New answers tagged laplace-transform
2
$$- \frac{\pi }{2}\int\limits_0^\infty {\frac{u}{{{s^2} + {u^2}}}\frac{{du}}{{{u^2} + 1}}}$$
$$ - \frac{\pi }{4}\int\limits_0^\infty {\frac{{dm}}{{{s^2} + m}}\frac{1}{{m + 1}}} $$
Now by partial fractions you can separate and get:
$$ - \frac{\pi }{4}\int\limits_0^\infty {\frac{{dm}}{{{s^2} + m}}\frac{1}{{m + 1}}} = - \frac{\pi }{4}\frac{1}{{{s^2} - ...
3
Denoting $\Psi(t) = \int_t^\infty \psi(\tau) \, d\tau$, we have by definition of $\mathcal L$, that
\begin{align*}
\mathcal L(\Psi)(s) &= \int_0^\infty \exp(-st)\Psi(t)\, dt\\
&= \int_0^\infty \exp(-st)\int_t^\infty \psi(\tau)\, d\tau\, dt\\
&= \int_0^\infty \int_t^\infty \exp(-st)\psi(\tau)\, d\tau\, dt\\
&= \int_0^\infty ...
1
OK, I think I have a systematic way of getting this ILT. Let's consider a periodic function $f(t)$ with period $T$. The Laplace transform of such a function is
$$\begin{align}\int_0^{\infty} dt \, f(t) e^{-s t} &= \sum_{k=0}^{\infty} \int_{k T}^{(k+1) T} dt \, f(t) e^{-s t}\\ &= \sum_{k=0}^{\infty} e^{-k s T} \int_0^T du \, f(u) e^{-s u} \\ &= ...
2
Here is a solution by the method of characteristics. So, we assume that the equation is
$$
u_t+u_x=\cos^2 u,\quad u(x,0)=u_0(x).
$$
Consider the characteristics defined by
$$
\frac{dx}{ds}=1,\quad x(0)=\tau,\\
\frac{dt}{ds}=1,\quad t(0)=0,\\
\frac{du}{ds}=\cos^2 u,\quad u(0)=u_0(\tau).
$$
Obviously, from first two equations
$$
x=s+\tau,\quad t=s\implies ...
2
ILTs like this may be evaluated using the residue theorem: the ILT is, in the this case, the sum of the residues at the poles to the left of the vertical line of integration. Here, the poles at $s_{\pm}=\pm 4 i$ are double poles.
The residue of a double pole $s=s_0$ of a function of the form
$$f(s) = \frac{p(s)}{q(s)^2}$$
is given by the expression
...
2
Observe that $$\int \frac s{(s^2+16)^2}ds=-\frac1{s^2+16}+C\iff \frac{d \left(\frac1{s^2+16}\right)}{ds} =- \frac s{(s^2+16)^2}$$
and from this, $$\mathscr{L}^{-1}\left\{ \frac1{(s^2+16)} \right\}=\frac{\sin4t}4$$
So, $$\mathscr{L} \left\{ t^1\cdot\frac{ \sin 4t}4 \right\}=(-1)^1\frac{d^1}{ds^1}\left(\frac1{s^2+16}\right)=+\frac{2s}{(s^2+16)^2} $$
0
I think this is about Z-Transformation.
In order to make the overall transfer function causal, we rewrite the given transfer function as
$$ \dfrac{K}{(a+bz^{-1})(x+yz)} = \dfrac{Kz^{-1}}{(a+bz^{-1})(y+xz^{-1})}. $$
Next, do:
$$ \dfrac{Kz^{-1}}{(a+bz^{-1})(y+xz^{-1})} = \dfrac{A}{a+bz^{-1}} + \dfrac{B}{y+xz^{-1}} $$
It is easy to see that:
$$
aA + yB = ...
1
If you're looking for tracking error, you can use Mason's formula to find the transfer function between $\Theta_{ref}$ and $\Theta$. First find the forward path gains. There is only one forward path:
$$F_1 = K_{2}*\frac{1}{N}*K_1\frac{1}{s}*N*\frac{1}{s}=\frac{K_1K_2}{s^2}$$
Now find the loop gains. there are two loops, the inner loop ($L_1$) and the outer ...
1
For every integrator block (you have two), you have one state ($\omega$ and $\theta$). Then you can find the relations between all the states, and construct the matrices for the state space representations. From A, B, C, D, you can find the transfer function for your system. If it is type 0, you will have a steady state error for step inputs, otherwise the ...
4
Google the Square wave function $\mathbf{sqw}(x)=(-1)^{\mathbf{floor}(x)}$ to find out it is a periodic function which is obviously piecewise and by using the L.T. rules we can see that $$\mathcal{L}\{\mathbf{sqw}(x)\}=\frac{1}{s}\tanh(s/2)=\frac{e^s-1}{s(e^s+1)}$$ So since $$\mathcal{L}\{1\}=\frac{1}{s}$$ I think you can do the rest.
1
Note that $\dfrac{\mathrm e^{-st}}t=\displaystyle\int_s^\infty\mathrm e^{-ut}\mathrm du$ hence, using Fubini theorem,
$$\int_0^\infty\mathrm e^{-st}\frac{x(t)}t\mathrm dt=\int_0^\infty x(t)\left(\int_s^\infty\mathrm e^{-ut}\mathrm du\right)\mathrm dt=\int_s^\infty\left(\int_0^\infty\mathrm e^{-ut}x(t)\mathrm dt\right)\mathrm du,
$$
that is, considering ...
6
See my solution to this problem. The analysis is identical, save for the denominator.
$$\oint_C ds \,e^{-a \sqrt{s (s+b)}}\, e^{s t}$$
where $C$ is the following contour pictured below:
This one's a bit odd because we are removing the branch point singularities by subtracting the inner contour $C_i$ from the outer contour $C_o$, i.e. $C=C_o+C_i$.
The ...
0
I have done it, you can see if it's useful to you, I.m not sure so if you look out the mistake please fix it..
$\begin{array}{l}
F(s) = \frac{{2{{({s^2} + 4s + 5)}^2} + 4}}{{{{({s^2} + 4s + 5)}^2}}} = 2 + \frac{{40}}{{{{{\rm{[}}{{(s + 2)}^2} + 1{\rm{]}}}^2}}} \\
{L^{ - 1}}{\rm{[2] = }}2\delta (t) \\
{\rm{\{ }}\int_0^\infty {2.{e^{ - st}}\delta (t)dt ...
0
I did by let u = at, so the period T = 2pi
$\begin{array}{l}
f(t) = f(t + \frac{{2\pi }}{a}) = \frac{{\sin at}}{{|\sin at|}} \\
Let:f(u) = (u + 2\pi ) = \frac{{\sin u}}{{|\sin u|}} \\
F(s) = L{\rm{[f(u)]}} = \frac{1}{{1 - {e^{ - 2\pi s}}}}\int_o^{2\pi } {{e^{ - st}}\frac{{\sin {a^2}u}}{{|\sin {a^2}u|}}du} \\
= \frac{1}{{1 - {e^{ - 2\pi s}}}}{\rm{\{ ...
1
Note that $T = 2 \pi/a$, and
$$\frac{\sin{a t}}{|\sin{a t}|} = \begin{cases} \\ +1 & a\, t \in (0,\pi) \\ -1 & a \,t \in (\pi,2 \pi)\end{cases}$$
Then
$$\begin{align}\int_0^{2 \pi/a} dt\, \frac{\sin{a t}}{|\sin{a t}|} \, e^{-s t} &= \int_0^{\pi/a} dt \, e^{-s t} - \int_{\pi/a}^{2 \pi/a} dt \, e^{-s t}\\ &= \frac{1-e^{-\pi s/a}}{s} ...
0
The problem reduces to $$2+\frac{40}{(s^2+4s+5)^2}$$
Let us denote the Laplace Transform of $f(t)$ as $L\left(f(t)\right)=F(s)$
and the inverse as $L^{-1}\left(F(s)\right)=f(t)$
For $L^{-1}\frac1{(s^2+4s+5)^2},$
$$\text{If }L\left(f(t)\right)=\frac{s-a}{\{(s-a)^2+b^2\}^2},$$
$$L\left(\frac {f(t)}t\right)=\int_s^\infty \frac{s-a}{\{(s-a)^2+b^2\}^2} ds$$
...
2
Let $F(s)$ denote the fraction in the post, hence $F(s)=2+40\frac1{(s^2+4s+5)^2}$. The $2$ part of $F(s)$ is the Laplace transform of twice the Dirac measure at $0$. The fraction $\frac1{s^2+4s+5}$ is a linear combination of $\frac1{s+2\pm\mathrm i}$ hence it is the Laplace transform of a linear combination of the functions $t\mapsto\exp(-(2\pm\mathrm i)t)$ ...
2
Here is how we compute the LT of $\sin^4{t}$ directly. As mentioned in @Mhenni's solution, write as $\sin^4{t} = (e^{i t}-e^{-i t})^4/(2 i)^4$
$$F(u) = \frac{1}{16} \int_0^{\infty} dt \, (e^{i 4 t} - 4 e^{i 2 t} + 6 - 4 e^{-i 2 t} + e^{-i 4 t}) e^{-u t}$$
Evaluate separately and combine judiciously:
$$F(u) = \frac18 \left [ \frac{u}{u^2+16} - 4 ...
4
Recalling the Laplace transform of a function $f(x)$
$$ F(s)=\int_{0}^{\infty} f(x)e^{-sx}dx .$$
Let
$$G(u)=\frac{1}{u^3} \implies g(u)=\frac{u^2}{2!}, $$
and
$$ f(u)= \sin(u)^4 \implies F(u)=\frac{24}{u(u^2+4)(u^2+16)}. $$
Now,
$$ \int_0^\infty \frac{\sin^4 x}{x^3} \, dx= \frac{24}{2}\int_0^\infty \frac{u^2}{u(u^2+4)(u^2+16)} \, dx = \ln(2)$$
...
2
The partial fraction expansion is:
$$\displaystyle \frac{s+3}{(s-2) (s+1)^2} = -\frac{5}{9(s+1)}-\frac{2}{3(s+1)^2}+ \frac{5}{9(s-2)}$$
The inverse Laplace Transform can now be done for each of those terms. We get:
$$\displaystyle \frac{1}{9} e^{-t} (-6 t+5 e^{3 t}-5)$$
Note that this is a somewhat easier way to display it. Of course, if you want to line ...
0
Hint:
$y(x)=f(x)+\int_0^{x+l}y(z)g(x-z)~dz$
$y(x)=f(x)+\int_x^{-l}y(x-t)g(t)~d(x-t)$
$y(x)=f(x)-\int_x^{-l}y(x-t)g(t)~dt$
$y(x)=f(x)+\int_{-l}^xy(x-t)g(t)~dt$
0
$\mathcal{L}_{t\to s}\left\{\int_0^te^{it}\dfrac{dt}{\sqrt{2\pi t}}\right\}=\dfrac{\mathcal{L}_{t\to s}\left\{\dfrac{e^{it}}{\sqrt{2\pi t}}\right\}}{s}=\dfrac{\mathcal{L}_{t\to s-i}\left\{\dfrac{1}{\sqrt{2\pi t}}\right\}}{s}=\dfrac{1}{s\sqrt{2(s-i)}}$
-1
$\mathcal{L}^{-1}_{t\to x}\left\{\int_t^\infty\dfrac{e^{-u}}{u}du\right\}=\dfrac{\mathcal{L}^{-1}_{t\to x}\left\{\dfrac{e^{-u}}{u}\right\}}{x}=\dfrac{H(x-1)}{x}$
0
I have done it:
$$\begin{array}{l}
X(s) = \frac{1}{{{s^2} - 9s + 20}} = \frac{1}{{(s - 4)(s - 5)}} = \frac{{ - 1}}{{s - 4}} + \frac{1}{{s - 5}} \\
= > x(t) = - {e^{4t}} + {e^{5t}} \\
\end{array}$$
0
One wants to write $f$ as the Laplace transform of $g$, that is, to find $g$ such that $f(t)=\int\limits_0^{+\infty}\mathrm e^{-tx}g(x)\mathrm dx$ for every $t\geqslant0$. The change of variable $u=tx$ indicates that $f(t)=\int\limits_1^{+\infty}\mathrm e^{-tx}x^{-1}\mathrm dx$.
Thus, the function $g$ defined by $g(x)=x^{-1}\mathbf 1_{x\geqslant1}$ for every ...
0
I think you mean the Laplace transform, not its inverse. Rewrite $f$ as
$$f(t) = \int_1^{\infty} dy \frac{e^{-t y}}{y} $$
The the LT of $f$, $\hat{f}(s)$, is
$$\begin{align}\hat{f}(s) &= \int_0^{\infty} dt \, f(t) e^{-s t}\\ &= \int_0^{\infty} dt \,\int_1^{\infty} dy \frac{e^{-t y}}{y} e^{-s t} \\ &=\int_1^{\infty} \frac{dy}{y} \: ...
1
Let $f(t)=\sum\limits_{n=0}^\infty a_nt^n$ ,
Then $\sum\limits_{n=0}^\infty na_nt^{n-1}=\sum\limits_{n=0}^\infty a_n(kt)^n$
$\sum\limits_{n=1}^\infty na_nt^{n-1}-\sum\limits_{n=0}^\infty k^na_nt^n=0$
$\sum\limits_{n=1}^\infty na_nt^{n-1}-\sum\limits_{n=1}^\infty k^{n-1}a_{n-1}t^{n-1}=0$
$\sum\limits_{n=1}^\infty(na_n-k^{n-1}a_{n-1})t^{n-1}=0$
...
5
To actually evaluate the inverse Laplace transform integral, we must perform an integration in the complex plane. Many times, this means using the theorems of complex analysis, notably Cauchy's residue theorem. Because many students are introduced to the Laplace transform before they take complex analysis - many times, such students won't ever take a ...
1
In general, think of a "transform" as a tool for converting a problem into some new form that's easier to solve. Suppose you have some difficult problem, and you have a transform that converts it into a different problem that you already know how to solve. If you can also construct a "reverse" transform, this then gives you a three-step process for solving ...
0
In many cases, a Laplace transform (or a Fourier transform for that matter) transforms a partial differential equation into an ordinary differential equation. It also transforms an ordinary differential equation into an algebraic equation. The thinking goes that if the we can find a function whose transform is the solution to that algebraic equation, then ...
2
Assuming $a>0$, get
$$\begin{align}\int_0^{\infty} dt \frac{f(t)}{1-e^{-a t}} e^{-s t} &=\int_0^{\infty} dt\, f(t) \sum_{k=0}^{\infty} e^{-(s + a k)t} \\ &= \sum_{k=0}^{\infty} \hat{f}(s+k a) \end{align}$$
where we reversed the order of summation and integration, and
$$\hat{f}(s) = \int_0^{\infty} dt \,f(t) e^{-s t}$$
1
As much as I can possibly guess at what you are doing (your notation is very confusing), it seems you are trying to find the inverse LT of
$$E(z) = \frac{5}{z} + \frac{1}{z^2} + \frac{2}{z^4}$$
where
$$E(z) = \int_0^{\infty} dt \, e(t) \, e^{-z t}$$
You want to use the easily verified formula
$$\mathcal{L}(t^k) = \frac{k!}{z^{k+1}}$$
It seems you have ...
2
$L\{f'(t)\}=\int_0^\infty e^{-st}f'(t)dt$
Integrating by parts we have,
$L\{f'(t)\}=e^{-st}f(t)|_0^\infty+s\int_0^\infty e^{-st}f(t)dt$
$L\{f'(t)\}=e^{-s(\infty)}f(\infty)-e^{-s(0)}f(0)+sL\{f(t)\}$
If $e^{-st}$ grows more rapidly than $f(t)$, we have $e^{-st}f(t)\to0$ when $t\to\infty$
$L\{f'(t)\}=sL\{f(t)\}-f(0)$
Since $f(0)=0$, this reduces to
...
3
The equation for $u(x,t)$ is
$$
\partial_t u = \partial_{xx}u - u^3 + u + B\left(u_0-\frac{1}{L}\int_{0}^{L} u\;dx\right)
$$
This is a non-linear integro-differential equation (if not for the $u^3$, it would be linear). Laplace transform techniques don't get you very far with this equation, because the Laplace transform of $u^3$ is an integral in the ...
0
The LT of a convolution of two functions is the product of the individual LTs. If you recall, we found that
$$\hat{f_T}(p) = \frac{2 - e^{-p T}}{p}$$
Thus, the LT of the above convolution is simply
$$\hat{h_{S*T}}(p) = \left ( \frac{2 - e^{-p S}}{p} \right )\left ( \frac{2 - e^{-p T}}{p} \right )$$
2
The issue here is continuity, since you typically cannot hope to recover the behavior of a function at a point which is not a point of continuity from an integral transform. If you consider continuous functions which decay fast enough (for example, subexponentially decaying functions) then you can prove uniqueness fairly easily by a standard calculation. ...
0
Another way to look at this is to see that there is a complete LT, plus some extra. To wit:
$$\hat{f}(s) = \int_0^{\infty} dt \, e^{-s t} + \int_0^{T} dt \, e^{-s t} $$
This is easy to evaluate:
$$\hat{f}(s) = \frac{1}{s} + \frac{1}{s} \left ( 1-e^{-s T}\right) = \frac{2-e^{-s T}}{s}$$
0
For piece-wise defined functions, such as your example, you either apply the definition, as Mhenni solution shows (and that is the easiest for this example) or if you want a more general approach you need to read about the step function also known as Heaviside function. Here is a brief description of what is involved.
For $c\ge0$ define the step function ...
2
Here is it
$$\int_{0}^{\infty}f_T(t)e^{-st}dt=\int_{0}^{T}2\,e^{-st}dt + \int_{T}^{\infty}1\,e^{-st}dt $$
$$ = 2\int_{0}^{T}e^{-st}dt + \int_{T}^{\infty} e^{-st}dt $$
$$ = 2 \frac{e^{-st}}{-s}\Big|_{0}^{T} + \frac{e^{-st}}{-s}\Big|_{T}^{\infty}=\dots\,. $$
I think you can finish it now. Note that you need to assume $Re(s)>0$.
1
It is the professor's typo. The function $2te^{-t}$ has a Laplace transform $\frac{2}{(s+1)^2}$
1
Let $F(s)$ denote the (one-sided) Laplace transform of $f(t)$, a continuous function defined on $[0,+\infty)$. Then the initial value theorem says that
$$f(0)=\lim_{s\rightarrow +\infty}sF(s)$$
and the final value theorem says that, if all the poles of $F$ are in the open left hand plane, then
$$\lim_{t\rightarrow+\infty}f(t)=\lim_{s\rightarrow ...
1
If $G(s)$ is the laplace transform of $g$, then $F(\omega)=G(i\omega)$ is the fourier transform of $g$, if the fourier transform exists. You can easily see that by comparing the formulas for the two transforms - in the case of the laplace transform the kernel is $e^{-s}$ whereas for the fourier transform it's $e^{-i\omega}$. (Since you usually assume that ...
2
Hints:
a). is correct
b.) is incorrect - think about the $t$ term and that being $(t-2)$ in the exponential (you got the Heaviside portion correct, so it is close).
It should be:
$$\displaystyle e^{9 (t-2)} H(t-2)$$
1
The appearance of the Heaviside $\theta$ is due to imposing a condition of convergence on the circular arc section of the contour, $T_R$ as you call it. When $t \gt 0$, substitute $p = R e^{i \phi}$ and we have
$$\int_{T_R} dp \frac{e^{p t}}{p} = i R \int_{\pi/2}^{3 \pi/2} d\phi\, e^{i \phi} \frac{e^{R t \cos{\phi}} e^{i R t \sin{\phi}}}{i R e^{i \phi}}$$
...
1
This Laplace transform does not exist. Why?
From Boyce and DiPrima's Elementary Differential Equations:
Theorem 6.1.2:
Suppose that:
$f$ is piecewise continuous on the interval $0\le t\le A$ for any positive $A$
$|f(t)| \le Ke^{at}$ when $t\ge M$. In this inequality, $K$, $a$, and $M$ are real constants, $K$, $M$ necessarily positive.
...
0
Well the formal definition of the Laplace transform is:
$$\mathcal{L}\left\{f(t)\right\}=\int_0^\infty e^{-st}f(t)dt$$
Here, your function is $f(t)=e^{(t-1)^2}$ so your Laplace transform is:
$$F(s)=\int_0^\infty e^{-st}e^{(t-1)^2}dt=\int_0^\infty e^{(t-1)^2-st}dt$$
Work this one out and you'll find what you want
Edit: As @anorton noted, this integral is ...
2
It seems correct. But note that we don't necessarily have $$\mathcal{L}(f(t)g(t))=\mathcal{L}(f(t))\times\mathcal{L}(g(t)) $$
0
$s^2-4 s+5$ has zeroes at $s_{\pm} = 2 \pm i$. The ILT $f(t)$ is simply the sum of the residues of
$$\frac{2 s+1}{s^2-4 s+5} e^{s t}$$
at these poles. Then
$$f(t) = \frac{2 s_+ + 1}{2 s_+-4} e^{s_+ t} + \frac{2 s_- + 1}{2 s_--4} e^{s_- t}$$
Expanding this a bit:
$$f(t) = e^{2 t} \left [ \left (1-i \frac52 \right ) (\cos{t}+i \sin{t}) + \left(1+i ...
0
Remember, the Laplace transform tables have the following kinds of terms
$$
\frac{s-a}{(s-a)^2+b^2},\frac{b}{(s-a)^2+b^2}
$$
Your term with an $s$ in the numerator has to look like $s-2$ in order to use that form. So your problem is at this step:
$$
\frac{2s}{(s-2)^2 + 1}+ \frac{1}{(s-2)^2 + 1} = \frac{2(s-2)+4}{(s-2)^2 + 1}+ \frac{1}{(s-2)^2 + 1} = ...
4
Caveat: One "can't use the identity table of Laplace transforms" when the function one is dealing with is not in the list. The function $f$ is NOT in the list. Hence the only option is to go back to the definition...
By definition, $Lf(s)=\displaystyle\int_0^\infty\mathrm e^{-st}f(t)\mathrm dt=\int_0^\pi\mathrm e^{-st}\sin(t)\mathrm dt$ for every ...
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