Question

What is the value of \(\int_{(n-1)T}^{nT} x(t)dt\) according to trapezoidal rule?

(a) \([\frac{x(nT)-x[(n-1)T]}{2}]T\)

(b) \([\frac{x(nT)+x[(n-1)T]}{2}]T\)

(c) \([\frac{x(nT)-x[(n+1)T]}{2}]T\)

(d) \([\frac{x(nT)+x[(n+1)T]}{2}]T\)

I had been asked this question in an international level competition.

My query is from Bilinear Transformations topic in section Digital Filters Design of Digital Signal Processing

Answer 1

Correct choice is (b) \([\frac{x(nT)+x[(n-1)T]}{2}]T\)

The best explanation: The given integral is approximated by the trapezoidal rule. This rule states that if T is small, the area (integral) can be approximated by the mean height of x(t) between the two limits and then multiplying by the width. That is

\(\int_{(n-1)T}^{nT} x(t)dt=[\frac{x(nT)+x[(n-1)T]}{2}]T\)