# Tag Info

0

Short answer: yes Long answer: For any 2 functions $f(t)$ and $g(t)$, the Laplace tranform of the convolution is the product of Laplace transforms $$\mathcal{L} \{ f(t) * g(t) \} = \mathcal{L}\{f(t)\} \mathcal{L}\{g(t)\}$$ It makes sense that this would hold no matter how many times we do the convolution

0

\begin{align} F(s)&=\frac{s}{(s - \frac{1}{2})^{2} + 4}\\ &=\frac{\frac{1}{2}}{(s - \frac{1}{2})^{2} + 4} + \frac{s-\frac{1}{2}}{(s - \frac{1}{2})^{2}+ 4}\\ \\ f(t) &= (1/4)e^{t/2} (\sin(2 t) + 4\cos(2 t)) \end{align} Credit to Chappers, Oliver, and Mattos for guiding me to the above solution

-1

This can be used to solve differential equations having a derivative term multiplied by $t$. e.g. : $$t\frac{\mathrm dy}{\mathrm dt}=t.$$

1

The transform is equal to $$\frac{2}{(s-1)^2+\sqrt{7}^2}+\frac{2(s-1)}{(s-1)^2+\sqrt{7}^2}$$ which turn into cosine and sine transforms combined with the first shift theorem.

0

The solution is $$y(x,t) = \left( e^{-\frac{b}{a}x}y_l + e^{-\frac{b}{a}x}\frac{c}{b}\left( 1 - e^{b(t+\frac{x}{a})} \right)\right)\sigma\left(t + \frac{x}{a}\right) - \frac{c}{b}\left(1-e^{bt}\right) - \mathcal{L}^{-1}\{ F(x,s) \},$$ where $$F(x,s) = \frac{1}{as}e^{\frac{s-b}{a}x} \int\limits_0^x e^{\frac{b-s}{a}\xi}f(\xi) \text{d}\xi,$$ with $y_l := ... 0 I think I have a solution. Define the Laplace transform as \begin{equation*} \mathcal{L}_t[f(t)](s)=\int^{\infty}_{0}f(t)e^{-st}dt. \end{equation*} Applying this to both sides of the equation gives \begin{equation*} \mathcal{L}_t[y''(t)+3y(t)](s)=\mathcal{L}_t[\sin(t)](s)\\ \Rightarrow \mathcal{L}_t[y''(t)](s)+3(\mathcal{L}_t[y(t)](s)). \end{equation*} ... 2 $$\int_a^b{f(x)dx}=\int_a^{b}{f(x)[H(x-a)-H(x-b)]dx}=\int_0^{\infty}{f(x)[H(x-a)-H(x-b)]dx}$$ 0 Here is another way to proceed. We will use the identity $$\mathscr{L}\{\log (t)\}(s)=\frac{-\gamma-\log(s)}{s}$$ where$\mathscr{L}\{\log (t)\}(s)$is the Laplace Transform of$\log (t)$and where$\gamma$is the Euler-Mascheroni constant. We will prove this identity at the end of this writing. Now, taking the Laplace Transform of$(1-\cos(t))/t$... 3 Notice that $$\frac{1-\cos{u}}{u} = \int_0^1 \sin{u} \, da.$$ Interchanging the order of integration, $$\mathcal{L}\left( \frac{1-\cos{u}}{u} \right)(s) = \int_0^1 \int_0^{\infty} e^{-su} \cos{au} \, du \, da.$$ Now, we know the Laplace transform of$\cos{au}$, or at least we can get it using complex exponentials or integration by parts: it's ... 2 You may simplify the derivation considerably by noticing that $$A(t) + B(t) = A_0 \implies \hat{A}(s) + \hat{B}(s) = \frac{A_0}{s}$$ Then $$s \hat{B}(s) = k_1 \hat{A}(s) - k_2 \hat{B}(s) = \frac{k_1 A_0}{s} - (k_1+k_2) \hat{B}(s)$$ Therefore $$\hat{B}(s) = \frac{k_1 A_0}{s (s+k_1+k_2)}$$ 1 The Gamma Function$\Gamma$is defined as $$\Gamma(z) = \int_0^{\infty} t^{z-1}e^{-t}dt$$ for$\text{Re}\{z\}>0$. We can write this integral representation as a Laplace Transform by letting$t \to st. Then, we have \begin{align} \Gamma(z) &= \int_0^{\infty} (st)^{z-1}e^{-st}sdt\\\\ &=s^z\int_0^{\infty} t^{z-1}e^{-st}dt \end{align} The ... 1 $$s Y'(s) +Y(s) = \frac{d}{ds} (s Y(s)) = -\frac{a}{s^2} \implies sY(s) = \frac{a}{s} + C$$ Thus, $$Y(s) = \frac{a}{s^2} + \frac{C}{s}$$ Inverting the LT, we have $$y(t) = a t +C$$ Information ony(0)$would determine$C$. 2 The idea is to undo the operations you may find in the transform using the properties of the Laplace transform. The Laplace transform of the logarithm times Heaviside is essentially a logarithm divided by$s$. The transform of a derivative is essentially the transform of the function times$s. And finally the transform of a function multiplied by an ... 2 One approach is to integrate by parts twice \begin{align} &\int_0^\infty e^{-\alpha x}\sin(\beta x)\,\mathrm{d}x\tag{1}\\ &=-\frac1\alpha\int_0^\infty\sin(\beta x)\,\mathrm{d}e^{-\alpha x}\tag{2}\\ &=\frac\beta\alpha\int_0^\infty e^{-\alpha x}\cos(\beta x)\,\mathrm{d}x\tag{3}\\ &=-\frac\beta{\alpha^2}\int_0^\infty\cos(\beta ... 2 Maybe the easiest (or at least most economic) way to solve this is using the fact that \sin(\beta x)=\Im[{e^{i\beta x}}]: I=\Im\left[\int_0^{\infty}{e^{i\beta x- ax}}\right]=-\Im\left[\frac{1}{i\beta-a}\right]=\frac{\beta}{\beta^2+a^2} $$1 For inverse Laplace transforms, we require that \gamma be larger than the real part of any pole of the Laplace transform, in this case, 1/s^2. There is good reason for this. Recall that the Laplace transform is defined as$$\int_0^{\infty} dt \, f(t) e^{-s t} $$which always converges when t \gt 0, right? Wrong. One must consider any exponential ... 1 We can build such a function as follows: As we want f not to be of exponential order, we will choose f such that$$ f(n) = n\exp(nx), \qquad n \in \mathbb N $$Then f is not of exponential order. To make f absolutely integrable, define a_n := (1+n)^{-2}\exp(-2nx), then f(n)a_n \to 0 and define f by$$ f(x) := \begin{cases} n\exp(nx)\frac ... 1 You don't really ''know'' it is satisfied, you more like demand that it has to be satisfied. If E.g. you solve a differential equation by using the laplace transform, then the solution you get should only be defined for positives$. This is not always stated explicitly because either the professor assumes you know that is the case or the function you found ... 0 By rationalizing the denominator, we have$\mathcal{L}^{-1}\left\{\dfrac{s^\frac{3}{2}-a-bs}{s^\frac{3}{2}+a+bs}\right\}=\mathcal{L}^{-1}\left\{\dfrac{(s^\frac{3}{2}-a-bs)^2}{(s^\frac{3}{2}+a+bs)(s^\frac{3}{2}-a-bs)}\right\}=\mathcal{L}^{-1}\left\{\dfrac{s^3-2s^\frac{3}{2}(bs+a)+b^2s^2+2abs+a^2}{s^3-b^2s^2-2abs-a^2}\right\}$... 0 What you could do is to introduce a new variable$\bar{x}(t) := x(t) + \frac{b_1}{b_0}$. This would enable you to express the transfer function by means of$\bar{X}(s)$, at least: $$\frac{\bar{X}(s)}{Y(s)} = \frac{a_0s+a_1}{b_0}.$$ 1 Note$t=(t-2)+2. Then, \begin{align} \mathscr{L}\{tu(t-2)\}&=\mathscr{L}\{(t-2+2)u(t-2)\}\\ &=\mathscr{L}\{(t-2)u(t-2)\}+2\mathscr{L}\{u(t-2)\} \end{align} Can you finish? 1 Error in finding the laplace transform. You should have $$\mathcal{Y}(s) = \frac{1}{s^2+4s+5}\mathcal{U}(s).$$ 2 HINT: Use Werner Formula , $$2\sin A\cos B=\sin(A+B)+\sin(A-B)$$ and we know, $$L\{\sin(at)\}=\dfrac a{s^2+a^2}$$ 1 When using\mathcal{L}^{-1}\left\{{f(s-a)}\right\}= e^{at}\mathcal{L}^{-1}\left\{{f(s)}\right\}$, You have replaced$s-a$with$s$in the denominator of your transform, but not the numerator. Add$+2-2$to the numerator and use linearity of$\mathcal{L}^{-1}$to split the transform and you should get the book's answer. 1 If the problem given was to solve$y''(t)-y'(t)-6y(t)=0$subject to initial conditions$y(0)=1$and$y'(0)=-1$, using the Laplace transform, then we write$Y(s) =\mathscr{L}\{y(t)\}(s)and \begin{align} \mathscr{L}\{y''(t)-y'(t)-6y(t)\}(s)&=\left(s^2Y(s)-sy(0)-y'(0)\right)\\ &-\left(sY(s)-y(0)\right)\\ &-6Y(s)\\\\ &=(s^2-s-6)Y(s)-(s-2)\\ ... 0 For the differential equation y'' - y' -6y = 0 the Laplace transform is \begin{align} ( s^2 y(s) - y'(0) - s y(0) ) - (s y(s) - y(0)) - 6 y(s) = 0 \end{align} which leads to (s^2 - s - 6) y(s) = y'(0) + y(0) (s-1) and \begin{align} y(s) &= \frac{y'(0) - y(0)}{s^2 - s -6} + \frac{y(0) \, s}{s^2 - s - 6} \\ &= \frac{y'(0) - y(0)}{5} \left( ... 0 It is not suggested to solve by Laplace transform. \dfrac{\partial y}{\partial t}=\dfrac{\partial y^n}{\partial x} \dfrac{\partial y}{\partial t}=ny^{n-1}\dfrac{\partial y}{\partial x} Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example: \dfrac{dt}{ds}=1 , letting t(0)=0 , we have t=s \dfrac{dy}{ds}=0 , ... 0 Define the Laplace transform as \begin{align} f(s) = \int_{0}^{\infty} e^{-s t} \, y(t) \, dt \end{align} then for the differential equation \begin{align} y'' - 3 y' -4 y = -4 t + \delta(t), \end{align} where y(0) = -2 and y'(0) = -1 then \begin{align} s^{2} f(s) - y'(0) - s y(s) - 3s f(s) + 3 y(0) - 4 f(s) = - \frac{4}{s^{2}} + 1 \end{align} which ... 1 You just need to compute a convolution. If X is a random variable having pdf: f_X(x) = \frac{1}{2}e^{\frac{1-x}{2}}\cdot \mathbb{1}_{[1,+\infty)}(x) $$then, for every t>-\frac{1}{2},$$ \mathbb{E}[e^{-tX}]=\frac{e^{-t}}{1+2t} $$as wanted. 0 strong textYou can write$$ \begin{align} F^{\prime}(s)&=[\ln(s^{2}+a^{2})-\ln(s^{2}+b^{2})]^\prime\\ &=\frac{2s+2a}{s^{2}+a^{2}}-\frac{2s+2b}{s^{2}+b^{2}}\\ &=\frac{2s}{s^{2}+a^{2}}+\frac{2a}{s^{2}+a^{2}}-\frac{2s}{s^{2}+b^{2}}-\frac{2b}{s^{2}+b^{2}}\\ &= 2(\cos at)+2(\sin at)-2(\cos bt)-2(\sin bt) \end{align} $$2 \mathcal{L}^{-1}\left\{\ln\dfrac{s^2+a^2}{s^2+b^2}\right\} =\mathcal{L}^{-1}\left\{\int_s^\infty\left(\dfrac{2s}{s^2+a^2}-\dfrac{2s}{s^2+b^2}\right)ds\right\} =\dfrac{1}{t}\mathcal{L}^{-1}\left\{\dfrac{2s}{s^2+a^2}-\dfrac{2s}{s^2+b^2}\right\} =\dfrac{2\cos at-2\cos bt}{t} 0 For practical purposes, there is no closed form solution to to this problem. A closed form solution would require you to diagonalize the coefficient matrix in a closed form. Diagonalization of an n \times n matrix corresponds to solving a polynomial equation of degree n. For n=2 there is a familiar formula, the quadratic formula. For n=3 and n=4 ... 0 This is certainly solvable. It is a fourth order linear system$$\dot{x}=Ax+Bu$$with state vector$$x=\left[\matrix{C_1 & C_2 & C_3 & C_e}\right]^T$$input u=C_{inf} and matrices A, B given by$$A=\left[\matrix{-k_{12}-k_{13}-k_{10} & k_{21} & k_{31} & 0\\ k_{12} & -k_{21} & 0 & 0 \\ k_{13} & 0 &-k_{31} ... 0 This should be solvable using matrix differential equation methods: basically the idea is to write the equations in the form $$\mathbf{c}'(t) = \mathbf{K} \, \mathbf{c}(t) + \mathbf{b},$$ where\mathbf{c}(t)$is the vector$(C_1,C_2,C_3,C_e)$,$\mathbf{K}$is a constant matrix with components made from$k_{ij}$, to reproduce your equations and ... 0 This is far easier using the definition of the inverse transform, i.e., $$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{s^3+1}$$ If you know complex analysis, then you know that this integral is simply the sum of the residues of the poles of the integrand. The poles are at$s=e^{i \pi/3}$,$s=e^{i \pi}$, and$s=e^{i 5 \pi/3} = e^{-i ...

0

If you use taylor expansion for this fraction it gives 1-$S^3$+$S^6$ +....

2

If $X \sim U([0,\frac 12])$, that is $X$ is uniformly distributed on $[0,\frac 12]$, we have \begin{align*} \def\E{\mathbf E} \E[e^{tX}] &= 2\int_0^{\frac 12} \exp(tx)\, dx\\ &= \frac 2t \exp(tx)\bigl|_0^{\frac 12}\\ &= \frac{\exp(t/2) - 1}{t/2} \end{align*} So the density with respect to Lebesgue measure on $\mathbf ... 0 You made a few mistakes when you integrated, like the appearance of an extra$\frac{1}{2}$coefficient, and forgetting to include a negative when you integrate$\int^\infty_0{e^{-t(b+s)}}\mathrm{dt}.$After integration you should get $$\frac{e^{t(b-s)}}{2(b-s)}-\frac{e^{-t(b+s)}}{2(b+s)} \Bigg|_0^\infty$$ The next move (I believe) is that you assume ... 0 The limit: $$\lim_{t \to +\infty} \frac {e^{t(b-s)}}{2(b-s)}$$ exists if$\operatorname{Re}(s) > b$. This might be more obvious if you write it as: $$\lim_{t \to +\infty} \frac {e^{-t(s-b)}}{2(b-s)}$$ 0 (I'm understanding here that your$U_1(t)$here is a Heaviside function that transitions at$t=1$.) It's true to say that Heaviside functions are "like a switch" -- they start at one value, 0, and then they abruptly transition to the value of 1. You probably understand how to graph continuous functions, like$f(t) = t$, very well, and so only the ... 1 We have $$t*e^{3t}=\int_0^t (t-\tau)e^{3\tau}\,d\tau=\frac{1}{9}(e^{3t}-1-3t)$$ and that seems to be the same one gets doing partial fraction decomposition on$1/(s^2(s-3))$, $$\frac{1}{s^2(s-3)}=\frac{1}{9}\Bigl(\frac{1}{s-3}-\frac{3}{s^2}-\frac{1}{s}\Bigr),$$ and then inverse Laplace transform, yielding$\frac{1}{9}(e^{3t}-1-3t)again. So, everything ... 2 $$\mathcal{L}^{-1} \left\{ \frac{\frac{5s}{4} + \frac{13}{4}}{s^2+5s+8} \right\}$$ $$= \mathcal{L}^{-1} \left\{{\frac {\frac{5s}4 + \frac 5 4\frac 5 2 - \frac 5 4\frac 5 2 + \frac {13}4}{\left({s + \frac 5 2}\right)^2 + \frac 7 4}}\right\}$$ You'll need: $$\mathcal{L}^{-1}\left\{{F(s + \alpha)}\right\} = e^{-\alpha t}\mathcal L^{-1} \left\{ ... 4 This is going to be a little involved. Begin with the representation$$\int_{-1}^1 du \frac{u}{\sqrt{1-u^2}} \sin{r u} = \pi J_1(r) $$which may be obtained by differentiating the well-known relation$$\int_{-1}^1 du \frac{\cos{r u}}{\sqrt{1-u^2}} = \pi J_0(r) $$Then$$\begin{align}\int_{-\infty}^{\infty} dy \frac{J_1 \left ( \sqrt{x^2+y^2} \right ... 0 We have to find the steady state error $$\lim_{t\rightarrow\infty}(r(t)-y(t))$$ According to the final value theorem $$\lim_{t\rightarrow\infty}(r(t)-y(t))=\lim_{s\rightarrow 0}\left[s(R(s)-Y(s))\right]$$ The Laplace transform of the error is given by $$R(s)-Y(s)=R(s)-\frac{F(s)G(s)}{1+F(s)G(s)}R(s)=\frac{1}{1+F(s)G(s)}R(s)$$ Hence, for a step reference ... 0 From the Laplace transform, we obtain $$s^2Y(s) - 1 - 4Y(s) = e^{-s}\iff Y(s) = \frac{1+e^{-s}}{s^2-4}$$ as you have found. You can do this with tables and convolution theorems by writing the RHS as $$\frac{1}{s^2 - 4} + \frac{e^{-s}}{s^2 - 4}\tag{1}$$ which I will leave this method to you. However, I will present an alternate method via the use of the ... -1 I think your answer is wrong. Note \begin{eqnarray} y(t)&=&L^{-1}\{\frac{1+e^{-s}}{s^2-4}\}\\ &=&L^{-1}\{\frac{1}{s^2-4}\}+L^{-1}\{\frac{e^{-s}}{s^2-4}\}\\ &=&\frac{1}{2}\sinh(2t)+\frac{1}{2}u_1(t)\sinh(2(t-1)) \end{eqnarray} Here we usedL^{-1}\{e^{-as}F(s)\}=u_a(t)f(t-a)$1 For$u(t)$,$U(s) = \frac{s+1}{(s+2)^2}$, therefore$\lim_{s \to 0} s U(s) = 0$. However, for$y(t)$the Final Value Theorem does not hold, because$y(t)$does not converge. For FVT to work,$\lim_{t \to \infty} y(t)$must exists, i.e. finite. But this is not the case since the system is not stable, i.e. it has roots at right half plane. Therefore$\lim_{t ...

1

$J_0(x)$ is bounded on $\mathbb R$, so the integral defining the Laplace transform will converge absolutely for $\text{Re}(s) > 0$. From the asymptotics $J_0(x) = \sqrt{2/(\pi x)} \sin(x + \pi/4) + O(x^{-3/2})$ as $x \to \infty$, the integral diverges for $\text{Re}(s) < 0$. For $\text{Re}(s) = 0$, say $s = i\omega$, it converges (conditionally) ...

Top 50 recent answers are included