# Why a high number of harmonics means a lower bitrate?

I am studying on a book (I'm still at basics) and I haven't understood why a signal with a lower number of harmonics implies a low bit rate. I have this figure:

It clearly shows how a signal with a high number of harmonics means that the bits are easier to "recognize". The bits in the last figure are 01100010. But it also shows the relationships between the number of harmonics and the bit rate:

My question is: shouldn't it be the opposite? A signal with many harmonics could represent more bits than a signal with few harmonics. A signal with N harmonics is able to carry N bits in a certain period of time T, while a signal with just one harmonic can carry just 1 bit in T nano seconds. But there is something wrong in my argument, could someone clarify this?

• There appears to be some context missing. Which book are you studying? Perhaps someone familiar with it can help. – user_1818839 Mar 5 '13 at 18:58
• We need more context to answer this, but what I think its saying is, Given a fixed channel bandwidth, say 2400 Hz, you can carry more harmonics of a lower bps signal. – The Photon Mar 5 '13 at 19:00
• I think indeed some context is missing. As I see it, this is a perfect example of showing that you can't reliably encode bits at a given bitrate if the channel bandwidth is too small. Generally bitsPerSecond is 2×channelBandwidth. If you disregard that rule of thumb and use a too low bandwidth, then you get the top three silly images. In my opinion the first harmonic is at an entirely wrong frequency (way too low). I don't think this chapter is particulary clear. – jippie Mar 5 '13 at 19:12
• Sorry I said book because I am missing the term. It's a collection of slides, not properly a book. As for the context, it's just saying to show the relationship between bit rate and harmonics, after showing Fourier serie and showing how a signal can be interpreted as a sequence of bits. Do you need more information? – Ramy Al Zuhouri Mar 5 '13 at 19:25
• @RamyAlZuhouri, without more context, the text written explaining the table, I don't think we can explain what the table means. Therefore "it's difficult to tell what is being asked here." – The Photon Mar 5 '13 at 20:07

The table merely says that given a channel with a fixed bandwidth, if you transmit a signal through it (input can contain infinite harmonics), at the output only the first harmonics will get through, where a high baud rate (high fundamental frequency) will get less harmonics through than a low baud rate signal. In other words, a low baud rate signal will come out the other side relatively unaffected by the channel, while a high baud rate signal can be severely distorted (even completely filtered out).

This is important because if the signal gets too distorted, the original bits cannot be reliably recovered due to intersymbol interference and the increased relevance of noise perturbations.

So a low number of harmonics does not imply a low bit rate. Quite the opposite, a low bit rate implies a high number of harmonics going through the channel.

The general criterion for finding baud rate vs channel bandwidth limits are given by the Nyquist Criterion. Note that, although related, this is different than the sampling theorem mentioned in another answer (Shannon-Nyquist), because you are not trying to reconstruct an original "bandwidth limited signal" in the analog sense from a set of samples, it is a matter of determining the original symbols, which is related to threshold decisions, which can be based on sampling, transmitted power (areas), etc.

It is unclear what you mean by "implying" a particular bit rate, and what exactly you envision bit rate meaning.

If you wish to accurately measure a signal by sampling it at regular points in time, then the minimum necessary sample rate (not bit rate) is dictated by the frequency content of that signal. In fact, the absolute minimum sample rate is twice per cycle of the highest frequency of interest. This is often referred to as the Nyquist rate.

Any arbitrary repetitive signal can be decomposed into a sum of sines with specific amplitudes and phase shifts. The period of these sines is the basic repetition period of the composite signal divided by all positive integers. These are called the "harmonics" of the signal, with the first harmonic having the base period/1 sometimes called the "fundamental".

Therefore, sampling a repetitive signal is the same as sampling all the sines that are added up to make that signal. The Nyquist rate applies to each of those. Therefore, more harmonics, which means the sine components will to go to higher frequency, requires a faster sampling rate. It is the highest frquency of interest that dictates the minimum sampling rate. More harmonics means a higher frequency of interest.

The more harmonics, the more channel bandwidth is required to transmit a particular baud rate.

If the channel bandwidth is limited, then baud rate is available bandwidth divided by highest harmonic, which clearly sets up the inverse relationship. (Bit rate is baud rate adjusted for coding.)