FOURIER TRANSFORMS
Hence Fourier transform of does not exist. Example 2 Find Fourier Sine transform of i. ii. Solution: i. By definition we have.
Fourier Series and Fourier Transform
j d dw. X(jw) = FT(tx(t)). FT(tx(t)) = j d dw. X(jw). Example 12: Obtain the F.T. of the signal e?atu(t) and plot its magnitude and phase spectrum. SOLUTION: x
EE 261 – The Fourier Transform and its Applications
Fourier series to find explicit solutions. This work raised hard and far reaching questions that led in different directions. It was gradually realized.
Chapter 1 The Fourier Transform
01-Mar-2010 Example 1 Find the Fourier transform of f(t) = exp(?
fourier transforms and their applications
are known then finite Fourier cosine transform is used. Heat Conduction. Example 22: Solve the differential equation. . .
Fourier transform techniques 1 The Fourier transform
The Fourier transform is beneficial in differential equations because it can reformulate them as problems which are easier to solve.
Fourier Transform Solutions of PDEs In this chapter we show how
Several new concepts such as the ”Fourier integral representation” and ”Fourier transform” of a function are introduced as an extension of the Fourier series
6 Fourier transform
Note that this is the D'Alembert formula. Question 110: Solve the integral equation: f(x) + 1. 2?. ? +?.
Finite Fourier transform for solving potential and steady-state
13-May-2017 Many boundary value problems can be solved by means of integral transformations such as the Laplace transform function
FOURIER TRANSFORMS
where is any differentiable function. Example 4 Show that Fourier sine and cosine transforms of are and respectively. Solution: By definition. Putting.
Lecture 8: Fourier transforms - Scholars at Harvard
Fourier transforms 1 Strings To understand sound we need to know more than just which notes are played – we need the shape of the notes If a string were a pure infinitely thin oscillator with no damping it would produce pure notes
Problems and solutions for Fourier transforms and -functions
Problems and solutions for Fourier transforms and -functions 1 Prove the following results for Fourier transforms where F T represents the Fourier transform and F T [f(x)] = F(k): a) If f(x) is symmetric (or antisymmetric) so is F(k): i e if f(x) = f( x) then F(k) = F( k) b) If f(x) is real F (k) = F( k)
1 Fourier Transform - University of Toronto Department of
We introduce the concept of Fourier transforms This extends the Fourier method for nite intervals to in nite domains In this section we will derive the Fourier transform and its basic properties 1 1 Heuristic Derivation of Fourier Transforms 1 1 1 Complex Full Fourier Series Recall that DeMoivre formula implies that sin( ) = ei i ei
Fourier Series and Fourier Transform
2.1 INTRODUCTION
Fourier series is used to get frequency spectrum of a time-domain signal, when signal is a periodic function of
time. We have seen that the sum of two sinusoids is periodic provided their frequencies are integer multiple
of a fundamental frequency,w0.2.2 TRIGONOMETRIC FOURIER SERIES
Consider a signalx(t), a sum of sine and cosine function whose frequencies are integral multiple ofw0
x(t) =a0+a1cos(w0t)+a2cos(2w0t)+··· b1sin(w0t)+b2sin(2w0t)+···
x(t) =a0+¥å n =1(ancos(nw0t)+bnsin(nw0t))(1) a0,a1,...,b1,b2,...are constants andw0is the fundamental frequency.
Evaluation of Fourier Coefficients
To evaluatea0we shall integrate both sides of eqn. (1) over one period(t0,t0+T)ofx(t)at an arbitrary timet0 t 0+T? t0x(t)dt=t
0+T? t 0a0dt+¥å
n=1a nt 0+T? t0cos(nw0t)dt+¥å
n=1b nt 0+T? t0sin(nw0t)dt
Since?t0+T
t0cos(nw0dt) =0
t 0+T? t0sin(nw0dt) =0
a 0=1Tt 0+T? t0x(t)dt(2)
To evaluateanandbn, we use the following result:
t 0+T? t0cos(nw0t)cos(mw0t)dt=?0m?=n
T/2m=n?=0
9496•Basic System Analysis
Multiply eqn. (1) by sin(mw0t)and integrate over one period t 0+T? t0x(t)sin(mw0t)dt=a0t
0+T? t0sin(mw0t)dt+¥å
n=1a nt 0+T? t0cos(nw0t)sin(mw0t)dt+
n=1b nt 0+T? t0sin(mw0t)sin(nw0t)dt
b n=2 Tt 0+T? t0x(t)sin(nw0t)dt(4)
Example 1:
-3-2-1 -1.01.0 1 230-Fig. 2.1.
T→ -1 to 1T=2w0=px(t) =t,-1 a 0=1 21
-1t dt=14(1-1) =0 a n=0 b n=1 -1tsinpntdt=?-tcospnt np-cospntnp? 1 -1 -1 b n=-2 npcosnp=2p? -(-1)nn? b 1b2b3b4b5b6
2 p-22p23p-24p25p-2···6p x(t) =¥å n=12 p? -(-1)nn? sinnpt 2 p? sinpt-12sin 2pt+13sin 3pt-14sin 4pt+···? Fourier Series and Fourier Transform•97
Example 2:
-2π 1.0 2π 4π 6π
0t Fig. 2.2.
x(t) =t 2pT=2pw0=2pT=1
a 0=1 T2p? 0 x(t)dt=14p2? 12t2? 2p 0 =12 a n=2 4p22p?
0 tcosntdt=12p2? tsintn+sinntn? 2p 0 1 2p2? 2psin 2npn+sin 2npn?
=0 b n=2 4p22p?
0 tsinntdt=-12p2? tcosntn+cosntn? 2p 0 -1 2p2? 2pcos 2npn+cos 2npn-1n?
b n=-1 np x(t) =1 2+¥å
n=1? -1np? sinnt=12+1p¥å n=11ncos(nt+p/2) 1 2-1p? sint+sin 2t2+sin 3t3+···? Example 3:
-T/2-T/4T/4A x(t) t T/2 Fig. 2.3.Rectangular waveform
98•Basic System Analysis
Figure shows a periodic rectangular waveform which is symmetrical to the vertical axis. Obtain its F.S.
representation. x(t) =a0+¥å n=1(ancosnw0t+bnsinnw0t) x(t) =a0+¥å n=1a ncos(nw0t)bn=0 x(t) =0 for-T 2 +Afor-T 4 0 for T 4 a 0=1 TT/4? -T/4Adt=A2 a n=2 TT/4? -T/4Acos(nw0t)dt=2ATnw0? sinnw0T4+sinnw0T4? a n=4A 2pnsin?np2?
=2Apnsin?np2? w 0=2pT a 1=4A 2p=2Ap
a 2=0 a 3=2A 3psin3p2=2A3p(-1) =-2A3p
x(t) =A 2+2Ap?
cosw0t-13cos 3w0t+15cos 5w0t+···? Example 4:Find the trigonometric Fourier series for the periodic signalx(t). 1.0 0 1-1-3-5-7-9x(t)
3 5 7 911t
T Fig. 2.4.
Fourier Series and Fourier Transform•99
SOLUTION:
b n=0x(t) =?1-11T[2-2] =0?w0=2pT=2p4=p2
a n=2 T? ?1 -1cos(nw0t)dt+3 1 cos(nw0t)dt? 22pn?
2sinpn2?
sin3np2-sinnp2?? 1 np? 3sinnp2-sin3np2?
sin3np2=sin? p+np2? =-sinnp2 a n=4 npsin?np2? a n=? ?0n=even 4npn=1,5,9,13
-4 npn=3,7,11,15 x(t) =4 pcos?p2t? -43pcos?3p2t? +45pcos?5p2t?
-47pcos?7p2t? x(t) =4 p? cos?p2t? -13cos?3p2t? +15cos?5p2t? Example 5:Find the F.S.C. for the continuous-time periodic signal with fundamental freq.w0=p -1.51.5 0 1x(t) 32 54
Fig. 2.5.
100•Basic System Analysis
SOLUTION:
T=2p w0=2,w0=p a 0=an=0
b n=1 0 1.5sinnptdt-2
1 1.5sinnptdt
1.5 np? [-cosnp+1]+[cos2np-cosnp]? b n=3 np[1-cosnp] x(t) =3 p? 2sinpt+23sin3pt+25sin5pt+···?
6 p? sinpt+13sin3pt+15sin5pt+···? C 0=1 2? ?1 0 1.5dt-1.52
1 dt? ?=0 OR By using complex exponential Fourier series
C n=1 2? ?1 0 1.5e-jnptdt-1.52
1 e -jnptdt? C n=3 -4jnp? ?e-jnpt? ?1 -e-jnpt 0? ?21? -3 4jnp?e-jnp-1-e-j2np+e-jnp?
3 2jnp?1-e-jnp?=32jnp[1-cosnp]
x(t) =¥å n=-¥C ne-jnpt n=-¥3 2jnp?1-e-jnp?ejnpt
n=-¥3 2jnp?ejnpt-ejnptcospn?
102•Basic System Analysis
forn=1 A 2pp 0 sintsintdt=A2pp 0 (1-cos2t)dt A 2p[p] =A2
Whennis even
=A 2p? 2n+1-21-n?
=2Ap(1-n2) Example 7:
-2-3-1 -2 2 1 2 T3 0x(t) a b t Fig. 2.7.
SOLUTION:
T=2w0=2p
T=p x(t) =?2t-1Point (a)(-1,-2)
-1t dt=14(1-1) =0 a n=0 b n=1 -1tsinpntdt=?-tcospnt np-cospntnp? 1 -1 -1 b n=-2 npcosnp=2p? -(-1)nn? b
1b2b3b4b5b6
2 p-22p23p-24p25p-2···6p x(t) =¥å n=12 p? -(-1)nn? sinnpt 2 p? sinpt-12sin 2pt+13sin 3pt-14sin 4pt+···?Fourier Series and Fourier Transform•97
Example 2:
-2π 1.02π 4π 6π
0tFig. 2.2.
x(t) =t2pT=2pw0=2pT=1
a 0=1 T2p? 0 x(t)dt=14p2? 12t2? 2p 0 =12 a n=24p22p?
0 tcosntdt=12p2? tsintn+sinntn? 2p 0 1 2p2?2psin 2npn+sin 2npn?
=0 b n=24p22p?
0 tsinntdt=-12p2? tcosntn+cosntn? 2p 0 -1 2p2?2pcos 2npn+cos 2npn-1n?
b n=-1 np x(t) =12+¥å
n=1? -1np? sinnt=12+1p¥å n=11ncos(nt+p/2) 1 2-1p? sint+sin 2t2+sin 3t3+···?Example 3:
-T/2-T/4T/4A x(t) t T/2Fig. 2.3.Rectangular waveform
98•Basic System Analysis
Figure shows a periodic rectangular waveform which is symmetrical to the vertical axis. Obtain its F.S.
representation. x(t) =a0+¥å n=1(ancosnw0t+bnsinnw0t) x(t) =a0+¥å n=1a ncos(nw0t)bn=0 x(t) =0 for-T2 +Afor-T 4 0 for T 4 a 0=1 TT/4? -T/4Adt=A2 a n=2 TT/4? -T/4Acos(nw0t)dt=2ATnw0? sinnw0T4+sinnw0T4? a n=4A 2pnsin?np2?
=2Apnsin?np2? w 0=2pT a 1=4A 2p=2Ap
a 2=0 a 3=2A 3psin3p2=2A3p(-1) =-2A3p
x(t) =A 2+2Ap?
cosw0t-13cos 3w0t+15cos 5w0t+···? Example 4:Find the trigonometric Fourier series for the periodic signalx(t). 1.0 0 1-1-3-5-7-9x(t)
3 5 7 911t
T Fig. 2.4.
Fourier Series and Fourier Transform•99
SOLUTION:
b n=0x(t) =?1-11T[2-2] =0?w0=2pT=2p4=p2
a n=2 T? ?1 -1cos(nw0t)dt+3 1 cos(nw0t)dt? 22pn?
2sinpn2?
sin3np2-sinnp2?? 1 np? 3sinnp2-sin3np2?
sin3np2=sin? p+np2? =-sinnp2 a n=4 npsin?np2? a n=? ?0n=even 4npn=1,5,9,13
-4 npn=3,7,11,15 x(t) =4 pcos?p2t? -43pcos?3p2t? +45pcos?5p2t?
-47pcos?7p2t? x(t) =4 p? cos?p2t? -13cos?3p2t? +15cos?5p2t? Example 5:Find the F.S.C. for the continuous-time periodic signal with fundamental freq.w0=p -1.51.5 0 1x(t) 32 54
Fig. 2.5.
100•Basic System Analysis
SOLUTION:
T=2p w0=2,w0=p a 0=an=0
b n=1 0 1.5sinnptdt-2
1 1.5sinnptdt
1.5 np? [-cosnp+1]+[cos2np-cosnp]? b n=3 np[1-cosnp] x(t) =3 p? 2sinpt+23sin3pt+25sin5pt+···?
6 p? sinpt+13sin3pt+15sin5pt+···? C 0=1 2? ?1 0 1.5dt-1.52
1 dt? ?=0 OR By using complex exponential Fourier series
C n=1 2? ?1 0 1.5e-jnptdt-1.52
1 e -jnptdt? C n=3 -4jnp? ?e-jnpt? ?1 -e-jnpt 0? ?21? -3 4jnp?e-jnp-1-e-j2np+e-jnp?
3 2jnp?1-e-jnp?=32jnp[1-cosnp]
x(t) =¥å n=-¥C ne-jnpt n=-¥3 2jnp?1-e-jnp?ejnpt
n=-¥3 2jnp?ejnpt-ejnptcospn?
102•Basic System Analysis
forn=1 A 2pp 0 sintsintdt=A2pp 0 (1-cos2t)dt A 2p[p] =A2
Whennis even
=A 2p? 2n+1-21-n?
=2Ap(1-n2) Example 7:
-2-3-1 -2 2 1 2 T3 0x(t) a b t Fig. 2.7.
SOLUTION:
T=2w0=2p
T=p x(t) =?2t-1Point (a)(-1,-2)
4 0 for T 4 a 0=1 TT/4? -T/4Adt=A2 a n=2 TT/4? -T/4Acos(nw0t)dt=2ATnw0? sinnw0T4+sinnw0T4? a n=4A 2pnsin?np2?
=2Apnsin?np2? w 0=2pT a 1=4A 2p=2Ap
a 2=0 a 3=2A 3psin3p2=2A3p(-1) =-2A3p
x(t) =A 2+2Ap?
cosw0t-13cos 3w0t+15cos 5w0t+···? Example 4:Find the trigonometric Fourier series for the periodic signalx(t). 1.0 0 1-1-3-5-7-9x(t)
3 5 7 911t
T Fig. 2.4.
Fourier Series and Fourier Transform•99
SOLUTION:
b n=0x(t) =?1-11T[2-2] =0?w0=2pT=2p4=p2
a n=2 T? ?1 -1cos(nw0t)dt+3 1 cos(nw0t)dt? 22pn?
2sinpn2?
sin3np2-sinnp2?? 1 np? 3sinnp2-sin3np2?
sin3np2=sin? p+np2? =-sinnp2 a n=4 npsin?np2? a n=? ?0n=even 4npn=1,5,9,13
-4 npn=3,7,11,15 x(t) =4 pcos?p2t? -43pcos?3p2t? +45pcos?5p2t?
-47pcos?7p2t? x(t) =4 p? cos?p2t? -13cos?3p2t? +15cos?5p2t? Example 5:Find the F.S.C. for the continuous-time periodic signal with fundamental freq.w0=p -1.51.5 0 1x(t) 32 54
Fig. 2.5.
100•Basic System Analysis
SOLUTION:
T=2p w0=2,w0=p a 0=an=0
b n=1 0 1.5sinnptdt-2
1 1.5sinnptdt
1.5 np? [-cosnp+1]+[cos2np-cosnp]? b n=3 np[1-cosnp] x(t) =3 p? 2sinpt+23sin3pt+25sin5pt+···?
6 p? sinpt+13sin3pt+15sin5pt+···? C 0=1 2? ?1 0 1.5dt-1.52
1 dt? ?=0 OR By using complex exponential Fourier series
C n=1 2? ?1 0 1.5e-jnptdt-1.52
1 e -jnptdt? C n=3 -4jnp? ?e-jnpt? ?1 -e-jnpt 0? ?21? -3 4jnp?e-jnp-1-e-j2np+e-jnp?
3 2jnp?1-e-jnp?=32jnp[1-cosnp]
x(t) =¥å n=-¥C ne-jnpt n=-¥3 2jnp?1-e-jnp?ejnpt
n=-¥3 2jnp?ejnpt-ejnptcospn?
102•Basic System Analysis
forn=1 A 2pp 0 sintsintdt=A2pp 0 (1-cos2t)dt A 2p[p] =A2
Whennis even
=A 2p? 2n+1-21-n?
=2Ap(1-n2) Example 7:
-2-3-1 -2 2 1 2 T3 0x(t) a b t Fig. 2.7.
SOLUTION:
T=2w0=2p
T=p x(t) =?2t-1Point (a)(-1,-2)
4 a 0=1 TT/4? -T/4Adt=A2 a n=2 TT/4? -T/4Acos(nw0t)dt=2ATnw0? sinnw0T4+sinnw0T4? a n=4A 2pnsin?np2?
=2Apnsin?np2? w 0=2pT a 1=4A 2p=2Ap
a 2=0 a 3=2A 3psin3p2=2A3p(-1) =-2A3p
x(t) =A 2+2Ap?
cosw0t-13cos 3w0t+15cos 5w0t+···? Example 4:Find the trigonometric Fourier series for the periodic signalx(t). 1.0 0 1-1-3-5-7-9x(t)
3 5 7 911t
T Fig. 2.4.
Fourier Series and Fourier Transform•99
SOLUTION:
b n=0x(t) =?1-11T[2-2] =0?w0=2pT=2p4=p2
a n=2 T? ?1 -1cos(nw0t)dt+3 1 cos(nw0t)dt? 22pn?2pnsin?np2?
=2Apnsin?np2? w 0=2pT a 1=4A2p=2Ap
a 2=0 a 3=2A3psin3p2=2A3p(-1) =-2A3p
x(t) =A2+2Ap?
cosw0t-13cos 3w0t+15cos 5w0t+···? Example 4:Find the trigonometric Fourier series for the periodic signalx(t). 1.00 1-1-3-5-7-9x(t)
3 5 7 911t
TFig. 2.4.
Fourier Series and Fourier Transform•99
SOLUTION:
b n=0x(t) =?1-12sinpn2?
sin3np2-sinnp2?? 1 np?3sinnp2-sin3np2?
sin3np2=sin? p+np2? =-sinnp2 a n=4 npsin?np2? a n=? ?0n=even4npn=1,5,9,13
-4 npn=3,7,11,15 x(t) =4 pcos?p2t? -43pcos?3p2t? +45pcos?5p2t?-47pcos?7p2t? x(t) =4 p? cos?p2t? -13cos?3p2t? +15cos?5p2t? Example 5:Find the F.S.C. for the continuous-time periodic signal with fundamental freq.w0=p -1.51.5 0 1x(t) 32 54
Fig. 2.5.
100•Basic System Analysis
SOLUTION:
T=2p w0=2,w0=p a0=an=0
b n=1 01.5sinnptdt-2
11.5sinnptdt
1.5 np? [-cosnp+1]+[cos2np-cosnp]? b n=3 np[1-cosnp] x(t) =3 p?2sinpt+23sin3pt+25sin5pt+···?
6 p? sinpt+13sin3pt+15sin5pt+···? C 0=1 2? ?1 01.5dt-1.52
1 dt? ?=0 ORBy using complex exponential Fourier series
C n=1 2? ?1 01.5e-jnptdt-1.52
1 e -jnptdt? C n=3 -4jnp? ?e-jnpt? ?1 -e-jnpt 0? ?21? -34jnp?e-jnp-1-e-j2np+e-jnp?
32jnp?1-e-jnp?=32jnp[1-cosnp]
x(t) =¥å n=-¥C ne-jnpt n=-¥32jnp?1-e-jnp?ejnpt
n=-¥32jnp?ejnpt-ejnptcospn?
102•Basic System Analysis
forn=1 A 2pp 0 sintsintdt=A2pp 0 (1-cos2t)dt A2p[p] =A2
Whennis even
=A 2p?2n+1-21-n?
=2Ap(1-n2)Example 7:
-2-3-1 -2 2 1 2 T3 0x(t) a b tFig. 2.7.
SOLUTION:
T=2w0=2p
T=p x(t) =?2t-1Point (b)(1,2)
y-(-2) =2-(-2)1-(-1)(x-(-1))
y+2=42(x+1)
y+2=2x+2 y=2x x(t) =2tSince function is an odd function
a n=0,a0=1 T1 -12tdt=12×0=0 b n=2 T1 -1tsin(npt)dt=2T? ?-tcosnpt np? ?1-1+1 n2p2cosnpt? ?1-1?104•Basic System Analysis
2.3 CONVERGENCE OF FOURIER SERIES - DIRICHLET CONDITIONS
Existence of Fourier Series: The conditions under which a periodic signal can be represented by an F.S.
are known as Dirichlet conditions. F.P.→Fundamental Period (1) The functionx(t)has only a finite number of maxima and minima, if any within theF.P. (2) The functionx(t)has only a finite number of discontinuities, if any within theF.P. (3) The functionx(t)is absolutely integrable over one period, that is T 0? ?x(t)? ?dt<¥2.4 PROPERTIES OF CONTINUOUS FOURIER SERIES
(1) Linearity:Ifx1(t)andx2(t)are two periodic signals with periodTwith F.S.C.CnandDnthen F.C. of linear combination ofx1(t)andx2(t)are given byFS[Ax1(t)+Bx2(t)] =ACn+BDn
Proof: Ifz(t) =Ax1(t)+Bx2(t)
a n=1 Tt 0+T? t0[Ax1(t)+Bx2(t)]e-jnw0t=AT?
T x1(t)e-jnw0tdt+BT?
T x2(t)e-jnw0tdt
a n=ACn+BDn (2) Time shifting:If the F.S.C. ofx(t)areCnthen the F.C. of the shifted signalx(t-t0)areFS[x(t-t0)] =e-jnw0t0Cn
Lett-t0=t
dt=dt B n=1 T? T x(t-t0)e-jnw0tdt 1 T? T x(t)e-jnw0(t0+t)dt=1T? T x(t)e-jnw0tdt·e-jnw0t0 B n=e-jnw0t0·Cnquotesdbs_dbs14.pdfusesText_20[PDF] fox news misinformation statistics 2018
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