Question

Let y[n] denote the convolution of a casual sequence h[n] and x[n], where $h\left[n\right]={\left(\frac{1}{4}\right)}^{n}u\left[n\right]$ and $x\left[n\right]$. If $y\left[0\right]=1/2$ and $y\left[1\right]=1$ then $x\left[1\right]$ equal to __________

• A. 0.875

Answer 1

The correct option is A 0.875
Given x[n] is a causal sequence

$\therefore x\left[n\right]$ will be zero for n < 0.

$y\left[n\right]={\sum }_{k=-\infty }^{\infty }x\left[k\right]h[n-k]$

and $h\left[n\right]={\left(\frac{1}{4}\right)}^{n}u\left[n\right]$

$h\left[n\right]=\{\begin{array}{cc}{\left(\frac{1}{4}\right)}^n;& n\ge 0\\ 0;& n<0\end{array}$

$\therefore h\left[0\right]=1,h\left[1\right]=\frac{1}{4},h\left[2\right]=\frac{1}{16}$

$y\left[0\right]=x\left[0\right]h\left[0\right]+x\left[1\right]h[-1]+...$

$\frac{1}{2}=x\left[0\right]\left(1\right)+x\left[1\right]\times 0$

$\therefore x\left[0\right]=\frac{1}{2}$

$y\left[1\right]=x\left[0\right]h\left[1\right]+x\left[1\right]h\left[0\right]+x\left[2\right]h[-1]+...$

$1=\left(\frac{1}{2}\right)\left(\frac{1}{4}\right)+x\left[1\right]\left[1\right]+x\left[2\right]\left[0\right]$

$1-\frac{1}{8}=x\left[1\right]$

$\therefore x\left[1\right]=\frac{7}{8}=0.875$