- #1

- 19

- 0

I know that d/dt { u(t) } is q(t)...

now is it correct to think that d/dt {(u(-2-t) + u(t-2)} = q(-2-t) + q(t-2) ?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter ColdStart
- Start date

- #1

- 19

- 0

I know that d/dt { u(t) } is q(t)...

now is it correct to think that d/dt {(u(-2-t) + u(t-2)} = q(-2-t) + q(t-2) ?

- #2

tiny-tim

Science Advisor

Homework Helper

- 25,832

- 251

I know that d/dt { u(t) } is q(t)...

now is it correct to think that d/dt {(u(-2-t) + u(t-2)} = q(-2-t) + q(t-2) ?

Nope … try again, using the chain rule (with g = -2-t) …

what do you get?

- #3

- 19

- 0

ok so:

d/dt {(u(-2-t) + u(t-2)} = d/dt{ u(-(t+2)) + u(t-2)} = -q(t+2) + q(t-2) ?

d/dt {(u(-2-t) + u(t-2)} = d/dt{ u(-(t+2)) + u(t-2)} = -q(t+2) + q(t-2) ?

- #4

tiny-tim

Science Advisor

Homework Helper

- 25,832

- 251

ok so:

d/dt {(u(-2-t) + u(t-2)} = d/dt{ u(-(t+2)) + u(t-2)} = -q(t+2) + q(t-2) ?

Yes ,

- #5

- 19

- 0

- #6

tiny-tim

Science Advisor

Homework Helper

- 25,832

- 251

No, in your first post you had …

q(-2-t) + q(t-2) ?

with no minus in front of the q.

(the book result would be the same if q is an odd function)

- #7

- 19

- 0

cool thanks!

- #8

- 338

- 0

But the function u(t) is not continuous at t=0.

Share: