Staffordshire University
Faculty of Computing, Engineering and Technology
August 2005
Signal Processing
Page 1
PARAMETERS AND DEFINITIONS
CONTENTS
Basic Signal Parameters
Wavelength and Velocity
Prefix And Powers
Decibels
Staffordshire University
Faculty of Computing, Engineering and Technology
Signal Processing
August 2005
Page 2
PARAMETERS AND DEFINITIONS
BASIC SIGNAL PARAMETERS
This section will discuss:
Peak and Peak-to-peak
Average, Root Mean Square (RMS), Mean Square value
Average Power, Normalised Average Power
Decibels, dB, dBm, dBW
Noise Power Spectral Density
Some basic signal parameters are discussed using the periodic waveform illustrated below.
Peak
Peak-to-Peak
Periodic Time T
The Peak-to-Peak amplitude is the maximum amplitude excursion of the signal.
The Peak value is half the peak to peak value.
Peak-to-Peak and peak values may also be to assigned to other parameters. For example peak
frequency deviation in frequency modulation (FM) and frequency shift keying (FSK)
modulation.
The Periodic Time, T, is the time for a periodic signal to go through one complete cycle.
The Frequency, f, of a signal is the number of cycles per second; measured in Hz (Hertz).
The frequency f and periodic time T are related by:
f 
1
Hz
T
Staffordshire University
Faculty of Computing, Engineering and Technology
Signal Processing
August 2005
Page 3
For example a signal with a periodic time T of 1 millisecond has a frequency of 1000 Hz or 1
kHz.
Signals, which convey information, such as speech and data, are non-periodic, i.e. they have
random or non-predictable amplitudes and frequencies. An information-bearing signal may
have a maximum peak-to-peak value and a maximum frequency content.
In this module we will generally use ‘Cosine’ signals. We may write a signal as
V(t) = Vcost
V is the peak value, volts.
 is the angular frequency , radians per second
Also note  = 2f =
2
T
AVERAGE VALUE
The average value of any function of time, f(t), over an interval t1 to t2 is:
Average 
This is saying Average =
t2
1
f (t )dt
t 2  t1 t1
AREA
BASE
v(t)
AREA
t1
t2
BASE
t2 - t1
Staffordshire University
Faculty of Computing, Engineering and Technology
August 2005
Signal Processing
Page 4
For a periodic signal, with periodic time T,
Average 
T
AREA
1
=
f
(
t
)
dt
BASE
T 0
AREA
enables us to determine some averages very easily
BASE
‘by inspection’, and to check others that we may determine by using the equation.
The observation that Average =
The relationship between AREA, BASE and Average is illustrated below.
AREA = Base X Height
Height =
AREA
h = Average
AREA
= Average
BASE
BASE
Example
The average of the waveform shown below is easily found from inspection.

E
0
0
BASE = T
AREA of the signal from 0 to T = E + zero
Average =
AREA E
=
BASE
T
t
Staffordshire University
Faculty of Computing, Engineering and Technology
August 2005
Signal Processing
Page 5
ROOT MEAN SQUARE, RMS
To remember, and calculate, write the function backwards:
f(t)2.
S
Square the function, f(t)
M
Find the mean or average of f(t)2, ie
R
Find the square root,
T
1
f (t ) 2 dt

T0
T
Ie RMS =
1
f (t ) 2 dt

T0
Example
Using the previous example, and solving by inspection for the RMS
We had

E
0
0
BASE = T
t
Step 1, Square the function to give f(t)2.

E2
0
0
BASE = T
t
Staffordshire University
Faculty of Computing, Engineering and Technology
August 2005
Signal Processing
Page 6
Step 2, Find the Mean or Average
AREA of the signal from 0 to T = E2 + zero
Average =
AREA
=
BASE
RMS = E
E 2  0
E 2
=
T
T

T
The Root Mean Square or RMS value of a sinusoidal signal is given by:
VRMS =
V peak
2
 0.7071 V peak
MEAN SQUARE
The Mean Square value of a signal is the RMS squared, i.e.:


MS  RMS 2 ie we proceed as for RMS but don’t find the square root.
T
1
MS =  f (t ) 2 dt
T0
For our previous example,
E 2
MS =
T
The Mean Square value is related to the normalised average power in the signal, as will bed
discussed later
AVERAGE POWER
One way to express power is Power =
Voltage 2
, where R is the load resistance, ohms.
Re sis tan ce
Staffordshire University
Faculty of Computing, Engineering and Technology
August 2005
Signal Processing
Page 7
v(t ) 2
R
For functions of time, the power function
P(t) =
Hence Average Power,
PAV =
Note, the integral ,
PAV =
T
1
P(t ) dt
T 0
T
1
Anythingdt , finds the average of ‘Anything’.
T 0
T
VRMS 2
MeanSquare
1 v(t ) 2
=
=
dt
R
R
T 0 R
NORMALISED AVERAGE POWER
The Normalized Average Power is obtained by Setting R = 1 ohm, i.e. normalizing the
power to a 1ohm load resistance. Thus
Normalized Average Power =
VRMS 
R  1 ohm
2
 VRMS  = Mean Square Value
2
Thus, Normalized Average Power is equivalent to the mean square value.
Normalised Average Power (NAvP) is a very important quantity in signal processing. The
relationship below is very useful for converting from one form to another.
NAvP = Mean Square = (RMS)2.
Power levels in a system are usually measured in dBm (i.e. the power in decibels referred to 1
milliwatt).
A power meter for RF signals gives a reading of power in dBm referred (usually) to 50 ohm
or 75 ohms, i.e. the power meter is calibrated assuming a load resistance of 50 ohm (or 75
ohm).
Staffordshire University
Faculty of Computing, Engineering and Technology
Signal Processing
August 2005
Page 8
WAVELENGTH AND VELOCITY
The wavelength, , of a signal is the distance in metres, between successive peak of the
signal.
The velocity of propagation of a signal, Vp, is the distance travelled by a signal per second.
Velocity, wavelength and frequency are related by:
Vp  f metres per sec ond .
In ‘free space’ the velocity of an electromagnetic signal, (eg. A radio frequency or optical
signal, light) is denoted by c, the velocity of light, where:
c ~ 3 x 10 8 metres per second
A useful way to think of this is 300 metres per microsecond.
Radio waves in free space are assumed to travel at c metres per second.
A radio frequency signal at 1000 MHz (1 GHz) in free space will have a wavelength  of:
8
c 3 x10
   9  0.3metres
f
10
Wavelength  = 30cm at 1GHz
Generally signals are not in free space and the actual velocity of propagation, Vp, is less than
the free space velocity, c.
The velocity of radio signals through the atmosphere is only slightly less than c. The velocity
of a radio frequency signal in a co-axial cable is about 2 x 108 metres per second, i.e about
two-thirds the free space velocity.
The velocity of an RF signal depends on frequency. Thus an information signal, within a
band of frequencies from say FL to FU may suffer from Dispersion or Group Delay
Distortion.
Dispersion is due to the different propagation velocities for difference frequencies in the
same medium. For example if FL travels faster than FU, the signals which start with the
correct timing and phase, travel at different velocities and one component arrives before
another (i.e. the relative timing and phase is changed) causing a distortion called group delay
distortion. Group delay equalizers are required to compensate for this type of distortion.
PREFIX AND POWERS
Staffordshire University
Faculty of Computing, Engineering and Technology
August 2005
Signal Processing
Page 9
A wide range of parameter values (numbers) are encountered in electronics and
telecommunications. It is convenient to express values using powers and prefix’s as shown
below.
Power
Prefix
Symbol
Frequency
Hz
Power
Watts
Voltage
Volts
Current
Amps
Resistance
Ohms
Capacitance
Farads
Inductance
Henrys
109
Giga
G
GHz
GW*
GV*
GA*
G*
GF*
GH*
106
Mega
M
MHz
MW
MV*
MA*
M
MF*
MH*
103
Kilo
k
kHz
kW
kV*
kA
k
kF*
kH*
100=1
-
-
Hz
W
V
A

F*
H*
10-3
Milli
M
mHz
mW
mV
mA
m
mF
mH
10-6
Micro

Hz*
W
V
A
*
F
H
10-9
Nano
n
nHz*
nW
nV
nA
n*
nF
nH
10-12
Pico
P
pHz*
pW
pV
pA
pF
pH
p*
* denotes impractical values. For example GigaFarad and GigaHenry values are completely
impractical.
Staffordshire University
Faculty of Computing, Engineering and Technology
August 2005
Signal Processing
Page 10
DECIBELS – dB – dBm - dBW
Signals can have a wide range of possible values.
In communications for example, signal powers can range from nanowatts (10-12 watts) or less
to kilowatts (103 watts) or more.
This is a range of 1 to 1015 or:
1 to 1,000,000,000,000,000.
Gains (or losses) in communications can range from gains of less than 10-20 (eg the path
‘gain’ of a satellite) to more than 106.
The decibel ‘unit’ allows these types of values to expressed in a more manageable way and
also makes the arithmetic easier.
All ‘decibels’ are relative units and are determined by the relationship
XdB = 10log10(Power Ratio)
Decibels in units of dB
The unit dB expresses the output power of a system relative to the input, ie it is used to
express the gain, or loss, for example of an amplifier
Gains And Loss
Consider an amplifier, with a power gain G, input power PIN and output power POUT.
PIN
POUT
Power Gain G
The gain G as a ratio is G =
The gain expressed in dB is:
POUT
PIN
GdB = 10 log10 
POUT
PIN

Note: that the units are in decibels, dB, which expresses the output relative to the input.
Some gain ratios and the approximate corresponding gain in dB are shown below:
Staffordshire University
Faculty of Computing, Engineering and Technology
August 2005
Signal Processing
Page 11
G(ratio)
GdB
1
2
10
20
100
1000
0dB
3dB
10dB
13dB
20dB
30dB
Some important ‘rules of thumb’ can be deduced.
1)
If the gain increases by a factor of 10, the gain in dB increases by 10dB (we add 10
dB).
i.e.
2)
G1 x 10 = G1dB + 10dB
If the gain increases by a factor of 2, the gain in dB increases by (approximately) 3dB
(we add 3dB).
i.e.
G1 x 2 = G1dB + 3dB
Note, a factor of 2 actually adds 3.0103 dB. A close approximation is to add 3 dB.
Thus a gain of 10 (or 10dB) increased by a factor of 10, to give a gain of 100, gives a gain of
20dB, (10dB + 10dB).
A gain of 20 is 10 x 2 ie (10dB + 3dB) or 13dB.
Consider a gain of 400.
We know a gain of 100 is 20dB.
100 x 2 = 200 gives 20dB + 3dB = 23dB (add 3dB).
200 x 2 = 400 gives 23dB +3dB = 26dB (add 3dB).
Thus a gain of 400 is 26dB
Consider now a cable or element with a loss, i.e. power out is less than power in.
PIN
Loss
Power Loss = L
P
The power gain G is still G = OUT and
PIN
POUT
Power Gain = G
GdB = 10 log10 
POUT
PIN

Staffordshire University
Faculty of Computing, Engineering and Technology
August 2005
Signal Processing
Page 12
Since POUT is less than PIN, GdB will be negative, (logs’ of numbers between 0 and 1 give a
negative result). For example a Gain of 0.1 gives:
GdB = 10 log10 (0.1) = -10dB.
The power loss L is related to the power gain (ratios) by
L=
1  PIN 


G  POUT 
 P 
 LdB = 10 log10  IN 
 POUT 
For example if the gain is 0.1 then the loss is 10.
i.e. LdB = 10 log10 (10) = 10dB (loss)
Thus a gain of –10dB is the same as a loss of +10dB.
Power in decibels, dBm and dBW
Power is an absolute value, decibels are relative values. (The gain in dB is the output relative
to the input).
The dBm unit expresses the power of a signal, relative to 1 milliwatt, measured in a standard
load resistance (50 or 75 or 600). For radio frequencies the standard resistance is
usually 50 or 75.
When measuring power – it is important to make sure the power meter is calibrated with the
same load resistance as the system you are measuring.
For example a power meter with the specification “dBm relative to 1mW in 50 ohm” should
either not be used for a 75 ohm system, or the results will need to be interpreted (calculated) for
the 75 ohm system.
A power meter with a 50 ohm input impedance should not be used to measure the power from a
source with a higher impedance, say 600 ohms, because this will ‘load’ the source and may
damage the source.
Suppose we have a power P(mW).
The power in dBm is:
 PmW  
PdBm  10 log 10 
,
 1mW 
gives the power relative to 1 mW measured in a load resistance RL ohms.
Staffordshire University
Faculty of Computing, Engineering and Technology
Signal Processing
August 2005
Page 13
A power of 100 mW is a power of 20 dBm.
[Strictly this measures that the power is 20dB above 1mW, ie 20dB relative to 1 mW].
The power dBW is similar to dBm, except that the power is measured relative to 1 watt.
Thus, power in dBW is:
 Pwatts 
 in RL.
PdBW = 10 log10 
1
watt


Some values of powers in watts and milliwatts, and the approximate (rule of thumb)
corresponding powers in dBm and dBW are tabulated below:
Power (Watt)
Power (mW)
Power dBm
Power dBW
0.001
1
0dBm
-30dBW
0.01
10
10dBm
-20dBW
0.05
20
13dBm
-17dBW
0.1
100
20dBm
-10dBW
1
1000
30dBm
+0dBW
10
10000
40dBm
+10dBW
20
1
43dBm
+13dBW
100
1 x 105
50dBm
+20dBW
x 104
A power of 1 microwatt (10-6 watts) is - 30dBm or - 60dBW.
Note again, if we increase the power by a factor of 10, we add 10dB. If we increase by a
factor of 2, we add (approx.) 3dB.
The units for power are either dBm or dBW. We cannot ‘say’ a power of 20dB, nor can we
say a gain of 20dBm.
Staffordshire University
Faculty of Computing, Engineering and Technology
August 2005
Signal Processing
Page 14
Application
One of benefits of ‘decibels’ is that we can values, rather than multiply.
Consider the system below, with the parameter values shown.
PIN
G1
PIN = 1mW
G1 = 10
G2
G3
G4
G2 = ½
L=2
G3 = 20
G4 = 100
POUT
As ‘ratios’ POUT = PIN G1 G2 G3 G4
POUT = (1mW)(10)(½)(20)(100) = 10,000mW
or 40dBm
Converting to ‘dB’
PIN = 0dB
G1=10dB
G2=-3dB
G3=13dB
G4=20dB
POUT dBm = 10 log10 (PIN. G1 G2 G3 G4.)
POUT dBm = 10 log10 (PIN (mW)) + log10 G1 + 10 log10 G2 + 10 log10 G3 + 10 log10 G4
= 0dBm + 10dB + (-3dB) + 13dB + 20dB
POUT dBm = 40dBm
Measuring powers in dBm and gains in dB, and adding, is much easier than multiplying.
Staffordshire University
Faculty of Computing, Engineering and Technology
August 2005
Signal Processing
Page 15
EXERCISE
Q1
a)
The average value for a continuous-time function, f(t), over an interval T is
given by:
1
 f (t )dt
T
T
Average value =
o
For the function f(t) = V + V cos t:
b)
(i)
Sketch the waveform of f(t)
(ii)
Determine the RMS and Normalized Average Power.
Determine the RMS and normalized average power for the signals below.
Show clearly how you arrived at your answer.
i)
A pulse waveform.
+V

0V
T
c)
Determine the average and normalised average power for the signal below.
Solve by equation and inspection methods.
VA
0
B
A
- VB
T
Staffordshire University
Faculty of Computing, Engineering and Technology
Signal Processing
Q2
August 2005
Page 16
For the system below, express the power gain of the amplifier and the input and
output power in dB and dBm respectively.
Input Power = 1 mW
Output Power ?
Power Gain
G = 80