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Engineering and PhysicsUniversity of Central Oklahoma
Dr. Mohamed Bingabr
Ch2Signals and Signal Space
ENGR 4323/5323Digital and Analog Communication
Outline
• Size of a Signal• Classification of Signals• Useful Signals and Signal Operations• Signals Versus Vectors• Correlation of Signals• Orthogonal Signal Sets• Trigonometric and Exponential Fourier Series
Signal Energy and Power
Energy Signal
Power Signal
Energy
Power
𝐸𝐸𝑔𝑔 = �−∞
∞𝑔𝑔(𝑡𝑡) 2𝑑𝑑𝑡𝑡
𝑃𝑃𝑔𝑔 = lim𝑇𝑇→∞
1𝑇𝑇�−𝑇𝑇/2
𝑇𝑇/2|𝑔𝑔 𝑡𝑡 |2𝑑𝑑𝑡𝑡
Signal Classification
1. Continuous-time and discrete-time signals2. Analog and digital signals3. Periodic and aperiodic signals4. Energy and power signals5. Deterministic and probabilistic signals6. Causal and non-causal7. Even and Odd signals
Analog continuous Digital continuous
Analog Discrete Digital Discrete
PeriodicDeterministic Aperiodic Probabilistic
Useful Signal Operation
Time Shifting Time Scaling Time Inversion
1-1
12
t
x(t)
Useful Signals
Unit impulse Signal
Unit step function u(t)𝑑𝑑𝑑𝑑(𝑡𝑡)𝑑𝑑𝑡𝑡
= 𝛿𝛿(𝑡𝑡)
Signals Versus Vectors
By sampling, a continuous signal g(t) can be represented as vector g.
g = [ g(t1) g(t2) … g(tn)]
Vector ApproximationTo approximate vector g using another vector x then we need to choose c that will minimize the error e.
g = cx + e
Remember: Dot product: <g, x> = ||g||.||x|| cos θ
θ θ θ
Signals Versus Vectors
Value of c that minimizes the error
Signal Approximationg(t) = cx(t) + e(t)
x(t)
Correlation of Signals
Two vectors are similar if the angle between them is small.
Correlation coefficient
Note: Similarity between vectors or signals does not depend on the length of the vectors or the strength of the signals.
Example
ρ1=1ρ2=1ρ3=-1
ρ4=0.961ρ5=0.628ρ6=0
Which of the signals g1(t), g2(t), …, g6(t) are similar to x(t)?
Cross-Correlation Function
2
g(t-τ)
13 4
Autocorrelation Function
g(t)
g(t-τ)
Orthogonal Signal Sets
g = c1x1 + c2x2 + c3x3
Orthogonal Vector Space
g(t) = c1x1(t)+ c2x2(t) + … + cNxN(t)Orthogonal Signal Space
Parseval’s Theorem
Trigonometric Fourier Series
( ) ( ) ( )∑ ∑∞
=
∞
=
++=1 1
000 2sin2cosn n
nn tnfbtnfaatx ππ
( ) ( )∫=0
00
2cos2
Tn dttnftx
Ta π
( ) ( )∫=0
00
2sin2
Tn dttnftx
Tb π
( )∫=00
01
T
dttxT
a
Example
• Fundamental periodT0 = π
• Fundamental frequencyf0 = 1/T0 = 1/π Hzω0 = 2π/T0 = 2 rad/s
( ) ( ) ( )
( )
( )
. as amplitudein decrease and 1618 504.0 2sin2
1612 504.0 2cos2
504.0121
2sin2cos
202
202
20
20
10
∞→
+
==
+
==
≈
−−==
++=
∫
∫
∫
∑
−
−
−−
∞
=
nban
ndtnteb
ndtntea
edtea
ntbntaatf
nn
t
n
t
n
t
nnn
π
π
ππ
π
π
ππ0 π−π
1e-t/2
f(t)
( ) ( ) ( )
+
++
+= ∑∞
=122 2sin
16182cos
16121504.0
nnt
nnnt
ntf
To what value does the FS converge at the point of discontinuity?
00 aC =22
nnn baC +=
−= −
n
nn a
b1tanθ
Compact Trigonometric Fourier Series
We can use the trigonometric identitya cos(x) + b sin(x) = c cos(x + θ)
to find the compact trigonometric Fourier series
C0, Cn, and θn are related to the trigonometric coefficients an and bn as:
( ) ( )∑∞
=
++=1
00 2cosn
nn tnfCCtx θπdc component nth harmonics
Role of Amplitude in Shaping Waveform
( ) ( )∑∞
=
++=1
00 2cosn
nn tnfCCtx θπ
Role of the Phase in Shaping a Periodic Signal
( ) ( )∑∞
=
++=1
00 2cosn
nn tnfCCtx θπ
Compact Trigonometric
• Fundamental periodT0 = π
• Fundamental frequencyf0 = 1/T0 = 1/π Hzω0 = 2π/T0 = 2 rad/s
( ) ( )
nab
nbaC
aCn
nb
na
a
ntCCtf
n
nn
nnn
o
n
n
nnn
4tantan
1612504.0
504.01618 504.0
1612 504.0
504.0
2cos
11
2
22
0
2
2
0
10
−−
∞
=
−=
−=
+=+=
==
+
=
+
=
≈
−+= ∑
θ
θ
0 π−π
1e-t/2
f(t)
( ) ( )∑∞
=
−−+
+=1
1
24tan2cos
1612504.0504.0
nnnt
ntf
• The amplitude spectrum of x(t) is defined as the plot of the magnitudes |Cn| versus ω
• The phase spectrum of x(t) is defined as the plot of the angles versus ω
• This results in line spectra• Bandwidth the difference between the
highest and lowest frequencies of the spectral components of a signal.
Line Spectra of x(t)
)( nn CphaseC =∠
Line Spectra
nn
CC
n
n
4tan1612504.0 504.0
1
20
−−=
+==
θ0 π−π
1e-t/2
f(t)
( ) ( )∑∞
=
−−+
+=1
1
24tan2cos
1612504.0504.0
nnnt
ntf
f(t)=0.504 + 0.244 cos(2t-75.96o) + 0.125 cos(4t-82.87o) +0.084 cos(6t-85.24o) + 0.063 cos(8t-86.24o) + …
0.504
0.244
0.1250.084
0.063
Cn
ω0 2 4 6 8 10
ω
θn
-π/2
( ) ∑∞
−∞=
=n
ntfjneDtx 02π
D-n = Dn*
( ) ,....2,1,0 , 102 ±±== ∫ − ndtetx
TD
oT
ntfj
on
π
Exponential Fourier Series
To find Dn multiply both side by and then integrate over a full period, m =1,2,…,n,…∞
ntfje 02π−
Dn is a complex quantity in general Dn=|Dn|ejθ
Even Odd
|Dn|=|D-n| Dn = - D-n
D0 is called the constant or dc component of x(t)
• The line spectra for the exponential form has negative frequencies because of the mathematical nature of the complex exponent.
Line Spectra in the Exponential Form
Cn = 2 |Dn| n ≠ 0
Dn = Cn
...)2cos()cos()(
...||||
||||...)(
2021010
2221
012
2
001
0102
+++++=
++
++++= −−−
−−−
θωθω
ωθωθ
ωθωθ
tCtCCtx
eeDeeD
DeeDeeDtxtjjtjj
tjjtjj
D0= C0
D-n = - Cn
Example
• Fundamental periodT0 = 2π
• Fundamental frequencyf0 = 1/T0 = 1/2π Hzω0 = 2π/T0 = 1 rad/s
π/2−π/2
1f(t)
−π π 2π−2π
Find the exponential Fourier Series for the square-pulse periodic signal.
=−≠
=
=
=
==
= ∫−
−
,15,11,7,3,15,11,7,3 allfor 0
odd /1even 0
21
)2/sinc(5.02/sin
21
0
2/
2/
nn
nnn
D
D
nn
n
dteD
n
n
jntn
πθ
π
πππ
π
π
π
|Dn|
Dn
1 1
1 1
Exponential Line Spectra
Example
=−≠
=
=
=
,15,11,7,3,15,11,7,3 allfor 0
odd 2
even 021
0
nn
nn
nC
C
n
n
πθ
π
π/2−π/2
1f(t)
−π π 2π−2π
The compact trigonometric Fourier Series coefficients for the square-pulse periodic signal.
[ ]∑∞
=
−
−−++=
1
2/)1(
21)1(cos2
21)(
n
nntn
tf ππ
• Fundamental frequencyf0 = 1/T0 = 1/2π Hzω0 = 2π/T0 = 1 rad/s
{ }( ) { }
( )( )
000
sincos
Im2Re2
cDaCb
CaDDDjb
DDDa
nnn
nnn
nnnk
nnnn
==−=
=−=−=
=+=
−
−
θθ
Relationships between the Coefficients of the Different Forms
000
1
22
2
tan
DaCDDC
ab
baC
nn
nn
n
nn
nnn
==∠=
=
−=
+=
−
θ
θ
( )( )
000
5.05.0
5.0
5.0
CaDeCCD
jbaDDjbaD
njnnnn
nnnn
nnn
===∠=
+==
−=∗
−
θθ
ExampleFind the exponential Fourier Series and sketch the corresponding spectra for the impulse train shown below. From this result sketch the trigonometric spectrum and write the trigonometric Fourier Series.Solution
+=
====
=
=
∑
∑
∞
=
∞
−∞=
10
0
000
0
0
0
)cos(211)(
/1||/2||2
1)(
/1
0
0
0
nT
nn
n
tjnT
n
tnT
t
TDCTDC
eT
t
TD
ωδ
δ ω
2T0T0-T0-2T0
)(0
tTδ
1- The function g(t) is absolutely integrable over one period.
∫𝑇𝑇0 𝑔𝑔 𝑡𝑡 𝑑𝑑𝑡𝑡 < ∞
2- The function g(t) can have only a finite number of maxima, minima, and discontinuities in one period.
Dirichlet Conditions for FS Convergence
• Let x(t) be a periodic signal with period T• The average power P of the signal is defined as
• Expressing the signal as
it is also
( ) ∑∞
=
++=1
00 )cos(n
nn tnCCtx θω
∑∞
=
+=1
220 2
nnDDP∑
∞
=
+=1
220 5.0
nnCCP
Parseval’s Theorem
∫−=2/
2/
2)(1 T
Tdttx
TP
Numerical Calculation of Fourier Series
( ) ,....2,1,0 , 10 ±±== ∫ − ndtetg
TD
oT
tjn
on
ω
𝐷𝐷𝑛𝑛 = lim𝑇𝑇𝑠𝑠→0
1𝑇𝑇0
�𝑘𝑘=0
𝑁𝑁0−1
𝑔𝑔 𝑘𝑘𝑇𝑇𝑠𝑠 𝑒𝑒−𝑗𝑗𝑛𝑛𝜔𝜔0𝑘𝑘𝑇𝑇𝑠𝑠 𝑇𝑇𝑠𝑠
𝐷𝐷𝑛𝑛 = lim𝑇𝑇𝑠𝑠→0
1𝑁𝑁0
�𝑘𝑘=0
𝑁𝑁0−1
𝑔𝑔 𝑘𝑘𝑇𝑇𝑠𝑠 𝑒𝑒−𝑗𝑗𝑛𝑛Ω0𝑘𝑘
𝐷𝐷𝑛𝑛 =1𝑁𝑁0
�𝑘𝑘=0
𝑁𝑁0−1
𝑔𝑔 𝑘𝑘𝑇𝑇𝑠𝑠 𝑒𝑒−𝑗𝑗𝑛𝑛Ω0𝑘𝑘
Ω0 = 𝜔𝜔0𝑇𝑇𝑠𝑠
𝑁𝑁0 =𝑇𝑇0𝑇𝑇𝑠𝑠
Matlab Exercise Page 80
T0 = pi; N0 = 256; Ts = T0/N0;t = 0: Ts : Ts*(N0-1); g = exp(-t/2); g(1) = 0.604;Dn = fft(g)/N0;[Dnangle, Dnmag] = cart2pol( real(Dn), imag(Dn));k =0:length(Dn)-1; k = 2*k;subplot(211), stem(k, Dnmag)subplot(212), stem(k,Dnangle)
0 π−π
1e-t/2
g(t)
T0 = πf0 = 1/T0 = 1/π Hzω0 = 2π/T0 = 2 rad/s
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