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 mentionedI 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

What is the output y(n) when a signal x(n)=n*u(n)is passed through a a

1.	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 mentionedI 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» 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) =\(\frac{n(n+1)}{2}\)

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