Question

Two discrete-time signals $x\left[n\right]$ and $h\left[n\right]$ are both non-zero only for n =0, 1, 2 and are zero otherwise. It is given that $x\left[0\right]=1,$ $x\left[1\right]=2,$ $x\left[2\right]=1,$ $h\left[0\right]=1.$
Let $y\left[n\right]$ be the linear convolution of $x\left[n\right]$ and $h\left[n\right].$
Given that $y\left[1\right]=3$ and $y\left[2\right]=4,$ the value of the expression $\left[10y\right[3]+y[4\left]\right]$ is ________.

• A. 31

Answer 1

The correct option is A 31

x/h	1	a	b
1	1	a	b
2	2	2a	2b
1	1	a	b

$x\left[n\right]=\{1,2,1\}$

$h\left[n\right]=\{1,a,b\}$

$y\left[n\right]=\{A,3,4,B,C\}$

$y\left[0\right]=1;y\left[1\right]=2+a;y\left[2\right]=1+2a+b;$

$y\left[3\right]=a+2b;y\left[4\right]=b$

Given, $y\left[1\right]=2+a=3\Rightarrow a=1$

Given, $y\left[2\right]=1+2a+b=4\Rightarrow b=1$

So, $y\left[3\right]=a+2b=3$

and $y\left[4\right]=b=1$

So, $10y\left[3\right]+y\left[4\right]=10\times 3+1=31$