Data Services Unit 2 Signals
Signals
1.
2.
Introduction to Signals ..................................................................................................... 2
Scientific Notation ............................................................................................................. 3
2.1
To write a number in scientific notation: ...................................................................... 3
3. Engineering notation ........................................................................................................ 4
3.1
Engineering notation prefix ........................................................................................... 5
4. Frequency ............................................................................................................................ 5
5. Definition of a Sinusoidal Signal .................................................................................... 7
6. Cosine Wave ........................................................................................................................ 9
7. Time Domain and Frequency Domain Representations .......................................... 10
8. Filtering ............................................................................................................................. 12
8.1
Low pass filter .............................................................................................................. 12
8.2
High pass filter............................................................................................................. 13
8.3
Band pass filter ............................................................................................................ 14
8.4
Band stop filter ............................................................................................................ 14
9. Fourier Analysis ............................................................................................................... 15
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Data Services Unit 2 Signals
1. Introduction to Signals
There are various forms of electrical signals that are encountered in communications systems.
Examples include:
Continuous analogue






Discrete digital
Square wave
Sinusoidal
Noise Signal
Pulsed Signal
The continuous analogue signal varies continuously between a minimum level and a maximum
level. It could be representative of a speech or audio signal for example.
The discrete digital signal assumes a finite number of logic levels. A 2 level signal is shown in
the example above. Typically a high voltage corresponds to a logic 1 level and a low voltage as
a logic 0 level. The digital signal could be data from a computer or it might be a digital
representation of an analogue signal that has been passed through an analogue to digital
converter.
The square wave is a 2 level signal that has a 1 0 1 0 1 0 ….. sequence. It could be used as a test
signal or it could represent a data clock.
A sinusoidal signal has a specific frequency and amplitude. It is often used for testing analogue
systems.
A noise signal is random in nature. This signal is found as an interference. It tends to derate the
performance of a system.
A pulsed signal is similar to a square wave except that the duration of the high and low levels
are not equal. This type of signal can be used to produce sampling of analogue signals.
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2. Scientific Notation
Do you know this number, 300,000,000 m/sec.?
It's the Speed of light !
Do you recognize this number, 0.000 000 000 753 kg?
This is the mass of a dust particle!
Scientists have developed a shorter method to express very large numbers. This method is called
scientific notation. Scientific Notation is based on powers of the base number 10.
The number 123,000,000,000 in scientific notation is written as:
The first number 1.23 is called the coefficient. It must be greater than or equal to 1 and less than 10.
The second number is called the base . It must always be 10 in scientific notation. The base number
10 is always written in exponent form. In the number 1.23 x 1011 the number 11 is referred to as the
exponent or power of ten.
2.1 To write a number in scientific notation:
Put the decimal after the first digit and drop the zeroes.
In the number 123,000,000,000 The coefficient will be 1.23
To find the exponent count the number of places from the decimal to the end of the number.
In 123,000,000,000 there are 11 places. Therefore we write 123,000,000,000 as:
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Exponents are often expressed using other notations. The number 123,000,000,000 can also be
written as:
1.23E+11 or as 1.23 X 10^11
For small numbers we use a similar approach. Numbers smaller than 1 will have a negative
exponent. A millionth of a second is:
0.000001 sec. or 1.0E-6 or 1.0^-6 or
3. Engineering notation
Taken from www.purplemath.com
"Engineering" notation is very similar to scientific notation, except that the power on ten can only
be a multiple of three. In this way, numbers are always stated in terms of thousands, millions,
billions, etc. For instance, 13,460,972 is thirteen million and some. In the newspaper, it would
probably be abbreviated as "13.5 million". In engineering notation, you would move the decimal
point six places to get 13.460972 × 106. Once you get used to this notation, you recognize that 106
means "millions", so you would see right away that this is around 13.5 million. Every time a
newspaper refers to some number of millions or billions or trillions, rather than writing out the
whole number with all the zeroes, it is, in effect, using engineering notation.
Express 472,690,128,340 in engineering notation.
This is a twelve-digit number. You need to move the decimal point from the end of the number
toward the beginning of the number, but you must move it in steps of three decimal places. In this
case, you must move the decimal point to between the 2 and the 6, because this will leave nine
digits (and nine is a multiple of 3) after the decimal point, and no more than three digits before the
decimal point. Then the answer is:
472.690128340 × 109, or 472.7 billions.
Express 83,201 in engineering notation.
You need to move the decimal point over to the left in sets of three digits. You can't move the
decimal point any further than to the left of the 2, which is three places, so the answer is:
83.201 × 103, or 83.201 thousands.
Express 0.000 063 8 in engineering notation.
You need to move the decimal point over in sets of three. If you move the decimal point to the right
three places, you'll be left with "0.0638", which won't do. If you move the decimal point to the right
nine places, you'll get "63800", which is too many digits. So move the decimal point six places.
Since this started out as a small number, the power on 10 will be negative, so the answer is:
Copyright © Elizabeth Stapel 2000-2007 All Rights Reserved
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63.8 × 10–6, or 63.8 millionths.
Express 0.397 53 in engineering notation.
Move the decimal point to the right three places. Since this started as a small number, the power on
10 will be negative:
397.53 × 10–3, or 397.53 thousandths.
You should notice that, in engineering notation, it is perfectly okay to have more than one digit to
the left of the decimal point -- in fact, you should expect to have something other than always only
one digit. Just make sure that the power on 10 is a multiple of three.
3.1 Engineering notation prefix
Prefix
Symbol
giga
G
Value
109
Example
Gigahertz GHz)
mega
M
106
megavolt (MV)
kilo
k
103
kilometre (km)
milli
m
10-3
milligram (mg)
micro
μ
10-6
Micrometre (mm)
nano
n
10-9
Nanosecond (ns)
pico
p
10-12
picofarad (pf)
Examples
Write the following in S.I Units using a preferred prefix:
(a) 6000m
(b) 0.005V
(c) 0.0003s
6000m
0.005V
0.0003s
= 6 × 103 m = 6km
= 5 × 10-3V = 5mV
= 300 × 10-6 s = 300μs
In each case the number is written in engineering notation then the
power of ten is replaced by the corresponding prefix.
4. Frequency
Frequency = For a periodic function, the number of cycles or events per unit time.
Hertz = The SI unit of frequency, equal to one cycle per second.
Frequency =
1
Time
1 Second = 1 Hertz
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1 Hertz
1,000 Hertz = 1 kHertz
1,000,000 Hertz = 1,000 kHertz = 1MHertz
1,000,000,000 Hertz = 1,000,000 kHertz = 1,000 MHertz = 1 GHertz
1,000,000,000,000 Hertz = 1,000,000,000 kHertz = 1,000,000 MHertz = 1,000 GHertz = 1 THertz
1 kHertz = 1 kilo Hertz
1 MHertz = 1 Mega Hertz
1 GHertz = 1 Giga Hertz
1 THertz = 1 Tera Hertz
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5. Definition of a Sinusoidal Signal
Figure 5-1 Sine wave of various frequencies. Low at the top and high at the bottom.
A continuous analogue sinusoidal signal is defined by:
s(t )  A sin(2 f1t )  A sin(1t )
The amplitude of this sinusoidal signal or tone is A and the frequency is f1.
The frequency is specified in cycles per second or Hertz. The frequency can also be specified in
radians per second or rads. This is the term 1 where:
1  2 f1t
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1 kHz Tone of amplitude 10 V
10
8
6
Amplitude (V)
4
2
0
-2
-4
-6
-8
-10
0
0.5
1
Time ms
1.5
2
This plots is an example of a single tone of amplitude A=10, and of frequency f1=1 kHz. Note that
the period of this signal is 1 ms.
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6. Cosine Wave
A cosine wave is a signal waveform with a shape identical to that of a sine wave, except each point
on the cosine wave occurs exactly 1/4 cycle earlier than the corresponding point on the sine wave.
A cosine wave and its corresponding sine wave have the same frequency, but the cosine wave leads
the sine wave by 90 degrees of phase. The cosine wave is known as an even function because the
curve is symmetrical about the vertical axis. A sine wave is an odd function because the curve is
skew-symmetrical about the vertical axis.
S(t) = A Cos(2πf1t)
A more general expression is
S(t) = A Cos(2πf1t - τ)
In terms of a phase angle this can be written
S(t) = A Cos(2πf1t + φ)
Where φ = -2π τ/T1
A numerically negative value of φ results in a lagging phase angle, and a positive value a leading
phase angle.
For a Sine wave
τ = T1/4
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7. Time Domain and Frequency Domain Representations
The previous representation of the analogue tone is said to be a time domain representation. It
corresponds to the plot of the amplitude of the signal as a function of time. This plot can be
obtained by computer simulation, or it could also be obtained by using a cathode ray oscilloscope
(CRO) to display a real sinusoidal signal.
Another way to view this signal is in the frequency domain. This plots the amplitude of each
frequency component of a signal against a horizontal axis of frequency rather than time. The
advantage of the frequency domain representation is that it can display clearly the various frequency
components that exist in a signal. The frequency domain representation or spectrum of the single
tone is shown in the diagram below.
AV
Frequency (Hz)
f1
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Consider the following analogue signal:
s(t )  3 sin( 2f1t )  sin( 2 (3 f1 )t )
This signal consists of the addition of 2 frequency components. The time domain and frequency
domain representations of this signal are shown below.
1 kHz Tone + of 3 kHz Tone of amplitude 3 V and 1 V resp
3
2
Amplitude (V)
1
0
-1
-2
-3
0
0.5
1
Time ms
1.5
2
3V
1V
f1
3f1
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Frequency (Hz)
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Data Services Unit 2 Signals
8. Filtering
The process of filtering is to allow various frequencies within a signal to pass through the filter and
to block others. An example of a filter is a camera filter that allows certain light frequencies only to
pass through the camera lens. Typical filter responses are:

Low-pass

High-pass

Band-pass

Band-stop
8.1 Low pass filter
A low-pass filter allows low frequencies through, but removes or attenuates sufficiently medium
and high frequencies. The characteristics of a filter can be shown in an amplitude response plot of
the filter. This is a plot of the amplitude that will occur at the output of the filter for a range of input
frequencies. This is called the filter frequency response. The frequency response is obtained by
applying a sinusoidal signal to the input of the filter and plotting the output level for various
frequencies.
Sinusoidal
Generator
Variable
Filter
Sinusoidal
Output
Digital
Voltmeter
Measuring the frequency response of a filter
Amplitude
Frequency
Low-pass filter frequency response
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Sometimes a filter is assumed to be ideal for the sake of simplifying the analysis of a system. The
ideal low-pass filter for example, passes all frequencies up to its cut-off frequency and completely
rejects all frequencies above this frequency.
Amplitude
Cut-off frequency
Frequency
Ideal Low-pass filter frequency response
Composite Signal
s(t )  3 sin( 2f1t )  sin( 2 (3 f1 )t )
If the composite signal previously analysed consisting of the 2 frequency components of 1 kHz and
3 kHz is connected to the input of an ideal low-pass filter of cut-off frequency 2 kHz or 2f1, then the
output signal will be a single 1 kHz frequency component whose amplitude is 3 V. The output
would therefore be:
s(t )  3sin(2 f1t )
8.2 High pass filter
Ideal filter response
Amplitude
Stops frequencies
below the cutoff
frequency
Allows frequencies above
the cutoff frequency to pass
Frequency
Cutoff frequency
Ideal High-pass filter frequency response
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8.3 Band pass filter
Stops frequencies
either side of fstart
and fstop
Amplitude
Allows freqs
above fstart and
below fstop to pass
Ideal filter response
frequency
fstart
fstop
Ideal Band-pass filter frequency response
8.4 Band stop filter
Allows freqs
below fstart and
above fstop to pass
Amplitude
Stops freqs in
between
Ideal filter response
frequency
fstart
fstop
Ideal Band-Stop filter frequency response
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9. Fourier Analysis
A mathematical process known as Fourier Analysis shows that any waveform can be treated as if it
is made up of a number of sine waves of different frequencies, amplitudes and phase relationships.
For example a square wave which has amplitude Vm and frequency f is made up of a series of sine
waves with frequencies, f, 3f, 5f etc. and can be described by the following equation:
Vm
t
V (t )  4
Vm  1
1
1

sin(
2

f
t
)

sin(
2

3
f
t
)

sin( 2 5 f ot )  .........

o
o
 1
3
5

Where …indicates that the sum goes on for ever.
This can also be written as
V (t )  4
Vm
1
sin( 2kfot )

k odd , k 1

k

A closer approximation to a square wave is obtained by adding in more and more terms and can be
seen in the diagram.
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Data Services Unit 2 Signals
-0.5
41
36
31
21
26
36
41
36
41
36
41
31
26
16
11
21
31
26
21
1
41
7 terms to sin13x/13
0
-0.2
-0.4
-0.6
-0.8
-1
6
6 terms to sin11x / 11
1
0.8
0.6
0.4
0.2
36
31
26
21
16
6
1
11
5 terms to sin 9x/9
1
0.8
0.6
0.4
0.2
16
-1
11
-0.5
sin x+sin3x/3+sin5x/5+sin 7x/7
6
41
36
31
26
21
16
11
6
1
0
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
1
sin x + 1/3 sin 3x + 1/5 sin 5x
0.5
8 terms to sin15x/15
1
0.8
31
26
21
16
11
6
-0.2
-0.4
-0.6
1
41
36
31
26
21
16
11
6
0.6
0.4
0.2
0
1
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
11
-0.6
-0.8
-1
-1
0
-0.2
-0.4
-0.6
-0.8
-1
6
41
36
31
26
21
16
11
6
1
0
0.4
0.2
0
-0.2
-0.4
1
0.5
1
sin x + 1/3 sin3x
1
0.8
0.6
16
sin x
1
-0.8
-1
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A triangular wave can be described by
V (t )  8
Vm  1
1
1

sin(
2

f
t
)

sin(
2

3
f
t
)

sin(
2

5
f
t
)

.........


o
o
o
 2 1
9
25

Or
V (t )  8
Vm
1
sin( 2kfo t )
2 k  odd , k 1 2

k

Vm
t
A sawtooth wave is described by
V(t) = 2 * Vm/ (Sin t - 1/2 Sin 2t + 1/3 Sin 3t - 1/4 Sin 4t + 1/5 Sin 5t + ….)
V (t )  2
Vm  1
1
1
1

 sin( 2f ot )  sin( 2 2 f ot )  sin( 2 3 f ot )  sin( 2 4 f ot )......... 
 1
2
3
4

The diagrams above show the time representation of a signal, or it shows the time domain of the
signal i.e. how the voltage of the signal varies with time. Alternatively we can produce a frequency
diagram of the signal, also called a diagram in the frequency domain, or a frequency spectrum of
the signal. The frequency spectrum of a square wave is shown below. For a simple signal, such as a
square wave both the time and frequency domain diagrams give all of the information about the
signal. For a more complex waveform, such as music, speech or a digitised signal the frequency
domain diagram will give only a very rough impression of the signal in the time domain.
A square wave has an amplitude of 10 volts and a frequency of 1 kHz
Vm
t
In the time domain
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V (t )  4
10  1
1
1

 sin( 2 1000t )  sin( 2 3000t )  sin( 2 5000t )  ......... 
 1
3
5

V (t )  12.73sin( 21000t )  4.24 sin( 2 3000t )  2.54 sin( 2 5000t )  .........
Vm in Volts
12.73 V
4.24 V
2.54 V
1.81 V
f in kHz
In the frequency domain
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