# Conversion between DTFT in radians/sample to DTFT in cycles/sample

I have found that most commonly the DTFT is defined as: $$X(\omega) = \sum_{n=-\infty}^{\infty} x[n]e^{-j \omega n}$$.

However the class I am taking frequently uses the DTFT expressed in "normalized continuous frequency of discrete signals" units i.e. cycles/sample. I am trying to understand how to convert a DTFT pair using one units to other units.

For example, the DTFT pair with $$\omega$$ [radians/sample]:

$$x[n] = 1 \Leftrightarrow X(\omega) = 2 \pi \sum_{k=-\infty}^{\infty} \delta( \omega-2 \pi k)$$

My understanding that this pair with $$\mu$$ [cycles/sample] is given by: $$x[n] = 1 \Leftrightarrow X(\mu)= \sum_{k=-\infty}^{\infty} \delta( \mu -k)$$

I have honestly struggled to find many resources which make any mention of the latter DTFT.

You can use the following :

Consider the discrete time sequence $$x[n]$$ and the DTFT $$X(\omega)$$ related by: $$X(\omega) = \sum_k x[n] e^{-j \omega n} \longleftrightarrow x[n] = \frac{1}{2\pi} \int_{-\pi}^{\pi} X(\omega) e^{j\omega n} d\omega$$

Then define $$X'(\mu)$$ as the normalized DTFT, which is related to $$X(\omega)$$ as $$X'(\mu) = X(2\pi \mu) = \sum_k x[n] e^{-j 2\pi \mu n}$$ and then $$x[n] = \int_{-0.5}^{0.5} X'(\mu) e^{j 2\pi \mu n} d\mu$$

Applying this to your example case: $$x[n] = 1 \longleftrightarrow X(\omega) = 2\pi \sum_r \delta(\omega - 2\pi r)$$ then since $$X'(\mu) = X(2\pi \mu)$$ we have

$$X'(\mu) = 2\pi \sum_r \delta(2\pi \mu - 2\pi r) = 2\pi \sum_r \delta(2\pi [\mu - r]) = 2\pi \sum_r \frac{1}{2\pi} \delta(\mu - r)$$
$$X'(\mu) = \sum_r \delta(\mu - r)$$

as expected, where we've used the property that $$\delta(ax) = \frac{1}{|a|} \delta(x)$$.

• Thanks, it was the division by $2 \pi$ that was throwing me off. Perfect answer. – Filip Oct 15 '18 at 13:36