In mathematics, the Laplace transform, named after its inventor Pierre-Simon Laplace (), is an integral transform that converts a function of a real variable
t
{\displaystyle t}
(often time) to a function of a complex variable
s
{\displaystyle s}
(complex frequency). The transform has many applications in science and engineering because it is a tool for solving differential equations. In particular, it transforms linear differential equations into algebraic equations and convolution into multiplication.For suitable functions f, the Laplace transform is the integral
This is my answer to the ODE (I think it's correct) via Laplace transform, but I'm more concerned about whether or not my explanation for part b is correct or not?
Any help would be greatly appreciated!!
I will omit the theorem and its proof here, since it would mean a lot of typing. But the relevant part of the proof of the theorem is that we are considering the set ##H## of functions consisting of ##\psi_\lambda(x)=e^{-\lambda x}## for ##x\geq 0## and ##\lambda\geq 0##. We extend the...
For this problem,
The solution is,
However, I'm confused by the partial fraction decomposition of ##\frac{2}{s^4(s^2 + 1)}##
I never done that sort of thing before. However, I think it would be done like this (Please correct me if I am wrong, the algebra is crazy here).
##\frac{2}{s^4(s^2 +...
I tried to solve this as follows
$$f(t)=(u(t)-u(t-2\pi))\sin{t}$$
$$=u(t)\sin{t}-u(t-2\pi)\sin{t}$$
$$\mathcal{L}(f(t))=e^{0\cdot s}\mathcal{sin{t}}-e^{-2pi s}\mathcal{L}(\sin{(t+2\pi)})$$
$$=\frac{1-e^{-2\pi s}}{s^2+1}$$
where I used the fact that ##\sin{(t+2\pi)}=\sin{t}##.
Then I looked...
I am interested specifically in solving this problem by factoring the quadratic term into complex linear factors.
$$s^2+4=0$$
$$\implies s=\pm 2i$$
$$\frac{5s+6}{(s-2i)(s+2i)(s-2)}=\frac{A}{s-2i}+\frac{B}{s+2i}+\frac{C}{s-2}$$
We can solve for ##C## using the cover-up method with ##s=2## to...
I'm following the intuition behind getting the zero's and poles of a damped cosine function with this video
At around 11:50, he shows some graphics pertaining to multiplying the probing function with the impulse response, but the graphics don't seem correct.
For example, in the B+B' graphic...
Here follows the theorem and proof:
Questions:
1. I do not understand the following part "...and hence, in view of the preceding calculation, ##\int_0^\infty \int_0^\infty |e^{-st}f(\tau)g(t-\tau)|dtd\tau## converges".
We know that ##\mathcal{L}\big(f(t)\big)## and...
The following three results are used in the proof of the theorem I have a question about.
Now follows the theorem and its proof I have a question about.
I do not understand why ##e^{-st}f(t)\rvert_0^\infty=-f(0)##. By Lemma 2, this is only possible if ##f## is of exponential type of order...
I recently acquainted myself with Laplace transform, and it appears that it has some relations with phasor analysis. This observation stems from the fact that while in Laplace transform, we have ##s = \sigma + j \omega## as the variable, in phasor analysis, we just use ##j\omega,## apparently...
Trying to model friction of a linear motor in the process of creating a state space model of my system. I've found it easy to model friction solely as viscous friction in the form b * x_dot, where b is the coefficient of viscous friction (N/m/s) and x_dot represents the motor linear velocity...
The implementations for the two filters in simulink are as follow:
For the first filter:
For the second one:
The obtained results have values of 10^-12, while the expected results should be between 10^-3 - 10.
Since it's the first time when I try t implement a tf with variable coefficients I...
Could someone check whether my proof for this simple theorem is correct? I get to the result, but with the feeling of having done something very wrong :)
$$\mathcal{L} \{f(ct)\}=\int_{0}^{\infty}e^{-st}f(ct)dt \ \rightarrow ct=u, \ dt=\frac{1}{c}du, \
\mathcal{L}...
Hi,
I found Laplace transform of this Dirac delta function which is ##F(s) = e^{-st}## since ##\int_{\infty}^{-\infty} f(t) \delta (t-a) dt = f(a)##
and that ##\delta(x) = 0## if ##x \neq 0##
Then the Mellin transform
##f(t) = \frac{1}{2 \pi i} \int_{\gamma - i \omega}^{\gamma +i \omega}...
Lets consider very simple equation ##x''(t)=0## for ##x(0)=0##, ##x'(0)=0##. By employing Laplace transform I will get
s^2X(s)=0 where ##X(s)## is Laplace transform of ##x(t)##. Why then this is equivalent to
X(s)=0
why we do not consider ##s=0##?
\mathcal{L}^{-1}[\frac{e^{-5s}}{s^2-4}]=Res[e^{-5s}\frac{1}{s^2-4}e^{st},s=2]+Res[e^{-5s}\frac{1}{s^2-4}e^{st},s=-2]
From that I am getting
f(t)=\frac{1}{4}e^{2(t-5)}-\frac{1}{4}e^{-2(t-5)}. And this is not correct. Result should be
f(t)=\theta(t-5)(\frac{1}{4}e^{2(t-5)}-\frac{1}{4}e^{-2(t-5)})...
The Laplace transform gives information about the exponential components in a function, as well as oscillatory components. To do so there is a need for the complex plane (complex exponentials).
I get why the MGF of a distribution is very useful (moment extraction and classification of the...
So, I know the direct definition of the Laplace Transform:
$$ \mathcal{L}\{f(t) \} = \int_0^\infty e^{-st}f(t)dt$$
So when I plug in:
$$\frac{\cos(t)}{t}$$
I get a divergent integral.
however:https://www.wolframalpha.com/input/?i=+Laplace+transform+cos%28t%29%2F%28t%29
is supposed to be the...
My attempt at finding this was via convolution theorem, where we take F(s) = 1/s^2 and G(s) = e^(-sx^2/2). Then to use convolution we need to find the inverses of those transforms. From a table of Laplace transforms we know that f(t) = t. But I am sort of struggling with e^(-sx^2/2). My 'guess'...
I am interested in modeling a battery charging/discharging. I am starting off with a simple model using a voltage source in series with a parallel RC branch which is in series with a resistor. I will be measuring the open circuit voltage between the last series resistor and the bottom of the...
hi guys
i am facing a little problem calculating this Laplace transform ## \mathscr{L}(\frac{e^{\alpha t}}{t})## , when calculate it using the method of the inverse Laplace transform its equal to
$$ ln{\frac{1}{s-\alpha}}$$
but then when i try to use the theorem
$$...
I've included the problem statement and a bit about the function but my main issue is with the equation after "then" and the one with the red asterisk. I don't understand why the Laplace transform for a u(t)*e^(-t/4) isn't (1/s)*(1/(s+1/4)). The book I am reading says it's(1/(s+1/4)).
Homework Statement:: Why is the heaviside function in the inverse laplace transform of 1?
Relevant Equations:: N/A
This is a small segment of a larger problem I've been working on, and in my book it gives the transform of 1 as 1/s and vice versa. But as I've looked online for help in figuring...
Suppose I'm solving
$$y''(t) = x''(t)$$ where $$x(t)$$ is the ramp function. Then, by taking the Laplace transform of both sides, I need to know $x'(0)$ which is discontinuous. What is the appropriate technique to use here?
Was just practicing some problems on the Fundamentals of Electric Circuits, and came across this question.
I understand I will have to transform to the s domain circuit, which looks something like this:
Then doing nodal analysis, I will get the following for the first segement
(10/s-V1)/1 =...
I struggle to find an appropriate inverse Laplace transform of the following
$$F(p)= 2^n a^n \frac{p^{n-1}}{(p+a)^{2n}}, \quad a>0.$$
WolframAlpha gives as an answer
$$f(t)= 2^n a^n t^n \frac{_1F_1 (2n;n+1;-at)}{\Gamma(n+1)}, \quad (_1F_1 - \text{confluent hypergeometric function})$$
which...
Hi again,
The previous problem was done using y′′(t)+2y′(t)+10y(t)=10 with with intial condition y(0⁻)=0.
In the following case, I'm using an initial condition and setting the right hand side equal to zero.
Find y(t) for the following differential equation with intial condition y(0⁻)=4...
Hi,
This thread is an extension of this discussion where @DrClaude helped me. I thought that it'd be better to separate this question.
I couldn't find any other way to post my work other than as images so if any of the embedded images are not clear, just click on them. It'd make them clearer...
Hi,
I 'm trying to find the Laplace transform of the following expression.
I used the following conversion formulas.
I think "1" is equivalent to unit step function who Laplace transform is 1/s.
I ended up with the following final Laplace transform.
Is my final result correct? Thank you...
I used partial fraction method first as:
1/s(s^2+w^2)=A/s+Bs+C/(s^2+w^2)
I found A=1/w^2
B=-1
C=0
1/s(s^2+w^2)=1/sw^2- s/s^2 +w^2
Taking invers laplace i get
1/w2 - coswt
But the ans is not correct kindly help.
We would need to recognise that the integral in the equation is a convolution integral, which has Laplace Transform: $\displaystyle \mathcal{L}\,\left\{ \int_0^t{ f\left( u \right) \,g\left( t - u \right) \,\mathrm{d}u } \right\} = F\left( s \right) \,G\left( s \right) $.
In this case...
Since this is of the form $\displaystyle \frac{f\left( t \right)}{t} $ we should use $\displaystyle \mathcal{L}\,\left\{ \frac{f\left( t \right) }{t} \right\} = \int_s^{\infty}{F\left( u \right) \,\mathrm{d}u } $.
Here $\displaystyle f\left( t \right) = \cosh{\left( 4\,t \right) } - 1 $ and so...
Upon taking the Laplace Transform of the equation we have
$\displaystyle \begin{align*} s^2\,Y\left( s \right) - s\,y\left( 0 \right) - y'\left( 0 \right) + 4\,Y\left( s \right) &= -\frac{8\,\mathrm{e}^{-6\,s}}{s} \\
s^2 \,Y\left( s \right) - 2\,s - 0 + 4\,Y\left( s \right) &=...
So I have completed (a) as this (original on the left):
I have then went onto (b) and I have equated T(s)=Z(s) as follows:
and due to
hence
Does this look correct to you smarter people?
Thanks in advance! All replies are welcome :)
Problem: Find a (limited?) solution to the diff eq.
At the end of the solution, when you transform \frac{-1}{s+1} + \frac{2}{s-3}
why doesn't it become -e^{-t} + 2e^{3t} , t>0 ?
So the task is to solve the following integral with laplace transform.
Since t>0 we can multiply both sides with heaviside stepfunction (lets call it \theta(t)).
What I am unsure about is what happens with the integral part and how do we inpret the resulting expression?
What will it result...
Summary: A 1963 paper by Michael Wertheim uses a Laplace transform in spherical coordinates. How is the resulting equation obtained?
In 1963, Michael Wertheim published a paper (relevant page attached here), where he presented the following equation (Eq. 1):
$$ y(\bar{r}) = 1 + n...
Hi,
I was trying to see if the following differential equation could be solved using Laplace transform; its solution is y=x^4/16.
You can see below that I'm not able to proceed because I don't know the Laplace pair of xy^(1/2).
Is it possible to solve the above equation using Laplace...
So I could just try using the definition by taking the limit as T goes to infinity of ∫ from 0 to T of that entire function but that would be a mess. I tried breaking it down into separate pieces and seeing if I could use anything from the table but I honestly have no clue I'm really stuck. I'd...
In Mathematical Methods in the Physical Sciences by Mary Boas, the author defines the Laplace transform as...
$${L(f)=}\int_0^\infty{f(t)}e^{-pt}{dt=F(p)}$$
The author then states that "...since we integrate from 0 to ##\infty##, ##{L(f)}## is the same no matter how ##{f(t)}## is defined for...
I have used Laplace transform during my EE studies to solve differential equations and in control system analysis, but we were taught that as a tool kit to make the math easier. The physical meaning was never explained. I know basic time and frequency domain concepts (thanks to Fourier series)...