Why does more bandwidth means higher bit rate in digital transmission?

I'll start with what I do know:

Shannon Law gives the theoretical upper limit [tex]C_{noisy}=B*log_{2}(1+\frac{S}{N})[/tex]
if S = N, then C = B
As N→∞, C→0
As N→0, C→∞

Nyquist Formula says approximately how many levels are needed to achieve this limit [tex]C_{noiseless}=2*B*log_{2}M[/tex]
(If you do not use enough logic levels you can not approach the shannon limit, but by using more and more levels you will not exceed the shannon limit)


My problem is that I'm having a hard time understanding why bandwidth relates to bit rate at all. To me it seems like the upper limit of the frequency that can be sent down the channel is the important factor.

Here's a very simplified example: No noise at all, 2 logic levels (0V and 5V), no modulation, and a bandwidth of 300 Hz (30 Hz - 330 Hz). It will have a Shannon Limit of ∞, and a Nyquist Limit of 600bps. Also assume that the channel is a perfect filter so anything outside of the bandwidth is completely dissipated. As I double the bandwidth, I double the bit rate etc.

But why is this? For two level digital transmission With a bandwidth of 300 Hz (30 Hz - 330 Hz), the digital signal of "0V's" and "5V's" will be a (roughly) square wave. This square wave will have the harmonics below 30 Hz and above 330 Hz dissipated, so it will not be perfectly square. If it has a fundamental frequency at the minimum 30 Hz, (so the "0V's" and "5V's" are switching 30 times a second), then there will be a good amount of harmonics and a nice square wave. If it has a fundamental frequency at the max 330 Hz, the signal will be a pure sine wave as there are no higher order harmonics to make it square. However, as there is no noise the receiver will still be able to discriminate the zeros from the ones. In the first case the bit rate will be 60 bps, as the "0V's" and "5V's" are switching 30 times a second. In the second case the bit rate will be a maximum of 660bps, (if the threshold switching voltage of the receiver is exactly 2.5V), and slightly less if the threshold voltage is different.

However this differs from the expected answer of 600 bps for the upper limit. In my explanation it is the upper limit of the channel frequency that matters, not the difference between the upper and lower limit (bandwidth). Can somebody please explain what have I misunderstood?

Also when my logic is applied to the same example but using FSK modulation (frequency shift keying), I get the same problem.

If a zero is expressed as a 30 Hz carrier frequency, a one is expressed as a 330 Hz carrier frequency, and the modulation signal is 330 Hz, then the max bit rate is 660 bps.

Again, can somebody please clear up my misunderstanding?

Also why use a square wave in the first place? Why cant we just send sine waves and design the receivers to have a switching threshold voltage exactly in the middle between the max and min value of the sin wave? This way the signal would take up much less bandwidth.

Thanks for reading!
Last edited:


Science Advisor
Gold Member
Hi Craig

I'm not an expert in this field

to answer your topic title first ...
Why does more bandwidth means higher bit rate in digital transmission?
A larger bandwidth allows for a higher bitrate, it doesn't produce it
You mite have a 1 GBPS bandwidth system but only running 100MBPS, you are just not making full use of the available bandwidth

My problem is that I'm having a hard time understanding why bandwidth relates to bit rate at all.
BUT a higher bitrate demands a higher bandwidth...
increasing bitrate increases the frequency therefore a higher bandwidth is required

keeping the same bitrate, greater bandwidth allows for a better quality received signal



Science Advisor
Homework Helper
I think the basic thing you are missing, is that to transmit any information (as opposed to just a carrier) you need some form of modulation (amplitude, frequency, phase, whatever).

The limited bandwidth means the signal takes a finite amount of time to change from one state to another, and settle down again. If you try to force the signal to change faster than the bandwidth permits, the receiver will pick up scrambled data.

When you talk about "a good amount of harmonics and a nice square wave," you can't just ignore the difference from a "perfect" square wave. The difference is what limits the data transmission rate!

In your example of a 300Hz carrier and the channel with bandwidth 30-330Hz, think (or simulate) what happens when you start or stop the carrier wave. You can't just "jump" into a perfect sine wave. You will get some sort of starting and ending transient for about half a cycle. The same applies if you want to "instantly" change the frequency, phase, etc. Again, that is what limits the data transmission rate.


Science Advisor


Gold Member
Bandwidth affect how fast a signal can change. Lets just talk about switched DC pulses or steps. Assume an RC filter limits the bandwidth. The RC filter limits the risetime, so the signal cannot change quickly. A DC step will take R*C seconds to rise to ~63% of it's final value. If you switch at higher speed, the resulting amplitude will be really low, so you become more sensitive to noise. Noise causes bit errors. In this case, the highest frequency tranported is essentially the same as the bandwidth.

In a modulated signal the same priciples hold. For example, if you AM modulate a 1MHz signal with a 1Khz sinewave, you see side tones at +-1Khz. So, you need 2KHz bandwidth to transmit the 1KHz signal. If you narrow the bandwidth before reception, the 1KHz modulation becomes lower in magitude. In this case, the frequencies between the highest and lowest frequency sidebands are required.

When you switch a signal quickly, you require more bandwidth to transmit the resulting frequencies. The faster you switch and the faster the rise/falltime, the wider the bandwidth required. If you switch 1MHz, then the requirement is 1MHz plus and minus some bandwidth.

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