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pemodelan fisis dengan matlab
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PERCOBAAN 2
RESPON WAKTU FUNGSI TRANSFER
Bila D<0
+( + ) + =
( + ) + =Cari respon waktu dari fungsi transfer
1. ( ) =2. ( ) =
3. ( ) =4. ( ) =
A. Berdasar model matematik diatas cari responnya bila sistem di beri input
a) Impulse
b) Step
c) Ramp
d) Sinus = 2
Gunakan tabel Laplace dengan mengubah fungsi transfer ke domain waktu
B. Berdasarkan Model tersebut diatas cari responnya dengan input yang sama menggunakan Progam M-File dan Simulink
Catatan ambil waktu sampling 0,2 detik
C. Cari grafik error sistem fungsi waktu
Jawab
A. Respon Fungsi Transfer
1. ( ) =( )( ) = + 3+ 5 + 4 > 0
( ) = + 3( + 4)( + 1)+ 3( + 4)( + 1) = ( + 4) + ( + 1)
= ( + 1) + ( + 4)= ( + ) + ( + 4 )
+ 4 = 3+ = 1
3 = 2= 23= 1 − 23= 13
( ) =13( + 4) +
23( + 1)= 0.333( + 4) + 0.667( + 1)
( ) =13( + 4) +
23( + 1) ∗ 1
( ) = +a. Imput Impuls ( 1 )
( ) = + 3+ 5 + 4( ) = ( )( )*1
= + 3( + 4)( + 1)= 0.333( + 4) + 0.667( + 1)
( ) = +
b. ( )
( ) = 1 ∗ + 3+ 5 + 4= + 3+ 5 + 4= + 3( + 4)( + 1)= + ( + 1) + ( + 4)= 0.75 − 0.667( + 1) − 0.0833( + 4)
( ) = . − . − .
c. Imput Ramp ( )
( ) = 1 ∗ + 3+ 5 + 4= + 3( + 5 + 4) = + + + 4 + + 1= − 0.6875 + 0.75 + 0.0208( + 4) + 0.6667( + 1)
= −0.6875 + 0.75 + 0.0208( + 4) + 0.6667( + 1)= −0.6875 ∗ + 0.75 ∗ 1 + 0.0208 ∗ 1( + 4)
+ 0.6667 ∗ 1( + 1)( ) = – . + . + . + .
d. Imput Sinus ( ) dimana =( ) = + 3+ 5 + 4 ∗ 2( + 4)
= 2 + 6+ 5 + 8 + 20 + 16
= 0.033+ 4 + 0.2667+ 1 + −0.15 − 0.1( − 2 ) + −0.15 + 0.1( + 2 )= 0.033+ 4 + 0.2667+ 1 + −0.15 − 0.1( − 2 ) + −0.15 + 0.1( + 2 )= 0.033+ 4 + 0.2667+ 1 + −0.35 + 0.4+ 4= 0.033+ 4 + 0.2667+ 1 − 0.3 + 4 + 0.4 2+ 4
( ) = . + . + . + .
2. ( ) =( ) = + 2+ 6 + 10 ∗ 1
= + 2( + 3) + 1= + 3( + 3) + 1 − 1( + 3) + 1
( ) = −a. Imput Impuls (1)
( ) = + 2+ 6 + 10 ∗ 1( ) = + 2( + 3) + 1
Y(s) = −
b. ( )
( )( ) = + 2+ 6 + 10( ) = + 2+ 6 + 10 ∗ 1
= + 2+ 6 + 10+ 2+ 6 + 10 = + ++ 6 + 10+ 2+ 6 + 10 = [ ( + 6 + 10)]+ [ ( + )]+ 6 + 10
[ ( + 6 + 10)]+ [ ( + )] = + 2+ 6 + 10 + + = + 2
( + ) + (6 + ) + 10 = + 210 = 2
= 210 = 15+ = 0
= −= − 15
6 + = 1
(6 ∗ 15) + = 1= 1 − 65
= − 15
( ) =15 + −15 − 15+ 6 + 10
= 15 ∗ 1 − 15 ∗ + 1( + 3) + 1= 15 ∗ 1 − 15 + 3( + 3) + 1 − 2( + 3) + 1= 15 ∗ 1 − 15 + 3( + 3) + 1 − 2 ∗ 1( + 3) + 1= 15 ∗ 1 − 15 [ − 2 ]
( ) = [1 − ( − 2 )] = − ∗ − 2 ∗
c. Imput Ramp ( )
( )( ) = + 2+ 6 + 10( ) = + 2+ 6 + 10 ∗ 1
= + 2+ 6 + 10 = + + +( + 3) + 1= −0.02 + 0.2 + 0.01 + 0.07− (−3 + ) + 0.01 − 0.07− (−3 − )= −0.02 + 0.2+ [(0.01 + 0.07 )( + 3 + )] + [(0.01 − 0.07 )( + 3 − )]( + 3) + (1)= −0.02 ∗ + 0.2 ∗ 1 + 0.02 + 0.06 − 0.14( + 3) + 1
= −0.02 ∗ + 0.2 ∗ 1 + 0.02 − 0.08( + 3) + 1= −0.02 ∗ + 0.2 ∗ 1 + 0.02( + 3) + 1 − 0.08( + 3) + 1= −0.02 ∗ + 0.2 ∗ 1 + 0.02 ∗ + 3( + 3) + 1
− 0.02 ∗ 3( + 3) + 1 − 0.08 ∗ 1( + 3) + 1= −0.02 ∗ + 0.2 ∗ 1 + 0.02 ∗ + 3( + 3) + 1
− 0.06 ∗ 1( + 3) + 1 − 0.08 ∗ 1( + 3) + 1( ) = −0.02 + 0.2+ 0.02 cos − 0.06 sin − 0.08 sin
= − . + . + . − .
d. Imput Sinus ( ) dimana =( )( ) = + 2+ 6 + 10( ) = + 2+ 6 + 10 ∗ 2+ 4
= 2 + 4+ 6 + 14 + 24 + 40= −0.0333 – 0.1− 2 + −0.0333 + 0.1+ 2 + 0.0333 + 0.1− (−3 + )
+ 0.0333 – 0.1− (−3 − )= [(−0.0333− 0.1 )( + 2 )] + [(−0.0333 + 0.1 )( − 2 ) ]+ 4+ [(0.0333 + 0.1 )( + 3 + )] + [(0.0333 − 0.1 )( + 3 − )]( + 3) + 1= −0.0666 + 0.4+ 4 + 0.0666 + 0.1998 − 0.2( + 3) + 1= −0.0666 + 0.4+ 4 + 0.0666 − 0.0002( + 3) + 1
= −0.0666+ 4 + 0.4+ 4 + 0.0666( + 3) + 1 + −0.0002( + 3) + 1= −0.0002 ∗ 1( + 3) + 1 + −0.0666 ∗ + 4
+ 0.42 ∗ 2+ 4+ 0.0666 + 3( + 3) + 1 − 3 ∗ 1( + 3) + 1( ) = (−0.0666 cos 2 ) + (0.2 sin 2 )+ [0.0666( cos − 3 sin )] +(−0.0002 sin )= (− . ) + ( . ) + . −.
3. ( ) =( ) = 1+ 3 + 5( ) = 1+ 3 + 5
= −0.3015− (−1.5 + 1.6583 ) + 0.3015− (−1.5 − 1.6583 )= (−0.3015 )( + 1.5 + 1.6583 )
( + 1.5) + (1.6583)+ (0.3015 )( + 1.5 − 1.6583 )( + 1.5) + (1.6583)= −0.3015 − 0.45225 + 0.49998( + 1.5) + (1.6583)+ 0.3015 + 0.45225 + 0.49998( + 1.5) + (1.6583)= 0.99996( + 1.5) + (1.6583)= 0.999961.6583 ∗ 1.6583( + 1.5) + (1.6583)= 0.603 ∗ 1.6583( + 1.5) + (1.6583)
( ) = . . .a. Imput Impuls (1)
( ) = 1+ 3 + 5( ) = 1+ 3 + 5 ∗ 1
= −0.3015− (−1.5 + 1.6583 ) + 0.3015− (−1.5 − 1.6583 )= [(−0.3015 )( + 1.5 + 1.6583 )]
( + 1.5) + (1.6583)+ (−0.3015 )( + 1.5 + 1.6583 )
( + 1.5) + (1.6583)
= −0.3015 − 0.45225 + 0.49998( + 1.5) + (1.6583)+ 0.3015 + 0.45225 + 0.49998( + 1.5) + (1.6583)= 0.99996( + 1.5) + (1.6583)= 0.999961.6583 ∗ 1.6583( + 1.5) + (1.6583)= 0.603 ∗ 1.6583( + 1.5) + (1.6583)
( ) = . . .b. ( )
( ) = 1+ 3 + 5= 1+ 3 + 5 ∗ 1
= 1+ 3 + 5= −0.1 + 0.0905− (−1.5 + 1.6583 ) + −0.1 − 0.0905− (−1.5 − 1.6583 ) + 0.2
= [(−0.1 + 0.0905 )( + 1.5 + 1.6583 )]( + 1.5) + (1.6583)+ [(−0.1 − 0.0905 )( + 1.5 − 1.6583 )]( + 1.5) + (1.6583) + 0.2
= −0.2 − 0.3 − 0.3( + 1.5) + (1.6583) + 0.2
= −0.2 − 0.6( + 1.5) + (1.6583) + 0.2
= −0.2 ∗ + 1.5( + 1.5) + (1.6583)− −0.2 ∗ 1.5( + 1.5) + (1.6583)− 0.61.6583 ∗ 1.6583( + 1.5) + (1.6583) + 0.2 ∗ 1
( ) = −0.2 . cos 1.6583 + 0.2 . 1.6583− 0.3618 . sin 1.6583 + 0.2= − . . . − . . . + .
c. Imput Ramp ( )
( ) = 1+ 3 + 5( ) = 1+ 3 + 5 ∗ 1
= 1+ 3 + 5= 0.06 + 0.006− (−1.5 + 1.6583 ) + 0.06 − 0.006− (−1.5 − 1.6583 )
+ −0.12 + 0.2
= 0.06 + 0.006− (−1.5 + 1.6583 ) + 0.06 − 0.006− (−1.5 − 1.6583 ) + −0.12 + 0.2
= [( 0.06 + 0.006 )( + 1.5 + 1.6583 )]( + 1.5) + (1.6583)
+ [( 0.06 − 0.006 )( + 1.5 − 1.6583 )]( + 1.5) + (1.6583)+ −0.12 ∗ 1 + 0.2 ∗ 1
= 0.12 + 0.18 − 0.0199( + 1.5) + (1.6583) + −0.12 ∗ 1 + 0.2 ∗ 1
= 0.12 + 0.1601( + 1.5) + (1.6583) + −0.12 ∗ 1 + 0.2 ∗ 1
= 0.12( + 1.5) + (1.6583) + 0.1601( + 1.5) + (1.6583)+ −0.12 ∗ 1 + 0.2 ∗ 1
= 0.12 + 1.5( + 1.5) + (1.6583) − 1.51.6583 ∗ 1.6583( + 1.5) + (1.6583)+ 0.16011.6583 ∗ 1.6583( + 1.5) + (1.6583) + −0.12 ∗ 1+ 0.2 ∗ 1
( ) = 0.12( . cos 1.6583− 0.9045 . sin 1.6583 )+ 0.0965 . sin 1.6583 − 0.12 + 0.2= . . . − . . . −. + .
d. Imput Sinus ( ) dimana =
( ) = 1+ 3 + 5( ) = 1+ 3 + 5 ∗ 2+ 4
= 2+ 3 + 9 + 12 + 20= −0.0811 – 0.0135− 2 + −0.0811 + 0.0135+ 2
+ 0.0811 – 0.0570− (−1.5 + 1.6583 ) + 0.0811 + 0.0570− (−1.5 − 1.6583 )= ( −0.0811− 0.0135 )( + 2 )
+ 4+ ( −0.0811 + 0.0135 )( − 2 )+ 4
+ [(0.0811 − 0.0570 )( + 1.5 + 1.6583 )]( + 1.5) + (1.6583)+ [(0.0811 + 0.0570 )( + 1.5 − 1.6583 )]( + 1.5) + (1.6583)= −0.1622 + 0.0540+ 4 + 0.1622 + 0.2433 + 0.18904( + 1.5) + (1.6583)= −0.1622 + 0.0540+ 4 + 0.1622 + 0.43234( + 1.5) + (1.6583)= −0.1622+ 4 + 0.0540+ 4 + 0.1622( + 1.5) + (1.6583)
+ 0.43234( + 1.5) + (1.6583)
= −0.1622 ∗ + 4 + 0.05402 ∗ 2+ 4+ 0.1622 + 1.5( + 1.5) + (1.6583) − 1.51.6583∗ 1.6583( + 1.5) + (1.6583)+ 0.432341.6583 ∗ 1.6583( + 1.5) + (1.6583)
( )= −0.1622 cos 2+ 0.0270 sin 2+ 0.1622( . cos 1.6583− 0.9045 . sin 1.6583 ) + 0.2607 . sin 1.6583
=− . +. +. . . + . . .
4. ( ) =( ) = 32 + 6( ) = 32 + 6
= 32 + 6= 32 ∙ 1( + 3)
( ) =a. Imput Impuls (1)
( ) = 32 + 6( ) = *1
= 32 + 6= 32 ∙ 1( + 3)
( ) =
b. ( )
( ) = 32 + 6( ) = 32 + 6 ∗ 1
= 32 + 6= 0.5 − 0.5+ 3
( ) = 0.5 ∗ 1 − 0.5 ∗ 1+ 3( ) = . − .
c. Imput Ramp ( )
( ) = 32 + 6( ) = 32 + 6 ∗ 1
= 32 + 6 = + + + 3= −0.1667 + 0.5 + 0.1667+ 3= −0.1667 + 0.5 + 0.1667+ 3= −0.1667 ∗ 1 + 0.5 ∗ 1 + 0.1667 ∗ 1+ 3
( ) = − . + . + .
d. Imput Sinus ( ) dimana =( ) = 32 + 6( ) = 32 + 6 ∗ 4+ 4
= 122 + 6 + 8 + 24= 0.4615+ 3 + −0.2308 – 0.3462− 2 + −0.2308 + 0.3462+ 2= 0.4615+ 3
+ (−0.2308 – 0.3462 )( + 2 )+ 4
+ (−0.2308 + 0.3462 )( − 2 )+ 4= 0.4615+ 3 + −0.4615 + 1.3848+ 4= 0.4615+ 3 + −0.4615+ 4 + 1.3848+ 4= 0.4615 ∗ 1+ 3 + −0.4615 ∗ + 4 + 1.38482 ∗ 2+ 4( ) = . − . + .