Digital Transmission Fundamentals
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Digital Transmission Fundamentals Chapter 3. Communication Networks Leon-Garcia, Widjaja The Questions ● Digital representation ● Why digital transmission ● Digital representation of analog signals ● Characterization of communication channels ● Fundamental limits ● Line coding ● Properties of transmission media ● Error detection-correction Block vs Stream communication ● Block: – Text, pictures, software – Total time of communication ● Stream: – Internet radio, streaming video, midi – Restrictions on rate of arrival of packets Block information L D=t + p R Reducing delay ● Decrease propagation delay – Increase speed of light ;-) – Optimize routing ● Avoid satellite channels ● Increase transmission rate – The subject of intense research ● Reduce data length – Compression. Secret to compression ● In text – Some letters or letter pairs are far more common – “ae” is more common than “zq” – We may encode “ae” with 3 bits and “zq” with 16 ● In images, sound, etc – Do not transmit non perceivable information (lossy) – Exploit statistical correlations – Use different encoding Streaming ● For analog signals, the most important quantity is bandwidth (of the signal) – A measure of how fast the signal changes – The unit is the Hz – It is the frequency of the highest frequency component Streaming ● Voice (telephony) – 4 kHz is considered enough – We need 8,000 samples per second – This is called Pulse Code Modulation ● Music – 20 kHZ is enough, but use 22 kHz – We need 44,000 sample per second Streaming ● PCM – A stream of 8-bit (or 16 or ...) samples ● Differential PCM – Transmit difference from previous sample. ● Adaptive DPCM – Adapts to variation of voice level ● Linear Predictive Methods – Predict next value and transmit the difference only Why Digital Comunication ● Many common signals are analog ● Yet almost all communication is digital now ● Reasons – Cheaper – Ability to restore signal after degradation – Advanced routing Analog vs Digital ● Analog: if we transmit .8 Volt and receive .81 Volt the receiver does not know if it is noise – Noise is impossible (to a great extend) to remove ● Digital: If we transmit 1 Volt and receive .8 Volt we know it is 1 Volt – All the noise enters the system upon digitization. – Can have re-generators at regular intervals – Only if the noise is strong and signal weak we lose information (kind of) Channel Bandwidth ● Amplitude response function – The amplification (or attenuation) of every frequency component going through the channel ● Bandwidth is the frequency after which the components have much more significant attenuation. – Frequency after which the ARF dips Channel Bandwidth A(f) f Sawtooth components Intermediate components Signal to Noise Ratio ● SNR, one of the most useful quantities ● Ratio of (average) signal power to (average) noise power ● Power and variance of a signal are sometimes indistinguishable for electrical engineers ● Commonly reported in decibell (db) – It is a logarithmic scale Signal to Noise ratio 2 2 σ n =E {n } 1 P= n2 R Signal to Noise ratio P SNR= sig Pnoise SNR db=10 log10 SNR Why is SNR Important? ● Imagine we had no noise... ● We could take the entire wikipedia, say 1 TB ● Treat the whole thing as a number ● Transmit it as 0.10111001101.... Volt ● We need zero bandwidth for this! Noise and Signal Multilevel Signals Multilevel Signals ● The noise should be small enough to allow us to distinguish the levels – Increase the power of the signal – Decrease the noise – Lower the bit rate (if we transmit every bit for a longer time we can average the noise out) Example: Telephone lines ● They have 1% noise (amplitude) ● SNR is 10,000:1 ● SNR in db is 10*log 10000 = 10*4 = 40 Channel Capacity ● Depends on: – Bandwidth – SNR ● Studied by Claude E. Shannon, father of communication theory – Information Theory (1948) ● It is a probabilistic theory ● Provides upper limit for the amount of data the receiver can meaningfully determine Channel Capacity C=W log2(1+SNR) Example Telephone lines ● Bandwidth 3.4 kHz ● SNR 10,000 ● Capacity is: – 3400 * log(10000)/log(2) = 3400*4*3.3 = 45,178 Fourier Transform Signals x(t)=∑ αk cos(2 π k f o t+ϕk ) x(t)=∑ αk sin (2π k f ot +ϕ k ) x(t)=∑ αk cos(2 π k f o t)+¿ ∑ βk sin(2π k f o t) Signal Bandwidth ● It is the frequency of the highest frequency component... but ● Components of very high frequency with miniscule weight are always present ● A better definition is: – Bandwidth is the frequency range that contains 99% of the power of the signal Bandwidth: Examples 1 0 1 2 3 4 kHz 1 2 3 4 Sampling ● We can approximate an analog signal with a set of discrete measurements ● Under certain conditions the original analog signal can be reconstructed exactly ● The “Sampling Theorem” provides the preconditions – The bandwidth of the signal can be no more than half the sampling rate Sampling x(t) x(nT ) xr (t)=∑ xn s(t−nT) xn= x(nT ) n s(t ) Sampling xr =∑ xn s(t−nT) n x 1 x3 s(t ) x 1 x3 Sampling ● We can do linear interpolation with this little triangle ● We can do a smoother interpolation with a “spline” ● The “perfect” interpolation is the sampling function Sampling Function sin (2π W t ) s(t )= 2π W t 1 W = T Sampling Function ● Works perfectly... in theory ● It dies off very slowly (1/t) ● Cannot be implemented in practice ● We use various approximations ● Typically we window it with a Gaussian Sampling Function Quantization ● In modern applications samples are digitized ● Digital signals can be transmitted error-free (almost) ● All the error sneaks in during digitization ● Simplest is the uniform quantizer Quantization Error en= yn− xn 2 σ x SNR= 2 σ e Quantization Error y −Δ Δ 2 2 x Quantization Error 2 σ 2=Δ e 12 V Δ= 2m−1 Power of the Signal ● It is related to V – Signal ranges between -V...V ● V is chosen so that the signal is almost always between -V...V ● If we assume that the pdf of the signal is Gaussian – Signal is almost always within 4 standard deviations Power of the Signal V =4 σ x -V V SNR V 2 2 σ x 16 SNR = 2 = 2 σ e Δ 12 3 V 2 = 4 Δ2 2(m−1) 3 22 3 2(m−1) = V = 2 4 V 2 4 3 = 4m 16 SNR db 3 SNR db = 10log(4)m + 10log( ) 16 = 6m−7.27 3 = 10 log 4m 16 SNR - Example ● For 8 bit per sample (telephones) ● SNR = 48 – 7.27 = 40.73 db ● Every extra bit adds 6db to the SNR – An extra bit halves the quantization error – So it reduces the power of the quantization error by 4 – The log of 4 is about 6 Non-uniform Quantization ● Depending on the application it might not be optimal to have all intervals same size ● For audio we prefer larger intervals for larger signal magnitude ● This may increase – The SNR for a particular application – The perception of quality by humans Communication Channels ● We usually assume channels to be – Linear – Time invariant ● Neither is perfectly true but works in practice. ● Summarized as: – The response to the sum of two signals is equal to the sum of the responses. – The response does not change with time. – The response to a sine is a shifted scaled sine Linear Time Invariant x1(t )→ y1(t) x2(t)→ y2(t) x1(t )+ x2(t)→ y1(t)+ y2(t) x1 (t +τ)→ y1(t+ τ) Attenuation ● The ratio of power going in over the power coming out ● If we have attenuation 3db per kilometer – In one kilometer the power if halved – In 15km the power is down about 30,000 times! – Voltage is down about 170 times Amplitude Response Function & Fourier x(t) = ∑ αk cos(2π k f o t) k y(t ) = ∑ A(k f o )ak cos(2 π k f o t + ϕ (k f o)) k Delay ϕ(f ) = −2π f τ cos(2π k f o t−2π f τ) = cos(2 π k f o(t −τ)) Finding the Frequency Response ● We could send one cosine after another through the channel and measure the response ● Or we could send the sum of all the cosines at once – It is linear time invariant after all ● This sum is called impulse or Dirac function or Delta function Impulse Response Function ● It is the response of the channel to the impulse function ● It is mathematically equivalent to the amplitude and phase responses of the channel – The one can be obtained from the other through a Fourier transform Example: Low-pass channel A(f )=1 −W ≤f ≤W ϕ(f )=2π f τ h(t)=s(t− τ) sin(2 π W t) s(t ) = 2π W t The Nyquist Signaling Rate ● Let p(t) be the response to a pulse that appears at the other side of the channel ● What is a good shape for p(t) so that we recover the original signal most easily ● Reduce the inter-symbol interference. ● What is the pulse rate under these conditions? Nyquist Signaling Rate The Nyquist Signaling Rate The Nyquist Signaling Rate ● The peak of a pulse coincides with the zeros of all the other pulses ● This allows us to pack the pulses closely ● Pulse rate is twice the bandwidth ● The sampling function is not implementable ● There are very good approximations Shannon Channel Capacity Without noise Shannon Channel Capacity With little noise Shannon Channel Capacity More bits Shannon Channel Capacity Too many bits Shannon Channel Capacity C = W log2(1+SNR) Line Coding ● We want to transmit bits over a channel ● The obvious is to send 5V to transmit 1 and 0V to transmit 0 ● Sounds good, but can we do better ● What are the problems with this? – Bandwidth – Clocking – Polarity Return to Zero ● A transition 0, 5, 0 Volt is an one ● A constant 0 Volt is a zero ● Self clocking is easy (as long as we do not have too many zeros in a row) Non-Return to Zero ● Consecutive 1s have no intermediate zero in between ● Fewer transitions than RZ DC component ● This simplest scheme is called Unipolar NRZ ● Biggest problem: – DC component – Wastes energy – Very low frequencies are tricky ● Most transmission lines cut-off DC component ●