Above youtube video mentions a tabular method for finding convolution in case of discrete time.

But i am bit confused? Is this tabular method only applicable to discrete time or is it also valid for continous time?

My own understanding is that this method is only valid for discrete time,please correct me if i am wrong


1 Answer 1


To the extent a Riemann sum is considered a method for computing the integration of a continuous time signal, the tabular method also finds the convolution of a continuous time signal by using samples of that signal. Both methods are approximations with decreasing error as the sampling rate is increased.

The summary below compares the formulas for convolution in continuous and discrete time. The "tabular" method is an approach to do the Discrete Convolution as described by the equation below. Below that I provide further details showing how integration in continuous time is related to summation in discrete time.

Continuous Time Convolution

$$x(t)*y(t) \equiv \int_{\tau=-\infty}^{\infty}x(\tau)y(t-\tau)d\tau$$

Discrete Time Convolution

$$x[nT] * y[nT] \equiv \sum_{k=-\infty}^{\infty}x[kT]y[nT-kT]\Delta T$$

Discrete Convolution

$$x[n] * y[n] \equiv \sum_{k=-\infty}^{\infty}x[k]y[n-k]$$

With the difference that Discrete Time Convolution is given in units that match the Continuous Time Convolution (such as time in seconds), and Discrete Convolution is normalized time given in samples. With that we see that in order for the tabular method to approximate the Continuous Time Convolution result, the Discrete Convolution result must be multiplied by $\Delta T$. As the sampling rate is increased ($\Delta T$ gets smaller), the approximation error gets smaller.

Further Background on Riemann Sum and Integration

As demonstrated in the graphic below, using the "area under the curve" interpretation of a definite integral we can see intuitively why integration in continuous time is similar to summation in discrete time.

integration and summation

And specifically the definite integral is the limit of the Riemann sum as $\Delta T$ approaches $0$:

$$\int_{t_1}^{t_2}x(t)dt = \lim_{N\rightarrow\infty}\sum_{n=0}^{N-1}x[nT]\Delta T$$

For $N$ samples over the time duration $T=t_2-t_1$, where $\Delta T= \frac{T}{N}$.


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