Question

Problem #5 Calculate the Laplace Transform of f(t) if 1 + 2 e-t/2 + 5 cos(t/3) + 6 sin(t/4). f(t)s hen obtain the positive value of the parameter of the Laplace Transform for which the value of the transforn sequal to 1, round-off the number you have just found to three figures and present it below (20 points): (your numerical result for the value of the parameter must be written here)

Answer 1

$f(t)=1+2e^{-\frac{t}{2}}+5\cos \left(\frac{t}{3}\right)+6\sin \left(\frac{t}{4}\right)$

take laplace

$F(s)=L\left\{1+2e^{-\frac{t}{2}}+5\cos \left(\frac{t}{3}\right)+6\sin \left(\frac{t}{4}\right)\right\}$

$F(s)=L\left\{1\right\}+2L\left\{e^{-\frac{1}{2}t}\right\}+5L\left\{\cos \left(\frac{t}{3}\right)\right\}+6L\left\{\sin \left(\frac{t}{4}\right)\right\}$

apply laplace rule $L\left\{a\right\}=\frac{a}{s}$

hence ${\color{Blue} L\left\{1\right\}=\frac{1}{s}}$

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apply laplace rule  $L\left\{e^{at}\right\}=\frac{1}{s-a}$

$L\left\{e^{-\frac{1}{2}t}\right\}=\frac{1}{s+\frac{1}{2}}$

${\color{Blue} L\left\{e^{-\frac{1}{2}t}\right\}= \frac{2}{2s+1}}$

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apply laplace rule  $L\left\{\cos \left(at\right)\right\}=\frac{s}{s^2+a^2}$

$L\left\{\cos \left(\frac{t}{3}\right)\right\}=\frac{s}{s^2+\left(\frac{1}{3}\right)^2}$

${\color{Blue} L\left\{\cos \left(\frac{t}{3}\right)\right\}= \frac{9s}{9s^2+1}}$

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apply laplace rule  $L\left\{\sin \left(at\right)\right\}=\frac{a}{s^2+a^2}$

$L\left\{\sin \left(\frac{t}{4}\right)\right\}=\frac{\frac{1}{4}}{s^2+\left(\frac{1}{4}\right)^2}$

${\color{Blue} L\left\{\sin \left(\frac{t}{4}\right)\right\}= \frac{4}{16s^2+1}}$

$F(s)=\frac{1}{s}+2\cdot \frac{2}{2s+1}+5\cdot \frac{9s}{9s^2+1}+6\cdot \frac{4}{16s^2+1}$

${\color{Red} F(s)=\frac{1}{s}+\frac{4}{2s+1}+\frac{45s}{9s^2+1}+\frac{24}{16s^2+1}}$