elec 360: signals and systemsmbingabr/signals_systems/sigsys...engineering and physics university of...
TRANSCRIPT
![Page 1: ELEC 360: Signals and Systemsmbingabr/Signals_Systems/SigSys...Engineering and Physics University of Central Oklahoma Dr. Mohamed Bingabr Ch6 Continuous-Time Signal Analysis ENGR 3323:](https://reader034.vdocuments.us/reader034/viewer/2022042911/5f429fef1f53732d2c107158/html5/thumbnails/1.jpg)
Engineering and PhysicsUniversity of Central Oklahoma
Dr. Mohamed Bingabr
Ch6Continuous-Time Signal Analysis
ENGR 3323: Signals and Systems
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Outline
• Introduction• Fourier Series (FS) Representation of
Periodic Signals.• Trigonometric and Exponential Form of FS.• Gibbs Phenomenon.• Parseval’s Theorem.• Simplifications Through Signal Symmetry.• LTIC System Response to Periodic Inputs.
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Sinusoidal Wave and phasex(t) = Asin(ωt) = Asin(2π50t)
A
t
T0 = 20 msec
x(t)
A
ttd = 2.5 msec
x(t-0.0025)= Asin(2π50[t-0.0025])= Asin(2π50t-0.25π)= Asin(2π50t-45o)
Time delay td = 2.5 msec correspond to phase shift θ = 45o
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Representation of Quantity using Basis
• Any number can be represented as a linear sum of the basis number {1, 10, 100, 1000}
Ex: 10437 =10(1000) + 4(100) + 3(10) +7(1)
• Any 3-D vector can be represented as a linear sum of the basis vectors {[1 0 0], [0 1 0], [0 0 1]}
Ex: [2 4 5]= 2 [1 0 0] + 4[0 1 0]+ 5[0 0 1]
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Basis Functions for Time Signal• Any periodic signal x(t) with fundamental frequency ω0 can be represented by a linear sum of the basis functions {1, cos(ω0t), cos(2ω0t),…, cos(nω0t), sin(ω0t), sin(2ω0t),…, sin(nω0t)}
Ex:x(t) =1+ cos(2πt)+ 2cos(2 π2t)+ 0.5sin(2π3t)+ 3sin(2πt)
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x(t) =1+ cos(2πt)+ 2cos(2 π2t)+ 3sin(2πt)+ 0.5sin(2π3t)
+ +
+ =
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Purpose of the Fourier Series (FS)
FS is used to find the frequency components and their strengths for a given periodic signal x(t).
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The Three forms of Fourier Series
• Trigonometric Form
• Compact Trigonometric (Polar) Form.
• Complex Exponential Form.
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Trigonometric Form
• It is simply a linear combination of sines and cosines at multiples of its fundamental frequency, f0=1/T.
• a0 counts for any dc offset in x(t).• a0, an, and bn are called the trigonometric Fourier
Series Coefficients. • The nth harmonic frequency is nf0.
( ) ( ) ( )∑ ∑∞
=
∞
=
++=1 1
000 2sin2cosn n
nn ntfbntfaatx ππ
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Trigonometric Form• How to evaluate the Fourier Series Coefficients
(FSC) of x(t)?
( )∫=00
01
T
dttxT
a
To find a0 integrate both side of the equation over a full period
( ) ( ) ( )∑ ∑∞
=
∞
=
++=1 1
000 2sin2cosn n
nn ntfbntfaatx ππ
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Trigonometric Form
( ) ( ) ( )∑ ∑∞
=
∞
=
++=1 1
000 2sin2cosn n
nn ntfbntfaatx ππ
( ) ( )∫=0
00
2cos2
Tn dtntftx
Ta π
To find an multiply both side by cos(2πmf0t) and then integrate over a full period, m =1,2,…,n,…∞
To find bn multiply both side by sin(2πmf0t) and then integrate over a full period, m =1,2,…,n,…∞
( ) ( )∫=0
00
2sin2
Tn dtntftx
Tb π
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Example
• Fundamental periodT0 = π
• Fundamental frequencyf0 = 1/T0 = 1/π Hzω0 = 2π/T0 = 2 rad/s
( ) ( ) ( )
( )
( )
. as amplitudein decrease and 1618 504.0 2sin21612 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)
( ) ( ) ( )( )
+
++= ∑
∞
=12 2sin42cos
16121504.0
nntnnt
ntf
To what value does the FS converge at the point of discontinuity?
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Simplifications Through Signal Symmetry
• If x (t) is EVEN: It must contain Cosine Terms and it may contain DC. Hence bn = 0.
• If x(t) is ODD: It must contain ONLY Sines Terms. Hence a0 = an = 0.
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A periodic signal x(t), has a Fourier series if it satisfies the following conditions:1. x(t) is absolutely integrable over any period,
namely
2. x(t) has only a finite number of maxima and minima over any period
3. x(t) has only a finite number of discontinuitiesover any period
Dirichlet Conditions
∫ ∞<0
)(T
dttx
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• Using single sinusoid,
• are related to the trigonometric coefficients anand bn as:
( )
( )∑∞
=
++=1 harmonicnth
0component dc
0 2cosn
nn ntfCCtx θπ
00 aC =
nnC θ and ,
and22nnn baC +=
−= −
n
nn a
b1tanθ
Compact Trigonometric Form
The above relationships are obtained from the trigonometric identity
a cos(x) + b sin(x) = c cos(x + θ)
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Role of Amplitude in Shaping Waveform
( ) ( )∑∞
=
++=1
00 2cosn
nn ntfCCtx θπ
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Role of the Phase in Shaping a Periodic Signal
( ) ( )∑∞
=
++=1
00 2cosn
nn ntfCCtx θπ
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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
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• 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 =∠
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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
0 2 4 6
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• x(t) can be expressed as
( ) ∑∞
−∞=
=n
ntfjneDtx 02π
D-n = Dn*
( ) ,....2,1,0 , 102 ±±== ∫ − ndtetx
TD
oT
ntfj
on
π
Exponential Form
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)
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• The line spectra for the exponential form has negative frequencies because of the mathematical nature of the complex exponent.
Line Spectra of x(t) in the Exponential Form
Cn = 2|Dn| Cn = Dn
...)2cos()cos()(
...||||
||||...)(
2021010
2221
012
2
001
0102
+++++=
++
++++= −−−
−−−
θωθω
ωθωθ
ωθωθ
tCtCCtx
eeDeeD
DeeDeeDtxtjjtjj
tjjtjj
C0 = D0
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Example
τ/2−τ/2
1f(t)
−Το Το
Find the exponential Fourier Series for the square-pulse periodic signal.
Το/2−Το/2
𝐷𝐷𝑛𝑛 =1𝑇𝑇0
�−𝜏𝜏/2
𝜏𝜏/2
𝑓𝑓(𝑡𝑡)𝑒𝑒−𝑗𝑗𝜔𝜔𝑜𝑜𝑛𝑛𝑛𝑛𝑑𝑑𝑡𝑡
𝐷𝐷𝑛𝑛 =𝜏𝜏𝑇𝑇0𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
𝑠𝑠𝜔𝜔𝑜𝑜𝜏𝜏2
If To = 2π and τ = π then
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τ/2−τ/2
1f(t)
−Το ΤοΤο/2−Το/2
𝐷𝐷𝑛𝑛 =𝜏𝜏𝑇𝑇0𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
𝑠𝑠𝜔𝜔𝑜𝑜𝜏𝜏2
𝜏𝜏 = 0.2𝜋𝜋
𝑇𝑇0 = 𝜋𝜋
𝜏𝜏 = 0.1𝜋𝜋
𝑇𝑇0 = 𝜋𝜋
𝜏𝜏 = 0.1𝜋𝜋
𝑇𝑇0 = 0.5𝜋𝜋
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Exponential Line Spectra
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Example
=−≠
=
=
=
,15,11,7,3,15,11,7,3 allfor 0
odd 2
even 021
0
nn
nn
nC
C
n
n
πθ
ππ/2−π/2
1f(t)
−π π 2π−2π
The compact trigonometric Fourier Series coefficients for the square-pulse periodic signal.
[ ]∑∞
=
−
−−++=
oddn
nntn
tx1
2/)1(
21)1(cos2
21)( π
π
Does the Fourier series converge to x(t) at every point?
𝐷𝐷𝑛𝑛 =𝜏𝜏𝑇𝑇0𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
𝑠𝑠𝜔𝜔𝑜𝑜𝜏𝜏2
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3( )x t 9 ( )x t
Gibbs Phenomenon
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21( )x t 45 ( )x t
overshoot: about 9 % of the signal magnitude (present even if )N →∞
Gibbs Phenomenon – Cont’d
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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.
2T0T0-T0-2T0
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
)(0
tTδ
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( )( )
000
5.05.0
5.05.0
CaDeCCD
jbaDDjbaD
njnnnn
nnnn
nnn
===∠=
+==
−=∗
−
θθ
Relationships between the Coefficients of the Different Forms
{ }( ) { }
( )( )
000
sincos
Im2Re2
cDaCb
CaDDDjb
DDDa
nnn
nnn
nnnk
nnnn
==−=
=−=−=
=+=
−
−
θθ
000
1
22
2
tan
DaCDDC
ab
baC
nn
nn
n
nn
nnn
==∠=
=
−=
+=
−
θ
θ
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• 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
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LTIC System Response to Periodic Inputs
H(s)H(jω)
tje 0ω tjejH 0)( 0ωω
A periodic signal x(t) with period T0 can be expressed astjn
nneDtx 0)( ω∑
∞
−∞=
=
For a linear system
H(s)H(jω)
tjn
nneDtx 0)( ω∑
∞
−∞=
= ∑∞
−∞=
=n
tjnneDjnHty 0)()( 0
ωω
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Fourier Series Analysis of DC Power Supply
A full-wave rectifier is used to obtain a dc signal from a sinusoid sin(t). The rectified signal x(t) is applied to the input of a lowpass RC filter, which suppress the time-varying component and yields a dc component with some residual ripple. Find the filter output y(t). Find also the dc output and the rms value of the ripple voltage.
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05.0rms ripple
0025.0136)41(
2
||2
22
1
2
==
=+−
=
= ∑∞
=
ripple
ripple
n
nnripple
P
Pnn
D
DP
π
Ripple rms is only 5% of the input amplitude
20 /4 /2 ππ == DCPD
∑
∑
∑
∞
−∞=
∞
−∞=
∞
−∞=
+−=
=
+=
−=
−=
n
ntj
n
tjnn
n
ntj
n
enjn
ty
ejnHDty
jjH
en
tx
nD
22
0
22
2
)16)(41(2)(
)()(
131)(
)41(2)(
)41(2
0
π
ω
ωω
π
π
ω
Fourier Series Analysis of DC Power Supply
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clear allt=0:1/1000:3*pi;for i=1:100
n=i;yp=(2*exp(j*2*n*t))/(pi*(1-4*n^2)*(j*6*n+1));n=-i;yn=(2*exp(j*2*n*t))/(pi*(1-4*n^2)*(j*6*n+1));y(i,:)=yp+yn;
endyf = 2/pi + sum(y);plot(t,yf, t, (2/pi)*ones(1,length(yf)))axis([0 3*pi 0 1]);
Power=0;for n=1:50
Power(n) = abs(2/(pi*(1-4*n^2)*(j*6*n+1)));endTotalPower = 2*sum((Power.^2));figure; stem( Power(1,1:20));
This Matlab code will plot y(t) for -100 ≤ n ≤100 and find the ripple power according to the equations below
0025.0||2
)16)(41(2)(
1
2
22
==
+−=
∑
∑∞
=
∞
−∞=
nnripple
n
ntj
DP
enjn
tyπ
Fourier Series Analysis of DC Power Supply