are spoiled by extraneous noises.

For situations where information only is to be con-

veyed then much lower S/N Ratios are possible. Under

some circumstances, a useable signal can be extracted

even when it has less power than the noise. Most peo-

ple can hear a conversation in a noisy crowd by using

the ability of the human auditory system to select those

sounds that it wants to hear and, simultaneously, to

reject others.

The function – rejecting noise – is an important one

as it enables useful information to be extracted from a

signal that, at first sight, appears to be dominated by

noise. It also has advantages in military applications

where jamming is present, as jamming can have similar

characteristics to noise.

When examining the waveform of a sinusoid with

noise then the easiest parameter to measure is the

amplitude of the signal compared to the amplitude of

the noise. However, Signal to Noise Ratio is a ratio of powers not amplitudes, and a different formula is used

when we have amplitudes:



E05-2

S/N Ratio = 20 Log Signal Voltage

Noise Voltage

In Figure One, where a signal of 7 V rms (or 9.9 V

peak) has 1 V rms (or 1.41 V peak) of noise then the

signal to noise ratio is:

S/N Ratio = 20 Log (7 ÷ 1)

= 20 Log 7

= 17 dB

Note that the same result would have been

obtained had we used both peak voltages:

S/N Ratio = 20 Log (9.9 ÷ 1.41)

= 20 Log 7

= 17 dB

Note that it is not possible to improve the S/N Ratio

using an amplifier, as this will not only amplify both sig-nal and noise by the same amount - but also add some

noise of its own. The end result is more noise and a

deterioration in S/N Ratio.

REDUCING INTERFERENCE

Standard means of reducing interference operate in

one of two ways. Firstly, to reduce the amount of

interference that is generated by the equipment that is

causing it and, secondly, by blocking the means of

entry of the interference into the equipment that is

being affected by it.

When you have access to any equipment that iscausing interference then it can be fitted with any of the

following:

• Suppressors: these aim to reduce the interference

at source and are usually capacitors for carbon

brushes or inductors for cables or, perhaps, a com-

bination of the two. (The ‘lump’ in a computer key-

board and monitor cable adds inductance to the

cable to reduce interference.)

• Screens: EM waves cannot pass through conduc-

tors so enclosing the interference-causing equip-

ment in a metal box wil l reduce radiated

interference. Any gaps in the metal screen might

allow some radiation to escape. Cables can also be

screened, using either foil or braid (for flexibility).

• Twisted pair cables: where cables run side-by-

side then pairs can be twisted together to reduce

radiation and cross-talk. This works because the

pair of cables will always have one positive and one

negative - when twisted, an adjacent cable will

alternately be near a positive and a negative. Over a long run of cable then the interference tends to

cancel out. Fibre-optic cables may be used instead


8 Jun 04E05 Electrical Signals.QXD

Figure 1: A Sinusoid of 7 V (rms) with 1 V (rms)of Noise (Clean Wave Shown in Grey)

THE SOURCE OF ELECTRICAL NOISE

The production of electrical noise is linked to the

basic nature of charge and energy. In general,

the energy that produces the noise comes from heat

(remember that room temperature is about

295 Kelvins). There are several ways that electrical

noise is produced and some of the important ones

are listed below:

Shot Noise: When an electric current flows then this

is a flow of electrical charge (usually electrons).

Because the flow consists of individual electrons,

there will be fluctuations in the number that arrive

each second, mill-second, micro-second, etc. This is

because the electrons do not all have the same

energy and so they travel at different speeds. The

arrival of each electron will cause a (small) increase

in the amount of charge. This produces a variation in

current flow that is random and, when it occurs in an

audio system, sounds like the hissing of falling rain.(Rain produces the same sound because of the noise

of impact of individual drops and fluctuations in the

numbers of rain drops arriving on ground.)

Thermal Noise. The electrons in a conductor have

thermal energy and this makes them move at ran-

dom. Although this movement will have an average

value of zero (because, for example, there will just

as many electrons going up as there are going

down) the numbers do not balance exactly due to

the random nature of the movement. This random

movement of charge is a current and it produces avoltage signal wherever there is some resistance. (V

= I × R ). This noise increases with temperature and

bandwidth.

Partition Noise: This type of noise is produced in

thermionic devices (e.g. Klystron, Travelling-Wave

Tube)) where the current can take one of a number of

routes (parallel paths) through the device. Fluctuations

in the division of current cause random noise.

Photon Noise: Similar to the shot noise of electric cur-

rent, this noise arises in thermal imagers and night

sights. When the light levels are very low then each

individual photon contributes to the output signal from

the detector. Random fluctuations in the numbers of

photons cause similar fluctuations in the signal.



E05-3

of copper - these are immune to electrical interfer-

ence.

To reduce the interference that gains entry to your

system then the following steps might be used:

• Equipment screening: your equipment can be fit-

ted with a metal case and, when necessary, particu-

larly sensitive parts of the equipment might be

installed inside smaller metal boxes within the main

box. EM Waves cannot penetrate a metal box - pro-

vided that there are no gaps. (Since the wires must

pass in and out of the box then perfect screening

cannot be achieved.)

• Directional Antennae: ordinary television aerials

usually point towards the transmitter - this means

that any interference from another direction is

reduced as the aerial does not receive it well. Thisapproach obviously does not work when the inter-

ference comes from the same direction as the

transmitter. Some modern antennae are able to

ignore signals that arrive from a particular direction -

they have a ‘steerable-null’ that can be directed at a

source of interference or jamming.

• Cable screening: your cables can be fitted with foil

or braid screens that can block the entry of interfer-

ing signals. Twisted-pair wiring may also be used as

the interference that it picks up tends to cancel out.

Using fibre-optics will give complete immunity to

interference.

Generally, any measurements made on a signal will

be more reliable, more accurate and more repeatable

when the signal to noise ratio increases. This can be as

a result of increasing the signal (e.g. be using more

power) or by decreasing the noise (e.g. by cooling

some or all of the circuits in use or using devices that

are less noisy.)

The amount of information that can be sent down

any communications channel increases as the S/NRatio improves. If you have used a dial-up modem then

you might have noticed how the data-rate varies from,

say, 49 k one day to 39 k the next: this is because the

quality (i.e. S/N Ratio) of the particular route through

the public, switched telephone-network (PSTN) was dif-

ferent on each day. The modems are capable of oper-

ating at a data rate of 56 k bits per second - but few

ever achieve it. Broadband connections use a different

transmission system and do not suffer from this prob-

lem of daily variation. However, they are affected by the

distance between the exchange and the computer, as

the signal gets weaker with distance and, after severalkilo-metres, eventually the signal-to-noise ratio is

degraded and full speed cannot be obtained.

One function of jamming is to degrade a communi-

cations channel by reducing its S/N Ratio. This includes

radar systems, where the radar echo carries informa-

tion about the target - when the S/N Ratio is poor then

the radar becomes progressively less accurate.

NOISE FACTOR

The worsening of S/N ratio that occurs when a signal

passes through any circuit is called its ‘Noise

Factor’. It is found by subtracting the output S/N ratio

from the input S/N ratio (all expressed in dB). Thus, if a

signal with a S/N ratio of 30 dB passes through an

amplifier and emerges with a S/N ratio of 24 dB then

the noise factor of the amplifier would be 30 - 24 =

6 dB. The best possible value is 0 dB - but this is not

achievable in practice.


E05 Electrical Signals.QXD8 Jun 04

THE SOURCE OF ELECTRICAL INTERFERENCE

Whenever there is an acceleration of electrical

charge then an electro-magnetic (EM) wave is

produced. Examples of this include the following:

• Switching on a lamp: this will cause the elec-

trons in the wiring to accelerate as the current

starts. The same effect occurs when the device

is switched off. This can often be heard as a

‘click’ from a nearby radio receiver.

• Electric motors: dc motors are especially bad

as every time a commutator segment passes a

carbon brush then the current stops and starts

(acceleration of electrons).

• Computer signals: these are often on/off

(binary) signals and, because they occur very

rapidly (e.g. 500 MHz) the acceleration of elec-

trons is large.

• Radio or radar transmitters: these are, of

course, designed to produce EM waves and,when these waves find their way into another cir-

cuit then they cause interference (e.g. when your

Hi-Fi picks up transmissions from taxies in the

road nearby).

Interference differs from noise in that it tends to be

more impulsive (i.e. suddenly increasing from nothing

to some large value and then returning to zero again)

whereas noise is random. Interference may occur at a

regular interval or frequency, depending on the device

that is causing it. When several data links or commu-

nications channels share a common cable-duct tomulti-way cable then it is possible for some of the sig-

nal on one to cross-over to an adjacent cable. Cables

that run parallel to each other can have this ‘cross-

talk’ produced by inductive or capacitive coupling

between the cables. As with noise, some forms of

jamming have similar characteristics to interference.



E05-4

BANDWIDTH

This is the difference between the highest and the

lowest frequency of the range of frequencies that

make up a signal or are used in a system. The average

human ear can hear frequencies ranging from about 20

Hz to about 20 kHz so its bandwidth is 20,000 - 20 or

19,800 Hz. This is usually rounded up to 20 kHz.

The bandwidth of a signal is related to its informa-

tion content, the rate at which information is transmitted

and the method used to encode the information. Some

systems use bandwidth less efficiently than others. A

teleprinter, limited by the rate at which the human oper-

ator can press the keys, can operate in a bandwidth of

about 200 Hz. Broadcast television requires a band-

width of around 5.5 MHz, stereo-FM - which sends

audio frequencies between 20 Hz and 15 kHz - uses

250 kHz bandwidth when transmitted (the apparent

‘waste’ of bandwidth is partly responsible for the low

background noise).

The bandwidth of the system indicates the range of frequencies that it accepts and the range of noise and

interference frequencies to which it is susceptible. An

interfering frequency of 600 Hz might not have much

effect on a teleprinter system, because it is outside its

bandwidth, but it would certainly be able to affect a tele-

vision.

To limit the effects of noise and interference then

most systems are designed so that they reject signals

that lie outside the bandwidth of their ‘normal’ signals.

CHANNEL WIDTH

Many communications systems (e.g. television, tele-phone and radio) use transmission links (cable,

radio, fibre-optic) that have limited bandwidth. For

example, the FM (VHF) radio band has a total band-

width, for all signals, of about 25 MHz. The available

bandwidth must be shared out between all the users,

with ‘guard-bands’ sometimes used between each

channel, so that there is a gap between the frequenciy

band used by one channel and that used by the next, to

avoid ‘adjacent channel interference’.

In the band allocated for stereo-VHF, from around 85

MHz to 110 Mhz, there is 25 MHz of bandwidth. This

would allow for about 4 TV channels (6 MHz each) or

about 100 FM-Stereo channels (250 kHz each). Since

the signals can travel hundreds of miles then, once a

radio channel is allocated, no other station may use that

range of frequencies unless its transmitter is located at a

distance significantly greater than the range of the other

station on that frequency. This makes it important to try

to minimise the bandwidth of each channel to enable the

maximum number of radio stations. Local radio stations

minimise this problem by transmitting on much lower

power than the national stations.

Telephone signals are restricted to a total band-

width of 4 kHz (300 - 3400 in use, 0 - 300 and 3400 -

4000 as ‘Guard Bands’ and AM radio signals arrerestricted to a bandwidth of 4.5 kHz, even though the

range of audible frequencies is 20 kHz. This bandwidth

restriction is necessary as it allows more channels

within the allowed band but it does mean that some fre-

quencies are lost from the original sound.

Compression: the signal may be processed before

transmission to identify and reduce redundant informa-

tion. This can be performed by a powerful computer: a

television picture (usually 5.5 MHz) can be reduced to

a bandwidth of around 50 kHz to give a fuzzy, jerky pic-ture that can be sent through a 56 k modem, on the

Internet. A PC, equipped with a Pentium III CPU is



NOISE POWER

The amount of noise depends on a number of fac-

tors, but temperature is usually the significant

one. Other factors include bandwidth, component

design, the materials used for construction and cir-

cuit design.

For thermal noise, the effective noise power (in

Watts) from any resistance at a temperature ‘T’

Kelvin over a bandwidth ‘B’ Hz is:

Noise Power = kTB Watts

The symbol ‘k’ is Boltzman’s Constant - a num-

ber that relates to the amount of energy in a thermal

system. It has a value of 1.38 × 10 –23

JK –1

(Joules

per Kelvin). A temperature in degrees Celsius can

be converted to Kelvins by adding 273.

Example: the antenna of a ground-surveillance

points towards the ground, which has a temperature

of 300 K. If it is designed to work with a receiver thathas a bandwidth of 1 MHz then the thermal noise

power in the antenna would be:

Noise Power = 1.38 × 10 –23

× 300 × 106

= 4.15 × 10 –15

W

The antenna of an air-defence radar points

upwards - where the temperature is much lower

(e.g. 250 K) this means that this part of the noise

would be lower.

Any received radar echo would have to compete

with this noise. Increasing the amount of amplifica-tion would bring no benefits because it would

amplify the noise as much as the signal.

As the signal passes through the various circuits

in a radar, radio, amplifier or whatever, then any

noise generated there is added to the signal. The

result is that the signal to noise ratio (S/N Ratio)

gets worse as the signal proceeds through the sys-

tem.

Noise Factor : if a signal enters a system with a

S/N Ratio of 40 dB and leaves it with a S/N Ratio of

36 dB then the noise factor (sometimes called ‘noise

figure’) is found by subtraction of the two ratios: 40

dB - 36 dB = 4 dB and is a measure of how much

worse the noise gets as the signal passes through.

The best possible factor is 0 dB, when the S/N Ratio

is unchanged



E05-5

required, at the other end, to decode the compressed

data at sufficient speed to produce a moving picture.

Now that broadband is commonly available, videos

compressed using a technique called ‘MPEG’, can betransmitted with data rates of around 128 - 256 kilo-bits

per second. These videos have significantly better

quality than those that could be used with ordinary

modems.

Digital television and DVD movies are also com-

pressed so that more information may be transmitted or

recorded.

NOISE DISTRIBUTION

Although noise is random and, therefore, unpre-

dictable, it is no less unpredictable than other ran-

dom events, We know, for example, that tossing a coinis random - but we also know that about 50% of the

results will be heads. In other words, we can predict the

likelihood of a certain outcome. If we throw two dice

then the chance of a double-six is 1/36 and the chance

of throwing a ‘seven’ is 1/6, for example, but this does

not tell us which throws will result in a double-six or add

up to seven.

If there is an element of random noise in our system

then we can measure its power or Voltage and, from

that, calculate the chance that the noise will exceed

certain limits. It turns out that the distribution, or spread,

of noise values is shaped like a ‘bell’ - the same distrib-

ution as that of many other random events. This is illus-

trated in Figure Two, which is the distribution of noise

voltages for a noise of 1 Watt in a resistor of 1 Ω. This

corresponds to an ‘average’ voltage of 1 V rms.

From the curve, you should be able to see that a

noise voltage of greater than +3V or less than –3V is

very unlikely - in fact the chance of the noise voltage

exceeding those limits is less than 0.5%. Other results

that emerge from an analysis of this curve is that the

noise lies between ±1 Volt for 68% of the time and

between ±2 Volts for 95% of the time. The chance that

the noise would exceed four-times the ‘rms’ value is

less than one in ten-thousand. A radar receiver, listening for echoes from potential

targets, would be receiving these echoes against a

background of noise. If a signal is detected that is twice

the rms level of the noise then there is a 5% chance

that it is false - no echo, just the noise itself which, at

random times, can be bigger than its rms (average)

value. This implies that the radar might receive an aver-

age of five false echoes for every hundred pulses that it

transmitted. The operator would soon give up because

there would be too many false alarms.

Many types of radar receiver monitor both noise

and signal power and then use the above probabilities

to determine the probability that an echo originated

from a real target or was merely noise. A radar system

could monitor the noise level and set a ‘threshold’ (e.g.

five times the noise) with a suitably low probability of

being exceeded by noise. Any signals above that

threshold would be treated as likely targets. This is the

basic operating principle of the Constant, False-Alarm

Rate Radar (CFAR). The exact specifications of thresh-

old levels, etc., will depend on the design of the particu-

lar radar. Knowledge of these data would assist a jammer and, naturally, the specificat ions for military

radars are not released.

This feature of noise, that it obeys simple, statistical

theory, enables us to predict the error rate in a digital

communications channel. A digital (binary) signal is one

that has only two states, zero and one. If a ‘zero’ is rep-

resented by 0 Volts and a ‘one’ by 10 V then, in effect,

any signal bigger than 5 V is recognised as a ‘one’ and

eny signal less than 5 V is recognised as a ‘zero’. If the

rms (average) noise is, say, 1 Volt, then the noise

would have to exceed five-times the rms value to

change a ‘zero’ into a ‘one’ - hence causing an error.The chance that this might happen can be calculated

and comes to about one error in 200,000 bits. If the

noise in a digital system increases then the error rate

increases. However, the error rate remains fairly low for

reasonable values of signal to noise ratio.

EQUIVALENT INPUT NOISE POWER

It is easy to measure the noise at the output of an

amplifier because the noise will have been amplified

and, consequently, wil l be at its highest value.

However, one important parameter of an amplifier is

the smallest input signal that it can process. If we take

the power of the noise that comes from the output of an

amplifier and divide it by the gain of the amplifier then

the result represents the amount of input power that

would produce the output noise. This indicates the min-

imum possible value that could be used for the actual

signal. An input signal smaller than this would emerge

smaller than the noise - probably indetectable over the

noise.

For example, an amplifier that produced 10 mW of

noise at its output and had a gain of 20 dB (100×)

would have an equivalent input noise of 10 mW ÷ 100

or 100 µW. This would not be suitable for the first

amplifier in a radar receiver, where the echo wouldoften be much less than 1 µW.



Figure 2: Noise Distribution Curve



E05-6

THE POWER IN A PULSED SIGNAL

Many signals are intermittent or pulsed in nature

and their voltages and currents do not remain

constant. Previously, in dc and ac theory, methods

of calculating power have used either steady state

(dc) values or rms (ac) values of voltage and current

to calculate the power. For example, when a 12 V

battery supplies 1 A then the power is 12 W and

when a 12 V rms supply provides 1 A rms then the

power is also 12 W.

A lamp that was illuminated for 10 ms and then

switched off for another 10 ms would not appear to

flash because the human eye has ‘persistence of

vision’ that hides the short, dark period. However, the

brightness of the lamp would appear to be less than

usual - it would seem to be operating at half brightness

because it is on for only half the time.

Consider the pulsed signal shown in Figure Three.

When the signal is ‘on’ then it has 10 V and 0.5 A so

the power is 5 W. When the signal is ‘off’ then it haszero voltage and current so the power is zero. The

power when averaged over one cycle of the waveform,

is the effective power of the signal. In this case, the sig-

nal is on for 5 ms out of a possible 25 ms. The ratio of

these two times is called the Duty Cycle of the wave-

form and it is 5/25 or 1/5, 0.2 or 20%.

Duty Cycle: this is defined as the proportion of the

total time for which the pulse is supplying power. It only

applies to pulsed waveforms that switch on and off (e.g.

radar). The equation needed to calculate it is:

Duty Cycle = Time Switched ONTotal Time

During each week, a man might work for 37 hrs out

of a possible 7 × 24 = 168 hrs: his duty cycle is

37 ÷ 168 or about 22%. Duty Cycle is useful because,

for rectangular pulses like those used in computers and

radar, it links the peak power to the average power,

using the formula:

Average Power = Peak Power × Duty Cycle

(Duty Cycle may also be called ‘Duty Factor’)

In the example above, where a pulse with 5 W had

a duty cycle of 20% then the average power is 20% of

the peak power: in this case, 20% of 5 W or 1 W.

This technique is used in most electric cookers to

vary the amount of heat produced by the elements. On

a low setting, the element is switched on for a short

time and hten left off for a longer time. On a high set-

ting, the on-time is increased, increasing the duty-

cycle. The switch that performs this function often

makes an audible clicking noise, as it operates.

Radar transmitters, such as that used in Rapier sys-

tems, might transmit a pulse lasting a few micro-sec-

onds and then wait a few hundred micro-seconds for

an echo. For example, if the radar pulses are 30 kW,

last for 5 µs and are emitted at a rate of 10 000 pulses

per second then we can calculate:

Over one second:

Duty Cycle = Time OnTotal Time

= 10 000 × 5 µs

1 s

= 0.05

= 5%

Note that since there are ten thousand pulses emit-

ted each second and each pulse is five micro-seconds

then the ‘on’ time must be 50 000 µs out of each sec-

ond. To convert a decimal (0.05) to a percentage (5%)

then multiply by 100.

Alternatively, since there are 10 000 pulses per sec-ond then the pulse interval, or time between one pulse

and the next, must be 1 /10 000 second or 100 µs. The

pulse lasts for 5 µs out of a possible 100 µs so the duty

cycle can be calculated using:

Duty Cycle = Time On

Total Time

= Pulse Duration

Pulse Interval

= 5 µs

100 µs

= 0.05

= 5%

Once the duty cycle is known then the average

power can be calculated, as follows:


= 30 kW × 5%

= 1.5 kW

This average power represents the power drawn

from the electrical system that supplies the radar cir-

cuits that generate the transmitted signal. During trans-

mission of the pulse, these circuits generate asignificant amount of heat. During the time between

pulses - which is 95% of the time - the devices have



10 V

0 5 10 3015 2520 35 ms

0.5 A

25 ms

5 ms

Figure 3: A Pulsed Waveform



E05-7

time to cool. This allows them to generate large pulses

that are well beyond their continuous power rating.

Continuous-Wave radars emit what seems to be

much lower power than pulsed radars. However, when

the effects of duty-cycle are taken into account then thepower levels of the two types of radar are often similar

to each other.

DISTORTION

Asignal such as that shown in Figure Four is, clearly,

not a sine-wave. Nevertheless, it bears some

resemblance to a sine-wave. This signal was con-

structed by the simple addition of two sine-waves, with

the following properties:

• Wave One has an amplitude of five units, frequency

of one unit and phase of zero, at time zero.• Wave Two has an amplitude of one unit, frequency

of two units and phase of +90° at time zero.

These two waves are shown in Figure Five. You

can see from Figure Five that the smaller wave is nega-

tive during the positive peak of the larger signal and

negative (again) at the negative peak of the larger sig-

nal. When these two waves are are added then the

positive peaks of the total are smaller than the negative

peaks of the total. The end result is a distorted sine-

wave with one flattened peak and one narrow peak.

This sort of waveform often occurs during signal

processing, when the circuits through which the signal

passes are imperfect. or during transmission through a

cable or through the atmosphere. This change in the

shape of the wave is called ‘Distortion’ and it can take

many forms.

One consequence of distortion in signals, as they

pass through such circuits as amplifiers, is that it intro-duces frequencies in the output that were not present in

the original signal (input). In a radio communications

system, for example, if the operator were listening for a

singal on a frequency of 10 MHz and his radio received

a signal of 5 MHz, from another transmitter, then any

distortion in his receiver could generate a false signal of

10 MHz from the 5 MHz signal. The false signal might

either be mistaken for the real one or, more likely, mask

the real one and prevent its reception.

The distortion often generates a whole series of fre-

quencies, each being an integer (whole-number) multi-

ple of the basic frequency. These signals are called‘Harmonics’. Therefore, a signal that contained only 5

MHz could, if distorted, produce signals of 10, 15, 20

MHz, etc. Simple amplifiers can produce signals that

contain a few percent of distortion; high-quality ampli-

fiers might produce less than 0.01% distortion.

This simple example shows how one complex

waveform can be built up from a mixture of sine-waves.

It turns out that all waveforms are made up of mixtures

of sine-waves. It might not be immediately obvious

what the mixture actually is, but there are mathematical

processes that can be used to determine this. One rea-

son why the simple sine-wave is used to illustrate a sig-

nal is that all signals, of any shape and size, are made

up from a mixture of sine-waves.

Spectrum : the diagram of Figure Six shows

another way of illustrating the mixture of signals that

makes the wave-form of Figure Four. The two, vertical

bars show the relative amounts of each sine-wave that

is required to make the wave-form. This is a much sim-

pler representation than those shown in Figures Four

and Five. Note that the spectrum does not show details

of the phase relationship between the harmonics.



Figure 4: A Complex Wave, made from the Addition of Two Sine Waves

Figure 5: The Two Sine-Waves that Form theWaveform fo Figure Four

4

2

0

1 2

Figure 6:Spectrum of

theWaveform of Figure Four



E05-8

COMPONENTS OF A SIGNAL

Most real signals are not perfect sine waves and

consist of a mixture of sine-waves. A mathemati-

cian called Fourier developed a method by which any

periodic signal (one that has repeating elements, with a

fundamental frequency of repetition) can be broken

down into a number of sine waves. The sine waves

produced are the components or ‘harmonics’ of the

original waveform. In effect, sine waves are the buildingblocks of all other waveforms.

THE SQUARE-WAVE

The square-wave is shown in Figure Seven. The

wave is square when its two halves are of equal

size. (The pulsed waveform of Figure Three would be

called a ‘rectangular’ wave - not a square wave!) A

Fourier Analysis of a square wave reveals that it is

made up of an infinite number of sine waves. The low-

est frequency is the same as the repetition frequency of

the square wave (equal to 1/t) and the harmonics are

all odd-numbered multiples of that; the higher harmon-ics have ever decreasing amplitudes. The even-num-

bered harmonics have zero amplitude - which means

that they do not exist in the square-wave.

In simple terms, a square wave of basic frequency

1 kHz is made up of harmonics at 1 kHz, 3 kHz, 5 kHz,

7 kHz, 9 kHz, etc. This spectrum is illustrated in Figure

Eight. The dotted line joining the tops of the spectrum

lines is a hyperbola because the proportion of each

harmonic is dependant on the inverse of its multiplier.

Thus, the 3rd harmonic (3 kHz in this example) is one-

third of the amplitude (or 1/9th of the power) of the 1st

harmonic (the main frequency of 1 kHz).

Figure Nine, drawn using Excel, shows the result of

adding the first three, non-zero harmonics (frequencies

of 1×, 3× and 5× the fundamental frequency). Compare

this with the graph of Figure Seven, which has all the

harmonics included. One implication of these harmon-

ics is that a bandwidth of many times the basic fre-

quency is needed to transmit square waves without

significant distortion.

TRIANGULAR WAVE

When a waveform is assembled using only even-

numbered harmonics then it is triangular in form,

as shown in Figure Ten. Such waveforms are used

when scanning is required. The waveform is applied to

a drive system and it moves an object from left to right,

or up and down, as the voltage of the triangular wave

alters. Examples of this include:

• The waveform used to control the servo system that

produces the laser grid in Javelin and HVM sys-

tems.

• The waveform used to move the sub-reflector of theRapier tracking radar, when it is seraching for tar-

gets.

• The waveform used to scan a television to produce

the lines that form the picture.

As with any wave, if there is a requirement either to

amplify or process a signal in some other way then the

circuits used must be capable of processing a suffi-

ciently large number of harmonics to avoid distorting

the shape of the wave.



Figure 9: ‘Square’ Wave with First Three Harmonics

4

2

01 3 5 7 9 kHz

Figure 8: Spectrum of a Square Wave

Figure 7: A Square Wave

Figure 10: A Triangular Wave



E05-9

COMPLEX WAVEFORMS

The frequencies of the various harmonics that make

up any waveform are often based on a ‘fundamen-

tal’ or main frequency accompanied by a series of

higher frequencies (harmonics) that are whole-number multiples of that frequency. For the pulsed waveforsm,

that are often encountered in radar and digital systems,

these frequencies can be estimated in a relatively sim-

ple way:

• Fundamental Frequency (Lowest Frequency):

this is related to the duration of one cycle of the

waveform (e.g. the interval marked as ‘25 ms’ in

Figure Eleven). The lowest frequency is the recipro-

cal of this time: 1/(25 ms) or 40 Hz. Note that any

waveform that does not have equal and opposite

polarities will have a Zeroeth harmonic - which isequivalent to the dc value of its average amplitude

(the peak value multiplied by the duty cycle). In the

case of Figure Eleven, the average value would be

5/15 of 10 V or 2 V.

• Spacing Between Harmonics: this is always the

same as the fundamental. Note that some harmonics

might have zero amplitude (in other words they are not

required to produce the waveform). Thus, in Figure

Eleven, the harmonics are 40 Hz, 80 Hz, 120 Hz, 160

Hz, etc. There is usually - in theory - an infinite number

of harmonics, although their amplitude decreases

quite rapidly as the frequency increases. This means

that a circuit that passes the fundamental and some of

the lower harmonics will often give a satsifactory

approximation to the original signal, even though the

higher harmonics are missing.

• Bandwidth: this is the spread of frequencies that

contains most of the power in the signal. It is related

to the time of the smallest element in the waveform

(e.g. the interval marked as ‘5 ms’ in Figure Three).

The bandwidth is the reciprocal of this smallest

interval: in Figure Three, it is 1/(5 ms) or 200 Hz.

This means that the signal shown can retain its

information content within a bandwidth of 200 Hz.

Thus, for the waveform of Figure Three, the har-

monics beyond 200 Hz are present in relatively

small amounts and can, therefore, be ignored.

Consequently, as described above, the two fre-

quencies that are important for the waveform of Figure

Three are 1/(25 ms) or 40 Hz and 1/(5 ms) or 200 Hz.

The signal would, therefore, contain significant

amounts of frequencies of: Zero, 40, 80, 120, 160 and

200 Hz. Any higher frequencies will be present in small

amounts and can be ignored. The complete spectrum

of the wave of Figure Eleven is shown in Figure

Twelve. The levels of the higher harmonics rises and

falls (along the grey line) whilst steadily reducing (along

the dotted line).

Widening the bannwidth to include these higher har-monics will add only a small amount to the main signal

but will also add a large amount to the noise (because

noise is proportional to bancwidth). Consequently, in

many practical situations, the bandwidth is limited as

described.

Thus, the waveform of Figure Elven contains six,

different harmonics as follows:

• DC: zero frequency, +2 V.

• 40 Hz: fundamental frequency.

• 80 Hz: second harmonic (reducing amplitude

compared to the fundamental)

• 120 Hz: third harmonic (even less of this).

• 160 Hz: fourth harmonic.

• 200 Hz: fifth harmonic.

In general, when information about the timing of a

waveform is available - but it is frequency that is

required - the conversion between time and frequency

is very simple:

f = 1/t and t = 1/f

The two times that we have considered have been

the duration of one cycle of the waveform and the dura-tion of its smallest element.



200Hz

Spacing between linesis 40 Hz or 1/(Pulse Interval)

The first null occurs at afrequency of 1/(Pulse Duration)

400Hz

600Hz

800Hz

Freq.0

Figure 12: Full Spectrum of Harmonics of aPulsed Waveform of Pulse Duration 5 ms and

Pulse Interval of 25 ms.

10 V

0 5 10 3015 2520 35 ms

0.5 A

25 ms

5 ms

Figure 11 : A Pulsed Waveform



E05-10

ADDING HARMONICS TO MAKE A PULSE

The diagram of Figure Thirteen shows a graph from

an Excel spreadsheet where four sinusoids have

been added together to produce an approximation to a

pulsed waveform. The sinusoids form the first four in

the spectrum illustrated in Figure Twelve - representing

frequencies of zero, 40 Hz, 80 Hz, 120 Hz, etc., with

amplitudes decreasing as indicated in Figure Twelve.

You will note that the resulting pulses are not exactly

rectangular - because some harmonics are missing.

Also, the missing harmonics cause ripples in betweenthe pulses.

When more harmonics are added then the pulses

become narrower and the ripples reduce. Figure

Fourteen shows the wavefrom obtained when the first

eight harmonics are used. The pulse is narower, rises

to maximum in a shorter time and the ‘ripples’ or ‘side-

lobes’ are smaller.

Since all real signals have a limited bandwidth then

it is not possible to have a pulse that rises from, say,

Zero Volts to 5 five Volts in zero time - the rise must

always take some finite time. Nevertheless, we often

draw pulsed waveforms as if they had a rise-time of

zero, because it represents an ideal pulse. The rise-

time is linked to the bandwidth and wider bandwidths

allow for shorter rise-times. Since the pulse has curved

sides, with no clear beginning and end, practical mea-surements of rise-time are usually made be taking the

time between the 10% and 90% points on the wave-

form (since the zero and one-hundred points are on a

gentle curve and, therefore, difficult to measure with

precision).



Figure 14: Pulsed Wave, Produced by Adding Together the First Eight Harmonics of the Spectrum

Figure 13: Pulsed Wave, Produced by Adding Together the First Four Harmonics of the Spectrum



E05-11

BENEFITS OF DIGITAL SIGNALS

Digital siggnls are square-waves or pulsed waves.

When digital signals are received in the presence

of noise then, provided that the noise is much less than

half the peak height of the digital signal, the noise hasno effect. Compare the waveforms shown in Figures

Fifteen and Sixteen. Both show a signal that has has

one Volt (rms) of noise added to it. The analogue sig-

nal, Figure Fifteen, is of the same basic shape as the

noise. Once contaminated by noise then it cannot be

removed (a bit like putting lime in your lager - both are

liquids and there is no way of getting it out again). A

music signal, like that of Figure Fifteen, that contained

so much noise would be unuseable - the noise would

spoil the sound. A television picture, fromed from a sig-

nal with as much noise as Figure Fifteen would be un-

watchable.The digital signal is rectangular whereas the noise

is sinusoidal, as shown in Figure Sixteen. To extract

data from a digital signal, it is only necessary to know

whether the signal is up or down - i.e. above or below a

reference point. It is clear from the Figure that the pres-

ence of the noise has not impaired the ability to identify

the up and down parts of the digital signal. An almost

perfect signal can be easily extracted from the signal of

Figure Sixteen, simply by determining whether the sig-

nal is up or down. This resistance to small and medium

amounts of noise is one reason why digital signals are

replacing analogue signals in virtually every signal

application. Compare the picture quality of a digital tele-

vision (satellite) against an ordinary, terrestrial televi-

sion. You will easily observe that, on a terrestrial televi-

sion, there is significant twinlkling in areas that contain

uniform colour. This is caused by noise. The corre-

sponding digital picture does, actually, have noise but

the amount is so small that you probably will not be

able to see it.



Figure 15: A Sinusoid of 7 V (rms) with 1 V (rms)of Noise (Clean Wave Shown in Grey)

Figure 16: A Square of 10 V (peak) with 1 V (rms)of Noise (Clean Wave Shown in Grey)



E05-12





E05-13

FORMULAE & TERMS IN THIS HANDOUT

Power = V2

/ R

S/N Ratio = 10 Log Signal Power

Noise Power

S/N Ratio = 10 Log Square of Signal Voltage

Square of Noise Voltage

Bandwidth = Highest Freq - Lowest Freq

Noise Power = kTB Watts

(k = 1.38 × 10-23

JK-1

)

Noise Factor = Input S/N Ratio - Output S/N Ratio

Equivalent Input Noise = Ouput Noise Pwr ÷ Pwr Gain

Duty Cycle = Time Switched ON

Total Time

Duty Cycle = Pulse Duration

Pulse Interval

Duty Cycle = Pulse Duration × PRF


Sine Wave Values = A × Sin ( 2πf t )

Effective Bandwidth = 1

of a Pulse Pulse Duration

Fundamental Frequency = Pulse Repetition Frequency

Harmonic Spacing = 1

Pulse Repetition Frequency



SIGNAL TO NOISE RATIO AND DATA RATE

Shannon’s Law links the maximum number of

bits per second, without error, that can be sent

down a transmission channel to the bandwidth (B)

and the S/N ratio of the channel. The formula, using

logarithms (log) to base ten, is as follows:

Max Data-rate =3.33 × B × Log ( 1 +

S/N )

Examples:

1. A poor-quality telephone line has a bandwidth of

3 kHz and a S/N ration of 20 dB. Its maximum data-

rate is found by using the formula quoted:

20 dB has a numerical value of 100 times.

Max Data-Rate = 3.33 × 3000 × log ( 101 )=3.33 × 3000 × 2

= 19,980 bits per second (19 kb/s)

2. A good-quality telephone line has a bandwidth of

3.3 kHz and a S/N ratio of 45 dB. Its maximum data-

rate is found by using Shannon’s Law:

45 dB has a numerical value of 31,623 times.

Max Dat- Rate = 3.33 × 3300 × log (31,624 )

= 3.33 × 3300 × 4.5

= 49,450 bits per second (49 kb/s)

Shannon’s Law gives the theoretical maximum

data capacity of a line. Since the equipment that is

used to transmit and recieve the data is imperfect

then this figure cannot be achieved in practice.

If you use a modem to connect to the Internet

then you might have noticed that the conenction rate

varies from day to day. This is because your modem

probably gets allocated a different telephone route

each day (by the exhcnage) and, consequently, the

quality differs from day to day. The modem tests the

line at the start of the session to determine its maxi-

mum data-rate.



E05-14

SELF-TEST QUESTIONS

1. A external signal that enters your system and gets

mixed in with the signal that you are trying to receive is

called:

a. Shot noise.

b. Interference.

c. Grass

d. Partition noise.

2. A signal that has been produced by random effects

within your equipment and which gets mixed in with the

signal that you are trying to receive is called:

a. Interference.

b. Jamming.

c. Noise.

d. Harmonic.

3. A signal has a power of 10 W and there is 0.2 W of

noise mixed in with it. The Signal to Noise Ratio is:

a. 17 dB

b. 1.7 dB

c. 50 dB

d. 2 dB

4. A signal of 10 V has a noise of 1 V mixed in with it.The Signal to Noise Ratio is:

a. 10 dB

b. 20 dB

c. 11 dB

d. 9 dB

5. One difference between noise and interference is

that noise:

a. occurs at regular intervals.

b. is random.

c. is mostly positive.

d. is mostly negative.

6. Electrical equipment might be enclosed in a metal

case to reduce the effects of electrical:

a. Thermal noise.

b. Shot noise.

c. Interference.

d. Harmonics.

7. The bandwidth of a signal that contains frequencies

from dc (zero Hz) to 9 kHz is:

a. 4.5 kHz

b. 9 kHz

c. 18 kHz

d. 90 kHz

8. An amplifier with a noise factor of 6 dB is used to

amplify an input signal that has a Signal to Noise Ratio

of 56 dB. The Signal to Noise Ratio of the output signal

will be:

a. 62 dB

b. 56 dB

c. 50 dB

d. 6 dB

9. An amplifier receives a signal with a Signal to Noise

Ratio of 75 dB and produces an amplified output with a

Signal to Noise Ratio of 65 dB. The noise factor of the

amplifier is:

a. 75 dB

b. 65 dB

c. 140 dB

d. 10 dB

10. An amplifier has a power gain of 20 dB and, whenthere is no signal at its input, produces a output power

of 1 mW of noise. The equivalent input noise power is:

a. 10 µW

b. 50 µW

c. 100 mW

d. 20 mW



1 . I n t e r f e r e n c e i s a n e x t e r n a l s i g n a l ( b )

2 . N o i s e i s p r o d u c e d b y r a n d o m e f f e c t s ( c )

3 . S / N = 1 0 L o g ( 1 0 ÷ 0 . 2 ) = 1 7 d B ( a )

4 . S / N = 1 0 L o g ( V 2 2

÷ V 1 2 ) = 1 0 L o g 1 0 0 = 2 0 d B ( b )

5 . N o i s e i s r a n d o m . ( b )

6 . E x t e r n a l i n t e r f e r n c e c a n ’ t p e n e t r a t e m e t a l ( c )

7 . B ’ w i d t h = F m a x - F m i n = 9 k H z ( b )

8 . N o i s e F a c t o r i s s u b t r a c t e d f r o m I / p S / N ( c )

9 . N F = I n p u t S / N - O u t p u t S / N = 7 5 - 6 5 = 1 0 d B ( d )

1 0 . G a i n i s 2 0 d B ( x 1 0 0 ) , E I N = 1 m W ÷ 1 0 0 ( a )

Answers



E05-15

11. A system has an output noise voltage of 2 mV rms.

The probability that the noise voltage will lie outside the

range ± 6 mV (i.e. three times greater than the rms

value) is:

a. 0.5%

b. 32%

c. 68%

d. 99.5%

12. When the bandwidth of a system is doubled then

the thermal noise in the system will usually:

a. double in power.

b. halve in power.

c. remain the same

d. double in voltage.

13. A signal that is on for 15 ms and off for 60 ms has a

duty cycle of:

a. 20%

b. 25%

c. 15%

d. 60%

14. A signal has a peak power of 500 W and operates

on a duty cycle of 10%. Its average power is:

a. 10 W

b. 50 W

c. 5 kW

d. 500 W

15. A signal has an average power of 200 W and a duty

cycle of 1%. Its peak power would be:

a. 1 W

b. 200 W.

c. 20 kW

d. 2 W

16. A radar transmits pulses lasting 4 µs at a rate of

1 000 per second. The duty cycle of this radar is:

a. 0.4%

b. 0.2%

c. 2.5%

d. 4%

17. Using Fourier Analysis, a square wave can be bro-

ken down into:

a. a single harmonic frequency.

b. many even-numbered harmonics

c. many odd-numbered harmonics

d. both odd and even harmonics.

18. A pulse of duration 20 µs that repeats at a rate of

4 000 pulses per second has harmonics that are sepa-

rated by:

a. 4 kHz.

b. 50 kHz

c. 54 kHz

d. 20 kHz

19. A pulse of duration 2 µs that repeats at a rate of 1 500 pulses per second has an effective bandwidth of:

a. 1.5 kHz

b. 6.7 kHz

c. 500 kHz

d. 2 MHz

20. The effective bandwidth of a pulse of duration 10 µs

that repeats at a rate of 10 000 pulses per second has

a spectrum that contains about:

a. 10 harmonics.b. 100 harmonics.

c. even harmonics only.

d. odd harmonics only.



1 1 . C h a n c e o f e x c e e d i n g 3 x a v e r a g e i s 0 . 5 % ( a )

1 2 . N o i s e P w r = 4 k T B . P r o p o r t i o n a l t o ‘ B ’ ( a ) 1 3 . D C = 1 5 ÷ 7 5 ( T i m e O N ÷ T O T A L ) = 2 0 % ( a )

1 4 . A v e P w r = P k P w r × D C = 5 0 0 × 0 . 1 = 5 0 W ( b )

1 5 . P k P w r = A v e P w r ÷ D C = 2 0 0 ÷ 0 . 0 1 = 2 0 k W ( c )

1 6 . P u l s e I n t e r v a l = 1 / 1 0 0 0 = 1 m s . D C = 4 µ s ÷ 1 m s =

0 . 0 0 4 . T i m e s b y 1 0 0 t o g e t p e r c e n t a g e 0 . 4 % ( a )

1 7 . S q u a r e w a v e s h a v e m a n y O D D h a r m o n i c s ( c )

1 8 . S e p a r a t i o n o f h a r m o n i c s = P R F = 4 k H z ( a )

1 9 . E f f B / W = 1 / ( p u l s e d u r ’ n ) = 1 / ( 2 µ s ) = 5 0 0 k H z ( c )

2 0 . E f f B / W = 1 / ( 1 0 µ s ) = 1 0 0 k H z a n d t h e h a r m o n i c s

a r e s p a c e d a t i n t e r v a l s o f 1 0 k H z . T h e r e i s r o o m f o r

1 0 h a r m o n i c s o f 1 0 k H z i n a B / W o f 1 0 0 k H z ( a )

Answers



E05-16


8 Jun 04E05 Electrical Signals QXD

Teaching Objectives Comments

E.05.01 Describe the Basic Properties of Electrical Signals

E.05.01.01 Describe an electrical signal as a voltage or current that

represents information in a system.

E.05.01.02 State that the signal must be recognisable abo ve noise and

interference

E.05.02 Describe the Properties of Electrical Noise

E.05.02.01 Define noise as a random voltage or current produced by

natural processes in the equipment and surroundings.

Difficult to screen.

E.05.02.02 Define the meaning of signal to noise ratio (S/N Ratio)and calculate its value.

E.05.02.03 Describe the formation and properties of various types of

noise

Including shot, thermal, partition, photon.

E.05.02.04 Define the meaning of noise figure for a circuit and

calculate i ts value.

As input S/N – output S/N in dBs

E.05.02.05 Define the term equivalent input noise of a circuit and

calculate its value.

As output noise ÷ power gain

E.05.02.06 State that the average value of a noise voltage or current

is zero.

E.05.02.07 Describe methods of reducing the effects of noise. Can be reduced by averaging, limiting

bandwidth, careful design, cooling.

E.05.03 Describe the Properties of Electrical Interference

E.05.03.01 State that interference is a signal produced by other

electrical equipment and transfers by induction or

radiation.

E.05.03.02 State that interference is usually impulsive and non -

random.

Not necessarily reduced by averaging.

E.05.03.03 Describe methods of reducing interference. Screening, antenna orientation, re duction at

source

E.05.04 Describe the Spectrum of a Signal

E.05.04.01 State that a continuous sine wave consists of a single

frequency.

E.05.04.02 State that any periodic signal is comprised of a number of

sinusoids of different amplitudes, phases and frequencies.

Fourier Analysis, FFT

E.05.04.03 Describe the spectra of common waveforms. Including: square, triangular, pulse

E.05.04.04 State the bandwidth requirements of common signals Including: AM/FM Radio, Telephone, TV,

Digital

E.05.05 Calculate the power in a signal

E.05.05.01 Describe duty cycle as the fraction of the period for

which a pulse is active.

Duty Cycle = pulse duration ÷ pulse

intervalMay also be expressed as a percentage

E.05.05.02 Calculate the average power using peak pwr × DCyc And vice-versa

bst handout e05

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