Integration with a unit step function

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SUMMARY

The discussion centers on the integration of a function with a unit step function, specifically the Heaviside step function, denoted as u(t). Participants clarify that u(t) equals 1 for t ≥ 0 and 0 for t < 0, leading to the conclusion that the integral from -∞ to ∞ of g(t) multiplied by u(t) simplifies to the integral from 0 to ∞ of g(t). This understanding is crucial for correctly applying the unit step function in definite integrals.

PREREQUISITES
  • Understanding of the Heaviside step function (u(t))
  • Knowledge of definite integrals in calculus
  • Familiarity with the properties of integrals involving piecewise functions
  • Basic concepts of mathematical notation and limits
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  • Study the properties of the Heaviside step function in detail
  • Learn about the application of unit step functions in signal processing
  • Explore integration techniques involving piecewise functions
  • Investigate the use of Laplace transforms with unit step functions
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Students and professionals in mathematics, engineering, and physics who are dealing with integration involving unit step functions, particularly in the context of signal processing and control systems.

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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