Convolution of discrete and continuous time signals

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SUMMARY

Convolution of discrete-time signals with continuous-time signals is feasible and can be mathematically expressed. Given a discrete-time signal x(nT) and a continuous-time signal y(t), the convolution can be computed by treating the discrete signal as a series of impulses using the delta function. The resulting convolution is represented as x(t) ⊗ y(t) = ∑ x(nT) y(t - nT), assuming convergence of the series. This approach effectively combines the two signal types in a coherent manner.

PREREQUISITES
  • Understanding of discrete-time signals and continuous-time signals
  • Familiarity with convolution operations in signal processing
  • Knowledge of delta functions and their properties
  • Basic grasp of signal convergence concepts
NEXT STEPS
  • Study the properties of convolution in signal processing
  • Learn about delta functions and their applications in signal representation
  • Explore convergence criteria for series in signal analysis
  • Investigate practical applications of discrete and continuous signal convolution in systems
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Signal processing engineers, students in electrical engineering, and researchers working with mixed signal types in analysis and system design.

swuster
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Not a specific question per se but...

Is it possible to convolve a discrete-time signal with a continuous-time one?

if you have x(n) and y(t) can you calculate the convolution of x and y (say, by taking y(t) for t in the set of integers or by treating each x(n) as its value multiplied by the delta function) or is there something inherently wrong about this?
 
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A discrete-time signal x(nT), where T is the sampling period, is a special case of a continuous-time signal, as you can write

x(t)=\sum_{n}x(nT)\delta(t-nT)

so if y(t) is a continuous-time signal, the convolution is

x(t)\otimes y(t)=\sum_{n}x(nT)\delta(t-nT)\otimes y(t)=\sum_{n}x(nT)y(t-nT)

(you have to assume this converges).
 
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