- Course Instrumentation And Control Systems 1 (24.71 KiB) Viewed 19 times
Course: Instrumentation and Control Systems
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Course: Instrumentation and Control Systems
Course: Instrumentation and Control Systems
6 The inverse Laplace transform of F(s) = 1 + 4 -3 is S s-8 (Multiple answers are allowed) е Select one or more: f(t) = 6*1(t) - Sin(8t) 81.1(t) + sin(41) €31.1(t) f(t) = 6t - e81 + 4 e3!,120. f(t) = 6*1(t) - €81.1(t) + 4 €31.1(t) f(t) = 6 - 281 + 4 e3, 120. f(t) = 6 - e-81+ 4 e-31.120. f(t) = 6 - sin(8t)*e 8! + sin(41)* 23.120. f(t) = 6*1(t) - e -81.1(t) + 4 e-31* 1(t) f(t) = 6*1*1(t) - €87.1(t) + 4 * 3.1t) e
6 The inverse Laplace transform of F(s) = 1 + 4 -3 is S s-8 (Multiple answers are allowed) е Select one or more: f(t) = 6*1(t) - Sin(8t) 81.1(t) + sin(41) €31.1(t) f(t) = 6t - e81 + 4 e3!,120. f(t) = 6*1(t) - €81.1(t) + 4 €31.1(t) f(t) = 6 - 281 + 4 e3, 120. f(t) = 6 - e-81+ 4 e-31.120. f(t) = 6 - sin(8t)*e 8! + sin(41)* 23.120. f(t) = 6*1(t) - e -81.1(t) + 4 e-31* 1(t) f(t) = 6*1*1(t) - €87.1(t) + 4 * 3.1t) e
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