# Why does a high optical carrier feq. provide higher transmission rate?

1. Sep 7, 2009

### gsmsmldb

For example , optical carrier in the infra red region can provide higher
capacity than microwave.

2. Sep 7, 2009

### Born2bwire

It generally comes down to bandwidth, which is the size of frequency range that we can modulate our signal over. When we have a digital or analog signal, we encode the information across a small band of frequencies. The larger the band of frequency we encode across, the more information we can cram into the signal per unit of time. So let's say we choose a carrier signal of 1 MHz and another at 1 THz. A band of 500 KHz is chosen for our signal. That means we have spread our signal across .75-1.25 MHz and 1.00075-1.00125 THz. The 1 MHz means that we have taken up a much larger amount of frequency range in regards to the carrier frequency than the 1 THz. This is hard to do, first it restricts the number of other transmitters in the frequency range, the next one may have to be at 1.5 MHz or more to prevent overlap. The second reason is that now our signal's wavelength at 1 MHz is shifting by a large percentage with respect to the carrier frequency. If we make a resonant antenna at 1 MHz, we would need an antenna with a much larger bandwidth than the 1 THz signal to properly receive the signal.

So it comes down to the fact that higher frequencies allow us to use a larger amount of bandwidth to encode more information.

3. Sep 7, 2009

### gsmsmldb

Thanks Born2bwire! It's very clear.

If hypotetically speaking, just out of curiosity, if I use a 1 MHz carrier, and modulate it just to transmit one pulse. If I try to brute-force modulate to have a mega short duration, so the frequency spreading will be huge (e.g. occupying a huge bandwidth). Where if you were to plot the signal in Fourier space, the bandwidth even extends to negative domain. Physically speaking in practice, what do you expect one to see?

4. Sep 7, 2009

### Born2bwire

It's been a good while since I did signal processing so this may not be correct, what we do in practice is not the same as theory but... It is my recollection that when you modulate a signal, you move the signal up from the baseband of 0 Hz to X Hz, but you also get a mirror image at -X Hz. So, if your actual bandwidth is larger than 2X Hz, then the positive and negative frequency bandwidths will overlap and you will get aliasing. It would corrupt the received signal. You would either need to shift to a higher frequency or filter the signal before modulation (at which you would lose information which may or may not be salvagable).

5. Sep 7, 2009

### gsmsmldb

Cool, Thanks so much! I think you are correct. Going back to my signal processing book, there's a requirement where carrier freq must be at least greater than the bandwidth in order to avoid overlapping of the spectra. Otherwise information of the modulating signal will permanently be lost in the process of modulation. What field are you in btw?

6. Sep 7, 2009

### Born2bwire

I do computational electromagnetics. Signal processing is rather useful to know for my research area as it aids in the practical application of electromagnetics and the mathematical techniques can be very useful to us as well. Sampling, interpolation, anterpolation, FFT are just a few topics off the top of my head that I have seen used in popular fast solver techniques.

7. Jun 15, 2011

### fisico30

Hello Born2bwire,

can I ask you a clarification on the good answer you provided?
thanks
fisico32

8. Jun 15, 2011

### Born2bwire

What exactly do you wish to be clarified?

9. Jun 15, 2011

### sophiecentaur

Yes, there is a limit to the actual bandwidth of signals you can squeeze onto a carrier, whatever modulation.

As well as the 'absolute bandwidth' factor, there is also the 'fractional bandwidth' issue. It is difficult to modulate a carrier and to transmit it with a baseband signal which is a large fraction of its frequency. This is partly because of device limitations but also, of course, there are often channel bandwidth limitations because other users want to use nearby frequencies for communication. This may not be relevant for cable and optical fibres, of course but it is often more convenient to use frequency (or wavelength) multiplexing (several carriers with different modulation) than to squeeze all the information onto just one carrier (complete time multiplexing).

But, just to be awkward, I should point out that information and bandwidth are not the same thing. Shannon's Communication Theory tells us that the only thing that limits the amount of actual information that can be transmitted within a certain bandwidth is the signal to noise ratio. Improve the signal to noise ration in any channel and you can (theoretically) get more information through it - without limit.