tugas 5 deret fourier
DESCRIPTION
Deret fourierTRANSCRIPT
Pemrosesan sinyal
TUGAS 5
MOHAMMAD ALFIAN IRSYADUL IBAD
Tugas 5A
Diketahui :
A=−5+ j 2
B=5− j 2
C=−2− j 4
Hitunglah :
(i) C – B
(ii) 2A + 3B + 4C
(iii) C2 (A+B)
(iv) B Re[A] + A Im[B]
(v) [(A - A*)(B + B*)*]*
(vi) (1/C) – (1/B)*
(vii)(B+C) / (2BC)
Penyelesaian :
(i) C – B =−2− j 4−(5− j2 )
= −2− j 4−5+ j 2
= −7− j 2
(ii) 2A + 3B + 4C = 2 (−5+ j 2 )+3 (5− j2 )+4 (−2− j 4 )
=−10+ j 4+15− j 6−8− j16
=−3− j 18
(iii) C2 (A+B) = (−2− j 4 )2 (−5+ j2+5− j 2 )
= (−2− j 4 ) (−2− j 4 ) ( 0+0 )
= 0
(iv) Bℜ [ A ]+ A ℑ [ B ] = (5− j 2 ) (−5 )+ (−5+ j 2 ) (−2 )
= −25+ j10+10− j 4
= −15+ j6
(v) [ ( A−A¿ ) ( B+B¿)¿ ]¿ = [¿*
= ( j4 ×10 )¿
=− j40
(vi)1C
−( 1B )
¿
= 1
−2− j 4−( 1
5− j 2 )¿
=1
−2− j 4− 1
5+ j 2
=5+ j 2−(−2− j 4 )(−2− j4 ) (5+ j 2 )
=7+ j 6
−10− j24+8
=7+6
−2− j24
=(7+ j6 ) (−2+ j24 )
(−2− j24 ) (−2+ j 24 )
=−156+ j 158
580
= −0,268+ j 0,272
(vii)(B+C )2 BC
= 5− j2−2− j 4
2 [ (5− j 2 ) (−2− j 4 ) ]
=3− j6
2 (−10− j 20+ j 4−8 )
=3− j 6
2 (−18− j 16 )
= 3− j6
−36− j32
= (3− j6
−36− j 32)(−36+ j 32
−36+ j 32)
= −108+ j 96+ j216+192
1296+1024
= 84+ j312
2320
= 0,036+ j 0,134
Tugas 5B
1) Gunakan (1) sampai (4) untuk menghitung:
(i) ej2
(ii) e1+j2
(iii) cos(j2)
(iv) sin(j2)
2) Pada saat t = 0,5s, hitunglah : (d/dt) (3 cos 2t – j2 sin 3t)
Penyelesian :
1) ejᶿ = cosθ+¿ jsin θ ¿ ………….. (1) cosθ=12(e j θ+e− j θ) ………........ (3)
e-jᶿ = cosθ−¿ jsin θ ¿ .................. (2) sin θ=− j12(e j θ−e− j θ)………… (4)
(i) e j 2 =cosθ+ j sin θ
= cos2+ j sin2
=1+ j 0,034
(ii) e1+ j 2 =e cos2+ j sin 2
=e (1+ j 0,034)
=2,718+ j0,092
(iii) cos ( j2 ) = 12
(e j ( j 2)+e− j ( j 2 ))
= 12(e−2+e2)
= 3,762
(iv) sin ( j 2 ) = − j12
(e j ( j 2)+e− j ( j 2 ))
=− j12
(e−2+e2 )
= − j3,762
2)ddt
¿ = d ¿¿
= 2¿
= −6 sin 2 t− j 6cos3 t
Saat t = 0,5 sekon
t (0,5 ) =−6 sin 2 (0,5 )− j 6 cos3 (0,5 )
=−6 sin 1− j6 cos1,5
= −0,104− j6
Tugas 5C
o Isyarat x(t) didefinisikan sebagai berikut:
x(t) = 1+ sin ω0 t + 2 cos ω0t + cos (2 ω0t + /4)
oNyatakanlah x(t) dalam deret Fourier
Penyelesaian :
x (t )= ∑k=−∞
+∞
ak e jkω0 t
x(t) = 1+ sin ω0 t + 2 cos ω0t + cos (2 ω0t + /4)
= 1+ 1j 2
{e j ω0 t−e− j ω0 t }+{e j ω0 t+e− j ω0 t }+ 12
{e j(2ω0 t+π4 )−e
− j(2 ω0 t+π4 )}
=1+[1+ 1j 2 ] e j ω0 t +[1− 1
j 2 ]e− j ω0 t+ 12
ej π
4 e j 2 ω0t +12
e− j π
4 e− j 2 ω0 t
Maka diperoleh
x(t) = 1 k = 0
+[1+ 1j 2 ]e jω0 t
k = 1
+[1− 1j2 ]e− jω 0 t k = -1
+12
ej π
4 e j 2 ω0 t k = 2
+12
e− j π
4 e− j 2 ω 0 t k = -2
Dapat diketahui bahwa
a0=1
a1=[1+ 1j 2 ]=[1− j
12 ]
a−1=[1− 1j2 ]=[1+ j
12 ]
a2=12
ej π
4 =√24
(1+ j)
a−2=12
e− j π
4=√24
(1− j)
ak=0 untuk|k|>2