Question

For a unit step input u[n], a discrete- time LTI system produces an output signal $\left(2\delta \right[n+1]+\delta [n]+\delta [n-1\left]\right)$
Let y[n] be the output of the system for an input $\left[{\left(\frac{1}{2}\right)}^{n}u\right[n\left]\right]\cdot$ The value of y[0] is

Answer 1

$\text{The impulse response}h\left(n\right)=s\left(n\right)-s(n-1)$

$h\left(n\right)=2\delta (n+1)+\delta \left(n\right)+\delta (n-1)-2\delta \left(n\right)-\delta (n-1)-\delta (n-1-1)$

$=2\delta (n+1)+\delta \left(n\right)+\delta (n-1)-2\delta \left(n\right)\delta (n-1)-\delta (n-2)$

$h\left(n\right)=2\delta (n+1)-\delta \left(n\right)-\delta (n-2)$

$\text{For input}{x}_1\left(n\right)=\left(1/2{)}^{n}u\right(n)\text{then output}{y}_1\left(n\right)$

$=x_1\left(n\right)\ast h\left(n\right)$

$y_1\left(n\right)={\left[\frac{1}{2}\right]}^{n}u\left(n\right)\ast \left[2\delta \right(n+1)-\delta (n)-\delta (n-2)$

$y_1\left(n\right)={\left[\frac{1}{2}\right]}^{n}u\left(n\right)\ast 2\delta (n+1)-{\left[\frac{1}{2}\right]}^{n}u\left(n\right)\ast \delta \left(n\right)-{\left[\frac{1}{2}\right]}^{n}u\left(n\right)\ast \delta (n-2)$

$y_1\left(n\right)=2{\left[\frac{1}{2}\right]}^{n+1}u(n+1)-{\left[\frac{1}{2}\right]}^{n}u\left(n\right)-{\left[\frac{1}{2}\right]}^{n-2}u(n-2)$

$y_1\left(n\right){|}_{n=0}=2{\left[\frac{1}{2}\right]}^{1}u\left(1\right)-{\left[\frac{1}{2}\right]}^{0}u\left(0\right)-{\left[\frac{1}{2}\right]}^{-2}u(-2)$

$=1-1$

$y_1\left(0\right)=0$