Question

I'm dealing with some laplace transforms for the first time regarding trig functions mixed with Heaviside and am running into trouble; posted below momentarily.

Answer 1

\[\mathcal{L}\left\{\cos(2t)\mathcal{U}(t-\pi)\right\}\]

Answer 2

I'm confused about how to put the argument of the cosine function in terms of t-pi; I'm thinking of using trigonometric identities, but I can't find any looking through pages that deal with pi, only pi/2.

Answer 3

@midhun.madhu1987

Answer 4

@Kainui :|

Answer 5

I am not aware of this :(

So sorry

Answer 6

No problem, thank you nonetheless!

Answer 7

If you have something with pi/2 surely if you apply it twice you can get yourself to pi?

As for the Heaviside function is concerned, all that really does is means that from -infinity to pi the function is 0 and from pi to +infinity the function is normal. So just adjust your bounds of integration accordingly and play around with your identities.

Answer 8

period of cos(2t) is pi so cos(2t) = cos(2t \(\pm\) pi)

Answer 9

I know how the Heaviside function works and how the lower limit contribution disappears in the derivation for it; I'm just confused about how I should play around with those identities. I'm not sure how to move forward from what @ganeshie8 posted, but I'm thinking over it.

Answer 10

does this help

cos(2t) = cos(2t-2pi) = cos(2(t-pi))

Answer 11

Oh, yeah, hah, that makes sense, I understand that more than I do the first mention. I want to think over your first post for a minute, but I can see how the second one would work.

Answer 12

Alright, so from the two posts, I'm confused on two different accounts unless there was a typo in the first one. Changing the argument of a trig function changes its period, so-in agreement with the first post-\[\cos(2t)=\cos(2t \pm \pi)\]

But in the second post, it looks like the periodicity of it didn't change despite being multiplied by 2, and 2pi is still being used for the period.

Answer 13

adding / subtracting ANY natural number of `pi`s from the argument of cos(2t) wont change its value cuz cos(2t) repeats itself after every cycle of period, pi.

Answer 14

cos(2t + pi) = cos(2t + pi + pi) = cos(2t + pi + pi + ... )

Answer 15

Oh, yeah, derp.

Answer 16

Alright, so \[\mathcal{L}\left\{\cos(2t)\mathcal{U}(t-\pi)\right\}=\mathcal{L}\left\{\cos(2t-2\pi)\mathcal{U}(t-\pi)\right\}=\mathcal{L}\left\{\cos(2(t-\pi)\mathcal{U}(t-\pi)\right\}\]

Answer 17

From there, the identification\[a=\pi, \ f(t)=\cos(2t)\]

Answer 18

\[\mathcal{L}\left\{\cos(2t)\mathcal{U}(t-\pi)\right\}=\frac{se^{-\pi s}}{s^2+4}\]

Answer 19

I'm not sure what you mean, sorry. I get it now, though.

Answer 20

http://prntscr.com/58i841

Answer 21

nvm i see what you're doing now :)

Answer 22

ty!