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
𝐸𝑔=∫− ∞
∞
𝑔2 (𝑡 )𝑑𝑡 𝐸𝑔=∫− ∞
∞
|𝑔 (𝑡)|2𝑑𝑡
𝑃𝑔= lim𝑇 →∞
1𝑇 ∫
−𝑇 / 2
𝑇 / 2
𝑔2 (𝑡 )𝑑𝑡 𝑃𝑔= lim𝑇 →∞
1𝑇 ∫
−𝑇 / 2
𝑇 / 2
¿𝑔 (𝑡 )∨¿2𝑑𝑡 ¿
Signal Classification
1. Continuous-time and discrete-time signals 2. Analog and digital signals 3. Periodic and aperiodic signals 4. Energy and power signals 5. Deterministic and probabilistic signals 6. Causal and non-causal 7. 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
Useful Signals
Unit impulse Signal
Unit step function u(t)𝑑𝑢(𝑡)𝑑𝑡 =𝛿(𝑡 )
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
Dot product: <g, x> = ||g||.||x|| cos 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
1=12=13=-1
4=0.9615=0.6286=0
Which of the signals g1(t), g2(t), …, g6(t) are similar to x(t)?
Cross-Correlation Function
4
g(t-)
Autocorrelation Function
g(t)
g(t-)
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
∫0
00
2cos2
Tn dttnftx
Ta
∫0
00
2sin2
Tn dttnftx
Tb
∫00
01
T
dttxT
a
Example
• Fundamental periodT0 =
• Fundamental frequencyf0 = 1/T0 = 1/ Hzw0 = 2/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
0
1e-t/2
f(t)
122 2sin
16182cos
16121504.0
n
ntn
nntn
tf
To what value does the FS converge at the point of discontinuity?
1 harmonicnth
0
component dc
0 2cosn
nn tnfCCtx
00 aC 22
nnn baC
n
nn a
b1tan
Compact Trigonometric Fourier Series
We can use the trigonometric identity a cos(x) + b sin(x) = c cos(x + )to find the compact trigonometric Fourier series
C0, Cn, and θn are related to the trigonometric coefficients an and bn as:
Role of Amplitude in Shaping Waveform
1
00 2cosn
nn tnfCCtx
Role of the Phase in Shaping a Periodic Signal
1
00 2cosn
nn tnfCCtx
Compact Trigonometric
• Fundamental periodT0 =
• Fundamental frequencyf0 = 1/T0 = 1/ Hzw0 = 2/T0 = 2 rad/s
nab
nbaC
aCn
nb
na
a
ntCCtf
n
nn
nnn
o
n
n
nnn
4tantan
161
2504.0
504.01618 504.0
1612 504.0
504.0
2cos
11
2
22
0
2
2
0
10
0
1e-t/2
f(t)
1
1
24tan2cos
161
2504.0504.0n
nntn
tf
• The amplitude spectrum of x(t) is defined as the plot of the magnitudes |Cn| versus w
• The phase spectrum of x(t) is defined as the plot of the angles versus w
• 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
4tan161
2504.0 504.0
1
20
0
1e-t/2
f(t)
1
1
24tan2cos
161
2504.0504.0n
nntn
tf
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
w0 2 4 6 8 10
w
n
-/2
n
ntfjneDtx 02
D-n = Dn*
,....2,1,0 , 102 ∫ ndtetx
TD
oT
ntfj
on
Exponential Fourier Series
To find Dn multiply both side by and then integrate over a full period, m =1,2,…,n,…
ntfje 02
Dn is a complex quantity in general Dn=|Dn|ej
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
ww
ww
ww
tCtCCtx
eeDeeD
DeeDeeDtxtjjtjj
tjjtjj
D0= C0
D-n = - Cn
Example
• Fundamental periodT0 = 2
• Fundamental frequencyf0 = 1/T0 = 1/2 Hzw0 = 2/T0 = 1 rad/s
/2/2
1f(t)
22
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
|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
/2/2
1f(t)
22
The compact trigonometric Fourier Series coefficients for the square-pulse periodic signal.
1
2/)1(
21)1(cos2
21)(
n
nntn
tf
• Fundamental frequencyf0 = 1/T0 = 1/2 Hzw0 = 2/T0 = 1 rad/s
000
sincos
Im2Re2
cDaCb
CaDDDjb
DDDa
nnn
nnn
nnnk
nnnn
Relationships between the Coefficients of the Different Forms
000
1
22
2
tan
DaCD
DC
ab
baC
nn
nn
n
nn
nnn
000
5.05.0
5.0
5.0
CaDeCCD
jbaDD
jbaD
njnnnn
nnnn
nnn
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
w
w
2T0T0-T0-2T0
)(0
tT
1- The function g(t) is absolutely integrable over one period.
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 w
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 , 102 ∫ ndtetg
TD
oT
ntfj
on
𝐷𝑛= lim𝑇 𝑠→ 0
1𝑇 0
∑𝑘=0
𝑁 0 −1
𝑔 (𝑘𝑇 𝑠 )𝑒− 𝑗𝑛𝜔 0𝑘𝑇𝑠𝑇 𝑠
𝐷𝑛= lim𝑇 𝑠→ 0
1𝑁0
∑𝑘=0
𝑁 0 −1
𝑔 (𝑘𝑇 𝑠 )𝑒− 𝑗𝑛Ω 0𝑘
𝐷𝑛=1𝑁0
∑𝑘=0
𝑁 0 −1
𝑔 (𝑘𝑇 𝑠 )𝑒− 𝑗𝑛Ω 0𝑘
Ω 0=𝜔0𝑇 𝑠
𝑁 0=𝑇 0
𝑇 𝑠
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
1e-t/2
g(t)
T0 = f0 = 1/T0 = 1/ Hzw0 = 2/T0 = 2 rad/s