Repet F Series - syd.kth.se€¦ · Transform Methods and Calculus of Several Variables 6H3709 , 5...
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Transcript of Repet F Series - syd.kth.se€¦ · Transform Methods and Calculus of Several Variables 6H3709 , 5...
Transform Methods and Calculus of Several Variables 6H3709 , 5 p Lecturer: Armin Halilovic KTH, Campus Haninge E-mail: [email protected] , www.syd.kth.se/armin REPETITION before the exam PART 2, Transform Methods Fourier series: FS1. Find the Fourier series of f on the given interval:
a) 3, 0
( )1 0
( 2 ) ( )
tf t
tf t f t
ππ
π
− − < <⎧= ⎨ − < <⎩+ =
b) ( ) 5 2 ,( 2 ) ( )
f t t tf t f t
π ππ
= + − < <+ =
a) )55sin
33sin
1sin(42 L++++−
tttπ
b) )44sin
33sin
22sin
1sin(45 L+−+−+
tttt
FS2. A signal with period T= 2 is described by
⎩⎨⎧
<≤−<≤−+
=10,101,1
)(tttt
tf
)()2( tftf =+ Find the Fourier series of f(t) Solution:
T=2, π=π
=T2Ω .
The function is even ⇒ 0=nb and
∫ π−=1
0
cos)1(2 tdtntan
101
)2
(2)1(21
0
2
0 =⎥⎦
⎤⎢⎣
⎡−=−= ∫
ttdtta
=π−= ∫1
0
cos)1(2 tdtntan (Partial integration)
= 01sin)1(2 ⎥⎦⎤
⎢⎣⎡
ππ
−n
tnt + =ππ
∫1
0
sin2 dtn
tn
=⎥⎦⎤
⎢⎣⎡
ππ
−01cos20 22n
tn =⎥⎦⎤
⎢⎣⎡
π−π
− 22
1cos2n
n ⎥⎦
⎤⎢⎣
⎡ −−− 22
1)1(2πn
n
Answer:
tnn
Sn
n
f π⎥⎦
⎤⎢⎣
⎡π−−
−+= ∑∞
=
cos1)1()2(21
122
FS3.
A signal with period T= 2 is described by⎩⎨⎧
<≤−<≤−+
=ππ
ππtt
tttf
0,0,2
)(
)()2( tftf =+ π Led )(tS f denote the Fourier series of the function f. a) Sketch the graph of f(t) in the interval ]5,5[ ππ− and calculate )3( π−f and )3( π−fS b) Find the Fourier series of )(tf Suitable formulas:
)sincos(2
)(1
0 tnbtnaa
tSn
nnf Ω+Ω+= ∑∞
=
where 2 , ( ) ( )f t T f tTπ
Ω = + =
∫−
Ω=2
2
cos)(2T
Tn dttntf
Ta , ∫
−
Ω=2
2
sin)(2T
Tn dttntf
Tb
Solution: Answer: a)
ππ −=− )3(f and 22
)3()3()3( ππππ −
=−+−
=− +− ffS f
b) π2=T , 12==Ω
Tπ
== ∫−
2
2
0 )(2T
T
dttfT
a =∫−
π
ππdttf )(
22
∫−
+0
)2(1
π
ππ
dtt + ∫ −π
ππ 0
)(1 dtt = 0+ 2π =
2π
For 1≥n we have
=Ω= ∫−
2
2
cos)(2T
Tn dttntf
Ta =∫
−
π
ππdtnttf cos)(
22
∫−
+0
cos)2(1
π
ππ
dtntt + ∫ −π
ππ 0
cos)(1 dtntt = (Partial integration)
= π
π2
cos22n
n− + π
π2
cos1n
n− = π
π2
cos33n
n− = π2
)1(33n
n−−
=Ω= ∫−
2
2
sin)(2T
Tn dttntf
Tb =∫
−
π
ππdtnttf sin)(
22
∫−
+0
sin)2(1
π
ππ
dtntt + ∫ −π
ππ 0
sin)(1 dtntt = (Partial integration)
= n
nπcos1−− + n1 =
nnπcos− =
n
n)1(−−
Answer: b)
tnbtnaa
tSn
nnf Ω+Ω+= ∑∞
=
sincos2
)(1
0 = tnn
tnn
n
n
n
Ω−−
+Ω−−
+∑∞
=
sin)1(cos)1(334 1
2ππ
FS4. A) For the given function )(tf
ππ <≤−= tttf ,2)( )()2( tftf =+ π
i) determine a ) )11( πf , b) )
29( πf c) )
2103( πf
ii) Find the Fourier series of )(tf Answer: i) a) π− 2 b) π c) π−
ii)
ntn
Sf n
nsin)1(4 1
1
+∞
=
−= ∑
FS5. Find the amplitude phase form for the following functions a) )4sin(80)4cos(60)( tttg += b) )3sin(8)3cos(5)2sin(3)2cos(2)( ttttth +++= Answer: a) )13,534cos(100)( o−= ttg b) )99,573cos(89)31,562cos(13)( oo −+−= ttth FS6.a) Find the Fourier series of )(tf where
ππ <<−= tttf ,||)( )()2( tftf =+ π
b) Determine the sum ...71
51
31
11
2222 +++
Answer: a)
ntn
Sfn
n
cos)1)1((22 1
2∑∞
=
−−+=π
π
or
⎟⎠⎞
⎜⎝⎛ +++−= ...
55cos
33cos
1cos4
2)( 222
ttttSfπ
π
Substituting 0=t in the previous result gives
⎟⎠⎞
⎜⎝⎛ +++=
⇒⎟⎠⎞
⎜⎝⎛ +++=
⇒⎟⎠⎞
⎜⎝⎛ +++−=
⇒⎟⎠⎞
⎜⎝⎛ +++−=
...51
31
11
8
...51
31
114
2
...51
31
114
20
...51
31
114
2)0(
222
2
222
222
222
ππ
ππ
ππ
πf
b) 8
2π
FS7. For the following Fourier series
5+ ∑∞
=⎟⎠⎞
⎜⎝⎛ −
133
sin3
cosn n
ntn
nt
determine a) complex form b) amplitude- phase form
Answer: 33330 261,
261,
25
ni
nc
ni
ncc nn −=+== − , 0≠n
b) o561.71,3105 30 === nn n
AA α , 0≠n
FS8. a) Find the Fourier series of )(tf where
,06
02)(
⎭⎬⎫
⎩⎨⎧
<<<<−
=π
πtt
tf
)()2( tftf =+ π Answer:
ntn
Sf n
nsin))1(1(44
1−−+= ∑
∞
= π
FS9. a) Find the Fourier series of )(tf where
ππ <<−+= tttf ,1)( )()2( tftf =+ π
b) Determine the sum ...71
51
31
11
−+−
Answer:
a) ntn
Sf n
nsin)1(21 1
1
+∞
=
−+= ∑
⎟⎠⎞
⎜⎝⎛ −+−+= ...
33sin
22sin
1sin21)( ttttSf
Substituting 2π
=t in the previous result gives
⇒⎟⎠⎞
⎜⎝⎛ −+−=
⇒⎟⎠⎞
⎜⎝⎛ −+−=
⇒⎟⎠⎞
⎜⎝⎛ −+−+=+
⇒⎟⎠⎞
⎜⎝⎛ +−−−+=
...71
51
31
11
4
...71
51
31
112
2
...71
51
31
1121
21
...51
40
31
20
1121)0(
π
π
π
f
b) 4π
FS10. For the following Fourier series
2+ ∑∞
=⎟⎠⎞
⎜⎝⎛ −−
144
sin4
cosn n
ntnnt
determine a) complex form b) amplitude- phase form Answer:
a) 44440 281,
281,2
ni
nc
ni
ncc nn −
−=+
−== − 0≠n
b) o36.0.104,417,2 40 === nn n
AA α 0≠n
