# Tag Info

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There is a way to do this analytically using Fourier transforms. I will show this as soon as I can, but I promise you, it will be a lot messier than what I present below. GRAPHICAL METHOD To do this at the level of an undergrad signals class, you really need to draw a picture. But while it may be error-prone, there are checks along the way to see that ...

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It is a typo. Dividing by $iw$ corresponds to the Fourier transform of $f(x)$ integrated. For example, in control engineering an integrator block is labeled $1/s$ where $s$ is the complex frequency variable of the Laplace transform, which becomes the Fourier transform for $s = iw$. Is it a first edition book? ...

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Let $\varphi \in \mathcal D$, where $\mathcal D$ is the space of $\varphi \in C^\infty_0(\mathbb R)$ with the usual test function continuity notion. Consider $$\tilde \varphi(\xi) = \int_{\mathbb R} \frac{dx}{(2\pi)^{1/2}}e^{-i\xi x}\varphi(x)=\int_{\mathbb R} \frac{dx}{(2\pi)^{1/2}}\sum_{n=0}^\infty \frac{(-i\xi x)^n}{n!}\varphi(x);$$ since the integral ...

1

The Fourier transform has different definitions in different settings. Some have a $\frac{1}{\sqrt{2\pi}}$, some have it without the square root, some have no such factor. The inverse transform is also defined with these $2\pi$ factors so that no matter which convention you use a consistent result is obtained. Partly it is a question of units, i.e. ...

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Check out the Wikipedia page: There are several common conventions for defining the Fourier transform ... Wolfram Mathworld has at least one alternate definition. The problem is that different branches of mathematics and physics use different definitions for the Fourier Transform. Each definition gives a slightly different result. So perhaps we ...

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The coefficient of the transformed function depends on your definition of the Fourier transform and the inverse transform. For example, I could define the forward transform and inverse respectively as $$F(k) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty f(x) e^{-ikx}\ dx$$ $$f(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty F(k) e^{ikx}\ dx$$ But this ...

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Suppose $f$ is a periodic function on $\mathbb R$ with period $p$, with Fourier series $$f \sim \sum_{n\,\in\,\mathbb Z} c_n e^{2\pi inx/p}$$ The inner product $\langle f,g\rangle = \frac 1 p\int_0^p f(x)\,\overline{g(x)}\, dx$ (where $a\mapsto \overline{a}$ is complex conjugation) is used to determine the coefficients, which must satisfy $$c_n = ... 1 Partial work. Let define the following function:$$g(\xi)=\int_{\mathbb{R}}\exp\left(-\frac{x^2}{2a}\right)\exp(-i\xi x)\,\mathrm{d}x.$$Notice that:$$g'(\xi)=-i\int_{\mathbb{R}}x\exp\left(-\frac{x^2}{2a}\right)\exp(-i\xi x)\,\mathrm{d}x=-\frac{\xi}{a}g(\xi).$$I used derivation under the integral and then integration by parts. Moreover, using substitution, ... 1 Morera's Theorem and the related Cauchy's integral theorem and may be what you're seeking.$$\oint_\gamma f(z) dz = 0$$If f is continuous on an open set that contains \gamma, and satisfies the above integral, then f is holomorphic on that set (analytic). By extension, if the above integral vanishes for every possible \gamma, then f is analytic ... 1 You know that \bar{f}(t) corresponds to \bar{F}(-\omega). There are two (reasonable) options to determine the Fourier transform of e^{iat}\bar{f}(t). The first is to define a function$$g(t)=e^{-iat}f(t)\Longleftrightarrow G(\omega)=F(\omega+a)$$from which$$\bar{g}(t)=e^{iat}\bar{f}(t)\Longleftrightarrow \bar{G}(-\omega)=\bar{F}(-\omega+a) ...

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