Question

$X\left(z\right)=1=3z^{-1},$ $Y\left(z\right)=1+2z^{-2}$ are Z-transforms of two signals x[n], y[n] respectively. A linear time invariant system has the impulse response h[n] defined by these two signals as h[n] =x[n-1] * y[n] where * denotes discrete time convolution. Then the output of the system for the input $\delta [n-1]$

• A.
  does not satisfy any of the above three
• B.
  has Z-transform $z^{-1}X\left(z\right)Y\left(z\right)$
• C.
  equals $\delta [n-2]-3\delta [n-3]+2\delta [n-4]-6\delta [n-5]$
• D.
  has Z-transform $1-3z^{-1}+2z^{-2}-6z^{-3}$

Answer 1

The correct option is C

equals $\delta [n-2]-3\delta [n-3]+2\delta [n-4]-6\delta [n-5]$
$X\left(z\right)=(1-3z^{-1})$

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

$h\left[n\right]=x(n-1)\ast y\left[n\right]$

$\Rightarrow H\left[z\right]=z^{-1}X\left(z\right).Y\left(z\right)$

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

$\Rightarrow H\left[z\right]=z^{-1}(1+2z^{-2}-3z^{-1}-6z^{-3})$

$\Rightarrow H\left[z\right]=(z^{-1}-3z^{-2}+2z^{-3}-6z^{-4})$

when input,
$I\left(n\right)=\delta [n-1]then,I\left[z\right]=z^{-1}$

Therefore output,
$P\left(n\right)=h\left[n\right]\ast I\left[n\right]$

$\Rightarrow P\left(z\right)=H\left(z\right)I\left(z\right)$

$P\left(z\right)=(z^{-1}-3z^{-2}+2z^{-3}-6z^{-4})\times z^{-1}$

$\Rightarrow P\left(z\right)=z^{-2}-3z^{-3}+2z^{-4}-6z^{-5}$

$\therefore p\left(n\right)=\delta [n-2]-3\delta [n-3]+2\delta [n-4]-6\delta [n-5]$