[amp_mcq option1=”$${z \\over {z – {a^T}}}$$” option2=”$${z \\over {z + {a^T}}}$$” option3=”$${Z \\over {z – {a^{ – T}}}}$$” option4=”$${z \\over {z + {a^{ – T}}}}$$” correct=”option1″]<\/p>\n
The correct answer is A.<\/p>\n
The z-transform of a sequence $x[n]$ is defined as<\/p>\n
$$X(z) = \\sum_{n=-\\infty}^{\\infty} x[n] z^{-n}$$<\/p>\n
The z-transform of the unit impulse sequence $\\delta[n]$ is<\/p>\n
$$X(z) = \\sum_{n=-\\infty}^{\\infty} \\delta[n] z^{-n} = 1$$<\/p>\n
The z-transform of the shifted unit impulse sequence $\\delta[n-k]$ is<\/p>\n
$$X(z) = \\sum_{n=-\\infty}^{\\infty} \\delta[n-k] z^{-n} = z^{-k}$$<\/p>\n
Therefore, the z-transform of the time function<\/p>\n
$$\\sum_{k=0}^{\\infty} \\delta[n-k] = \\sum_{k=0}^{\\infty} z^{-k}$$<\/p>\n
is<\/p>\n
$$X(z) = \\frac{1}{1-z^{-1}} = \\frac{z}{z-1}$$<\/p>\n
which is option A.<\/p>\n
Option B is the z-transform of the time function $e^{-an}$. Option C is the z-transform of the time function $e^{an}$. Option D is the z-transform of the time function $e^{-an} u[n]$.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”$${z \\over {z – {a^T}}}$$” option2=”$${z \\over {z + {a^T}}}$$” option3=”$${Z \\over {z – {a^{ – T}}}}$$” option4=”$${z \\over {z + {a^{ – T}}}}$$” 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-55783","post","type-post","status-publish","format-standard","hentry","category-signal-processing","no-featured-image-padding"],"yoast_head":"\n