Question

The Fourier transform of h(n) is defined as $H\left(e^{j\omega }\right)$, where h(n) is the impulse response of the system whose input is
x(n) and output y(n). If $h\left(n\right)=3{\left(\frac{1}{2}\right)}^{n}u\left(n\right)-2{\left(\frac{-1}{3}\right)}^{n}u\left(n\right),$ then which of the following difference equations represent the system

• A. $y\left(n\right)=\frac{1}{6}y(n-1)+\frac{1}{6}y(n-2)+x\left(n\right)+2x(n-1)$
• B. $y\left(n\right)=\frac{5}{6}y(n-1)-\frac{1}{6}y(n-2)+x\left(n\right)$
• C. $y\left(n\right)=-\frac{1}{6}y(n-1)-\frac{1}{6}y(n-2)+2x(n-1)$
• D. $y\left(n\right)=\frac{-5}{6}y(n-1)-\frac{1}{6}y(n-2)+x\left(n\right)$

Answer 1

The correct option is A $y\left(n\right)=\frac{1}{6}y(n-1)+\frac{1}{6}y(n-2)+x\left(n\right)+2x(n-1)$

$h\left(n\right)=3{\left(\frac{1}{2}\right)}^{n}u\left(n\right)-2{(-\frac{1}{3})}^{n}u\left(n\right)$

Taking 'z' transform, we get

$H\left(z\right)=\frac{3}{(1-\frac{1}{2}{z}^{-1})}-\frac{2}{(1+\frac{1}{3}{z}^{-1})}$

$\frac{Y\left(z\right)}{X\left(z\right)}=\frac{3+z^{-1}-2+z^{-1}}{1-\frac{1}{6}{z}^{-1}-\frac{1}{6}{z}^{-2}}=\frac{1+2z^{-1}}{1-\frac{1}{6}{z}^{-1}-\frac{1}{6}{z}^{-2}}$

$Y\left(z\right)-\frac{1}{6}{z}^{-1}Y\left(z\right)-\frac{1}{6}{z}^{-2}Y\left(z\right)=X\left(z\right)+2z^{-1}X\left(z\right)$

Taking inverse 'z' transform

$y\left(n\right)-\frac{1}{6}y(n-1)-\frac{1}{6}y(n-2)=x\left(n\right)+2x(n-1)$


$y\left(n\right)=\frac{1}{6}y(n-1)+\frac{1}{6}y(n-2)+x\left(n\right)+2x(n-1)$