[amp_mcq option1=”0″ option2=”$${1 \\over 2}$$” option3=”1″ option4=”$${3 \\over 2}$$” correct=”option4″]<\/p>\n
The correct answer is $\\boxed{{1 \\over 2}}$.<\/p>\n
The convolution of two sequences $h[n]$ and $g[n]$ is defined as<\/p>\n
$$y[n] = \\sum_{k=-\\infty}^{\\infty} h[k] g[n-k]$$<\/p>\n
where $u[n]$ is the unit step function.<\/p>\n
In this case, $h[n] = {\\left( {\\frac{1}{2}} \\right)^n} u[n]$ is a causal sequence, which means that $h[n] = 0$ for $n < 0$.<\/p>\n
The convolution of a causal sequence with an arbitrary sequence is also causal. This means that $y[n] = 0$ for $n < 0$.<\/p>\n
We are given that $y[0] = 1$ and $y[1] = {\\frac{1}{2}}$. Using the definition of convolution, we can write<\/p>\n
$$1 = \\sum_{k=-\\infty}^{\\infty} h[k] g[-k]$$<\/p>\n
and<\/p>\n
$${\\frac{1}{2}} = \\sum_{k=-\\infty}^{\\infty} h[k] g[-k-1]$$<\/p>\n
Subtracting these two equations, we get<\/p>\n
$${\\frac{1}{2}} = \\sum_{k=-\\infty}^{\\infty} h[k] (g[-k-1] – g[-k])$$<\/p>\n
Since $h[k] = 0$ for $k < 0$, the only terms that contribute to the sum are $k = 0$ and $k = -1$. Therefore, we have<\/p>\n
$${\\frac{1}{2}} = h[0] g[-1] – h[-1] g[0]$$<\/p>\n
Solving for $g[1]$, we get<\/p>\n
$$g[1] = \\frac{h[0] + h[-1]}{{2}} = \\frac{1 + {\\frac{1}{2}}}{{2}} = {1 \\over 2}$$<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”0″ option2=”$${1 \\over 2}$$” option3=”1″ option4=”$${3 \\over 2}$$” correct=”option4″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[959],"tags":[],"class_list":["post-47823","post","type-post","status-publish","format-standard","hentry","category-signal-processing","no-featured-image-padding"],"yoast_head":"\n