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

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SUMMARY

The derivative of the unit step function, represented as u(t), with discontinuities at -2 and 2 is calculated using the chain rule. The correct expression for the derivative is d/dt {(u(-2-t) + u(t-2)} = -q(t+2) + q(t-2), where q(t) is the derivative of u(t). The confusion arose from differing representations in literature, highlighting the importance of careful notation in mathematical expressions.

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ColdStart
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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) ?
 
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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 \frac{du}{dt}=\delta (t) valid for all t?

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

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