# Why do digital signals have an infinite bandwidth

## Why do all digital signals have an infinite bandwidth?

Constructing a discontinuous voltage waveform as the sum of sine waves would require an infinite number of sine waves. A "perfect" digital signal immediately switches between VSS and VDD (or vice versa). Such instantaneous switching events would represent discontinuities in the waveform.

In practice, chips neither generate perfect digital signals at their outputs, nor do they need them at their inputs. Examining an output signal in a sufficiently good range will usually show that it is a little "smoothed", and most chips will tolerate input signals that are essentially "smoothed" provided they don't linger or hop around in the 1/4 range and 3/4 VDD. In fact, some chips are designed to intentionally smooth their output waveforms (sometimes by a programmable amount) and / or accept inputs that are even "muddy" smoothed to a point.

It's worth noting that while something like a perfect 1 Hz square wave can be expressed as the sum of continuous sine waves in the 1 Hz to 1 MHz range and beyond, it is very unlikely that a device capable of receiving of a 1 MHz signal, a 1 Hz rectangle perceives a wave with a continuous 1 MHz component. The 1 Hz square wave would contain, among other things, a 999,999 Hz component whose strength was 1 / 999,999 of the fundamental wave and a 1,000,001 Hz component whose strength was 1 / 1,000,001 of the fundamental wave. The device trying to receive a "1 MHz" signal would detect these and many other components to varying degrees; During each one second interval there were times when they were all in phase and times when about half were in phase and half out. The device would thus sense a variable amount of "1 MHz" signal - most likely sense a considerable amount near the moments when the input switched (because all the waves sensed would be in phase) and a much smaller amount at other times ( because the detected waves would have a mixture of phases). A really sharp 1 Hz square wave driving a strong antenna would therefore not cause continuous interference in a 1 MHz transmission, but would result in a 2 Hz "tick-tick-tick".

### Kaz

A square wave function is not discontinuous. It is piecemeal throughout. Indeed, continuity is one of the necessary criteria for the existence of a Fourier transform, IIRC.

### Super cat

@Kaz: While functions that are not continuous everywhere can be broken down into piece-wise continuous and non-piece-wise, I've usually heard both of these subcategories called "discontinuous". Although I could invent some functions whose value is defined everywhere, but which are not piecewise continuous (e.g. let f (x) 1 for rational values ​​of x and 0 for irrational values), I can't think of any such functions that I would consider "useful". There might be math functions that work for different areas of x had no value, but ...

### Super cat

I'm not sure if such things could be called "waveforms".