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0

Starting from $$H(z)=\frac{1-z^{-1}+z^{-2}-3z^{-3}}{z^{-2}(1-z^{-1})}$$ then \begin{align} H(z) &= \frac{z^{3}}{z^{3}} \cdot \frac{1-z^{-1}+z^{-2}-3z^{-3}}{z^{-2}(1-z^{-1})} \\ &= \frac{z^{3} - z^{2} + z - 3}{z-1} = \frac{z^{2} (z-1) + (z-1) -2}{ z-1} \\ &= 1 + z^{2} + \frac{2}{1-z} = 1 + z^{2} + 2 \, \sum_{n=0}^{\infty} z^{n} \\ &= 3 + 2 z + ...


3

In the form it is in right now, I do not think the Laplace transform is of use. However, you can rewrite it in such a way to use the Laplace transform: $$-y\sin x + y'\cos x = \cos^2 x \Longrightarrow \frac{d}{dx}(y\cos x) = \cos^2 x.$$ From here you can use a Laplace transform. You might want to use a double angle identity to simplify $\cos^2 x$. That ...


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Your function is $f(t)=t$ but not $f(t)=1$. You need to use the formula $$\mathcal{L}\{u_c(t) f(t-c)\} = e^{-cs}*F(s)$$ 3 times with $c=1,2, 3$ to get the answer. The answer derived by Leucippus looks correct.


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It might be helpful to consider it as three separate parts: $$h(t) = \left\{\begin{array}{l} 0,\, 0\leq t<\pi \\ 1,\, \pi\leq t<2\pi\\ 0, t\geq2\pi \end{array}\right.$$ The most systematic way I find to turn these into heaviside functions is to start with the top-most, then add heaviside times the function on the next step minus that of the ...


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The inverse Laplace transform considered is $$F(s)=e^{-3s}\frac{1+s}{s^3+2s^2+2s}$$ and can be reduced as follows. First notice that \begin{align} F(s) = \frac{e^{-3 s}}{s} \cdot \frac{s+1}{(s+1)^{2} + 1} = f_{1}(s) \cdot f_{2}(s) \end{align} where \begin{align} f_{1}(s) = \frac{e^{-3 s}}{s} \hspace{10mm} f_{2}(s) = \frac{s+1}{(s+1)^{2} + 1}. \end{align} ...


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Using the notation $u_{c}(t) = H(t-c)$, where $H(t)$ is the Heaviside step function, then $$g(t)= (t-1) u_1(t) - 2(t-2) u_2(t) + (t-3) u_3(t)$$ is evaluated as follows. Let $g(s) = \mathcal{L}\{g(t)\}$ such that: \begin{align} g(s) &= \int_{0}^{\infty} e^{-s t} \left[ (t-1) \, H(t-1) - 2 \, (t-2) \, H(t-2) + (t-3) \, H(t-3) \right] \, dt \\ &= ...


0

HINT: $$\mathscr{L}(tu(t))(s)=\int_0^{\infty}te^{-st}dt=\frac1{s^2}$$ Thus, $$\mathscr{L}((t-c)u(t-c))(s)=\int_c^{\infty}(t-c)e^{-st}dt=e^{-sc}\frac1{s^2}$$ NOTE: We derive the result for the Laplace Transform of $t$. To that end, we write $$\mathscr{L}(t)(s)=\int_0^{\infty}te^{-st}dt$$ Now, integrate by parts with $u=t$ and $v=-e^{-st}/s$. Then, ...


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You can reduce this problem to a more familiar one with the following trick. Consider the problem $$ q_t(x,t) = -au_x(x,t) + \frac{D}{2}q_{xx}(x,t) \quad (0 < t, 0 < x < 2L)\\ q(x,0) = 0, \quad aq(0,t) - \frac{D}{2}q_x(0,t) = f(t) = -aq(2L,t) - \frac{D}{2}q_x(2L,t) $$ and solve it with a standard method. The solution will have a specific symmetry ...


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The Fourier Transform of a spatial variable is no different mathematically from a Fourier Transform of a temporal variable. The mathematics is agnostic to parameter interpretation. For the Fourier Transform pair for the time-frequency domain are often written $$F(\omega) = \mathscr{F}(f)(k) = \int_{-\infty}^{\infty} f(t) e^{i \omega t} \, dt$$ $$f(t) = ...


1

If $f \in L^{2}[0,\infty)$, then $\mathscr{L}\{f\}$ is holomorphic in the right half plane where $\Re s > 0$. and the Laplace transform is square integrable on all vertical lines in the right half plane, with $$ \frac{1}{2\pi}\int_{-\infty}^{\infty}|\mathscr{L}\{f\}(v+iw)|^{2}dw \le \int_{0}^{\infty}|f(t)|^{2}dt,\;\;\; 0 < v < \infty. $$ The ...


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It is convenient to start with the complex Laplace transform. Fourier transform \begin{eqnarray*} \tilde{f}(\omega ) &=&\int_{-\infty }^{+\infty }dt\exp [i\omega t]f(t) \\ f(t) &=&\frac{1}{2\pi }\int_{-\infty }^{+\infty }d\omega \exp [-i\omega t]% \tilde{f}(\omega ) \end{eqnarray*} Complex Laplace transform $$ \hat{f}(z)=\int_{0}^{+\infty ...


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The denominator $1 + T_d s$ acts as a filter to prevent the D action from actually being a derivative. Pure derivatives are not physically implementable, because the gain would go to infinity at high frequencies. (Trace the Bode plot to verify!) The extra pole at $T_d$ limits the gain and allows the use of the D term in practice.


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Consider the following. \begin{align} \frac{2s+5}{s^2+6s+34} = \frac{2 s + 5}{ (s+3)^{2} + 5^{2}} = 2 \, \frac{(s+3)}{(s+3)^{2} + 5^{2}} - \frac{1}{5} \, \frac{5}{(s+3)^{2} + 5^{2}} \end{align} Now, for \begin{align} f(s) \doteqdot \int_{0}^{\infty} e^{-st } \, f(t) \, dt, \end{align} then \begin{align} \frac{2s+5}{s^2+6s+34} \doteqdot 2 e^{-3t} \, \cos(5 t) ...


0

Use the method of completing the square to see that the denominator is $(s+3)^2+5^2$. Then we have $$\frac{2s+5}{(s+3)^2+5^2}=2\frac{s+3}{(s+3)^2+5^2}-\frac{1}{5}\frac{5}{(s+3)^2+5^2}.$$ These functions are now in a standard form to be inverse-transformed.


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When you have taken the transform of the differentiated equation, you have missed off constant terms. For the transform of the derivative is $$sX(s)-x(0)$$ You should find that the relation between you 2 functions then causes the original constant to be present, as one should expect.


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Keep in mind that the Laplace transform takes the following form: $$F(s)= \int_{0}^{\infty} f(x) e^{-sx}dx. $$ Let us make use of the following set of properties : $$ \begin{align} \int_0^\infty F(u)g(u) \, du & = \int_0^\infty f(u)G(u) \, du \\[6pt] L[f(t)] & = F(s) \\[6pt] L[g(t)] & = G(s)\end{align} $$ Let $$ G(x)=\frac{1}{x^3} ...


3

Note how you have defined $\sinh(t)$. A factor of $\frac{1}{2}$ is missing, which when you correct for should give you the answer you're looking for.


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Let $f(t)=y(t+2)$. That way $f(0)=y(2)$ and $f'(0)=y'(2)$. Equivalently, $f(t-2)=y(t)$, $y'(t)=f'(t-2)$ and $y''(t)=f''(t-2)$. Then $$ f''(t-2)-3f'(t-2)+2f(t-2)=e^{-t},\;\;\; f(0)=1, f'(0)=0. $$ Substitute $t$ with $t+2$: $$ f''(t)-3f'(t)+2f(t) = e^{-t}e^{-2}, \;\;\; f(0)=1, f'(0)=0. $$


0

I cant understand where you get the $\delta$-function, too. According to my work: $$y''(t+2)-3y'(t+2)+2y(t+2)=e^{-(t+2)}\qquad y(0)=1\,,\,y'(0)=0$$ $$s^2\mathcal{L}[y(t+2)]-sy(0)-y'(0)-3(s\mathcal{L}[y(t+2)]-y(0))+2\mathcal{L}[y(t+2))]=e^{2s}\mathcal{L}[e^{-t}]$$ ...


1

You do not need to convert e^(-t) to a dirac-delta function. It is a dirac-delta function that is converted to e^(at) form to reduce the function. Secondly, as we are supposed to work with initial conditions, which are not given here. we have, t=2 => consider n=t-2 so that n=0. => t=n+2 now the given equation can be written as: ...


0

Firstly I can't see where you get the $\delta$-function from, please consider showing your working when you post a question here. You want to use the following properties of the Laplace transform: $$\mathcal{L}[f'(t)]=sF(s)-f(0)$$ $$\mathcal{L}[f''(t)]=s^2F(s)-sf(0)-f'(0)$$ where $F(s)$ denotes the Laplace transform of $f(t)$. Unfortunately, you only have ...


0

Since: $$ \cos(x)=\prod_{n\geq 0}\left(1-\frac{4x^2}{(2n+1)^2\pi^2}\right)\tag{1} $$ by considering the logarithmic derivative of both sides we have: $$ \tan(x) = \sum_{n\geq 0}\frac{8x}{(2n+1)^2\pi^2-4x^2}=\sum_{n\geq 0}\left(\frac{-1}{x-\frac{2n+1}{2}\pi}+\frac{-1}{x+\frac{2n+1}{2}\pi}\right)\tag{2}$$ so $\tan(x)$ does not have a Laplace transform, but, at ...


0

In general case we have: $$L(f^n)=s^nL(f)-s^{n-1}f(0)-\dots-sf^{n-2}(0)-f^{n-1}(0)$$ so $$L(y')=sL(y)-y(0)=sL(y)-1$$ then $$sL(y)-1+2L(y)=\frac{s}{s^2+9}$$ $$L(y)=\frac{s^2+s+9}{(s+2)(s^2+9)}$$ $$L(y)=\frac{\frac{2}{13}s+\frac{9}{13}}{s^2+9}+\frac{\frac{11}{13}}{s+2}$$ ...


1

The general formula is $$ L(f^{(n)})(s) = s^n L(f)(s) - \sum_{k=1}^n s^{n-k} f^{(k-1)}(0) $$ In particular, $L(y')(s) = s L(y)(s) - y(0)$.


1

Just note $$\frac{1}{s^2+a^2}-\frac{1}{s^2+b^2}=\frac{b^2-a^2}{(s^2+a^2)(s^2+b^2)}$$ Then, if $a^2\neq b^2$ and $ab\neq 0$, we have \begin{align*} \frac{1}{(s^2+a^2)(s^2+b^2)}&=\frac{\frac{1}{b^2-a^2}}{s^2+a^2}-\frac{\frac{1}{b^2-a^2}}{s^2+b^2}\\ ...


0

As you said, $$\mathcal{L}^{-1}\left(\frac{a}{s^2+a^2}\right)=\sin(at)\tag{1} $$ hence, assuming $a\neq b$ and $a,b\neq 0$: ...


1

Your partial fractions decomposition is off. You should get $$ \frac{1}{(s^2+a^2)(s^2+b^2)} = \frac{1}{b^2-a^2} \left( \frac{1}{s^2+a^2} - \frac{1}{s^2+b^2} \right), $$ so the inverse Laplace transform is $$ \frac{1}{b^2-a^2} \left( \frac1a \sin at - \frac1b \sin bt \right). $$


1

You also needn't use the Laplace transform here. Your Ansatz could be $Ae^{3x}$ and you would have $18Ae^{3x}+3Ae^{3x}-Ae^{3x}=e^{3x}$ $18A+3A-A=1$ which gives you the particular solution. The solution to the homogenous equation, as you know, will be a linear combination of $e^{r_1x}$ and $e^{r_2x}$. Then you can plug in the initial values. To me, this ...


3

In general, when you do Laplace transforms, you need to do partial fraction decomposition to separate all the terms: $$L(y)=\frac{1}{(s-3)^2}\cdot\frac{1}{2s-1}\cdot\frac{1}{s+1}$$ $$\frac{1}{(s-3)^2}\cdot\frac{1}{2s-1}\cdot\frac{1}{s+1}=\frac{A}{s-3}+\frac{B}{(s-3)^2}+\frac{C}{2s-1}+\frac{D}{s+1}$$ Clear the denominator: ...


0

Just another approach. The zeroes of $s^2+2s+5$ occur at $s=-1\pm 2i$, and: $$\text{Res}\left(\frac{s+2}{s^2+2s+5},s=-1\pm 2i\right)=\frac{1}{2}\pm \frac{i}{4},\tag{1}$$ so: $$ f(s) = \frac{s+2}{s^2+2s+5} = \frac{\frac{1}{2}+\frac{i}{4}}{s-(-1+2i)}+\frac{\frac{1}{2}-\frac{i}{4}}{s-(-1-2i)}\tag{2} $$ and since ...


1

I'm assuming there is a typo in the first equation. I will assume the second equation is right. Some standard Laplace transforms are $$\mathcal L[e^{-at}\cos(bt)]=\frac{s+a}{(s+a)^2+b^2},$$ and $$\mathcal L[e^{-at}\sin(bt)]=\frac{b}{(s+a)^2+b^2}.$$ You can write your Laplace transform as $$\mathcal ...



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