[amp_mcq option1=”esTF(s)” option2=”e-sTF(s)” option3=”$${{F\\left( s \\right)} \\over {1 + {e^{sT}}}}$$” option4=”$${{F\\left( s \\right)} \\over {1 – {e^{ – sT}}}}$$” correct=”option4″]<\/p>\n
The correct answer is $\\frac{F(s)}{1-e^{-sT}}$.<\/p>\n
The Laplace transform of a function $f(t)$ is defined as<\/p>\n
$$L[f(t)] = \\int_0^\\infty f(t) e^{-st} dt$$<\/p>\n
If $L[f(t)] = F(s)$, then the Laplace transform of $f(t-T)$ is<\/p>\n
$$L[f(t-T)] = \\int_0^\\infty f(t-T) e^{-st} dt = \\int_T^\\infty f(t) e^{-(s-T)t} dt = \\int_0^\\infty f(t) e^{-st} dt e^{-Ts} = F(s) e^{-Ts}$$<\/p>\n
Therefore, the Laplace transform of $f(t-T)$ is $\\frac{F(s)}{1-e^{-sT}}$.<\/p>\n
Option A is incorrect because it does not take into account the time delay of $T$. Option B is incorrect because it takes the time delay into account, but it does so incorrectly. Option C is incorrect because it is the Laplace transform of a shifted exponential function, not a delayed exponential function. Option D is correct because it is the Laplace transform of a delayed exponential function.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”esTF(s)” option2=”e-sTF(s)” option3=”$${{F\\left( s \\right)} \\over {1 + {e^{sT}}}}$$” option4=”$${{F\\left( s \\right)} \\over {1 – {e^{ – sT}}}}$$” 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-47390","post","type-post","status-publish","format-standard","hentry","category-signal-processing","no-featured-image-padding"],"yoast_head":"\n