Integration with a unit step function

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Homework Help Overview

The discussion revolves around the integration of a function involving a unit step function, specifically addressing the behavior of the unit step function u(t) over different intervals.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to clarify the definition of the unit step function u(t) and its implications for integration limits. There is a focus on whether u(t) is equal to 1 or 0 over specified intervals and how this affects the integration of another function g(t).

Discussion Status

There is an active exploration of different interpretations regarding the unit step function's definition. Some participants are questioning the original poster's understanding and suggesting alternative views on the behavior of u(t) across the defined intervals.

Contextual Notes

Participants are referencing an image for clarification, which is not visible in the text. There may be confusion stemming from the visual representation of the unit step function and its integration limits.

killerfish
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Hi,

i have a problem with integration a function with a unit step function.

Homework Statement


Given,

eafe.JPG


Refer to the image, i dun understand is that u(t) is equal to 1 from a definite integration from -\infty to \infty since u(t)=1 from -\infty to 0 and u(t)=0 from 0 to \infty.


Thanks.
 
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killerfish said:
Hi,

i have a problem with integration a function with a unit step function.

Homework Statement


Given,

View attachment 23734

Refer to the image, i dun understand is that u(t) is equal to 1 from a definite integration from -\infty to \infty since u(t)=1 from -\infty to 0 and u(t)=0 from 0 to \infty
since u(t) = 0 for all t < 0.


Thanks.

Isn't it the other way around like in your drawing? I.e., that u(t) = 0 for t < 0 and u(t) = 1 for 0 <= t < infinity?

That means that
\int_{-\infty}^{\infty} g(t) u(t)dt = \int_0^{\infty} g(t) dt
 
Mark44 said:
Isn't it the other way around like in your drawing? I.e., that u(t) = 0 for t < 0 and u(t) = 1 for 0 <= t < infinity?

That means that
\int_{-\infty}^{\infty} g(t) u(t)dt = \int_0^{\infty} g(t) dt

so if i have muliplication of few unit step function like in the image below,

faea.JPG


am i right this way?
 
Ayuh.
 

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