2. Fundamental of Transmissions Dr. Farahmand Updated: 2/9/2009

3. What is telecommunications?  Conveying information between two points or between one and multi-points  Transmitting information wirelessly is achieved via electromagnetic signals (E) Electric current flowing through a wire creates magnetic field around the wire An alternating electric current flowing through a wire creates electromagnetic waves Electromagnetic radiation is waves of energy These waves collectively called electromagnetic spectrum RX TX Medium

4. Signal Characteristics  Analog (continuous) or digital (discrete)  Periodic or aperiodic  Components of a periodic electromagnet wave signal Amplitude (maximum signal strength) – e.g., in V Frequency (rate at which the a periodic signal repeats itself) – expressed in Hz Phase (measure of relative position in time within a single period) – in deg or radian (2 = 360 = 1 period)

5. Sine Waves

6. Sound Wave Examples Each signal is represented by x(t) = sin (2  f.t) A dual tone signal with f1 and f2 is represented by x(t) = sin (2  f1.t) + sin (2  f2.t) f = 5Kz f = 1Kz

7. Periodic Signal Characteristics  The simplest signal is a sinusoidal wave  A sine wave can be expressed in time or space (wavelength) Wavelength is the distance the signal travels over a single cycle Wavelength is a function of speed and depends on the medium (signal velocity) Exact speed light through vacuum is 299,792,458 m/s f=v

8. Periodic Signal Characteristics  A signal can be made of many frequencies All frequencies are multiple integer of the fundamental frequency Spectrum of a signal identifies the range of frequencies the signal contains Absolute bandwidth is defined as: Highest_Freq – Lowest_Freq Bandwidth in general is defined as the frequency ranges where a signal has its most of energies  Signal data rate Information carrying capacity of a signal Expressed in bits per second (bps) Typically, the larger frequency larger  data rate Example 

9. Periodic Signal Characteristics  Consider the following signal Consists of two freq. component (f) and (3f) with BW = 2f f 3f BW What is the Max amplitude of this component? http://www.jhu.edu/~signals/listen-new/listen-newindex.htm Second harmonic

10. Periodic Signal Characteristics S(t)=sin(2ft) S(t)=1/3[sin(2f)t)] S(t)= 4/{sin(2pft) +1/3[sin(2f)t)]}

11. Frequency Domain Representation S(t)= 4/{sin(2ft) +1/3[sin(2f)t)]} frequency domain function for a single square pulse that has the value 1{s(t)=1} between –X/2 and X/2, and is 0 {s(t)=1} elsewhere Refer to NOTES!

12. Data Rate & Frequency  Example: What is f1? What is f2? Which case has larger data rate? (sending more bits per unit of time)  f1 = 2(1/10^-3)=2KHz  Case I data rate=one bit per (0.25msec)  4 Kbps  f2 = 1 KHz  data rate=2Kbps  Case 1 has higher data rate (bps)

13. Bandwidth and Data Rate  Case 1: Assume a signal has the following components: f, 3f, 5f ; f=10^6 cycles/sec What is the BW? What is the period? How often can we send a bit? What is the data rate? Express the signal equation in time domain  Case 2: Assume a signal has the following components: f, 3f, 5f; f=2x10^6 cycles/sec What is the BW? What is the period? How often can we send a bit? What is the data rate?  Case 3: Assume a signal has the following components: f, 3f ; f=2x10^6 cycles/sec What is the BW? What is the period? How often can we send a bit? What is the data rate? Express the signal equation in time domain BW=4MHz T=1usec 1 bit every 0.5usec Data rate=2*f=2bit/usec=2MHz BW=8MHz (5 x 2 - 2=8) T= ½ usec 1 bit every 0.25usec Data rate=2*f=2bit/0.5 usec=4MHz BW=4MHz T=0.5 usec 1 bit every 0.25usec Data rate=2*f=4bit/usec=4MHz Remember: Greater BW  larger cost but Lower BW  more distortion;

14. Nyquist Formula and Bandwidth  Assuming noise free system and assuming that only one bit is provided to represent the signal:  Nyquist’s formula states the limitation of the data rate due to the bandwidth: If the signal transmission rate is 2B, then a signal with frequency of less or equal B is required to carry this signal: TR(f)=2B  f  B If bandwidth is B (Hz)  the highest signal rate that can be carried is 2B (bps): f=B  TR(f)  B  Example: if the highest frequency is 4KHz (bandwidth) a sampling rate of 8 Kbps is required to carry the signal Remember: Channel Capacity = (number of bit) x (signal bandwidth)

15. Channel Capacity Nyquist’s formulation when multilevel signaling is present  channel capacity (C) is the tightest upper bound on the amount of information that can be reliably transmitted over a communications channel (max. allowable data rate)  What if the number of signal levels are more than 2 ( we use more than a single bit to represent the sate of the signal)?  C = Maximum theoretical Channel Capacity in bps  M = number of discrete signals (symbols) or voltage levels  n = number of bits per symbol Remember: More bits per symbol  more complexity! Example: Log2(8)=ln(8)/ln(2)=3

16. Channel Capacity Example:  Voice has a BW of 3100 Hz. calculate the channel capacity Assuming we use 2 signal levels Assuming we use 8 signal levels  channel capacity required to pass a voice signal:  Channel capacity (or Nyquist capacity) is 2 x 3100 cycles/sec = 6.1Kbps – note in this case one bit is being used to represent two distinct signal levels.  If we use 8 signal levels: channel capacity: 2x3100x3=18600 bps

17. S/N Ratio  The signal and noise powers S and N are measured in watts or volts^2, so the signal- to-noise ratio here is expressed as a power ratio, not in decibels (dB) Example: Assume signal strength is 2 dBm and noise strength is 5 mW. Calculate the SNR in dB. 2dBm  1.59 mW SNR = 10log(1.59/5)=-5dB

18. Signal Impairments Attenuation  Strength of a signal falls off with distance over transmission medium  Attenuation factors for guided media: Received signal must have sufficient strength so that circuitry in the receiver can interpret the signal Signal must maintain a level sufficiently higher than noise to be received without error Typically signal strength is reduced exponentially Expressed in dB Attenuation is greater at higher frequencies, causing distortion

19. Signal Impairments Attenuation Impacts  Lowers signal strength  Requires higher SNR  Can change as a function of frequency More of a problem in analog signal (less in digital) Higher frequencies attenuate faster Using equalization can improve – higher frequencies have stronger strength

20. Signal Impairments Delay Distortion  In bandlimited signals propagation velocity is different for different frequencies Highest near the center frequency Hence, bits arrive out of sequence  resulting in intersymbol interference  limiting the maximum bit rate!

21. Categories of Noise  Thermal Noise  Intermodulation noise  Crosstalk  Impulse Noise

22. Thermal Noise  Thermal noise due to agitation of electrons  Present in all electronic devices and transmission media  Cannot be eliminated  Function of temperature  Particularly significant for satellite communication When the signal is received it is very weak

23. Thermal Noise  Amount of thermal noise to be found in a bandwidth of 1Hz in any device or conductor is:  N 0 = noise power density in watts per 1 Hz of bandwidth  k = Boltzmann's constant = 1.3803 10 -23 J/K  T = temperature, in Kelvins (absolute temperature) – zero deg. C is 273.15  Expressed in dBW 10log(No/1W)

24. Thermal Noise  Noise is assumed to be independent of frequency  Thermal noise present in a bandwidth of B Hertz (in watts): or, in decibel-watts 

25. Thermal Noise (MATLAB Example) %MATLAB CODE: T= 10:1:1000; k= 1.3803*10^-23; B=10^6; No=k*T; N=k*T*B; N_in_dB=10*log10(N); semilogy(T,N_in_dB) title(‘Impact of temperature in generating thermal noise in dB’) xlabel(‘Temperature in Kelvin’) ylabel(‘Thermal Noise in dB’)

26. Other Types of Noise  Intermodulation noise – occurs if signals with different frequencies share the same medium Interference caused by a signal produced at a frequency that is the sum or difference of original frequencies  Crosstalk – unwanted coupling between signal paths  Impulse noise – irregular pulses or noise spikes Short duration and of relatively high amplitude Caused by external electromagnetic disturbances, or faults and flaws in the communications system Question: Assume the impulse noise is 10 msec. How many bits of DATA are corrupted if we are using a Modem operating at 64 Kbps with 1 Stop bit?

27. Other Types of Noise - Example Intermodulation noise Crosstalk Impulse noise

28. Channel Capacity with Noise and Error  An application of the channel capacity concept to an additive white Gaussian noise channel with B Hz bandwidth and signal-to-noise ratio S/N is the Shannon–Hartley theorem:  Establishing a relation between error rate, noise, signal strength, and BW  If the signal strength or BW increases, in the presence of noise, we can increase the channel capacity  Establishes the upper bound on achievable data rate (theoretical) Does not take into account impulse and attenuation Note: S/N is not in dB and it is log base 2!

29. Noise Impact on Channel Capacity  Presence of noise can corrupt the signal  Unwanted noise can cause more damage to signals at higher rate  For a given noise level, greater signal strength improves the ability to send signal Higher signal strength increases system nonlinearity  more intermodulation noise Also wider BW  more thermal noise into the system  increasing B can result in lower SNR

30. Example of Nyquist Formula and Shannon–Hartley Theorem  What is the wavelength associated with the highest energy level?  Calculate the BW of this signal.  Assuming the SNR = 24 dB, Calculate the maximum channel capacity.  Using the value of the channel capacity, calculate how many signal levels are required to generate this signal?  How many bits are required to send each signal level?  Express the mathematical expression of this signal in time domain.  What type of signal, more likely, is this? (TV, Visible light, AM, Microwave) – Next slide 3MHz 4MHz =10 8 m B=4-3=1 MHz SNR dB (24)  log -1 (24/10) 10 2.4 = 251 C=Blog 2 (1+S/N)=8Mbps C=2Blog 2 M  M=16 2 n =M  n=4 Signal Type: AM /4 /3x4 http://www.adec.edu/tag/spectrum.html

31. Radio-frequency spectrum: commercially exploited bands http://www.britannica.com/EBchecked/topic-art/585825/3697/Commercially-exploited-bands-of-the-radio-frequency-spectrum

32. Expression E b /N 0  Ratio of signal energy per bit to noise power density per Hertz R = 1/Tb; R = bit rate; Tb = time required to send one bit; S = Signal Power (1W = 1J/sec) Eb=S.Tb No = Thermal noise (W/Hz)  The bit error rate for digital data is a function of E b /N 0 Given a value for E b /N 0 to achieve a desired error rate, parameters of this formula can be selected As bit rate R increases, transmitted signal power must increase to maintain required E b /N 0 Note that as R increases power must increase as well to maintain signal quality

33. SNR & Expression E b /N 0  Using Thermal noise within the bandwidth of B Hertz (in watts): N=NoxB  Using Shannon’s Theorem – Channel Capacity in the presence of noise  The relation between SNR and Eb/No will be (R=C=Data rate)  C/B expressed in bps/Hz and called Spectral Density Q: What will be Eb/No if the spectral density is 6 bps/Hz???

34. Probability of Error Question: Assume we require Eb/No = 8.4 dB to achieve bit error rate of 10^-4. Assume temperature is 17oC and data rate is set to 2.4 Kbps. Calculate the required level of the received signal in W and dBW. 8.4 dB 10^-4

35. Probability of Error Question: Assume we require Eb/No = 8.4 dB to achieve bit error rate of 10^-4. Assume temperature is 17oC and data rate is set to 2.4 Kbps. Calculate the required level of the received signal in W and dBW. 8.4 dB 10^-4 8.4 dB  6.91 17oC  290oKelvin R=2400 bps K=1.38*10^-23 Eb/No=S/(KTR)  S=-161 dBW

36. Review: Power in Telecommunication Systems  Remember:  Example 1: if P2=2mW and P1 = 1mW  10log 10 (P2/P1)=3.01 dB  Example 2: if P2=1KW and P1=10W  20dB  What if dB is given and you must find P2/P1? P2/P1 = Antilog(dB/10) = 10 dB/10.  Example 3: if dB is +10 what is P2/P1? P2/P1 = Antilog(+10/10) = 10 +10/10 = 10

37. Colors and Wavelengths Color Red Orange Yellow Green Blue Violet Wavelength (nm) 780 - 622 622 - 597 597 - 577 577 - 492 492 - 455 455 - 390 Frequency (THz) 384 - 482 482 - 503 503 - 520 520 - 610 610 - 659 659 - 769 1 terahertz (THz) = 10^3 GHz = 10^6 MHz = 10^12 Hz, 1 nm = 10^-3 um = 10^-6 mm = 10^-9 m. The white light is a mixture of the colors of the visible spectra. Wavelengths is a common way of describing light waves. Wavelength = Speed of light in vacuum / Frequency. f=v

38. Colors and Wavelengths

40. References  Online calculator: http://www.std.com/~reinhold/BigNumCalc.html http://www.std.com/~reinhold/BigNumCalc.html  Wavelengths and lights http://www.usbyte.com/common/approximate _wavelength.htmhttp://www.usbyte.com/common/approximate _wavelength.htm & http://eosweb.larc.nasa.gov/EDDOCS/Wav elengths_for_Colors.html http://eosweb.larc.nasa.gov/EDDOCS/Wav elengths_for_Colors.html  Learn about decibel http://www.phys.unsw.edu.au/jw/dB.html http://www.phys.unsw.edu.au/jw/dB.html