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1

$$\ddot y+4y=\cos t-\cos t \cdot \theta(t-2\pi)$$ Because $y(t)$ is continuous at $t=2\pi$, we have $$\lim _{t\to2\pi^+}\ddot y(t)-\lim _{t\to2\pi^-}\ddot y(t)=\lim _{c\to 0}\left(\ddot y(2\pi+c)-\ddot y(2\pi-c)\right)$$ $$=-\cos (2\pi) \cdot \lim _{c\to 0}(\theta(c)-\theta(-c))=-1 \cdot (1-0)=-1$$


0

The transfer function of a dynamical linear system is defined as: $$G(s) = \frac{Y(s)}{U(s)},$$ where $U(s)$ is the Laplace transform of the input signal. The output signal can therefore be obtained just as follows: $$Y(s) = U(s) G(s) = \mathcal{L}_s [\cos{t}] \frac{1}{(s+1)^2},$$ and hence: \begin{align} \color{blue}{y(t)} = & \mathcal{L}^{-1} ...


2

First call: $$ G(u) = \mathcal{L}_uf'(t) = \int^\infty_0 f'(t) e^{-ut} \, \mathrm{d}t,$$ Integrate respect to $u$ from $0$ to $s$: $$\int_0^s G(u)\mathrm{d}u = \int^\infty_0 \int_0^s f'(t)e^{-ut} \mathrm{d}u \, \mathrm{d}t$$ $$=\int^\infty_0 \frac{f'(t)}{t}[1-e^{-st}] \, \mathrm{d}t=- \mathcal{L}\left[\frac{f'(t)}{t}\right]+\int^\infty_0 \frac{f'(t)}{t} ...


0

Since you mention $\sqrt{t}$ I assume that your original function is defined on $[0,\infty )$. Then the complex Laplace transform is \begin{equation*} L(z)=\int_{0}^{\infty }dt\exp [izt]f(t),\;{Im}z>0. \end{equation*} With $\theta (t)$ the Heaviside step function and setting $z=\omega +i\delta $ we can write \begin{equation*} L(z)=L(\omega +i\delta ...


3

Have you considered using the properties of the Laplace transform to simplify calculation? The one's I'm thinking of are: $$\mathcal{L}(e^{at}f(t))=F(s-a)$$ and, $$\mathcal{L}\left(\frac{d^n}{dt^n}f(t)\right)=s^nF(s)-\sum_{i=1}^ns^{i-1}f^{(n-i)}(0)$$ where $F(s)=\mathcal{L}(f(t))$ and $f^{(n)}$ is the n-th derivative of $f$. The $\frac{1}{n!}$ is a ...


1

The integral you've given calculates the inverse Laplace transform. To go from the time-domain to the Laplace-domain (i.e. perform the Laplace Transform), you shouldn't need any contour integration. The (unilateral) Laplace transform is given by $$ F(s) = \int_{0}^{\infty} {f(t)e^{-st}dt}$$ Unless you specifically want to practice integration, these are ...


0

For the case $1)$ you can use the following trick: Define $z(t) = y-y_0 $ such that $z(0) = 0$ and $z'(0) = y'_0 = 0$. Rewrite the problem for $z$: $$ z'' - 3z' + 2 z = \cos{\alpha t} - 2 \equiv f(t), \quad z(0)=z'(0) = 0, \quad t > 0. \tag{1}$$ Laplace-transform both sides to have: $$ Z(s) = \frac{1}{s^2-3 s+2} \, \mathcal{L}_s [f[t]] = ...


1

They're asking for $$\lim_{t \to 2 \pi^+} y'(t) - \lim_{t \to 2 \pi^-} y'(t)$$ and the same at $7 \pi$. However, you don't need the solution to calculate these, all you need to assume is that $y(t)$ and $y'(t)$ are bounded (or even just integrable). Rewrite the equation $$y''(t) = -4y'(t) - 13y(t) + \delta(t-2\pi)-\delta(t-7\pi)$$ and integrate both ...


0

You already proved that the solution on $(0,\pi)$ is $$y(t)=\frac{4}{17} \sin{t} - \frac{16}{17} \cos{t}+A\mathrm e^{-t/2}\cos t + B\mathrm e^{-t/2} \sin{t},$$ for some constants $A$ and $B$. Your next step should be to determine $(A,B)$ using the initial conditions $y(0)$ and $y'(0)$. You (nearly) already proved that the solution on $(\pi,\infty)$ is ...


1

Your equation is a 2nd order, constant coefficients, non-homogenous and linear ODE. Furthermore, the non-homogenous term is a piecewise function that, since $t > 0$ splits your domain into two subdomains, $I_1 = (0,\pi)$ and $I_2 = [\pi,\infty)$. This leads you to solve the equation for each of the subdomain. For doing that, follow, for example, this link ...


1

If the denominator is a quadratic, you should find suitable entries in any standard Laplace table: $$L^{-1}\left\{\frac{a}{s^2+a^2}\right\}=\sin at\ ,\quad L^{-1}\left\{\frac{s}{s^2+a^2}\right\}=\cos at\ .$$ You can convert your example into this form by completing the square, $$L^{-1}\left\{\frac{60+12s}{s^2 + 10s + 50}\right\} ...


1

Here, you have to use the result $\mathcal L[t\, p(t)] = -\dfrac{d}{ds} \mathcal L[p(t)]$. Using this: $\mathcal L[t \cos \omega t] = -\dfrac{d}{ds}\mathcal L[\cos \omega t] = -\dfrac{d}{ds} \dfrac{s}{s^2 + \omega^2} = \dfrac{s^2 - \omega^2}{(s^2 + \omega^2)^2}$ This can be split up as: $\dfrac{s^2 + \omega^2 - 2\omega^2}{(s^2 + \omega^2)^2} = ...


1

Would this be correct? As written, it is not (Wolfram Alpha). Also the time shift property applies to functions that are zero for a while, i.e., contain a Heaviside factor. There isn't one in your formula. So, if the formula is correct, you just have to calculate explicitly, reducing the transform to $\cos t$ and $t\cos t$. But if the first factor is ...


1

Indeed, $$ f(t) = L^{-1}\biggl[\dfrac{1}{(s^2 + 1)^2}\biggr] = L^{-1}\biggl[\dfrac{1}{s^2 + 1} - \dfrac{s^2}{(s^2 + 1)^2}\biggr] = \sin t - L^{-1}\biggl[\dfrac{s^2}{(s^2 + 1)^2}\biggr] \ (1) $$ But, $$ \dfrac{d}{ds}\biggl(\dfrac{s}{s^2 +1}\biggr) = \dfrac{1}{(s^2 + 1)^2} - \dfrac{s^2}{(s^2 + 1)^2} \quad \Rightarrow $$ $$ ...


2

If I understand correctly you are asking two questions. The first is about an apparent issue because you are not using the fact that $y_0 = y_{-1} + x_0$. The second is if you can solve difference equations using the z-transform. A few comments first: My discussion ignores issues of convergence and existence of the z-transform. I assume that the relevant ...


2

If you multiply two power series $\sum_{n=0}^{\infty}a_{n}z^{n}$ and $\sum_{n=0}^{\infty}b_{n}z^{n}$, then a type of convolution also appears because of trying to gather like powers $$ \sum_{n=0}^{\infty}a_{n}z^{n}\sum_{n=0}^{\infty}b_{n}z^{n} = \sum_{n=0}^{\infty}\left(\sum_{k=0}^{n}a_{k}b_{n-k}\right)z^{n} $$ You can think of a Laplace ...


0

Well convolutions are really DEFINED to make that identity true. Their use comes into play when you want to separate the elements of the equations that relate to the system itself, those that relate to the initial conditions of the system, and those that relate to external forces. For example, consider $$ ay''+by'+cy = g(x) $$ Transforming, $$a\big ...


2

The standard argument is: the Laplace transform of $f$ is $$F(s)= \int_0^\infty e^{-st} f(t)\,dt$$ For the delta-function at $0$ this gives $F(s)\equiv 1$. However, "constant in one domain is delta in the other domain" does not work in the other direction. In contrast to the Fourier transform (which is very similar to its inverse), the Laplace ...


0

Using Maple, I find the Laplace transform of the noncentral chi-square density with $\nu$ degrees of freedom and noncentrality parameter $\delta$ is $$ \left( 1+2\,s \right) ^{-\nu/2} \exp\left(-{\frac {\delta\,s}{1+2\,s}}\right) $$


0

If you mean the probability function for the chi-square distribution with $r$ degrees of freedom $$P_r(x)=\frac{x^{-1+r/2}e^{-x/2}}{2^{r/2}\Gamma(r/2)}$$ The Laplace transform is : $$\int_0^\infty P_r(x)e^{-sx}dx = \frac{(s+1/2)^{-r/2}\Gamma(r/2)}{2^{r/2}\Gamma(r/2)}= (2s+1)^{-r/2}$$


0

You can think of it as though the Fourier transform is giving you the "steady state" behavior after the system has been exposed to that sinusoid for an infinite amount of time. Hence there isn't much to say except the magnitude and phase because everything other than the input frequency will be gone by then. On the other hand, the Laplace transform ...


2

The value of $C$ is wrong. It should be $A=\frac{7}{4}, B=\frac{1}{2}, C=-\frac{3}{4}$


2

We get the following system: $$ \left( \begin{array}{l l} s & 1\\ 1 & s \end{array} \right) \cdot \left( \begin{array}{l} \bar X\\ \bar Y \end{array} \right) = \left( \begin{array}{c} 2 + \frac{1}{s^2 + 1}\\ \frac{s}{s^2+1} \end{array} \right) $$ Then you will have to multiply the left hand side with $$ \left( ...


0

$$ L( \frac{1}{a}cos( \frac{bt}{a} ) ) = \frac{as^2}{a^2s^2 + b^2} $$ and hence,your reasoning must surely be wrong. Think about change of variable from $s$ to $s/a$ .


1

$\hat{h}(s) = \frac{s}{a^2s^2+b^2} = {1 \over a^2} {s \over s^2 + ({b \over a})^2 }$. From this we get $h(t) = {1 \over a^2} \cos ({b \over a} t)$.


1

The generating function of Catalan's numbers $y_n=\frac 1{n+1}\binom{2n}n$ is $$\sum_{n\geqslant 0 }\frac1 {n+1}\binom{2n}nx^n=\frac{1-\sqrt{1-4x}}{2x}$$ Multplying by $x$ and differentitating gives that $$\sum_{n\geqslant 0}\binom {2n}nx^n=\frac{d}{dx}\frac{1-\sqrt{1-4x}}{2}=\frac{1}{\sqrt{1-4x}}$$ One can also try to prove this directly, by noting that ...



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