# How does sampling rate of $x[n]$ relate to sampling rate in frequency domain after DFT transformation?

I've got an analog signal $x(t)$ sampled at frequency $F_s$ to obtain samples:

$$x[n] = x(t) \bigg|_{t=n/F_s}$$

I transform this signal with DFT defined as:

$$X[k] = \sum_{n=0}^{N-1}x[n]e^{-i2\pi kn/N}$$

Is there such a thing as a sampling frequency / rate in the frequency domain? How does it relate to $F_s$?

• Yes -- you have $N$ spectrum samples over a frequency band of $F_s$ hertz, so you're sampling the spectrum at $N/F_s$ samples/hertz. – MBaz May 2 '18 at 19:31
• Sexy, just to be clear. What we mean by "$x[n]$" (with square brackets) is a sequence of numbers (that we call "samples"). Older DSP textbooks and some lit still use the notation "$x_n$" (or "$X_k$" for the output of the DFT). but because analog signals can be a vector of signals (e.g. multichannel audio), we use the notation "$x_m(t)$ to represent the different analog signals in time. then when we simultaneously sample those different analog signals we have samples: $$x_m[n] \triangleq x_m(t)\bigg|_{t=n/F_s}$$ we thing that notation is better than "$x_{n,m}$" for the samples. – robert bristow-johnson May 2 '18 at 22:38

A better way to think about it is the DFT is agnostic about the sampling rate. What matters is the number of sample points. If you look at the DFT definition, $F_s$ does not appear in it, but $N$ does twice.

The sampling rate can be used to figure out the duration of the sampling frame given the number of samples, or used to figure out how many sample points for a given duration. Once the DFT is taken on the sample, the bin number corresponds to the number of cycles per frame. The number of cycles per frame can be divided by the number of seconds per frame (the frame duration), or whatever unit of time is being used, to get the number of cycles per second (or time unit).

I think the equation the r b-j gave is better expressed as:

$$f = k \cdot \frac{F_s}{N}$$

Where the units are:

$$\frac{cycles}{timeunit} = \frac{cycles}{frame} \cdot \frac{\frac{samples}{timeunit}}{\frac{samples}{frame}}$$

The latter factor can be considered a conversion factor for the frequency.

When $k = N/2$:

$$f = \frac{N}{2} \cdot \frac{F_s}{N} = \frac{F_s}{2}$$

Which confirms r b-j's statement about the halfway bin corresponding to the Nyquist frequency.

Hope this helps.

Ced

• Good to have somebody show the dimensional analysis side:) Sometimes it's necessary to get the scaling correct and is always handy. – rrogers May 8 '18 at 21:31

And the essential answer to your question is that $X\left[\tfrac{N}{2}\right]$ is the DFT bin that corresponds to the Nyquist frequency, $\frac{F_s}{2}$.

So $X[k]$ is the DFT bin that corresponds to a frequency of $f = F_s \tfrac{k}{N}$.