What is the particular solution of the difference equation y(n)=\(\frac{5}{6}y(n-1)-\frac{1}{6}\)y(n-2)+x(n) when the forcing function x(n)=2^n, n≥0 and zero elsewhere?(a) \(\frac{1}{5}\) 2^n(b) \(\frac{5}{8}\) 2^n(c) \(\frac{8}{5}\) 2^n(d) \(\frac{5}{8}\) 2^-nThe question was posed to me by my school principal while I was bunking the class.My doubt stems from Discrete Time Systems Described by Difference Equations in chapter Discrete Time Signals and Systems of Digital Signal Processing

What is the particular solution of the difference equation y(n)=\(\fra

1.	What is the particular solution of the difference equation y(n)=\(\frac{5}{6}y(n-1)-\frac{1}{6}\)y(n-2)+x(n) when the forcing function x(n)=2^n, n≥0 and zero elsewhere?(a) \(\frac{1}{5}\) 2^n(b) \(\frac{5}{8}\) 2^n(c) \(\frac{8}{5}\) 2^n(d) \(\frac{5}{8}\) 2^-nThe question was posed to me by my school principal while I was bunking the class.My doubt stems from Discrete Time Systems Described by Difference Equations in chapter Discrete Time Signals and Systems of Digital Signal Processing
Answer» Correct choice is (c) \(\frac{8}{5}\) 2^n To explain I would say: The assumed solution of the difference equation to the forcing equation x(n), called the particular solution of the difference equation is yp(n)=Kx(n)=K2^nu(n) (where K is a scale factor) Upon substituting yp(n) into the difference equation, we obtain K2^nu(n)=\(\frac{5}{6}\)K2^n-1u(n-1)-\(\frac{1}{6}\) K2^n-2u(n-2)+2^nu(n) To DETERMINE K we must evaluate the above equation for any n>=2, so that no term vanishes. => 4K=\(\frac{5}{6}\)(2K)-\(\frac{1}{6}\) (K)+4 => K=\(\frac{8}{5}\) => yp(n)=(8/5) 2^n.

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