Question

What is the output y(n) when a signal x(n)=n*u(n)is passed through a accumulator system under the conditions that it is initially relaxed?

(a) \(\frac{n^2+n+1}{2}\)

(b) \(\frac{n(n+1)}{2}\)

(c) \(\frac{n^2+n+2}{2}\)

(d) None of the mentioned

I had been asked this question by my school principal while I was bunking the class.

I would like to ask this question from Discrete Time Systems topic in portion Discrete Time Signals and Systems of Digital Signal Processing

Answer 1

Right choice is (b) \(\frac{n(n+1)}{2}\)

The best explanation: Given that the system is initially relaxed, that is y(-1)=0

According to the equation of the accumulator,

y(n)=\(∑_{k=-∞}^n x(n)\)

=\(∑_{k=-∞}^{-1} x(n)+∑_{k=0}^n x(n)\)

=\(y(-1)+ ∑_{k=0}^n n*u(n)\)

=\(0+∑_{k=0}^n n\)(since u(n)=1 in 0 to n)

=\(\frac{n(n+1)}{2}\)