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

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?

• But then how do I transform 1/cos(b)? – Vitali Pom May 29 at 14:06
• Continuous or discrete? "n" implies discrete, "sinc" implies continuous. (Your function is the normalized sinc in disguise.) – Cedron Dawg May 29 at 14:23
• 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 at 15:21
• @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 at 17:09
• @MattL. Thanks, OP doesn't say. In that case, this should help: dsp.stackexchange.com/questions/48736/… – Cedron Dawg May 29 at 17:23

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