Derivative of unit step

  • Thread starter ColdStart
  • Start date
  • #1
19
0
ok letsa say i have d/dt {(u(-2-t) + u(t-2)}

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) ?
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,832
251
HI ColdStart! :wink:
ok letsa say i have d/dt {(u(-2-t) + u(t-2)}

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? :smile:
 
  • #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) ?
 
  • #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 :smile:, except it should still be -(t+2) inside the first q, shouldn't it? :wink:
 
  • #5
19
0
well then it turns out that its what i wrote in my first post, but in one book it shows -q(t+2)... thats why i got confused and was asking it here..
 
  • #6
tiny-tim
Science Advisor
Homework Helper
25,832
251
well then it turns out that its what i wrote in my first post, but in one book it shows -q(t+2)... thats why i got confused and was asking it here..

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
Is [tex]\frac{du}{dt}=\delta (t)[/tex] valid for all t?

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

Related Threads on Derivative of unit step

  • Last Post
Replies
6
Views
26K
  • Last Post
Replies
3
Views
4K
  • Last Post
Replies
1
Views
4K
Replies
1
Views
1K
Replies
2
Views
773
  • Last Post
Replies
3
Views
3K
Replies
1
Views
2K
  • Last Post
Replies
4
Views
47K
  • Last Post
Replies
2
Views
15K
Top