What is the Convolution of a Unit Step Function and an Exponential Function?

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SUMMARY

The convolution of a unit step function, h(t) = u(t), and an exponential function, x(t) = e-t, is calculated over the interval from 0 to t. The integral of e from 0 to t results in -e-t, which indicates a misunderstanding in the application of the convolution theorem. The discussion clarifies that this is a discrete convolution, emphasizing the need for proper limits and definitions in the convolution process.

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  • Understanding of convolution in signal processing
  • Familiarity with unit step functions
  • Knowledge of exponential functions and their properties
  • Basic calculus, specifically integration techniques
NEXT STEPS
  • Study the properties of discrete convolution in signal processing
  • Learn about the unit step function and its applications
  • Explore integration techniques for exponential functions
  • Review the convolution theorem and its implications in systems analysis
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Students and professionals in mathematics, engineering, and signal processing who are working with convolution operations and need to understand the interaction between step and exponential functions.

Quincy
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Homework Statement


h(t) = u(t) (the unit step function)

x (t) = e-t

The Attempt at a Solution



There is only one interval where the two functions overlap, and that's from 0 to t.

The integral from 0 to t of e-\tau d\tau = -e-t

Doesn't look right to me... what am I doing wrong?

EDIT: This is discrete convolution, by the way.
 
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