tugas matrek laplace

Nama : Khusnul Khotimah

NRP : 24 11 100 703

Tugas : Matematika Rekayasa

A. Selesaikan Transformasi Laplace Berikut.

1. L { 4 e2t + 6 Sin 2t + 2t6 } = L (4e2t ) + L ( 6 Sin 2t) + L ( 2t6 )

= 4L (e2t ) + 6L (Sin 2t ) + 2L ( t6)

= 4 ( 1/s-2 ) + 6 ( 1/S2+22) + 4 (6!/66+1) = {4/(S-2)} + {6/(S2+4)} + {2880/ 67}

2. L {e-5t Sin 5t}= L (e-5t) L (Sin 5t)

= (1/S+5) (5/S2+25)

= {5/(S+5)2 + 25}

= {5/ S2+10S+50}

3. L (8 Cos 10t) = 8 L (Cos 10t)

={ 8 S/(S2 + 102) }

= {8S/(S2+100) }

4. L dari d2 x/dt2 + 12 dx/dt + 6x = Sin 3t, beserta fungsi transfernya.

L (dx2/dt2) = s2 x (s) – sd2 x/dt2 (0) – dx/dt (0)

L (dx/dt) = 12x (s) – x (0)

L (x) = x (s)

L (y) = {3/(s2+9) }

f (s) adalah; xs (s2 + 12s +6) = {3/(s2 + 9) }Dapat dituliskan fungsi transfernya, (Output/Input) = 3/s4 +12s3 +15s2 +109s+54

B. Selesaikan Transformasi Laplace Balik Berikut.5. L-1 dari f (s) = {(s3 + 4s + 8)/ (s+2)2 }

Dengan perumusan yang telah ditentukan; b3 = (s+2)2 (s3 +4s+8)/(s+2)2 s = -2

= (-23 + 4. -2+ 8)

= 8

b2 = d/ds {(s+2) (s3+4s+8) /(s+2)2 } s = -2

= 2s2 + 4 s = -2

= 4

b3 = 1/2 d2/ds2 { s3+4s+8} s = -2

= 1/2 (4s) s = -2

= -4

f(s) = -4/(s+2)2 + 4/(s+2)2 + 8/(s+2)2

= 8/(s+2)2

f(t) = 8/s2+4s+4

= 2/s2+4s

= 2/s(s+4)

= 1/s(s+a)

=