Bandwidth and noise

Bandwidth basically means how fast your signal can change or how fast can you send out symbols. – Symbol is something you send out to represent bit(s) Noise means that although you sent 1 to me, I may receive something like 1+x, where x is the noise added by the media.

Ideal case If the bandwidth is infinite and absolutely no noise, how fast can you send/receive data?

Bandwidth If the media is of infinite bandwidth but with some noise, how fast can you send/receive data? Assuming that your device is fast enough.

Noise If there is absolutely no noise but the bandwidth is limited, how fast can you send/receive data? Assuming that your device is fine enough to tell the slightest differences of signal voltage.

Shannon’s Theorem C=B*log(1+S/N) – C is the capacity of the channel, B is the bandwidth of the channel, S is power of the signal and N is the power of the noise – Channel capacity means how many bits you can send out per second reliably

Shannon’s Theorem There is actually a very simple way to understand Shannon’s theorem – B means how fast can you send out symbols – S/N determines how many bits each symbol carries – why there is a log?

Limited Bandwidth Bandwidth is limited because of many reasons – The wire itself, if too long, is a capacitor and slows down voltage transition – In wireless transmissions, the whole spectrum shared by many communication parties and each can have only a limited chunk of it

Limited bandwidth The trouble is, we live in a world with limited bandwidth and some noise. Noise is easy to model. In mathematical languages, any signal can be viewed as the sum of a series of sine waves on different frequencies You pass the signal to a channel that can pass frequencies up to B, all sine waves on frequencies higher than B will be lost – you will receive a distorted signal

Nyquist Theorem If the bandwidth is limited to B, in the ideal case when there is no noise, how fast can you send/receive symbols? – Note that the channel capacity is infinity because each symbol can carry infinite number of bits Nyquist Theorem says that it only makes sense for you to send/receive symbols at a speed of 2B – if B is 4KHz, you send/receive 8K symbols per second – the baud rate is 8K per second. Why? If a signal is band-limited by BHz, by taking 2B samples per second, you can completely reconstruct it. Nothing more can be reconstructed, so no point of sending.

How modulation is done Given data bit streams, translate them into baseband waveforms Up-convert it to the carrier frequency

How demodulation is done Given the received waveform, down-convert it to the baseband waveform Translate the baseband waveform to bit streams

Wireless communications FDMA – Frequency division multiplexing TDMA – Time division multiplexing CDMA – Code division multiplexing

Wireless Demodulation You multiply the received signal with a sine wave. Then you pass it to a low pass filter. This is the baseband signal. Then you do phase tracking. Then you sample the wave, and get samples of the data. Then you figure out what the bits are.

Demodulation With noise, it’s all about guessing, because you don’t know what the noise is when this symbol is sent as noise is random. You may know some statistics of the noise, based on which you make your best guess. For example, let’s say 0 is 0 volt 1 is 5 volts. Suppose you know that very rarely the noise exceeds 2.5 volts. If you received a 2.2 volts, you would guess it to be 0 or 1? What is the chance that you got it right/wrong?

Maximum Likelihood Detection Detection – given a received signal, determine which of the possible original signals was sent. There are finite number of possible original signals (2 for the binary case – 0 or 1) Compute a likelihood value for every possible input, choose the one with largest likelihood – maximum likelihood detection

Maximum Likelihood Detection There are two inputs, x1 and x2. Noise is n. What you receive is y. If I sent x1, you receive y=x1 + n. If I sent x2, you receive y=x2+n. You don’t know what I sent and how large n is. You compute the likelihood of receiving y if I sent xi, Li (i=1,2). If L1 > L2, you say I sent x1. Else you say I sent x2. How to compute L1 and L2?

Maximum Likelihood Detection If n=0 always, y=x1 if I sent x1 and y=x2 if I sent x2. Of course x1 != x2. Will you make mistake in this case? What is the likelihood of y=x2 if I sent x1? If n is not always zero, we assume n follows some probability distribution. If it is Gaussian, the channel is called AWGN. Given y, the likelihood of x1 being sent is the likelihood that n=y-x1. Similarly, the likelihood of x2 being sent is the likelihood that n=y- x2. (likelihood is derived from probability, but likelihood could be taking some values that probability cannot take depending on how you define likelihood) So what you are doing is to compare the likelihood of n=y-x1 and n=y-x2. So the detection rule is if p(n=y-x1)/p(n=y-x2) > 1, output x1, else output x2. That’s all! Wait, what if you know that x1 is more likely to be sent than x2?

Wired Communication – Telephone Company Dial-up – 56kbps DSL – Digital Subscriber Line – ADSL: Asymmetric DSL, different upload and download bandwidth – Available bandwidth is about 1.1MHz, divided into 256 channels, one for voice, some unused or for control, the rest divided among upstream and downstream data. My DSL at Pittsburgh was 100kbps upstream and 768kbps downstream – How ADSL is set up. Fig The ADSL modem is 250 QAM modems operating at different frequencies. The actual QAM depends on the noise.

Wired Communications – The Cable TV Company Cable frequency allocation. Fig – Downstream channel bandwidth is 6MHz. If using QAM-64, how much of a speed we can get? – Upstream channel is worse so use QAM-4. – Upstream – stations contend for access (MAC layer issue, will be discusses later) – Downstream – no contention, from the head end to user – Shared medium, so some security is needed

Wired communication – Optical Backbone SONET