What is the derivative of a unit step function with discontinuities at -2 and 2?

  • Context: Undergrad 
  • Thread starter Thread starter ColdStart
  • Start date Start date
  • Tags Tags
    Derivative Unit
Click For Summary

Discussion Overview

The discussion focuses on the derivative of a unit step function with discontinuities at -2 and 2, exploring the application of the chain rule and the properties of the derivative in relation to the unit step function and the Dirac delta function.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant proposes that the derivative of the sum of unit step functions can be expressed as the sum of their derivatives, questioning if d/dt {(u(-2-t) + u(t-2)} equals q(-2-t) + q(t-2).
  • Another participant suggests using the chain rule to find the derivative, prompting a reevaluation of the expression.
  • A participant reformulates the derivative as d/dt{ u(-(t+2)) + u(t-2)} and questions if it equals -q(t+2) + q(t-2).
  • There is a discussion about the correct form of the derivative, with one participant noting a discrepancy between their expression and a book's notation, leading to confusion.
  • Another participant points out that the book's result would be consistent if q is considered an odd function.
  • A new question arises regarding the validity of the expression \(\frac{du}{dt}=\delta(t)\) for all t, highlighting concerns about the continuity of u(t) at t=0.

Areas of Agreement / Disagreement

Participants express differing views on the correct formulation of the derivative of the unit step function, and the discussion remains unresolved regarding the validity of certain expressions and the implications of continuity.

Contextual Notes

Participants reference specific mathematical properties and definitions, such as the chain rule and the nature of the Dirac delta function, which may depend on interpretations and assumptions about continuity and odd functions.

ColdStart
Messages
18
Reaction score
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) ?
 
Physics news on Phys.org
HI ColdStart! :wink:
ColdStart said:
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:
 
ok so:
d/dt {(u(-2-t) + u(t-2)} = d/dt{ u(-(t+2)) + u(t-2)} = -q(t+2) + q(t-2) ?
 
ColdStart said:
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:
 
well then it turns out that its what i wrote in my first post, but in one book it shows -q(t+2)... that's why i got confused and was asking it here..
 
ColdStart said:
well then it turns out that its what i wrote in my first post, but in one book it shows -q(t+2)... that's why i got confused and was asking it here..

No, in your first post you had …
ColdStart said:
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)
 
cool thanks!
 
Is [tex]\frac{du}{dt}=\delta (t)[/tex] valid for all t?

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

Similar threads

  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 33 ·
2
Replies
33
Views
4K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 19 ·
Replies
19
Views
5K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K