Question

What is the value of y(n)-y(n-1) in terms of input x(n)?

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

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

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

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

The question was posed to me in final exam.

My doubt is from Bilinear Transformations in portion Digital Filters Design of Digital Signal Processing

Answer 1

Right answer is (a) \([\frac{x(n)+x(n-1)}{2}]T\)

Best explanation: We know that the derivative equation is

dy(t)/dt=x(t)

On applying integrals both sides, we get

\(\int_{(n-1)T}^{nT}dy(t)=\int_{(n-1)T}^{nT} x(t)dt\)

=> y(nT)-y[(n-1)T]=\(\int_{(n-1)T}^{nT} x(t)dt\)

On applying trapezoidal rule on the right hand integral, we get

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

Since x(n) and y(n) are approximately equal to x(nT) and y(nT) respectively, the above equation can be written as

y(n)-y(n-1)=\([\frac{x(n)+x(n-1)}{2}]T\)