[amp_mcq option1=”0.5 – j0.25″ option2=”$${1 \\over {\\left( {0.5 + j0.25} \\right)}}$$” option3=”$${1 \\over {\\left( {0.5 – j0.25} \\right)}}$$” option4=”2 + j4″ correct=”option1″]<\/p>\n
The correct answer is $\\boxed{{1 \\over {\\left( {0.5 – j0.25} \\right)}}}$.<\/p>\n
Let $x[n]$ be a sequence with even symmetry, i.e., $x[n] = x[-n]$. The z-transform of $x[n]$ is given by<\/p>\n
$$X(z) = \\sum_{n=-\\infty}^{\\infty} x[n] z^{-n} = \\sum_{n=-\\infty}^{\\infty} x[-n] z^{n} = \\sum_{n=0}^{\\infty} x[n] z^{n}$$<\/p>\n
Therefore, the poles of $X(z)$ are the complex numbers $z$ such that $z^{-1}$ is a zero of $X(z)$.<\/p>\n
In this case, we are given that $0.5 + j0.25$ is a zero of $X(z)$. Therefore, $z^{-1} = 0.5 – j0.25$ is a pole of $X(z)$. The reciprocal of a pole is a zero, so $1 \/ (0.5 – j0.25)$ is also a zero of $X(z)$.<\/p>\n
The other options are not necessarily zeros of $X(z)$. For example, $0.5 + j0.25$ is not a zero of $X(z)$, since $z^{-1} = 0.5 + j0.25$ is not a pole of $X(z)$.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”0.5 – j0.25″ option2=”$${1 \\over {\\left( {0.5 + j0.25} \\right)}}$$” option3=”$${1 \\over {\\left( {0.5 – j0.25} \\right)}}$$” option4=”2 + j4″ correct=”option1″]<\/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-56945","post","type-post","status-publish","format-standard","hentry","category-signal-processing","no-featured-image-padding"],"yoast_head":"\n