# Spectrum and Spectrogram explanation

I'm reading this material but I don't understand this example:

1. "An example of a 94% overlap transform processing could be demonstrated by a 279 sample frame with 262 samples overlapped from each frame into the next. At 12.5 MHz sample rate, each spectrum frame would occupy 23.76 µs. Each spectrum would start 1.36µs after the previous one and could show a 10 MHz span." Could anybody please help to explain what is 10 MHz span and why why each spectrum could show a 10 MHz span?

2. The same problem with the following picture (I don't understand the example):

"An example of a 94% overlap transform processing could be demonstrated by a 279 sample frame with 262 samples overlapped from each frame into the next.

A spectrogram is really just a sequence of spectra. In this case, each individual spectrum is taken of 279 time samples (which I will call a frame). Each subsequent frame consists of the last 262 samples of the previous frame, plus the next 17 time samples.

That means percentage of common samples between subsequent frames is:

$$\frac{262}{279} = 93.9068 \%$$

At 12.5 MHz sample rate, each spectrum frame would occupy 23.76 µs.

We have 279 samples in each frame. The distance between time samples is

$$\Delta t = \frac{1}{12.5 \times 10^6} = 80 {\rm ns}$$

Add 279 of those together and we get

$$279 \times 80 {\rm ns} = 22.32 {\rm \mu s}$$

It looks like someone can't multiply: the number 23.76 µs is actually $297 \times 80 {\rm ns}$.

Each spectrum would start 1.36µs after the previous one...

Each frame starts 17 samples after the last. This is $17 \times 80 {\rm ns} = 1.36 {\rm \mu s}$

... and could show a 10 MHz span." Could anybody please help to explain what is 10 MHz span and why why each spectrum could show a 10 MHz span?

It's not clear what they are meaning by this. If you sample at 12.5MHz, then the total possible frequency span is 12.5MHz (if you include positive and negative frequencies). There may be some allowance for guard bands that reduces the usable bandwidth from the ideal.