[amp_mcq option1=”$$\\left| z \\right| < {5 \\over 6}$$\" option2=\"$$\\left| z \\right| > {5 \\over 6}$$” option3=”$${5 \\over 6} < \\left| z \\right| < {6 \\over 5}$$\" option4=\"$${6 \\over 5} < \\left| z \\right| < \\infty $$\" correct=\"option3\"]<\/p>\n
The correct answer is $\\boxed{{5 \\over 6} < \\left| z \\right| < {6 \\over 5}}$.<\/p>\n
The region of convergence of the z-transform of a sequence is the set of all values of $z$ for which the z-transform converges. The z-transform of the sequence ${\\left( {{5 \\over 6}} \\right)^n}u\\left( n \\right) – {\\left( {{6 \\over 5}} \\right)^n}u\\left( { – n – 1} \\right)$ is given by<\/p>\n
$$Z\\left{ {\\left( {{5 \\over 6}} \\right)^n}u\\left( n \\right) – {\\left( {{6 \\over 5}} \\right)^n}u\\left( { – n – 1} \\right) \\right} = {1 \\over 1 – {5 \\over 6}z^{-1}} – {1 \\over 1 – {6 \\over 5}z}$$<\/p>\n
The z-transform converges if and only if the denominator does not have any poles in the region of interest. The denominator has poles at $z = {6 \\over 5}$ and $z = {5 \\over 6}$. Therefore, the region of convergence is the set of all values of $z$ such that $z \\neq {6 \\over 5}$ and $z \\neq {5 \\over 6}$. This can be written as $${5 \\over 6} < \\left| z \\right| < {6 \\over 5}$$<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”$$\\left| z \\right| < {5 \\over 6}$$\" option2=\"$$\\left| z \\right| > {5 \\over 6}$$” option3=”$${5 \\over 6} < \\left| z \\right| < {6 \\over 5}$$\" option4=\"$${6 \\over 5} < \\left| z \\right| < \\infty $$\" correct=\"option3\"]\n<\/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-46079","post","type-post","status-publish","format-standard","hentry","category-signal-processing","no-featured-image-padding"],"yoast_head":"\n