pemodelan sistem dasar

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pemodelan fisis dengan matlab

Text of pemodelan sistem dasar

  • PERCOBAAN 2

    RESPON WAKTU FUNGSI TRANSFER

    Bila D 0

  • )1 +()4 +(3 + = )()4 +( +) +( =)4 +( + )1 +( =)1 +( + )4 +( = )1 +()4 +(3 +

    3 =4 +2 =31 = +

    32 =32 1 =31 =

    )1 +(766.0 + )4 +(333.0 =)1 +(32 + )4 +(31 = )(1 )1 +(32 +)4 +(31 = )(

    + = )() 1 ( slupmI tupmI .a

    )1 +(766.0 + )4 +(333.0 =)1 +()4 +(3 + =1*)()( = )(4 +5 +3 + = )(

  • + = )() ( .b

    )4 +( + )1 +( + =)1 +()4 +(3 + =4 +5 +3 + =4 +5 +3 + 1 = )()4 +(3380.0 )1 +(766.0 57.0 =

    . . . = )(( pmaR tupmI .c

    )

    )1 +(7666.0 + )4 +(8020.0 + 57.0 +5786.0 =1 + + 4 + + + = )4 +5 +(3 + =4 +5 +3 + 1 = )()1 +(1 7666.0 +)4 +(1 8020.0 +1 57.0 + 5786.0 =)1 +(7666.0 + )4 +(8020.0 + 57.0 + 5786.0 =

    . + . + . + . = )(( suniS tupmI .d

    = anamid )

    61 +02 +8 +5 +6 +2 =)4 +(2 4 +5 +3 + = )(

  • 4 +2 4.0 +4 + 3.0 1 +7662.0 + 4 +330.0 =4 +4.0 + 53.0 + 1 +7662.0 + 4 +330.0 = ) 2 +(1.0 + 51.0 + ) 2 (1.0 51.0 + 1 +7662.0 + 4 +330.0 =) 2 +(1.0 + 51.0 + ) 2 (1.0 51.0 + 1 +7662.0 + 4 +330.0 = . + . + . + . = )(

  • = )( .21 +)3 +(2 + =1 01 +6 +2 + = )(

    1 +)3 + (1 1 +)3 +(3 + = = )()1( slupmI tupmI .a

    1 +)3 +(2 + = )(1 01 +6 +2 + = )( = )s(Y

    01 +6 +2 + = )()() ( .b01 +6 +2 + =1 01 +6 +2 + = )(

    01 +6 + + + =01 +6 +2 +01 +6 +]) +([ + ])01 +6 +([ =01 +6 +2 +

    2 + = ]) +([ + ])01 +6 +([2 =012 + =01 +) +6( +) +(2 + = + +01 +6 +51 = 012 =0 = +

    =1 = +651 =

  • 1 = + )51 6(56 1 =

    51 =] 2 [ 51 1 51 =1 +)3 +(1 2 1 +)3 +(3 + 51 1 51 =1 +)3 +(2 1 +)3 +(3 + 51 1 51 =1 +)3 +(1 + 511 51 =01 +6 +51 51 +51 = )( 2 = ]) 2 ( 1[ = )(

    ( pmaR tupmI .c01 +6 +2 + = )()()

    1 +)3 +( + + + = 01 +6 +2 + =1 01 +6 +2 + = )() 3( 70.0 10.0 + ) + 3( 70.0 + 10.0 + 2.0 +20.0 =)1( +)3 +(]) 3 +() 70.0 10.0([ + ]) + 3 +() 70.0 + 10.0([ +2.0 + 20.0 =1 +)3 +(41.0 60.0 +20.0 +1 2.0 + 20.0 =

  • = 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)+ 10.08 1(+ 3)+ 1= 0.02 + 0.2 1+ 0.02 + 3(+ 3)+ 1

    0.06 1(+ 3)+ 10.08 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)+ 13 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)

  • )3856.1( +)5.1 +(89994.0 + 52254.0 5103.0 =)3856.1( +)5.1 +(89994.0 + 52254.0 + 5103.0 +)3856.1( +)5.1 +(69999.0 =)3856.1( +)5.1 +(3856.1 3856.169999.0 =)3856.1( +)5.1 +(3856.1 306.0 =

    . . . = )() ( .b

    2.0 + ) 3856.1 5.1( 5090.0 1.0 + ) 3856.1 + 5.1( 5090.0 + 1.0 =5 +3 +1 =1 5 +3 +1 =5 +3 +1 = )(2.0 + )3856.1( +)5.1 +(]) 3856.1 5.1 +() 5090.0 1.0([ +)3856.1( +)5.1 +(]) 3856.1 + 5.1 +() 5090.0 + 1.0([ =2.0 +)3856.1( +)5.1 +(3.0 3.0 2.0 =2.0 +)3856.1( +)5.1 +(6.0 2.0 =)3856.1( +)5.1 +(5.1 + 2.0 =

    )3856.1( +)5.1 +(5.1 2.01 2.0 +)3856.1( +)5.1 +(3856.1 3856.16.0

  • () = 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)3 +(1 23 =6 +23 =6 +23 = )(6 +23 = )( = )()1( slupmI tupmI .a

    )3 +(1 23 =6 +23 =1* = )(6 +23 = )( = )() ( .b

    3 +5.0 5.0 =6 +23 =1 6 +23 = )(6 +23 = )(3 +1 5.01 5.0 = )(

    . . = )(

  • 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() = . . + .

    PERCOBAAN 2

    RESPON WAKTU FUNGSI TRANSFER

    Bila D