Discrete-time convolution has the same fundamental relationship as continuous-time convolution. Which statement best describes this relationship?

Discrete-time convolution has the same fundamental relationship as continuous-time convolution. Which statement best describes this relationship?

• A. It yields a time-domain product of inputs
• B. It is equivalent to multiplication in time
• C. It yields the output as the sum of products of the input with time-shifted impulse response
• D. It cannot be used with non-periodic signals

Answer: (C)

In discrete-time convolution the output of an LTI system is built as a weighted sum of impulse responses that are shifted in time, with the weights taken from the input samples. Concretely, y[n] is formed by summing x[k] times h[n−k] over all k, which means each input sample x[k] contributes a shifted copy of the impulse response to the output, and all these contributions are added together. This mirrors the continuous-time case, where the output is the integral of x(τ) times h(t−τ); here the integral becomes a sum and the delay becomes an integer shift.

This description is why the statement is the best fit: it captures the superposition of time-shifted copies of the impulse response weighted by the input samples, which is exactly what the discrete-time convolution does. The alternative idea of simply multiplying the signals in time would miss the way the system’s memory (the impulse response) spreads and combines past input samples. Similarly, convolution is not limited to periodic signals; it applies to non-periodic signals as well, with periodic cases yielding a special kind of cyclic convolution.