frekuensi kompleks
DESCRIPTION
analisis and fourier methodeTRANSCRIPT
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teVtv t
m cos
tVtv m cos
0cos VVtv m
tt
m eVeVtv 0cos
Frekuensi Kompleks
Fungsi sinusoida teredam eksponensial
0,0
0
0
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2
tjeVtv 0
Frekuensi Kompleks
tVtv m cos
tjtj eet21cos
tjj
m
tjj
m
tjtj
m
eeVeeV
eeVtv
21
21
21
Fungsi sinusoida dengan sudut fase 0
stKetf )(
Rumus Euler
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3
tstseKeKtv 21
21
tjtjt
m
t
m
eeeV
teVtv
21
cos
tjj
m
tjj
m eeVeeVtv 21
21
jssjs *
121
Rumus Euler
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4
500
500 500sin2
02
0 100
*
12
1
2
jss
jsttv
jsetv
stv
t
63
63 106sin4
*
12
1
3
jss
jstetv t
js
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Fungsi Pemaksa Sinusoida Teredam
teVtv t
m cos
tjt
m
tjt
m
eeVtv
eeVtv
Re
Re
tjj
m eeVtv Re
stj
m
stj
m
eeIti
eeVtv
Re
Re
Fungsi Pemaksa Riil
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Fungsi Pemaksa
02 104cos60 tetv t
teIti t
m 4cos2
0104202 60Re104cos60 tjtt eetetv
sttjj Veeetvo
Re60Re 4210
oV 1060 42 js Fungsi Pemaksa Kompleks
stoe1060
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Respons dengan kuantitas kompleks
stIe mII
dtiCdt
diLRitv
1
i(t)
v(t)
3H
0.1 F
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dtidt
diitv 1032
ststststo Ies
sIeIee10
321060
Is
sIIo 10321060
s
o
sI
1032
1060
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9
42104232
1060
j
o
jI
oI 6,10637,5
ot teti 6,1064cos37,5 2
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sYdansZ
L teVtv t
m cos
teIti t
m cos
+
-
sL
I
+
-
V
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Sumber Tegangan
teVtv t
m cos
Diberikan kepada induktor L, maka respons arus yang timbul :
teIti t
m cos
ststj
m VeeeVtv ReRe
ststj
m IeeeIti ReRe
Rangkaian daerah waktu
Hub. Arus dan tegangan untuk induktor
dt
tdiLtv
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stst sLIeVe ReRe sLIV
Impedansi Kompleks sLI
VsZ
Admitansi sL
sY1
R L C
R sL 1/sC
1/R 1/sL sC
sZ
sY
Dengan cara yang sama
diperoleh
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-1-j2
i(t) -6+j12
60 100
s=-2+j4
Contoh Soal
A
jjjI
o
oo
6,10637,5
105
1060
211262
1060
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Soal Tentukan ekivalen
Thevenin
Vte t 10cos100 5
4H
20
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Respons alami dalam bidang s
tititi fn
tiR
L
tvs
t=0
sLR
VsI s
f
L
Rs
V
LsI s
f
1
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L
RsLsH
sV
sI f
1
0 jsAsIL
R
Dengan mentransformasikan respons alami ke daerah waktu
L
tR
LtR
Aeti
Aeti
n
n
Re
Fungsi Pemindah H(s) adalah perbandingan antara respons paksaan yang
dikehendaki dengan fungsi pemaksa.
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tiAeti fL
tR
.......
.......
42
312
ssss
ssssksH
V
sV
s
Jaringan tanpa
sumber bebas sI1sV
+ - sV2 ...42
422 tsts
n eAeAtv
transferfungsikutubkutubadalahss ,......, 42
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Untuk mencari respons alami ti n1
Menentukan kutub-kutub fungsi transfer sV
sIsH 1
Contoh Soal
x x’
2H
1
3H 21i 2i
Aii 1100 21
Sumber tegangan Vs dipasang antara x dan x’
2136
23
23612
2
1
ss
Vs
sss
Vsi
s
s
61
32
21
1
2
ss
s
V
sIsH
s
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6/2
1
tt BeAeti
Kondisi awal Ai 1101 BA11Persamaan Tambahan
Biidt
di
t
61
1221
0
1 -2A 2
11220102
diperoleh 6/2
1 38 tt eeti Dengan cara yang sama
6/2
2 12 tt eeti
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t=0
2H 6H
12
3
Ateti t
s 2cos
tv
Ri 1Li 2Li+
-
Cari respons lengkap v(t) untuk rangkaian berikut
tvtvtv nf
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126
121
31
ss
sIsV
31
23
ss
ss
I
sVsH
s
Respons alami
tt
n BeAetv 3
Mencari respons paksaan
21 1 jssIs
222
21213
jj
jjsHsIsV s
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22
osV 452875,1
ot
f tetv 452cos2875,1
ottt teBeAetv 452cos2875,13
Respons lengkap
Kondisi awal Vv 30
2
2875,130 BAv
Kondisi awal untuk dtdv /
BAdt
dv
t
32
1
2
22875,1
0
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BAdt
dv
t
3625,50
Ritv 3 213 LLs iiitv
dt
di
dt
di
dt
di
dt
dv LLs 213
90
tdt
dv
didapat 125,1 0 BA
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Respons Lengkap
ott teetv 452cos2875,1125,1 3
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RASIO TEGANGAN
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IDEAL OP-AMP
H(s) = - Zf/Z1
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Cari besarnya tiap komponen dari rangkaian jika
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