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Find the Discrete-time Fourier transform of $\frac{\cos(\frac{n\pi} 6)}{(n+3)\pi}$

I thought of making it to be a sinc, but at the bottom there is $n+3$ and if I replace $n+3$ then I don’t know how to get rid of $-\pi/2$ in the new $\cos((\pi/6)m-0.5\pi)$

Could someone through me a hint?

Matt L.
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Vitali Pom
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  • But then how do I transform 1/cos(b)? – Vitali Pom May 29 '20 at 14:06
  • Infinity. n is index and result is periodic. – Vitali Pom May 29 '20 at 14:14
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    Continuous or discrete? "n" implies discrete, "sinc" implies continuous. (Your function is the normalized sinc in disguise.) – Cedron Dawg May 29 '20 at 14:23
  • Discrete, sinc I meant - sampled. – Vitali Pom May 29 '20 at 14:45
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    Well, that clarifies nothing. There are versions of the sinc function: In the continuous domain, there is "sinc" and "normalized sinc". In the discrete domain, there is "discrete sinc", aka "alias sinc" and "Dirichlet kernel". Your function is a rescaled normalized sinc. The DFT is defined on a finite interval. So, if you are asking for the DFT of your function, the answer will depend on your interval definition. Your function is not periodic. The DFT of the continuous sinc is not going to be a "clean" answer. In the continuous realm, sinc and rectangular functions are Fourier pairs. – Cedron Dawg May 29 '20 at 15:21
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    @CedronDawg: The OP didn't mention the DFT. I suppose that the question is about the DTFT, and the DTFT of $\sin(n\omega_c)/(n\pi)$ is a well-known function. – Matt L. May 29 '20 at 17:09
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    @MattL. Thanks, OP doesn't say. In that case, this should help: https://dsp.stackexchange.com/questions/48736/dtft-of-1n-cdot-mathrmsinc – Cedron Dawg May 29 '20 at 17:23
  • Cease fire, I understood. I got cos() of something minus pi/2. It IS sinus of purely something. NOTE: that cos of pi is sinus of minus pi/2. – Vitali Pom May 29 '20 at 20:33

1 Answers1

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HINT:

$$\frac{\cos(n\pi /6)}{(n+3)\pi}=\frac{\cos[(n+3)\pi/6 -\pi/2]}{(n+3)\pi}=\frac{\sin[(n+3)\pi/6]}{(n+3)\pi}$$

Matt L.
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    Clear. I still wonder why I commented – Laurent Duval May 29 '20 at 16:47
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    @LaurentDuval: Sometimes it's a good thing that comments can be deleted, isn't it? Unlike in real life ... :) – Matt L. May 29 '20 at 17:10
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    It's not deleted in my mind. Like in real life – Laurent Duval May 29 '20 at 17:20
  • 1)x[t-t0]=“something”*X(something else). 2)cos((n-3)pi-p/2)=sinus((n-3)pi). – Vitali Pom May 29 '20 at 20:44
  • My point of confusion was that I thought that the unit circle of e is the same unit circle of cos and sin. NOT TRUE! There are 3 different unit circles and they are not being described anywhere. This is something you and I need to understand. I’m willing to add this description to Wikipedia, let me know if it’ll help you. If not, I will do it later. – Vitali Pom May 29 '20 at 20:47
  • @VitaliPom Not sure what you are talking about about three different unit circles. Nor does your last sentence make sense in the other comment.

    $$ e^{i\theta} = \cos( \theta ) + i \sin( \theta ) $$

    $$ \sin( \theta ) = \cos( \pi/2 - \theta ) = \cos( \theta - \pi/2 ) $$

    These articles of mine may help you out: https://www.dsprelated.com/showarticle/754.php and https://www.dsprelated.com/showarticle/1238.php

     – Cedron Dawg May 30 '20 at 00:54
  • Teta is (n -3)pi in the other comment, this exactly what I wrote. It's what being developed from the exercise and if you're refering to x[n-3] then it's just time shifting. All I'm looking is how to help to people with description in Wikipedia which will summarize the differences between e's circle, cos circle and sinus circle. Many people fall here down because sin & cos can be defined through the unit circle. – Vitali Pom May 30 '20 at 10:51
  • @VitaliPom Sine and cosine ARE defined by the unit circle. The unit circle exists in the real plane (usually denoted with (x,y)) and the complex plane (usually denoted with (real, imaginary)), but the trig functions are identically defined in each. There is a one-to-one mapping between the real and complex planes. The location of a point on the unit circle (either one) can be found by how far along the circumference it is. This distance is often denoted by $\theta$. So Euler's equation is actually a conversion from the distance along the circumference to Cartesian coordinates. – Cedron Dawg May 30 '20 at 11:23
  • Ok, thanks Cedron :)! – Vitali Pom May 30 '20 at 17:05