1.
Communication System
Ass. Prof. Ibrar Ullah
BSc (Electrical Engineering)
UET Peshawar
MSc (Communication & Electronics Engineering)
UET Peshawar
PhD (In Progress) Electronics Engineering
(Specialization in Wireless Communication)
MAJU Islamabad
E-Mail: ibrar@cecos.edu.pk
Ph: 03339051548 (0830 to 1300 hrs)
1
3.
Chapter-2
•
•
•
•
•
•
•
•
•
Signals and systems
Size of signal
Classification of signals
Signal operations
The unit impulse function
Correlation
Orthogonal signals
Trigonometric Fourier Series
Exponential Fourier Series
3
4.
Signals and systems
•
A signal is a any time-varying quantity of information or data.
•
Here a signal is represented by a function g(t) of the independent time
variable t. Only one-dimensional signals are considered here.
•
Signals are processed by systems.
•
"A system is composed of regularly interacting or interrelating groups of
activities/parts which, when taken together, form a new whole."
(from Wikipedia)
•
Here a system is an entity that processes an input signal g(t) to produce a
new output signal h(t).
4
5.
Size of Signals
The size of any entity is a number that indicates the largeness or strength of that entity
• Energy
The energy Eg of a signal g(t) can be calculated by the formula
For complex valued signal g(t) it can be written as
The energy is finite only, if
5
6.
Size of Signals (cont…)
• Power
The power Pg of a signal g(t) can be calculated by the formula
For complex valued signal g(t) it can be written as
The power represents the time average(mean) of the signal amplitude squared. It is
finite only if the signal is periodic or has statistical regularity.
6
7.
Size of Signals (cont…)
•
Examples for signals with finite energy (a) and finite power (b):
•
•
Remark:
The terms energy and power are not used in their conventional sense as
electrical energy or power, but only as a measure for the signal size.
7
8.
Example 2.1 Page: 17
•
Determine the suitable measures of the signals given below:
•
The signal (a)
→ 0 as t → ∞ Therefore, the suitable measure for this
signal is its energy
Eg
given by
∞
0
∞
−∞
−1
0
Eg = ∫ g 2 (t )dt = ∫ (2) 2 dt + ∫ 4e −t dt = 4 + 4 = 8
•
8
9.
Example 2.1 Page: 17 (Cont.)
The signal in the fig. Below does not --- to 0 as t ∞ . However it is periodic,
therefore its power exits.
9
13.
Example 2.2 (cont…)
And rms value
We can extent this result to a sum of any number of sinusoids with distinct frequencies.
13
14.
Example 2.2 (cont…)
Recall that
Therefore
The rms value is
14
15.
Classification of signals
1)
2)
3)
4)
5)
6)
Continuous-time and discrete-time signals
Analog and digital signals
Periodic and aperiodic signals
Energy and power signals
Deterministic and random signals
Causal vs. Non-causal signals
15
16.
Classification of signals
Continuous time (CT) and discrete time (DT) signals
CT signals take on real or complex values as a function of an independent
variable that ranges over the real numbers and are denoted as x(t).
DT signals take on real or complex values as a function of an independent
variable that ranges over the integers and are denoted as x[n].
Note the use of parentheses and square brackets to distinguish between CT
and DT signals.
16
17.
Classification of signals
Analog continuous time signal x(t)
Analog discrete time signal x[n]
17
18.
Classification of signals
Digital continuous time signal
Digital discrete time signal
18
19.
Classification of signals
periodic and aperiodic signals
Examples:
19
22.
Classification of signals
Energy and Power Signals
22
23.
Classification of signals
Remarks:
• A signal with finite energy has zero power.
• A signal can be either energy signal or power signal, not both.
• A signal can be neither energy nor power e.g. ramp signal
23
24.
Classification of signals
Deterministic and Random signals
•
A signal g(t) is called deterministic, if it is completely known and
can be described mathematically
•
A signal g(t) is called random, if it can be described only by
terms of probabilistic description, such as
– distribution
– mean value (The average or expected value)
– squared mean value (The expected value of the squared
error)
– standard deviation (The square root of the variance)
24
25.
Classification of signals
Causal vs. Non-causal signals
A causal signal is zero for t < 0 and an non-causal signal is
zero for t > 0 or
A causal signal is any signal that is zero prior to time zero.
Thus, if x(n) denotes the signal amplitude at time (sample) n,
the signal x is said to be causal if x(n)=0 for all n< 0
25
26.
Classification of signals
Right- and left-sided signals
A right-sided signal is zero for t < T and a left-sided signal is
zero for t > T where T can be positive or negative.
26
27.
Classification of signals
Even signals xe(t) and odd signals xo(t) are defined as
xe(t) = xe(−t) and
xo(−t) = −xo(t).
If the signal is even, it is composed of cosine waves. If the signal
is odd, it is composed out of sine waves. If the signal is neither
even nor odd, it is composed of both sine and cosine waves.
27
35.
Signals and Vectors
•
•
•
•
•
•
Analogy between Signals and Vectors
--A vector can be represented as a sum of its components
--A Signal can also be represented as a sum of its components
Component of a vector:
A vector is represented by bold-face type
Specified by its magnitude and its direction.
•
E.g
•
•
•
Vector x of magnitude | x | and Vector g of magnitude | g |
Let the component of vector g along x be cx
Geometrically this component is the projection of g on x
The component can be obtained by drawing a perpendicular from the
tip of g on x and expressed as
g = cx + e
35
36.
Component of a Vector
•
There are infinite ways to express g in terms of x
• g is represented in terms of x plus another vector which is called
the
error vector e
• If we approximate g by cx
36
37.
Component of a Vector (cont..)
The error in this approximation is the vector
e
e = g - cx
The error in the approximation in both cases for last figure
are
37
38.
Component of a Vector (cont..)
•
We can mathematically define the component of vector g along x
•
We take dot product (inner or scalar) of two vectors g and x as:
g.x = | g || x | cos θ
•
The length of vector by definition is
|x|² = x.x
•
The length of component of g along x is
| g | cos θ
Multiply both sides by | x |
c| x |
=
c | x | ² = | g | | x | cos θ =
g.x
38
39.
Component of a Vector (cont..)
Consider the first figure again and expression for c
• Let g and x are perpendicular (orthogonal)
• g has a zero component along x gives c = 0
• From equation g and x are orthogonal if the inner (scalar or
dot) product of two vectors is zero i.e.
g.x = 0
39
40.
Component of a Signal
• Vector component and orthogonality can be extended to signals
• Consider approximating a real signal g(t) in terms of another real
signal x(t)
The error e(t) in the approximation is given by
40
41.
Component of a Signal
• As energy is one possible measure of signal size.
• To minimize the effect of error signal we need to minimize its
size-----which is its energy over the interval
This is definite integral with
This is definite integral with
dummy variable t t
dummy variable
Hence function of cc(not t)
Hence function of (not t)
For some choice of c the energy is minimum
Necessary condition
Necessary condition
41
43.
Component of a Signal
Recall equation for two vectors
•• Remarkable similarity between behavior of
Remarkable similarity between behavior of
vectors and signals. Area under the product of two
vectors and signals. Area under the product of two
signals corresponds to the dot product of two
signals corresponds to the dot product of two
vectors
vectors
••The energy of the signal is the inner product of
The energy of the signal is the inner product of
signal with itself and corresponds to the vector
signal with itself and corresponds to the vector
length squared (which is the inner product of the
length squared (which is the inner product of the
vector with itself)
vector with itself)
43
44.
Component of a Signal
Consider the signal equation again:
t2
1
c=
∫ g (t ) x(t )dt
E x t1
• Signal g(t) contains a component cx(t)
• cx(t) is the projection of g(t) on x(t)
• If cx(t) = 0 ⇒ c = 0
⇒ signal g(t) and x(t) are orthogonal over the interval
[
t1 , t 2
]
44
45.
Example 2.5
Component of a Signal (cont..)
For the square signal g(t), find the component of g(t) of the form
sint or in other words approximate g(t) in terms of sint
g (t ) ≅ c sin t
0 ≤ t ≥ 2π
45
46.
Example 2.5 (cont…)
x(t ) = sin t
and
From equation for signals
c=
⇒
1
c=
π
t2
1
∫ g (t ) x(t )dt
E x t1
π
2π
4
1
∫ g (t ) sin tdt = π ∫ sin tdt + π − sin tdt = π
∫
o
0
2π
g (t ) ≅
4
sin t
π
46
47.
Orthogonality in complex signals
For complex functions of t over an interval
g (t ) ≅ cx(t )
t2
2
Ex = ∫ x(t ) dt
t1
Coefficient c and the error in this case is
e(t ) = g (t ) − cx(t )
t2
Ee = ∫ g (t ) − cx(t )
t1
2
47
48.
Orthogonality in complex signals
2
t2
Ee = ∫ g (t ) − cx(t )
t1
We know that:
(
)
u + v = ( u + v ) u + v = u v + u v + uv
2
t2
2
Ee = ∫ g (t )dt −
t1
1
Ex
t2
∗
∗
2
2
2
∗
1
∫ g (t ) x (t )dt + c Ex − Ex
t1
∗
∗
2
t2
g (t ) x ∗ (t )dt
∫
t1
48
49.
Orthogonality in complex signals
t
1 2
c=
g (t )x ∗ (t )dt
Ex ∫
t1
So, two complex functions are orthogonal over an interval, if
t2
∫ x (t ) x (t )dt = 0
1
∗
2
t1
or
t2
x1∗ (t ) x2 (t )dt = 0
∫
t1
49
50.
Energy of the sum of orthogonal
signals
• Sum of the two orthogonal vectors is equal to the sum of the
lengths of the squared of two vectors. z = x+y then
2
2
z = x + y
2
• Sum of the energy of two orthogonal signals is equal to
the sum of the energy of the two signals. If x(t) and y(t) are
orthogonal signals over the interval, [ t1 ,t 2 ] and if
z(t) = x(t)+ y(t) then
Ez = Ex + E y
50
51.
Correlation
Consider vectors again:
• Two vectors g and x are similar if g has a large component along x
OR
• If c has a large value, then the two vectors will be similar
c could be considered the quantitative measure of similarity between g
and x
But such a measure could be defective. The
But such a measure could be defective. The
amount of similarity should be independent of the
amount of similarity should be independent of the
lengths of g and x
lengths of g and x
51
52.
Correlation
Doubling g should not change the similarity between g and x
Doubling g doubles the value of c
Doubling g doubles the value of c
Doubling x halves the value of c
Doubling x halves the value of c
⇒
However:
However:
c is faulty measure
for similarity
• Similarity between the vectors is indicated by angle between the
vectors.
• The smaller the angle , the largest is the similarity, and vice versa
• Thus, a suitable measure would be c = cos θ , given by
n
g .x
cn = cos θ =
g x
Independent of the lengths of g
and x
52
53.
Correlation
g .x
cn = cos θ =
g x
This similarity measure cn is known as correlation co-efficient.
The magnitude of cn is never greater than unity
− 1 ≤ cn ≥ 1
•Same arguments for defining a similarity index (correlation
co-efficient) for signals
• consider signals over the entire time interval
• normalize c by normalizing the two signals to have unit
∞
1
energies.
c =
g (t ) x(t )dt
n
Eg Ex
∫
−∞
53
54.
Correlation
consider
g (t ) = kx(t )
If k is positive then:
cn = 1
Related signals-------Best friends
Related signals-------Best friends
Negative then:
c n = −1
Dissimilarity worst enemies
Dissimilarity worst enemies
If g(t) and x(t) are orthogonal then
cn = 0
Unrelated signals-------Strangers
Unrelated signals-------Strangers
54
55.
Example 2.6
Find the correlation co-efficient cn between the pulse x(t) and the
pulses g i (t ) =, i = 1,2,3,4,5,6
5
5
0
0
E x = ∫ x 2 (t )dt = ∫ dt = 5
cn =
1
Eg Ex
∞
∫ g (t ) x(t )dt
−∞
Similarly
E g1 = 5
5
1
⇒ cn =
∫ dt = 1
5× 5 0
Maximum possible similarity
Maximum possible similarity
55
56.
Example 2.6 (cont…)
5
5
0
cn =
E g 2 =1.25
0
E x = ∫ x 2 (t )dt = ∫ dt = 5
1
Eg Ex
∞
∫ g (t ) x(t )dt
−∞
5
1
⇒ cn =
∫ (0.5)dt = 1
1.25 × 5 0
Maximum possible similarity……independent of amplitude
Maximum possible similarity……independent of amplitude
56
57.
Example 2.6 (cont…)
5
5
0
0
E x = ∫ x 2 (t )dt = ∫ dt = 5 Similarly
cn =
1
Eg Ex
∞
∫ g (t ) x(t )dt
−∞
E g1 = 5
5
1
⇒ cn =
∫ (1)(−1)dt = −1
5× 5 0
57
58.
Example 2.6(cont…)
5
5
0
0
E x = ∫ x 2 (t )dt = ∫ dt = 5
T
E = ∫ (e
2
− at
0
Here
1
a=
5
T
) dt = ∫ e
E g 4 = 2.1617
0
T =5
− 2 at
1
dt =
(1 − e − 2 aT )
2a
5
−t
5
1
cn =
∫ e dt = 0.961
5 × 2.1617 0
Reaching Maximum similarity
Reaching Maximum similarity
58
61.
Trigonometric Fourier series
Consider a signal set:
{1, cos wot , cos 2wot........ cos nwot ,.... sin wot , sin 2wot.... sin nwot ,....}
•A sinusoid function with frequency nwo is called the nth harmonic
of the sinusoid of frequency w o when n is an integer.
• A sinusoid of frequency
wo
is called the fundamental
•This set is orthogonal over any interval of duration
because:
0
cos nw o t cos mw o tdt = T
∫
To
o 2
n≠ m
n= m≠ o
wo
n≠m
0
sin nw o t sin mw o tdt = T
∫
To
o 2
To = 2π
n=m≠o
61
62.
Trigonometric Fourier series
and
∫ sin nw
o
for all n and m
t cos mw o tdt = 0
To
The trigonometric set is a complete set.
Each signal g(t) can be described by a trigonometric Fourier
series over the interval To :
g ( t ) = a o + a 1 cos w o t + a 2 cos 2 w o t + ...
or
t 1 ≤ t ≤ t 1 + To
+ b1 sin w o t + b 2 sin 2 w o t + ...
∞
g ( t ) = a o + ∑ a n cos nw o t + b n sin nw o t
t 1 ≤ t ≤ t 1 + To
n =1
wn =
2π
To
62
63.
Trigonometric Fourier series
We determine the Fourier co-efficient
Cn =
∫
t 1 +T o
t1
∫
t1
2
an =
To
2
bn =
To
as:
g ( t ) cos nw o tdt
t 1 +T o
1
a0 =
To
ao ,an ,b n
cos 2 nw o tdt
t 1 +T o
∫ g ( t )dt
t1
t 1 +T o
∫ g ( t ) cos nw
o
tdt
t1
t 1 +T o
∫ g ( t ) sin nw
t1
o
tdt
n = 1, 2 , 3 ,......
n = 1, 2 , 3 ,......
63
64.
Compact Trigonometric Fourier series
Consider trigonometric Fourier series
g ( t ) = a o + a 1 cos w o t + a 2 cos 2 w o t + ...
t 1 ≤ t ≤ t 1 + To
+ b1 sin w o t + b 2 sin 2 w o t + ...
It contains sine and cosine terms of the same frequency. We
can represents the above equation in a single term of the same
frequency using the trigonometry identity
a n cos nw o t + b n sin nw o t = C n cos( nw o t + θ n )
Cn =
2
2
a n + bn
− bn
θ n = tan −1
a
n
Co = ao
64
65.
Compact Trigonometric Fourier series
∞
g ( t ) = C 0 + ∑ C n cos( nw o t + θ n )
t 1 ≤ t ≤ t 1 + To
n =1
65
66.
Example 2.7
Find the compact trigonometric Fourier series for the following
function
66
67.
Example 2.7
Solution:
We are required to represent g(t) by the trigonometric Fourier
series over the interval 0 ≤ t ≤ π and To = π
wo =
2π
= 2 rad
sec
To
Trigonometric form of Fourier series:
n =1
a0 ?, an ?, bn ?
π
g (t ) = ao + ∑ an cos 2nt + bn sin 2nt
0 ≤≤
t
∞
67
68.
Example 2.7
π
1 −t 2
a0 = ∫ e dt = 0.50
π 0
C o = ao
π
2 −t
2
a n = ∫ e 2 cos 2 ntdt = 0.504
2
π 0
1 + 16 n
2
2
Cn = an + bn
π
2 −t 2
8n
bn = ∫ e sin 2ntdt = 0.504
2
π 0
1 + 16n
Compact Fourier series is given by
n =1
π
g (t ) = C0 + ∑ Cn cos(nwot + θ n )
0 ≤≤
t
∞
68
69.
Example 2.7
Co = ao = 0.504
Cn = a + b = 0.504
2
n
2
n
64n 2
2
+
= 0.504(
)
2 2
2
(1 + 16n )
1 + 16n
4
(1 + 16n )
2 2
−b
θ n = tan −1 n = tan −1 ( − 4n ) = − tan 4n
a
n
+ 0.084 cos(6t − 85.24o ) + 0.063 cos(8t − 86.42o ) + .......
π
= 0.504 + 0.244 cos(2t − 75.96o ) + 1.25 cos(4t − 82.87 o )
0 ≤≤
t
)
π
(
2
cos 2nt − tan −1 4n
1 + 16n 2
n =1
⇒ g (t ) = 0.504 + 0.504∑
0 ≤≤
t
∞
69
70.
Example 2.7
n
0
1
2
3
4
5
6
7
Cn
0.504
0.244
0.125
0.084
0.063
0.054
0.042
0.063
Өn
0
-75.96
-82.87
-85.24
-86.42
-87.14
-87.61
-87.95
Amplitudes and phases for first seven harmonics
70
71.
Periodicity of the trigonometric Fourier
series
The co-efficient of the of the Fourier series are calculated for the
interval [ t1 , t1 + To ]
∞
φ ( t ) = C o + ∑ C n cos( nw o t + θ n )
for all t
n =1
∞
φ ( t + T 0 ) = C o + ∑ C n [cos( nw o ( t + T 0 ) + θ n ]
n =1
∞
= C + ∑ C cos( nwt + 2 n π + θ )
o
n
o
n
n =1
∞
= C + ∑ C cos( nwt + θ )
o
n
o
n
n =1
for all t
= φ (t )
71
72.
Periodicity of the trigonometric Fourier series
ao =
1
To
∫g (t )dt
To
an =
2
∫g (t ) cosnw otdt
To To
n= 1,2,3,……
bn =
2
∫g (t ) sinnw otdt
To To
n= 1,2,3,……
∫
Means integration over any interval of To
Means integration over any interval of To
To
72
73.
Fourier Spectrum
Consider the compact Fourier series
∞
g (t ) = C0 + ∑ Cn cos(nwot + θ n )
n =1
This equation can represents a periodic signal g(t) of
frequencies: 0(dc), wo ,2wo ,3wo ,....., nwo
Amplitudes: C0 , C1 , C2 , C3,......,Cn
Phases:
0, θ1 , θ 2 ,θ 3 ,.....θ n
73
74.
Fourier Spectrum
Frequency domain description of φ ( t )
cn
vs w
(Amplitude spectrum)
θ vs w (phase spectrum)
Time domain description of
φ(t )
74
75.
Fourier Spectrum
Consider the compact Fourier series
∞
g (t ) = C0 + ∑ Cn cos(nwot + θ n )
n =1
This equation can represents a periodic signal g(t) of
frequencies: 0(dc), wo ,2wo ,3wo ,....., nwo
Amplitudes: C0 , C1 , C2 , C3,......,Cn
Phases:
0, θ1 , θ 2 ,θ 3 ,.....θ n
75
76.
Fourier Spectrum
Frequency domain description of φ ( t )
cn
vs w
(Amplitude spectrum)
θ vs w (phase spectrum)
Time domain description of
φ(t )
76
77.
Example 2.8
Find the compact Fourier series for the periodic square wave
w(t) shown in figure and sketch amplitude and phase spectrum
Fourier series:
∞
w ( t ) = a o + ∑ a n cos nw o t + b n sin nw o t
W(t)=1 only over (-To/4, To/4)
W(t)=1 only over (-To/4, To/4)
and
and
n =1
1
a0 =
To
t 1 +T o
∫ g ( t )dt
t1
1
⇒ a0 =
To
To
∫
To
4
4
1
dt =
2
w(t)=0 over the remaining
w(t)=0 over the remaining
segment
segment
77
78.
Example 2.8
2
an =
To
To
4
∫ cos nw o tdt =
−T o
4
0
2
=
nπ
−2
nπ
2
bn =
To
To
∫
To
2
nπ
sin
nπ
2
n − even
n = 1 , 5 , 9 , 13...
n = 3 , 7 , 11 , 15...
4
4
sin ntdt = 0
⇒ bn = 0
All the sine terms are zero
All the sine terms are zero
78
79.
Example 2.8
w(t ) =
1 2
1
1
1
+ cos w o t − cos 3 w o t + cos 5 w o t − cos 7 w o t + ....
2 π
3
5
7
The series is already in compact form as there are no sine terms
The series is already in compact form as there are no sine terms
Except the alternating harmonics have negative amplitudes
Except the alternating harmonics have negative amplitudes
The negative sign can be accommodated by aaphase of π radians as
The negative sign can be accommodated by phase of
radians as
− cos x = cos( x − π )
Series can be expressed as:
w(t ) =
1 2
1
1
1
1
+ cos w o t + cos( 3 w o t − π ) + cos 5 w o t + cos( 7 w o t − π ) + cos 9 w o t + ....
2 π
3
5
7
9
79
80.
Example 2.8
1
Co =
2
0
−π
θn =
Cn
0
= 2
nπ
n − even
n − odd
for all n≠ 3,5,7,11,15,…..
for all n = 3,5,7,11,15,…..
We could plot amplitude and phase
We could plot amplitude and phase
spectra using these values….
spectra using these values….
In this special case ififwe allow Cnnto
In this special case we allow C to
take negative values we do not need aa
take negative values we do not need
phase of − π to account for sign.
phase of
to account for sign.
Means all phases are zero, so only
Means all phases are zero, so only
amplitude spectrum is enough
amplitude spectrum is enough
80
81.
Example 2.8
Consider figure
w o ( t ) = 2 ( w ( t ) − 0 .5 )
w(t ) =
4
1
1
1
cos w o t − cos 3 w o t + cos 5 w o t − cos 7 w o t + ....
π
3
5
7
81
84.
Example
Consider example 2.7 again, calculate exponential Fourier series
wo =
2π
= 2 rad
sec
To
To = π
∞
ϕ (t ) =
D n e j 2 nt
∑
n = −∞
1
Dn =
To
π
π
1 −t 2 − j 2 nt
− j 2 nt
∫To ϕ ( t )e dt = π ∫ e e dt
0
1
= ∫e
π 0
−(
1
+2 n ) t
2
dt =
84
85.
Example
and
0.504
=
1+ j 4 n
∞
1
ϕ ( t ) = 0.504 ∑
e j 2 nt
n = −∞ 1 + j 4 n
1
1
1
1+
e j 2t +
e j 4t +
e j 6 t + ...
1+ j 4
1+ j 8
1 + j 12
= 0.504
1
1
1
e− j 2t +
e−j 4t +
e − j 6 t + ...
1− j 4
1− j 8
1 + j 12
Dnnare complex
D are complex
Dnnand D-n are conjugates
D and D-n are conjugates
85
86.
Example
1
D n = D −n = C n
2
< D n = θ n and
thus
D n = D n e jθ n
< D −n = − θ n
and
D − n = D n e − jθ n
D o = 0.504
o
0.504
⇒ 0.122 e − j 75.96
1+ j 4
o
0.504
=
⇒ 0.122 e − j 75.96
1− j 4
D1 =
D −1
< D 1 = −75.96 o
< D −1 = 75.96 o
86
87.
Example
o
0.504
⇒ 0.625 e − j 82.87
1+ j 8
0.504
− j 82.87 o
=
⇒ 0.625 e
1− j 8
D2 =
< D 1 = −82.87 o
D −2
< D −1 = 82.87 o
And so on….
87
Be the first to comment