July 23rd, 2017, 04:54 AM  #11  
Member Joined: Jul 2017 From: europe Posts: 51 Thanks: 0  Quote:
OK, that's why I reduced my question to the following: are the quoted transformation correct? I just want obsessively to know whether they are correct or not. To know it as a fact, confirmed by Mathematician...  
July 23rd, 2017, 09:35 PM  #12  
Senior Member Joined: Jul 2012 From: DFW Area Posts: 635 Thanks: 96 Math Focus: Electrical Engineering Applications 
Hi DesertFox, Quote:
To answer your question: 1. Incorrect, wrong lower limit (to deal with a lower limit of 0 instead of $\infty$ for sinusoids, it helps to know about convolution). Also, $F(sin(t))$ is incorrect. See 3., below. 2. Correct, although $F(w)$ is usually written as $F(jw)$ to imply that it is usually complex, unlike $f(t)$, which is, of course, real. 3. Incorrect. The Dirac delta function, $\delta(w)$ is needed for the FT of perpetual sinusouds. 4. Incorrect. This is really confusing because it seems that the text is trying to take the inverse FT of a time domain function, rather than a (complex) frequency domain function. But it is hard to tell for sure. Another poster has asked if you have studied FS, and you replied that you have not. I also think that understanding FS before moving on to the FT is the right path to take. If you Google "Fourier Series", you should find some good references. To try to explain why it is a good idea to start with FS, consider that FS are used in analyzing the frequency content of periodic waveforms in terms of the harmonics of the waveform angular frequency ($w_0, 2w_0, 3w_0, \text{etc.}$). Note that mathematicians have shown that in a periodic waveform, only the fundamental frequency and its harmonics are present. (I can heuristically explain this in another post if requested, and it actually has application towards the FT, but I would rather not take focus away from FS now). The complex coefficients for the FS of a function $f(t)$ are given by: $\displaystyle \large c_n=\frac{1}{T} \int_{T/2}^{T/2} f(t)e^{jnw_ot} \ dt$ where $T$ is the period and $w_0$ is the fundamental angular frequency. The waveform is reconstructed as follows: $\displaystyle \large f(t)=\sum_{n=\infty}^{\infty} c_n e^{jnw_0t} \qquad$ Eqn. [1] So let's take the FS (find the $c_n$ values) of $\cos(t+\theta)$, so $w_0=1$: First: $\displaystyle \large \cos (t+\theta)=\frac{e^{j(t+\theta)}+e^{j(t+\theta)}}{2}=\frac{1}{2} \left (e^{jt}e^{j \theta}+e^{jt}e^{j \theta} \right )$ So: $\displaystyle \large c_n=\frac{1}{4\pi} \underbrace{\int_{\pi}^{\pi}e^{jt}e^{j \theta}e^{jnt} \ dt}_{\text{0 when } n \neq 1} + \frac{1}{4\pi} \underbrace{ \int_{\pi}^{\pi}e^{jt}e^{j \theta}e^{jnt} \ dt}_{\text{0 when } n \neq 1}$ $\displaystyle \large c_1=\frac{1}{4\pi} \int_{\pi}^{\pi}e^{j \theta} \ dt = \left. \frac{1}{4\pi} \ e^{j \theta} \ t \ \right _{\pi}^{\pi}=\frac{e^{j \theta}}{2}$ $\displaystyle \large c_{1}=\frac{1}{4\pi} \int_{\pi}^{\pi}e^{j \theta} \ dt = \left. \frac{1}{4\pi} \ e^{j \theta} \ t \ \right _{\pi}^{\pi}=\frac{e^{j \theta}}{2}$ Which is what we expect. I will leave it to you to show that these coefficients reconstruct $\cos(\theta)$ using Eqn [1] above. The point of working this example is to show that there are 2 non zero FS coefficients of a perpetual sinusoid and since we integrated over the period, the coefficients are well defined, and look like the incorrect results given for the FT of a similar signal in your text. But, for the FT where we integrate from $ \infty$ to $\infty$, perpetual sinusoids produce weighted Dirac delta functions as coefficients. That is why I said that the text was incorrect. Integrating from $0$ to $\infty$ produces further complications. I would like to point out one final thing: In 1., the text mentioned the reason for using a minus sign in $e^{jwt}$ in calculating the FT. I am not sure that I understood it. But from the FS calculation example for the sinusoid given above, if we want to extract the frequency information of a signal at, say, $e^{j2w_0t}$, we need to multiply by $e^{j2w_0t}$ to get to $e^0=1$ which will integrate to the coefficient, and to $e^{j4w_0t}$ which will integrate to 0. I think the reason for the minus sign is much clear here.  

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application, electromagnetism, fourier, transform 
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