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## Hot answers tagged laplace-transform

15

The Fourier and the Laplace transform obviously have many things in common. However, there are cases where only one of them can be used, or where it's more convenient to use one or the other. First of all, even though in the definitions you simply replace $s$ by $j\omega$ or vice versa to go from one transform to the other, this cannot generally be done ...

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The Fourier Transform is the Laplace Transform with the complex variable s restricted to be the imaginary axis on the s plane. For this reason the Fourier Transform only exists when the imaginary axis is within the region of convergence. The variable s is called a "complex frequency" as it is the frequency variable that can take on real ($\sigma$) and ...

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Both notations are common and correct. As pointed out by Yuri Nenakhov, the advantage of the argument $j\omega$ is that it coincides with the complex (Laplace transform) variable $s$ when its real-part is zero. Note that in the complex $s$-plane the frequency axis is the imaginary axis. So $j\omega$ has nothing to do with complex frequency (which makes no ...

9

Assume you have a $N\times M$ sized image. If you know take what is classically used, a square filter kernel, of let's say size $L\times L$, you'd need to convolve that with the picture – which gives you $N\times M$ pixels, each needing $L^2$ multiply-accumulates. So you end up with $A_{2D}=L^2MN$ operations. Now, if you can decompose that filter into an $... 8 Why is the fourier transform a special case of the laplace transform? The Laplace transform produces a 2D surface of complex values, while the Fourier transform produces a 1D line of complex values. The Fourier transform is what you get when you slice the Laplace transform along the jω axis. For instance, a simple lowpass filter$H(s)=\frac{1}{s+1}$has a ... 8 As pointed out by Peter K., it is important to distinguish between Laplace and Fourier transforms. The first few transform pairs in your question are Fourier transform pairs, whereas the last pair is a correspondence of the unilateral Laplace transform: $$F(s)=\int_{0}^{\infty}f(t)e^{-st}dt$$ In the last transform pair in your question the$\mathcal{F}$... 8 Given the approach started in the OP's Github code I have this suggestion: Observe that the unilateral Laplace Transform given as: $$X(s) = \int_0^\infty x(t)e^{-st}dt$$ Is just the Fourier Transform of a causal function with a weighting exponential: $$X(s) = \int_0^\infty x(t)e^{-(\sigma+j\omega)t}dt$$ $$X(s) = \int_0^\infty [e^{-\sigma t}x(t)]e^{-j\omega t}... 7 Matt is correct that the sign is convention. I think that there is a reason for it beyond that though. If we look at complex frequencies in the complex plane, they look like a constant vectors that rotate in one direction or another. Positive frequencies rotate counter-clockwise, negative frequencies rotate clockwise, and "0 Hz" frequencies don't rotate ... 7 The widespread use of the unilateral Laplace transform reflects the fact that in practice we often deal with causal systems and signals that have a defined starting time (usually chosen as t_0=0). The Fourier transform is mainly used for analyzing ideal signals and systems, such as ideal filters (e.g., low pass, high pass, etc.) and ideal signals such as ... 7 Re-writing the strain-stress equation$$ 0 = E_1Ɛ + \frac{η(E_1+E_2)}{E_2}\dot{Ɛ} - \frac{η}{E_2}\dot{σ} - σ $$for displacement/restoring force variables$$ 0 = E_1x + \frac{η(E_1+E_2)}{E_2}\dot{x} + m\frac{η}{E_2}\dddot{x} + m\ddot{x} $$and applying the Laplace transform, we arrive at the equation$$ 0 = E_1 + \frac{η(E_1+E_2)}{E_2}s + ms^2 + m\frac{η}{... 6 The Laplace Transform is more representative of real systems that have a starting point, which is why the integral starts at 0, and also why the unit step function is generally talked about alongside the Laplace Transform. With the Laplace Transform, we can examine the transient and steady-state behavior of a system. Using$e^{st}$instead of$e^{iwt}$... 6 You cannot make conclusions about the stability of a system by only considering its transfer function evaluated on the imaginary axis$s=j\omega$. Replacing$s$by$j\omega$in the transfer function only makes sense for a stable system, otherwise you get a function of$\omega$that does not describe the system, but another (stable) system. Let me explain ... 6 Laplace transforms can capture the transient behaviors of systems. Fourier transforms only capture the steady state behavior. Of course, Laplace transforms also require you to think in complex frequency spaces, which can be a bit awkward, and operate using algebraic formula rather than simply numbers. If you want to see the power distribution of a signal ... 5 The Laplace transform evaluated at$s=j\omega$is equal to the Fourier transform if its region of convergence (ROC) contains the imaginary axis. This is also true for the bilateral (two-sided) Laplace transform, so the function need not be one-sided. As for real and imaginary parts, since$s$is a complex variable, both the Laplace and the Fourier transform ... 5 The bi linear transform is the transform from the Laplace Transform Domain to the Z Transform. The Laplace Transform Domain is a regular plane. This transform transforms vertical lines in the Laplace domain into circles in the Z Domain. Hence the Fourier Vertical Line in Laplace Domain (The Y Vertical Lines) is transformed into the unit circle in the Z ... 5 In engineering practice, the complex inversion integral is hardly ever used. As an engineer, you will almost exclusively need to invert rational functions, and this can be done by partial fraction expansion and elementary inversions. So first I'll show you how to obtain the inverse Laplace transform by partial fraction expansion, then I'll explain the ... 5 The short answer is yes, if you have the Laplace or Z-transform of a function you do not need the Fourier transform. This is because the CFT is a special case of the Laplace transform and the DTFT is a special case of the Z transform. The Fourier transform is used to find the complex sinusoids that compose a function, whereas the Laplace transform finds ... 5 You're comparing the transforms of two different functions. You consider the Fourier transform of the function$x_1(t)=\cos(\omega_0 t)$, but you took the Laplace transform of the function$x_2(t)=\cos(\omega_0t)u(t)$, where$u(t)$is the unit step function: $$X_1(j\omega)=\int_{-\infty}^{\infty}x_1(t)e^{-j\omega t}dt\\ X_2(s)=\int_{0}^{\infty}x_2(t)e^{-st}... 5 The problem is that you took the derivative of the function$$\hat{x}_u(t)=2e^{-3t}-e^{-4t}\tag{1}$$whereas using the Laplace transform you implicitly assumed that x_u(t) equals zero for t<0:$$x_u(t)=\hat{x}_u(t)u(t)=(2e^{-3t}-e^{-4t})u(t)\tag{2}$$where u(t) is the unit step function. If you take the derivative of (2) then you'll get the ... 5 Let H(s) be a transfer function of the form$$H(s) = \frac{1}{s-p}$$where p, which is a pole of H(s), can be written as a complex number a+jb. Taking the inverse Laplace transform of H(s) gives the corresponding impulse response h(t) (that is, the output of your system when given \delta(t) as input). Noting \mathcal{L}^{-1} the inverse ... 5 Let s = \sigma + j\omega, the inverse Laplace transform of f(t+a) is given by$$f(t+a) = \frac{1}{2\pi j} \int_{\sigma-j\infty}^{\sigma+j\infty} F(s)e^{s(t+a)} \mathrm{d}s = \frac{1}{2\pi j} \int_{\sigma-j\infty}^{\sigma+j\infty} F(s)e^{sa}e^{st} \mathrm{d}s.$$Hence the bilateral Laplace transform of f(t+a) is F(s)e^{sa} where F(s) is the Laplace ... 5 First of all, it's important to understand that there is no single best way to transform a continuous-time system to a discrete-time system. The method you're using is called backward Euler method, and it is defined by the mapping$$s\leftarrow\frac{1-z^{-1}}{T}\tag{1}$$Note that in (1) you scale by 1/T, where T is the sampling interval (i.e., 1/T ... 5 It's natural consequence of applying a transform to a convolution relation. The output y(t) of an (continuous-time) LTI system is described by a convolution integral :$$y(t) = h(t)\star x(t) = \int_{-\infty}^{\infty} x(\tau) h(t-\tau) d\tau $$And when you apply a Fourier transform on this relation, it turns out to be a multiplication in the transform ... 5 This has to do with the way the Laplace transform and the \mathcal{Z}-transform are defined:$$\mathcal{L}\big\{x(t)\big\}=\int_{-\infty}^{\infty}x(t)e^{-st}dt\tag{1}\mathcal{Z}\big\{x[n]\big\}=\sum_{n=-\infty}^{\infty}x[n]z^{-n}\tag{2}$$Of course, the question remains why they are defined that way. The answer lies in the fact that in continuous time ... 5 Concerning your first question, both, the Laplace and the Fourier transform, are frequency domain representations of a function or signal. In the Fourier transform we deal with a real-valued frequency variable$\omega$, whereas in the Laplace transform we have a generally complex-valued independent variable (usually$s$), the imaginary part of which equals ... 4$X(j \omega)$(frequency response) is a Fourier transform of system's impulse response. It's actually a function of frequency ($\omega$) but usually is written as$X(j \omega)$because replacing$j \omega$in the formula with$s$will give you system's Laplace transform$X(s)$without any additional conversions. (This works in the opposite direction as well: ... 4 You are right that the (bilateral) Laplace transform can be interpreted as the Fourier transform of$e^{-\sigma t}f(t)$. However, I think that the significance of the Laplace transform only becomes clear when$s=\sigma+j\omega$is viewed as a complex variable because then we can study the analytic properties of the system function. E.g., electrical networks ... 4 The best intuitive description of Laplace transform I've ever seen: At first glance, it would appear that the strategy of the Laplace transform is the same as the Fourier transform: correlate the time domain signal with a set of basis functions to decompose the waveform. Not true! Even though the mathematics is much the same, the rationale behind the two ... 4 As already mentioned by other people, the bilinear transform is often used to map a continuous-time system described in the$s$-domain to a discrete-time system described in the$z$-domain. However, a bilinear transform is a more general tool that can also be used to transform a discrete-time system to another discrete-time system. Since you didn't give any ... 4 Strictly speaking you can't because without specifying the ROC, the inverse Laplace transform is generally not unique. However, in many contexts there is the implicit assumption of causality of the corresponding time function (i.e.,$x(t)=0$for$t<0\$), which is equivalent to stating that the ROC is a right half-plane.

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