Question

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

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

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

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

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

1 = $\left(\frac{1}{2}\right)\left(\frac{1}{4}\right)+x\left[1\right]\left[1\right]+x\left[2\right]\left[0\right]$
$1-\frac{1}{8}=x\left[1\right]$

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