New answers tagged

1

If $\mu(\{0\})<1$, then as $e^{-\theta x}<1$ for $\theta,x>0$ we have \begin{align} \hat\mu(\theta) &= \mu(\{0\}) + \int_{(0,\infty)} e^{-\theta x}\ \mathsf d\mu(x)\\ &< \mu(\{0\}) + \int_{(0,\infty)}\ \mathsf d\mu(x)\\ &=\mu(\{0\}) + \mu((0,\infty))\\ &=\mu([0,\infty))\\ &\leqslant 1. \end{align}


1

HINT $$ v'(t)=-g+c u^2(t)-2 c u(t) v (t)+c v^2(t) $$ This is a Riccati's Equation, of the form $$ v'(t) = q_0(t) + q_1(t) \, v(t) + q_2(t) \, v^2(t) $$ where $q_0(t)= -g+c u^2(t)\neq 0$, $q_1(t)=-2 c u(t)$ and $q_2(t)=c\neq 0$.


0

Obviously, if you find the common denominator, you obtain $$ \frac{8s +0.2}{(4s+5)(8s + 0.2)} + \frac{3(4s + 5)}{(8s + 0.2)(4s+5)} = \frac{20s + 15.2}{32s^2 + 40.8s + 1} $$ so pick polynomials $a,b$ naturally.


0

Oops. Some of what's below is wrong. The part about $\lim=0$ is fine, and the part about $\lim=\infty$ in the case $0\le\rho\le 1$ is fine. But the counterexample in the case $\lim=\infty$, $\rho>1$ has big problems. The construction is "soft" enough that for all we know we got $H(s)=\infty$. A lawyer might argue that there's nothing about $H(s)$ being ...


1

I and II give precise asymptotics for $H$ and $h$. However, the statement $I^{\prime}$ below is imprecise. With some additional assumptions, you may be able to make some statement with inequalities, but I doubt it will be as nice as the original equivalence. I$^{\prime}$: $\beta^{\rho}H(\beta)\rightarrow 0$ as $\beta\rightarrow 0$ Note that $H(\beta)\sim ...


0

If you mean its inverse in time domain, then it is $$u(t - a) f(t)$$ where $$u = \begin{cases} 1, & t \geq a \\ 0, & \text{ otherwise} \end{cases} $$ And yes, the convolution theorem still applies.


0

This is what you get when you have a unit step function. (Heaviside function). If $\int_0^{\infty} f(t) e^{-st} dt = F(s)$ then $\int_a^{\infty} f(t) e^{-st} dt = e^{-as} F(s)$


2

Schaum's outlines are a great resource and contain lots of exercises to work through. For the "Mathematics 1" categories listed, you could probably find those all in the beginning and advanced calculus outlines. For the "Mathematics 2" categories, Schaum's Outline of Differential Equations will cover part of it. For the rest, I'd recommend Applied Partial ...


0

Thanks guys for the help.The partial fraction method makes the most sense. I found out that what I posted was a particular case of Bernoulli equation, which you might find interesting. http://tutorial.math.lamar.edu/Classes/DE/Bernoulli.aspx We can make the substitution v=1/y Then dy/dx=(dy/dv)*(dv/dy) = (-v^-2)(dv/dx) The origninal equation becomes ...


0

$$\frac{\color{blue}{3(s^2-3s-2)}}{s(s+1)(3s+1)}=\frac{A}{s}+\frac{B}{s+1}+\frac{C}{3 s+1}=\frac{\color{blue}{A(s+1)(3s+1)+Bs(3s+1)+Cs(s+1)}}{s(s+1)(3s+1)}$$ Thus we must have $$ 3s^2-9s-6=(3A+3B+C) s^2+(4 A+B+C) s+A $$ and equating the same power we have $3A+3B+C=3$, $4 A+B+C=-9$ and $A=-6$, and then $B=3$ and $C=12$ that is $$ ...


0

$$x'(t)+5x(t)+2y(t)=e^{-t}\Longleftrightarrow$$ $$\mathcal{L}_t\left[x'(t)+5x(t)+2y(t)\right]_{(s)}=\mathcal{L}_t\left[e^{-t}\right]_{(s)}\Longleftrightarrow$$ $$sx(s)-x(0)+5x(s)+2y(s)=\frac{1}{1+s}\Longleftrightarrow$$ $$x(s)\left[s+5\right]=\frac{1}{1+s}+x(0)-2y(s)\Longleftrightarrow$$ $$x(s)=\frac{\frac{1}{1+s}+x(0)-2y(s)}{s+5}$$ ...


0

$$ f(t)=\frac{g(t)}{t}=\frac{1-\cos t}{t} $$ thus the Laplace transform is $$ F(s)=\int_s^\infty G(z)\,\mathrm dz=\int_s^\infty \left[\frac{1}{z}-\frac{z}{z^2+1}\right]\,\mathrm dz=\left[\log{z}-\log\left(\sqrt{z^2+1}\right)\right]_s^\infty=\log\left(\tfrac{\sqrt{s^2+1}}{s}\right) $$


3

Using a vector notation, the system reads $$\frac{d\mathbf x}{dt}=A\mathbf x+\mathbf b.$$ Take the Laplace transform, giving $$s\mathbf X-\mathbf x_0=A\mathbf X+\frac{\mathbf b}s,$$ solved by $$\mathbf X=(sI-A)^{-1}\left(\mathbf x_0+\frac{\mathbf b}s\right).$$ To get the response at infinity, multiply by $s$ ant take the limit at $0$, $$\mathbf ...


3

As Ian has pointed out, it is hard to deal with the laplace transform of the equation. You can solve the equation in this way: $$\frac{y'}y =y+1$$ $$\ln (|y|)=\int y\,dx +x $$ Again $$\frac{y'}{y+1}=y$$ $$\ln(|y+1|) = \int y \,dx$$ So by taking the difference: $$\ln \left( |\frac y{y+1}| \right) = x+C$$ $$\frac y{y+1} =\pm k e^x$$ $$y=\frac 1{1 \mp ...


1

Denote the Laplace transform of $x(t)$ and $y(t)$ by $X(s)$ and $Y(s)$. Taking Laplace transform of the coupled ODEs yields the following: \begin{align} sX & = -kX + gY + \dfrac{E}{s} \tag{1}\\ sY & = -kY - gX \tag{2} \end{align} where we use the assumption that we have zero initial conditions. Rearranging (2) to get an expression for $Y(s)$: ...


0

You just want to have a function that satisfies zero boundary conditions and the differential equation $y^{\prime\prime}+4y=\delta(t-\tau)$. In Laplace transform domain this is $-(0)-s(0)+s^2Y(s)+4Y(s)=e^{-s\tau}$. Then $$Y(s)=\frac{e^{-s\tau}}{s^2+4}$$ The inverse transform is $$G(t,\tau)=\frac12\sin(2(t-\tau))u(t-\tau)$$ You need to add $\cos(2t)$ to the ...


0

Use that, when $\Re[s]>0$: $$\mathcal{L}_t\left[t^n\right]_{(s)}=\int_{0}^{\infty}t^ne^{-st}\space\text{d}t=\frac{s^{n+1}}{\Gamma(1+n)}=\frac{s^{n+1}}{n!}$$ ...


0

Such a curve would have a polar representation $$r(\phi)=a+b\sin(n\,\phi)\qquad(-\infty<\phi<\infty)\ ,\tag{1}$$ whereby $a\gg b>0$, and $n>0$ (not necessarily an integer) denotes the number of full periods per one turn around the origin. The representation $(1)$ can be rewritten as a parametric representation as follows: ...


2

\begin{align} Y(s)&=\frac{s+6}{s^2+4s+5}+\frac{\mathrm e^{-3(s+2)}}{(s+2)(s^2+4s+5)}\\ &=\frac{s+2}{(s+2)^2+1}+\frac{4}{(s+2)^2+1}+\left[\frac{1}{s+2}-\frac{s+2}{(s+2)^2+1}\right]\mathrm e^{-3(s+2)}\\ \end{align} so you have $$ y(t)=\left[\cos t+4\sin t\right]\mathrm e^{-2t}+\left[H(t-3)-\cos (t-3)H(t-3)\right]\mathrm e^{-2t}$$


0

The Weierstrass and Laplace transformations are related by this well known theorem: Theorem. $f(t)$ is a Weierstrass-transformable generalized function with the strip of definition $\{s: \sigma_1 < \Re s < \sigma_2\}$ for $(\mathcal W\{f(t)\}) (s)$ if and only if $\mathrm e^{-t^2/4}f(t)$ is a Laplace-transformable generalized function with the strip ...


0

This is really confusing and your question is not well-posed. I assume $p$ is $s$, "PID" is a PID controller, and you are using inverting feedback. If your input is frequency dependent, nothing says this compensator is going to be able to regulate the plant, if your input is a constant setpoint set at zero on the other hand, you have a good chance with a ...


1

You are better off looking at the transfer function in the $s$-domain where it looks like $$ H(s) = \frac{1}{1+RCs} $$ From the Laplace transformation you know that $0\leq s<\infty$ and from the range of $s$ you get $0 < H \leq 1$. Your version of the transfer function is obtained when you use the substitution $s\mapsto j\omega$ which only works for a ...


2

From your related question we have $F(s)=H(s)\cdot G(s)=\frac{s}{s+\frac{1}{\tau}}\cdot\frac{A}{s}\tanh\left(\frac{sT}{4}\right)$ and then \begin{align} \color{blue}{F(s)}&=\frac{A}{s+\frac{1}{\tau}}\cdot\tanh\left(\frac{sT}{4}\right) =\frac{A}{s+\frac{1}{\tau}}\cdot\frac{\mathrm e^{\frac{Ts}{4}}-\mathrm e^{-\frac{Ts}{4}}}{\mathrm ...


0

Your circuit is a passive differentiator I think. There is one zero and one pole and the operation of the differentiation on the square wave (hence the spikes to be expected as your picture shows) occurs at frequencies below the pole frequency. You should be getting a Dirac delta function in the output for this, I think, after taking the inverse transform.


2

The transfer function is $$ H(s)=\frac{V_{\text{out}}(s)}{V_{\text{in}}(s)}=\frac{s}{s+\frac{1}{\tau}} $$ where $\tau=RC$. The input voltage is $$ V_{\text{in}}(s)=\frac{A}{s}\tanh\left(\frac{sT}{4}\right)=\frac{A}{s}\left(1-\mathrm e^{-\frac{Ts}{2}}\right)^2\sum_{k=0}^{\infty} \mathrm e^{-kTs} $$ and then $$ ...


0

Since the problem with your work is something to do with the RHS of the ODE, I'll focus on computing that side's transform only. $$\begin{align*} \mathcal{L}\left\{4\,\mathcal{U}(\pi-t)\cos t\right\}&=4\int_0^\infty \mathcal{U}(\pi-t)\cos t\,e^{-st}\,\mathrm{d}t\\[1ex] &=4\int_0^\pi \cos t\,e^{-st}\,\mathrm{d}t\\[1ex] ...


2

$$y'(t)=t+1\Longleftrightarrow$$ $$\mathcal{L}_t\left[y'(t)\right]_{(s)}=\mathcal{L}_t\left[t+1\right]_{(s)}\Longleftrightarrow$$ $$sy(s)-y(0)=\frac{1}{s^2}+\frac{1}{s}\Longleftrightarrow$$ Use $y(0)=0$: $$sy(s)-0=\frac{1}{s^2}+\frac{1}{s}\Longleftrightarrow$$ $$sy(s)=\frac{1}{s^2}+\frac{1}{s}\Longleftrightarrow$$ ...


2

$$\mathscr{L}^{-1}\left\{\frac1{s^2}+\frac1{s^3}\right\}=\mathscr{L}^{-1}\left\{\frac1{s^2}\right\}+\mathscr{L}^{-1}\left\{\frac1{s^3}\right\}=\mathscr{L}^{-1}\left\{\frac1{s^2}\right\}+\color{blue}{\frac{1}{2}}\mathscr{L}^{-1}\left\{\frac{\color{blue}{2}}{s^3}\right\}=t+\frac12t^2$$


4

Your inverse Laplace transform of $\frac1{s^3}$ is wrong, it should be $\frac{t^2}2$ as you expect. $$\mathcal L^{-1} \left\{ \frac1{s^{n+1}} \right\} = \frac{t^n}{\color{red}{n!}} $$


2

We can write the square wave function as $$ f(t)=A\sum_{k=0}^{\infty} \left[u\left(t-kT\right)-2u\left(t-\frac{2k+1}{2}T\right)+u\left(t-(k+1)T\right)\right] $$ where $u(t)$ is the Heaviside's function. So the Laplace transform of $f(t)$ is \begin{align} F(s)&=A\sum_{k=0}^{\infty} \frac{1}{s}\left[\mathrm e^{-kTs}-2\mathrm e^{-\frac{2k+1}{2}Ts}+\mathrm ...


0

Consider one method from where the proposer's work ends: Starting with $$\mathcal{L}_{s}^{-1}\left[k\left(\frac{1}{s+\frac{1}{cr}}\right)\left(1-\frac{2}{e^{as}+1}\right)\right]_{(t)} = k\,e^{-\frac{t}{cr}} - 2ckr \, \mathcal{L}_{s}^{-1}\left[\frac{1}{\left(1+e^{as}\right)\left(1+crs\right)}\right]_{(t)}$$ then \begin{align} ...


0

If $\mathcal L\{f\} = F(s)$ vanishes on an infinite sequence of points that are located at equal intervals along a line parallel to the real axis $$ F(s_0+n\sigma)=0\qquad (\sigma >0, n=1,\,2,\,\ldots) $$ $s_0$ being a point of convergence of $\mathcal L\{f\}$; then it follows that $f(t)$ is a nullfunction. So it follows that a Laplace transform ...


0

If $\mathcal L\{f\} = F(s)$ vanishes on an infinite sequence of points that are located at equal intervals along a line parallel to the real axis $$ F(s_0+n\sigma)=0\qquad (\sigma >0, n=1,\,2,\,\ldots) $$ $s_0$ being a point of convergence of $\mathcal L\{f\}$; then it follows that $f(t)$ is a nullfunction. So it follows that a Laplace transform ...


1

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} ...


0

Let $h(t)=(f*g)(t)$. We have $$\mathcal{L}\{f*g\} = \mathcal{L}\{f\} \cdot \mathcal{L}\{g\}$$ that is $$H(s)=F(s)G(s)= \frac{3}{s^2+9}\cdot\frac{1}{s+2} =-\frac{9}{13}\cdot\frac{s}{s^2+9}+\frac{6}{13}\cdot\frac{3}{s^2+9}+\frac{9}{13}\cdot\frac{1}{s+2} $$ and then $$h(t)=\mathcal L^{-1}\{F(s)G(s)\}=\left[ ...


1

Observe that if $\mathcal L\{f(t)\}=F(s)$, we have $\mathcal L\left\{\mathrm e^{-ht}f(t)\right\}=F(s+h)$ and $\mathcal L\{f(kt)\}=\frac{1}{k}F\left(\frac{s}{k}\right)$ for $k>0$ and then $$\mathcal L\left\{e^{-ht}f(kt)\right\}=\frac{1}{k}F\left(\frac{s+h}{k}\right)$$ so for $f(t)=\frac{a}{2\sqrt{\pi t^3}}\exp{(-a^2/4t)}$ with Laplcae transform ...


0

I really wonder why you put two pair of unnecessary brackets in your formula? Furthermore $c\phi$ is a constant or is $\phi$ time dependent? If the latter, you can set it to zero to determine the behavior of $X_{in}$ to $\theta$. But I will asume you mean $c\theta$? $$\ddot\theta(t) = a (b X_{in}(t) - bk\dot\theta(t) - \ddot\theta(t)) - c\theta(t) $$ ...


0

Oh, it turns out to be pretty easy. The equation simply says that in $x \in [0, c)$ it has the form $y''(x) - a y(x) + b = 0$ and in $x > c$: $y''(x) - a y(x) = 0$. Boundary conditions give trivial solution on $x > c$, so I can treat $\theta$-function on $[0, c)$ as a constant to obtain: $$y(x) = b\theta(c - x)(e^{-\sqrt a x} + 1)$$


3

Hint: you can use the fact that $$\mathcal{L}\{f'(t)\} = s\mathcal{L}\{f(t)\}-f(0)$$ and $$\mathcal{L}\{tf(t)\} = -\frac{d}{ds}\mathcal{L}\{f(t)\}$$ to get that $$\mathcal{L}\{tf'(t)\} = -\frac{d}{ds}\mathcal{L}\{f'(t)\} = -\frac{d}{ds}\big(s\mathcal{L}\{f(t)\}-f(0)\big) = - \mathcal{L}\{f(t)\} - s\frac{d}{ds}\mathcal{L}\{f(t)\}$$ but you really don't ...


0

Observe that $$ \big(f(t) * \delta(t-T)\big)= \int_{-\infty}^\infty f(\tau) \delta(t-T-\tau) \,\mathrm d\tau= \int_{-\infty}^\infty f(\tau) \delta(\tau-(t-T)) \,\mathrm d\tau = f(t-T) $$ So you have $$f(t) * \big(\delta(t-1)-\delta(t-1)\big)=f(t) * \delta(t-1)-f(t) * \delta(t-2)=f(t-1)-f(t-2)$$ where $f(t)=t\mathrm e^{2t}u(t)$ and then $$\color{blue}{ ...


2

$$ \begin{align} \mathcal L\left[ f\left( t-a\right) H\left( t-a\right) \right] &=\int_a^\infty\mathrm e^{-pt}f\left( t-a\right)\mathrm dt\\ &=\int_0^\infty\mathrm e^{-p(u+a)}f\left( u\right)\mathrm du\qquad (t-a=u)\\ &=\mathrm e^{-pa}\int_0^\infty\mathrm e^{-pt}f\left( t\right)\mathrm dt\qquad (t=u)\\ &=\mathrm e^{-pa}\mathcal L\left[ ...


6

Note that $$ (xy(x))'=xy'(x)+y(x) $$ so that your differential equation can be solved by one simple integration.


0

I think I've figured it out. I'm not sure what the norms about answering your own question are, but I post this in case it's helpful to others. 1) Define $v_n = a_n \epsilon_n = Expo(1/a_n)$. 2) Define $X = \sum_n v_n$ As $X$ is a sum of independent random variables, its PDF $f(x)$ is a convolution of those random variables, thus the Laplace transform of ...


0

I'll be a little bit straightforward. We have \begin{align} f(t) &= t e^{2 t} \theta_0(t),\\ g(t) &= \delta_1(t) - \delta_2(t), \end{align} where $\theta_x(t) = \mathbb{1}_{\left\{t\geq x\right\}}$ and $\delta_x(t) = \mathbb{1}_{\left\{t=x\right\}}$. Given that both functions are supported on $\mathbb{R}_{+}$, we define $$h(t) = f(t) \ast g(t),$$ ...


0

Why don't simply use the definition of the convolution of two distributions, $$(T\star S) (\varphi) = T_x (S_y (\varphi(x+y)))$$ Hence, take $S = \delta_1-\delta_2$ and you got $S_y(\varphi(x+y))=\varphi(x+1)-\varphi(x+2)$. Hence \begin{align*} (te^{2t} \star (\delta_1-\delta_2))(\varphi)& = ...


0

Why not a substitution? $y=x-ct \implies dy=-cdt$, and $$ \mathcal{L}[u(t,x)](s) = \frac{1}{c} \int_{-\infty}^x \mathrm{exp}\left(-\frac{s}{c}(x-y)\right)u_0(y)dy = \int_0^\infty \mathrm{exp}(-st)\,u_0(x-ct)\,dt,$$ So $u(t,x) = u_0(x-ct)$.


0

First, you need to parametrize your curve. You can check that $f(t)=5t$ for $t\in[0,2]$ and $f(t)=15-2.5t$ for $t\in(2,6]$. Then, the function can be written as $$f(t)=5t[u(t)-u(t-2)]+(15-2.5t)[u(t-2)-u(t-6)]$$ note that the function is recovering the value at $t=2$ if we take the convention $u(0)=1/2$. For the Laplace transform, you get two kind of terms: ...


1

The substitution leads to a definite integral $$ I = \int\limits_0^\infty e^{-sv} f(v) \, dv $$ which does not depend on $u$. The prior integral $$ I_0 = \int\limits_u^\infty e^{-s(t-u)} f(t-u) \, dt $$ seems to depend on $u$, but it is not as $I$ shows.


1

Case 1: $f(t)=\cos\left(2t-\frac{\pi}{3}\right) u\left(t\right)$ Observing that $\cos(\omega t -\phi)=\cos(\omega t)\cos\phi+\sin(\omega t)\sin\phi$ and that $\mathcal L\left\{\cos(\omega t)\right\}=\frac{s}{s^2+\omega^2}$ and $\mathcal L\left\{\sin(\omega t)\right\}=\frac{\omega}{s^2+\omega^2}$, we have \begin{align} \mathcal L\left\{\cos(\omega t ...


0

OK I got it. Whenever we take a derivative of something we must lose some information. Here the rule is a specific part of the general thing. If we say $L(tf(t))=-F'(s)$ then the most general form of F(s) is $F(s)=L(f(t))+C$ for arbitrary C. Thus, considering the previous equation $L(t^2f(t))=F''(s)$ $F(s)=L(f(t))+as+b$ ...



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