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1

You're right, the answer's wrong. Let's just solve for $Y(s)$ from the second equation, $$Y(s) = -\frac{X(s)}{s-2}.$$ Subbing this into the first equation gives, after rearranging, $$ X(s) = \frac{10s-20}{s(s-3)(s-9)}, $$ the partial fraction decomposition of which is $$ X(s) = \frac{35}{27(s-9)} - \frac{5}{9(s-3)} - \frac{20}{27s}. $$ Note that we could sub ...


1

Try to use the Heaviside step function, which is defined as follows: $$\mathscr U(t-a) = \cases{1 &$0 \leq t \leq a$\\1 & $t\geq a$}.$$ Then, as the function $f$ is defined for $t \geq 0$ we can write: $$f(t) = 2 +t\mathscr U(t-1) -(t+2)\mathscr U(t-2). $$ Now apply the theorem: $$\mathcal L(f(t-a)\mathscr U(t-a))(s) = e^{-at}\mathcal L(f(t))(s),$$ ...


1

You have $${\mathcal L}(f)(s) = \int_0^\infty e^{-sx} f(x)\, dx = \int_0^1 e^{-sx} f(x)\, dx + \int_1^2 e^{-sx} f(x)\, dx + \int_2^\infty e^{-sx} f(x)\, dx.$$ Can you do the rest?


0

We are given: $$y''+2y'+y=0; \;\;\;\;\;y'(0)=2\;\;\;\;\;\mbox{and}\;\;\;\;\;y(1)=2$$ I am going to start you off, but you have to show some work and fill in the details. Taking the Laplace transform yields: $$\mathscr{L}(y''+2y'+y) = (s^2 y(s) -s y(0) -y'(0)) +2 (s y(s) - y(0)) + y(s) = 0$$ Substituting the ICs (let $y(0) = a$) yields: $$(s^2 y(s) -s ...


1

We have $$y_1' = 2y_1 + y_2 \\ y_2' = -y_1 \\ y_1(0) = 0, y_2(0) = 1$$ Taking Laplace transforms, we have: $$sy_1(s) - y_1(0) = 2y_1(s) + y_2(s) \\ s y_2(s) - y_2(0) = -y_1(s)$$ Substituting ICs yields: $$sy_1(s) = 2y_1(s) + y_2(s) \\ s y_2(s) - 1 = -y_1(s)$$ You can put this into matrix form to solve if desired (but it is instructive to see where ...


0

No, it hasn't. There is no element of multiplication of polar coordinates or rectangular coordinates in Laplace transform.


0

The best video I have found for Laplace Transform was this: https://www.youtube.com/watch?v=zvbdoSeGAgI https://www.youtube.com/watch?v=hqOboV2jgVo MIT professor explains where it comes from, its analogue: Power Series and so on. It will take 13 minutes of your time but you will understand the concept completely!


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If you don't have access to books you can visit this site for a nice introduction. If you prefer videos you can always visit this site and watch the whole playlist. For a general overview read this wikipedia article.


1

Hint: $$H(s)=\frac8{s^4+4}=\frac{2}{s+1+i}+\frac{2}{s+1-i}+\frac{2}{s-1+i}+\frac{2}{s-1-i}$$ The infverse Laplace transform of $H(s)$ is: $$f(t)=2\exp(1+i)+2\exp(1-i)+2\exp(-1+i)+2\exp(-1-i)$$ $$=8\cos (t)\cosh(t)$$


0

Let $$(\mathcal{L}f)(s)=\int^{\infty}_{0}f(t)e^{-st}\,dt$$ Denote the Laplace operator applied to a real valued function $f$ which satisfies some regularity conditions such that the integral above exists. Note that ...


0

$$ \underbrace{s^2-10s+29 = (s-5)^2+2^2}_{\text{completing the square}} = t^2+2^2 $$ So $$ \frac{6s+9}{s^2-10s+29} = \frac{6(s-5)+39}{(s-5)^2+2^2} = \frac{6t+39}{t^2+2^2} = 6\frac{t}{t^2+2^2} + \frac{39}{2}\cdot\frac{2}{t^2+2^2} $$ Now look at each term separately.


0

We have $s^2-10s+29=(s-5)^2+2^2$, which implies that $$F(s)=\frac{6s+9}{s^2-10s+29}=\frac{6(s-5)+39}{(s-5)^2+2^2}.$$ Therefore, $$\mathcal{L}^{-1}(F(s))=6\mathcal{L}^{-1}\left(\frac{s-5}{(s-5)^2+2^2}\right) +\frac{39}{2}\mathcal{L}^{-1}\left(\frac{2}{(s-5)^2+2^2}\right)\\ =6e^{5t}\cos(5t)+\frac{39}{2}e^{5t}\sin(2t)$$ by here.


0

$$ F(s) = \dfrac{6s+9}{s^2-10s+29}=\dfrac{6s-30+39}{(s-5)^2+4}=\dfrac{6s-30}{(s-5)^2+4}+\dfrac{39}{(s-5)^2+4}=6\dfrac{s-5}{(s-5)^2+4}+\dfrac{39}{2}\dfrac{2}{(s-5)^2+4} $$ Then $$ \mathcal{L}^{-1}\{F(s)\} =6e^{5t}\cos{2t}+\frac{39}{2}e^{5t}\sin{2t} $$


2

To answer your question, the most important thing you are leaving out is Cauchy's theorem. You are defining the contour so that it encloses no poles of the LT and is single-valued along itself. Thus, the sum of the contributions along the contours is zero. You should end up with something like $$\begin{align}\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ...


1

The limit is the supremum $\xi$ of the support of $X$, that is, $$\xi=\sup\{x\mid P(X\geqslant x)\ne0\}=\inf\{x\mid P(X\leqslant x)=1\}.$$ To prove this, note that for every positive $\varepsilon$: $X\leqslant\xi$ almost surely hence $E(e^{pX})\leqslant e^{p\xi}$ $q=P(X\geqslant\xi-\varepsilon)$ is positive and $E(e^{pX})\geqslant qe^{p(\xi-\varepsilon)}$ ...


0

Hint. You may use partial fraction $$ \frac{s^2-4s-2}{\left(s^2+2\right)^2}=\frac{1}{2+s^2}-\frac{4 s+4}{\left(2+s^2\right)^2} $$ and then make use of a table.


2

Have a look in, for example, Apostol's Mathematical Analysis. The edition I looked at was old (5th, 1981), it was exercise 11.39. Here is the formal statement of the inverstion formula for Laplace transforms. Let $c>0$ be a positive number such that $\int_0^\infty |f(t)|e^{-c t}\,dt$ exists, and define $F(s)=\int_0^\infty e^{-s t}f(t)\,dt$ for $\Re ...


1

Use the fact that $$e^{At}(B\cosh Ct + D\sinh CT) = \tfrac{B+D}{2}e^{(A+C)t} + \tfrac{B-D}{2}e^{(A-C)t} $$ Compare this to your expression. The coefficients give you $$\left\{ \begin{array}{cc}\tfrac{B+D}{2}=\tfrac23 \\ \tfrac{B-D}{2}=\tfrac13 \end{array}\right. $$ The exponents give you $$\left\{ \begin{array}{cc}A+C=-4 \\ A-C=2 \end{array}\right. $$ From ...


0

I've solved this. One creates a system of equations out of this expression. And equate it to my simplified answer.


1

http://en.wikipedia.org/wiki/Laplace_transform#Properties_and_theorems Rule "Frequency integration", 7th row in the table. To answer the "how do I use it?", it would be helpful if you'd post your actual problem. Applying the rule to your problem yields $$f(t) = -\cos(5t) \ \ \ \Rightarrow \ \ \ F(\tilde{s}) = -\frac{\tilde{s}}{\tilde{s}^2+5^2}$$ Now ...


0

In the following $\langle f, \cdot \rangle$ denotes the linear functional on Schwartz space induced by $f$. By definition, for any Schwarts function $\varphi$ \begin{align*} \langle 1^\lor, \varphi \rangle=\langle 1, \varphi^\lor \rangle=&\int_\mathbb{R} \left(\int_\mathbb{R} e^{2\pi ixy}\varphi(y) dy\right)dx =\lim_{M\to\infty}\int_{-M}^M ...


1

First of all, the poles you are thinking of are at $s=-n^2 \pi^2$. ($s=0$ is a removable pole of the sinh terms.) Also note that, even though there are square roots in the sinh terms, we need not worry about taking branches because any odd phase behavior will cancel in the fraction. Therefore, what we have is a straightforward Bromwich contour encircling ...


1

First, let me fill in my comments regarding why the inverse Laplace transform of $$H(s)=\frac{R}{Ls+\frac{1}{Cs}+R}=\frac{Rs/L}{(s+R/2L)^2+1/LC-(R/2L)^2}=\frac{2as}{(s+a)^2+\omega^2}$$ (where I've used the definitions from my comments and theva's answer) can't quite be $$H(t)=\frac{a}{2}e^{-at}\cos(\omega t)u(t)-ae^{-at}\sin(\omega ...


0

Using partial fractions and Maple I am obtaining $$x \left( t \right) =c_{{2}}\sqrt [3]{-{\frac {k}{m}}} \left( a-\sqrt [ 3]{-{\frac {k}{m}}} \right) ^{-1} \left( -b+\sqrt [3]{-{\frac {k}{m}}} \right) ^{-1}{\frac {1}{\sqrt {\pi \,t}}}+ \left( \left( b-\sqrt [3] {-{\frac {k}{m}}} \right) \left( c_{{1}}+c_{{2}}{a}^{2} \right) {\it erfc} \left( -a\sqrt {t} ...


0

Two ways: 1) Observe first that $y [n]= \sum_{k=0}^{n} 2^k = (2^{n+1}-1)u[n]$ . Then $$Y(Z)= \sum_{n=0}^{\infty} z^{-n} (2^{n+1}-1)$$ (you should be able to go on from here) 2) Find the Z transform of $x[n]$ and apply the "accumulation" property (analogous to the integral for Fourier/Laplace transform).


1

Consider the integral \begin{align} J = \int_{0}^{\infty} e^{-a x} \, x^{\mu} \, dt = \frac{\Gamma(\mu+1)}{a^{\mu+1}}. \end{align} Let $\mu = i t$, where $i = \sqrt{-1}$, for which \begin{align} \int_{0}^{\infty} e^{-a x} \, x^{i t} \, dt &= \int_{0}^{\infty} e^{-a x} \, e^{i t \ln(x)} \, dt = \frac{\Gamma(i t+1)}{a^{it+1}} \end{align} From this it is ...


1

So I got the expression: $$H(t) = \frac{Rs}{L(s+\frac{R}{2L})^2+\frac{1}{C}-\frac{R^2}{4L}}$$ And then I divide everything by L and break things out until I get this: $$\frac{Rs}{L}\frac{1}{(s+a)^2+\omega^2}$$ where $ a=\frac{R}{2L}$ and $\omega^2=\frac{1}{CL}-(\frac{R}{2L})^2$ I can then rewrite it again as: ...


1

I'm thinking of solving this using convolution theorem but I'm unsure if it works for the product of 3 functions. You are looking at this being $F(s) p(s) q(s)$. But the product of two functions is also a function, so $F(s) z(s)$, it being two functions, is completely equivalent. As such, your method will work. Note that you could also take the ...


0

Note that the Laplace transform, transforms a function from the $t$-domain to the $s$-domain, now, notice that $\int\limits_0^1 {i(t)dt} $ is not a varying function of $t$, but a constant. Hence, its Laplace transform is equal to $\frac{1}{s}$ multiplied by itself (I suggest you use different notations for the dummy variable of the integral to avoid such ...


0

\begin{align}\int^{\infty}_{s}F(u)\,du&=\int^{\infty}_{s}\int^{\infty}_{0}f(\tau)e^{-u\tau}\,d\tau\,du\\&=\int^{\infty}_{0}f(\tau)\int^{\infty}_{s}e^{-u\tau}\,du\,d\tau\\&= ...


2

Really, you are asking for the inverse LT, which by definition is $$f(t) = \frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, e^{s t} \, \operatorname{sech}{a \sqrt{s}} $$ where $a=\sqrt{2} x$, and $c \gt 0$ is greater than the greatest real part of any pole of the integrand. Normally, I would take you through an integration contour in the complex plane ...



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