[amp_mcq option1=”2, 3″ option2=”$$\\frac{1}{2},3$$” option3=”$$\\frac{1}{2},\\frac{1}{3}$$” option4=”$$2,\\frac{1}{3}$$” correct=”option3″]<\/p>\n
The correct answer is $\\boxed{\\frac{1}{2},\\frac{1}{3}}$.<\/p>\n
The poles of a discrete-time system are the roots of the denominator of the system transfer function. In this case, the system transfer function is given by<\/p>\n
$$H(z) = \\frac{1}{1 – \\frac{1}{2}z^{-1} – \\frac{1}{3}z^{-2}}$$<\/p>\n
The poles of the system are the roots of the denominator polynomial, which is given by<\/p>\n
$$1 – \\frac{1}{2}z^{-1} – \\frac{1}{3}z^{-2} = (z – \\frac{1}{2})(z – \\frac{1}{3})$$<\/p>\n
Therefore, the poles of the system are located at $\\frac{1}{2}$ and $\\frac{1}{3}$.<\/p>\n
Option A is incorrect because the poles are not located at $2$ and $3$. Option B is incorrect because the poles are not located at $\\frac{1}{2}$ and $3$. Option C is incorrect because the poles are not located at $\\frac{1}{2}$ and $\\frac{1}{3}$.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”2, 3″ option2=”$$\\frac{1}{2},3$$” option3=”$$\\frac{1}{2},\\frac{1}{3}$$” option4=”$$2,\\frac{1}{3}$$” correct=”option3″]<\/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-50872","post","type-post","status-publish","format-standard","hentry","category-signal-processing","no-featured-image-padding"],"yoast_head":"\n