Determiner l′original de laplace F(p)=(1/((p^2 +p+1)^2 ))


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Question Number 144683 by lapache last updated on 27/Jun/21

$D e t e r m i n e r l^{'} o r i g i n a l d e l a p l a c e$ $F (p) = \frac{1}{{(p^{2} + p + 1)}^{2}}$
	Answered by Olaf_Thorendsen last updated on 27/Jun/21
	$F(p) = (1/((p^2 +p+1)^2 )) a = ((−1−i(√3))/2), b = ((−1+i(√3))/2) F(p) = (1/((p−a)^2 (p−b)^2 )) F(p) = −(1/3).(1/((p−a)^2 ))−(1/3)(1/((p−b)^2 )) +((2i)/( 3(√3))).(1/(p−a))−((2i)/(3(√3))).(1/(p−b)) f(t) = L^(−1) {F}(t) f(t) = −(1/3)te^(−at) −(1/3)te^(−bt) +((2i)/(3(√3)))e^(−at) −((2i)/(3(√3)))e^(−bt) f(t) = −(1/3)t(e^(−at) +e^(−bt) )+((2i)/(3(√3)))(e^(−at) −e^(−bt) ) f(t) = −(1/(3(√e)))t(e^(i(√3)t) +e^(−i(√3)t) )+((2i)/(3(√(3e))))(e^(i(√3)t) −e^(−i(√3)t) ) f(t) = −(2/(3(√e)))t.cos((√3)t)−(4/(3(√(3e))))sin((√3)t) f(t) = −(2/(3(√e)))[t.cos((√3)t)+(2/( (√3)))sin((√3)t)]$
	$F (p) = \frac{1}{{(p^{2} + p + 1)}^{2}}$ $a = \frac{- 1 - i \sqrt{3}}{2}, b = \frac{- 1 + i \sqrt{3}}{2}$ $F (p) = \frac{1}{{(p - a)}^{2} {(p - b)}^{2}}$ $F (p) = - \frac{1}{3} . \frac{1}{{(p - a)}^{2}} - \frac{1}{3} \frac{1}{{(p - b)}^{2}}$ $+ \frac{2 i}{3 \sqrt{3}} . \frac{1}{p - a} - \frac{2 i}{3 \sqrt{3}} . \frac{1}{p - b}$ $f (t) = L^{- 1} {F} (t)$ $f (t) = - \frac{1}{3} {t e}^{- a t} - \frac{1}{3} {t e}^{- b t} + \frac{2 i}{3 \sqrt{3}} e^{- a t} - \frac{2 i}{3 \sqrt{3}} e^{- b t}$ $f (t) = - \frac{1}{3} t (e^{- a t} + e^{- b t}) + \frac{2 i}{3 \sqrt{3}} (e^{- a t} - e^{- b t})$ $f (t) = - \frac{1}{3 \sqrt{e}} t (e^{i \sqrt{3} t} + e^{- i \sqrt{3} t}) + \frac{2 i}{3 \sqrt{3 e}} (e^{i \sqrt{3} t} - e^{- i \sqrt{3} t})$ $f (t) = - \frac{2}{3 \sqrt{e}} t . \cos (\sqrt{3} t) - \frac{4}{3 \sqrt{3 e}} \sin (\sqrt{3} t)$ $f (t) = - \frac{2}{3 \sqrt{e}} [t . \cos (\sqrt{3} t) + \frac{2}{\sqrt{3}} \sin (\sqrt{3} t)]$
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