Question

laplace transform of cos t+ cos 2t from Pi 13 years ago

Answer 1

What are we doing here exactly?

Answer 2

cos 2t =2cos^2 t-1

Answer 3

you will get a quadratic in t

Answer 4

solve for t

Answer 5

Huueefetcha!!

Answer 6

note that Pi<t<2Pi

Answer 7

how will the pi range affects the ans if i simply use formula to take L.T of the function?

Answer 8

for example if i get t=5pi/4 i need to convert it into an angle btw pi and -pi

Answer 9

where cos has same value

Answer 10

gt it?

Answer 11

ok i see

Answer 12

so i express cos 2t in t first?

Answer 13

u know the formula \[\text{L{cos (at)}}=\frac{s}{a^2+s^2}\]

Answer 14

yup

Answer 15

so without consdering the pi range i get s/(s^2+1)+s/(s^2+4)

Answer 16

den how do i continue?

Answer 17

I think you need the shifted step function in there, to truncate your cosines...
been a while since I did this..

Answer 18

how to do that? need to plot out the graph and see?

Answer 19

checking...

Answer 20

Laplace transform is defined as F(s)=L[f(t)] = ∫ e^(-st) f(t) dt
Let f(t)=g(t)+h(t)
where g(t)=cos t and h(t)=cos 2t
The integral becomes
H(s)=L[h(t)]=∫ e^(-st) cos(2t) dt

integrate by parts with
u=cos(2t) → du= -2sin(2t)
dv=e^(-st) → v= -(1/s)e^(-st)

∫ cos(2t)e^(-st) dt
= -(1/s)e^(-st)cos(2t) - (b/s) ∫ e^(-st)sin(2t) dt

to solve the integrals on right side integrate by parts with
u=sin(2t) → du= 2cos(2t)
dv=e^(-st) → v= -(1/s)e^(-st)

∫ cos(2t)e^(-st) dt
= -(1/s)e^(-st)cos(2t) - (2/s) [ -(1/s)e^(-st)sin(2t) + (2/s)∫ e^(-st)cos(2t) dt
<=>
∫ cos(2t)e^(-st) dt
= -(1/s)e^(-st)cos(2t) + (2/s²)e^(-st)sin(2t) - (4/s²)∫ e^(-st)cos(2t) dt

solve for ∫ cos(2t)e^(-st) dt
(1 + (4/s²)) ∫ cos(2t)e^(-st) dt
= -(1/s)e^(-st)cos(2t) + (4/s²)e^(-st)sin(2t)
<=>
((s² + 4)/s²) ∫ cos(2t)e^(-st) dt
= -(1/s)e^(-st)cos(2t) + (2/s²)e^(-st)sin(2t)
<=>
∫ cos(2t)e^(-st) dt
= [ s²/(s² + 4)]( -(1/s)e^(-st)cos(2t) + (2/s²)e^(-st)sin(2t))
= (-scos(2t) + 2sin(2t))e^(-st)/(s² + 4)

Now apply the limits to find L[h(t)]
\[L [ \cos(2t) ] = \int\limits_{\pi}^{2\pi} \cos(2t) e^{-st} dt\]

Similarly, you can find L[g(t)] = (-scos t+sin t) e^(-st)/(s²+1)
Find F(s) by adding the two solutions.
L[f(t)]=L[g(t)]+L[h(t)]