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  • 09/03/12

Chapter2 Presentation Transcript

  • 1. ECE 4790ELECTRICAL COMMUNICATIONS Fall 99 Dr. Bijan Mobasseri ECE Dept. Villanova University
  • 2. Policies and procedures 3 hours of lecture per week(MWF) 2 hours of lab per week(Wed. in CEER 118) Homework assigned and graded weekly Lab is due the same day 2 tests and one final Grade break down • 15% each test • 25% final • 25% labs • 20% homeworkCopyright1999 by BG Mobasseri 2
  • 3. Lab work Hands-on lab work is an integrated part of the course There will be about 10 experiments done using MATLAB with signal processing and COMM toolboxes Experiments, to the extent possible, parallel theoretical material Professional MATALB code is expectedCopyright1999 by BG Mobasseri 3
  • 4. Going online Send a blank message to ece4790_f99-subscribe@egroups.com You will then have access to all class notes, labs etc. Notes/labs are in MS Office format You can also participate in the online discussion group and, if you wish, chat roomCopyright1999 by BG Mobasseri 4
  • 5. Ethical standards This course will be online and make full use of internet. This convenience brings with it many responsibilities • Keep electronic class notes/material private • Keep all passwords/accounts to yourself • Do not exchange MATLAB codeCopyright1999 by BG Mobasseri 5
  • 6. Necessary Background This course requires, at a minimum, the following body of knowledge • Signal processing • Probability • MATLABCopyright1999 by BG Mobasseri 6
  • 7. Course Outlook Introduction Pulse Shaping • signals, channels • “best” pulse shape • bandwidth • interference • signal represent. • equalization Analog Modulation Digital Modulations • AM and FM Modem standards Source coding Spread Spectrum • Sampling, PAM Wireless • PCM, DM, DPCMCopyright1999 by BG Mobasseri 7
  • 8. SIGANLS AND CHANNELS IN COMMUNICATIONS AN INTRODUCTION
  • 9. A Block Diagram Information source user source source decoder encoder channel channel decoder encoder modulator demodulator channelCopyright1999 by BG Mobasseri 9
  • 10. Information source The source can be analog, or digital to begin with • Voice • Audio • Video • DataCopyright1999 by BG Mobasseri 10
  • 11. Source encoder Source encoder converts analog information to a binary stream of 1’s and 0’s 1 0 0 1 1 0 0 1 ... Source encoder PCM, DM, DPCM, LPCCopyright1999 by BG Mobasseri 11
  • 12. Channel encoder The binary stream must be converted to real pulses polar 10110 channel encoder on-offCopyright1999 by BG Mobasseri 12
  • 13. Modulator Signals need to be “modulated” for effective transmission 1 01 1 0 ModulatorCopyright1999 by BG Mobasseri 13
  • 14. Channel Channel is the “medium” through which signals propagate. Examples are: • Copper • Coax • Optical fiber • wirelessCopyright1999 by BG Mobasseri 14
  • 15. Signals and Systems Review
  • 16. Periodic vs. Nonperiodic A periodic signal satisfies the condition The smallest value of To for which this condition is met is called a period of g(t) g(t ) = g(t + T0 ) periodCopyright1999 by BG Mobasseri 16
  • 17. Deterministic vs. random A deterministic signal is a signal about which there is no uncertainty with respect to its value at any given time • exp(-t) • cos(100t)Copyright1999 by BG Mobasseri 17
  • 18. Energy and Power Consider the following + i(t) V(t) R Instantaneous power is given by - 2 v (t ) 2 p( t ) = = Ri(t ) RCopyright1999 by BG Mobasseri 18
  • 19. Energy Working with normalized load, R=1Ω 2 2 2 p(t ) = v(t ) = i(t ) = g(t ) Energy is then defined as T 2 E = lim ∫ g(t ) dt −T T→∞Copyright1999 by BG Mobasseri 19
  • 20. Average Power The instantaneous power is a function of time. An overall measure of signal power is its average power T 2 1 T→∞ 2T ∫ P = lim g( t ) dt −TCopyright1999 by BG Mobasseri 20
  • 21. Energy and Power of a Sinusoid Take m (t ) = A cos (2πfc t ) • Find the energy T T A2 cos2 (2πfc t )dt ⇒ ∞ 2 E = lim T→∞ ∫ −T T →∞ ∫ 1 44 2 4 4 g(t ) dt = lim −T 3 always>0Copyright1999 by BG Mobasseri 21
  • 22. Instantaneous Power Instantaneous power AMPLITUDE= 1, FREQ.=2 1 2 2 p(t ) = A cos ( 2πfc t ) 0.8 0.6 A2 A2 0.4 = + cos( 4πfc t ) 0.2 2 2 0 -0.2 INSTANT. POWER ORIGINAL SIGNAL -0.4 -0.6 -0.8 -1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 TIME(SEC)Copyright1999 by BG Mobasseri 22
  • 23. Average Power The average power of a sinusoid is T 2 T 1 1 T →∞ 2T ∫ T→∞ 2T ∫ P = lim g(t ) dt = lim A2 cos 2 (2πfc t )dt −T −T from 2 cos 2 x = 1 + cos(2x ),then T 1 A2 P = lim T →∞ 2T ∫ 2 (1 + cos(4πfct ))dt −T T T 1 A2 1 A2 A2 T→ ∞ 2T ∫ 2 T →∞ 2T ∫ 2 = lim + lim cos( 4πfc t )dt = −T 4 4 4 2 4 4 4 3 2 −T 1 averages to zeroCopyright1999 by BG Mobasseri 23
  • 24. Average Transmitted Power What is the peak signal amplitude in order to transmit 50KW? Assume antenna impedance of 75Ω. • Note the change in Pavg for non-unit ohm load ⎛ A2 ⎞ Pavg = ⎜ ⎟ R ⇒ A = 2RPavg = 2 × 75 × 50,000 ⎝2⎠ ⇒ A = 2,738voltsCopyright1999 by BG Mobasseri 24
  • 25. Energy Signals vs. Power Signals Do all signals have valid energy and power levels? • What is the energy of a sinusoid? • What is the power of a square pulse? In the first case, the answer is inf. In the second case the answer is 0.Copyright1999 by BG Mobasseri 25
  • 26. Energy Signals A signal is classified as an energy signal if it meets the following 0<E<infinity Time-limited signals, such as a square pulse, are examples of energy signalsCopyright1999 by BG Mobasseri 26
  • 27. Power Signals A power signal must satisfy 0<P<infinity Examples of power signals are sinusoidal functionsCopyright1999 by BG Mobasseri 27
  • 28. Example:energy signal Square pulse has finite energy but zero average power A E = A2 T T 1 T 2 1 T 2 P = lim ∫ g(t ) dt = lim ∫ A dt ⇒ 0 T→∞ 2T −T T→∞ 2T −T 123 fixedCopyright1999 by BG Mobasseri 28
  • 29. Example:power signal A sinusoid has infinite energy but finite power A ∞ E= ∫ A2 cos 2 (2πfc t ) dt ⇒ ∞ −∞ but T 1 ∞ cos (2πfc t )dt ⇒ ⇒ finite T→∞ 2T ∫ 2 Pavg = lim −T ∞Copyright1999 by BG Mobasseri 29
  • 30. SUMUP Energy and power signals are mutually exclusive: • Energy signals have zero avg. power • Power signals have infinite energy • There are signals that are neither energy or power?. Can you think of one?Copyright1999 by BG Mobasseri 30
  • 31. DEFINING BANDWIDTH
  • 32. WHAT IS BANDWIDTH? In a nutshell, bandwidth is the “highest” frequency contained in a signal. We can identify at least 5 definitions for bandwidth • absolute • 3-dB • zero crossing • equivalent noise • RMSCopyright1999 by BG Mobasseri 32
  • 33. ABSOLUTE BANDWIDTH The highest frequency Spectrum -W W fCopyright1999 by BG Mobasseri 33
  • 34. 3-dB BANDWIDTH The frequency where frequency response drops to .707 of its peak Spectrum W fCopyright1999 by BG Mobasseri 34
  • 35. FIRST ZERO CROSSING BANDWIDTH The frequency where spectrum first goes to zero is called zero crossing bandwidth. 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 -4 -3 -2 -1 0 1 2 3 4Copyright1999 by BG Mobasseri 35
  • 36. EQUIVALENT NOISE BANDWIDTH Bandwidth which contains the same power as an equivalent bandlimited white noise 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0Copyright1999 by BG Mobasseri 36 -4 -3 -2 -1 0 1 2 3 4
  • 37. RMS BANDWIDTH RMS bandwidth is related to the second moment of the amplitude spectrum 1 ∞ 2 ⎡∫−∞ f G( f ) ⎤ 2 2 Wrms = ⎢ ∞ 2 ⎥ ⎣ ∫−∞ G( f ) ⎦ This measures the tightness of the spectrum around its meanCopyright1999 by BG Mobasseri 37
  • 38. RMS BANDWIDTH OF A SQUARE PULSE Take a square pulse of duration 0.01 sec. Its spectrum is a sinc SQUARE PULSE SPECTRUM (WIDTH=0.01 SEC.) 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0Copyright1999 by BG Mobasseri 38 0 100 200 300 400 500 600 700 800 FREQUENCY(Hz)
  • 39. RMS BANDWIDTH The RMS bandwidth can be numerically computed using the following MATLAB code W rms= 35.34 HzCopyright1999 by BG Mobasseri 39
  • 40. Bandwidth of Real Signals This is the spectrum a 3 sec. clip sampled at 8KHz 0 -10 -20 -30 -40 -50 -60 -70 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 FrequencyCopyright1999 by BG Mobasseri 40
  • 41. Gate Function Gate function is one of the most versatile pulse shapes in comm. It is pulse of amplitude A and width T A -T/2 T/2Copyright1999 by BG Mobasseri 41
  • 42. rect function Gate function is defined based on a function called rect ⎧ 1 1 1 − <t< rect(t) = ⎨ 2 2 ⎪ ⎩0 otherwiseCopyright1999 by BG Mobasseri 42
  • 43. Expression for gate Based on rect we can write t g(t ) = Arect ⎛ ⎞ ⎝T ⎠ Note that for -T/2<t<T/2, the argument of rect is inside {-1/2,1/2} therefore rect is 1 and g(t)=ACopyright1999 by BG Mobasseri 43
  • 44. Generalizing gate Let’ say we want a pulse with amplitude A centered at t=to and width T T A t=toCopyright1999 by BG Mobasseri 44
  • 45. Arriving at an Expression What we have a is a rect function shifted to the right by to Shift to the right of f(t) by to is written by f(t-to) Therefore ⎛t − to ⎞ g(t ) = Arect ⎝ T ⎠Copyright1999 by BG Mobasseri 45
  • 46. gate function in the Fourier Domain The Fourier transform of a gate function is a sinc as follows sinc(t) 1 0.8 g(t ) = Arect ⎛t ⎞ ⎝T ⎠ 0.6 0.4 G( f ) = ATsinc( fT ) 0.2 Zero crossing 0 -0.2 -0.4 -3 -2 -1 0 1 2 3 TIME(t)Copyright1999 by BG Mobasseri 46
  • 47. Zero Crossing Zero crossings of a sinc is very significant. ZC occurs at integer values of sinc argument ⎧ x =0 1 sin c(x) = ⎨ ⎩ x = ±1,±2,L 0Copyright1999 by BG Mobasseri 47
  • 48. Some Numbers What is the frequency content of a 1 msec. square pulse of amplitude .5v? • We have A=0.5 and T=1ms g(t ) = 0.5rect(1000t) G( f ) = 5 × 10 −4 sinc(10 −3 f ) • First zero crossing at f=1000Hz obtained by setting 10^-3f=1Copyright1999 by BG Mobasseri 48
  • 49. RF Pulse RF(radio frequency) pulse is at the heart of all digital communication systems. RF pulse is a short burst of energy, expressed by a sinusoidal functionCopyright1999 by BG Mobasseri 49
  • 50. Modeling RF Pulse An RF pulse is a cosine wave that is truncated on both sides This effect can be modeled by “gating”the cosine waveCopyright1999 by BG Mobasseri 50
  • 51. Mathematically Speaking Call the RF pulse g(t), then t g(t ) = Arect ⎛ ⎞ cos(2πfc t ) ⎝T ⎠ This is in effect the modulated version of the original gate functionCopyright1999 by BG Mobasseri 51
  • 52. Spectrum of the RF Pulse:basic rule We resort to the following g1 (t )g2 (t ) ⇔ G1 ( f ) * G2 ( f ) Meaning, the Fourier transform of the product is the convolution of individual transformsCopyright1999 by BG Mobasseri 52
  • 53. RF Pulse Spectrum: Result We now have to identify each term AT gate → g1 (t ) ⇔ sin c(Tf ) 2 cosine → g 2 (t ) ⇔ [δ ( f − fc ) + δ ( f + fc )] Then, the RF pulse spectrum, G(f) AT G( f ) = ⎛ sin c(Tf )⎞ * [δ ( f − fc ) + δ ( f + fc )] ⎝ 2 ⎠ AT AT = sin c(T ( f − fc )) + sin c(T ( f + fc )) 2 2Copyright1999 by BG Mobasseri 53
  • 54. Interpretation The spectrum of the RF pulse are two sincs, one at f=- fc and the other at f=+fc RF PULSE SPECTRUM:two sincs at pulse freq -4 0 4 FREQUENCY (HZ)Copyright1999 by BG Mobasseri 54
  • 55. Actual Spectrum RF PULSE SPECTRUM, fc=1200Hz, duration= 5 msec 50 45 40 35 Bandwidth=400Hz 30 25 20 5 msec 15 10 5 0 0 400 800 1200 1600 2000 2400 FREQUENCY (HZ)Copyright1999 by BG Mobasseri 55
  • 56. Baseband and Bandpass Signals and Channels
  • 57. Definitions:Baseband The raw message signal is referred to as baseband, or low freq. signal Spectrum of a baseband audio signal 0.045 0.04 0.035 0.03 0.025 0.02 0.015 0.01 0.005 0 0 500 1000 1500 2000 2500 3000 3500 4000 FREQUENCY(HZ)Copyright1999 by BG Mobasseri 57
  • 58. Definitions:Bandpass When a baseband signal m(t) is modulated, we get a bandpass signal The bandpass signal is formed by the following operation(modulation) m (t )cos( 2πfc t )Copyright1999 by BG Mobasseri 58
  • 59. Bandpass Example:AM 3 BANDPASS BASEBAND 2m (t )cos( 2πfc t ) 1 0 -1 -2 -3 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2Copyright1999 by BG Mobasseri 59
  • 60. Digital Bandpass Baseband and bandpass concepts apply equally well to digital signals 1 1 1 1 0 RF pulsesCopyright1999 by BG Mobasseri 60
  • 61. Baseband vs. Bandpass Spectrum Creating a bandpass signal is the same as modulation process. We have the following 1 m (t )cos( 2πfc t ) ⇔ [ M( f − fc ) + M ( f + fc )] 2 InterpretationCopyright1999 by BG Mobasseri 61
  • 62. Showing the Contrast baseband frequencyBandwidth doubles bandpass bandwidthCopyright1999 by BG Mobasseri 62
  • 63. SIGNAL REPRESENTATION How to write an expression for signals
  • 64. Introduction We need a formalism to follow a signal as it propagates through a channel. To this end, we have to learn a few concepts including Hilbert Transform, analytic signals and complex envelopeCopyright1999 by BG Mobasseri 64
  • 65. Hilbert Transform Hilbert transform is an operation that affects the phase of a signal +90 H(f) f -90 |H(f)|=1 Phase responseCopyright1999 by BG Mobasseri 65
  • 66. More precisely ⎧ − jπ ⎪1e 2 f ≥ 0 H( f ) = ⎨ π ⎪ 2j ⎩1e f < 0 Equivalently H ( f ) = − jsgn( f )sgn( f ), or signum function, extracts the sign of its argument ⎧f >0 ⎪ 1sgn( f ) = ⎨ f = 0 0 ⎪ f <0 ⎩−1Copyright1999 by BG Mobasseri 66
  • 67. HT notation The HT of g(t) is denoted by HT ( g(t )) = g(t ) ˆ In the frequency domain, ˆ G( f ) = − j sgn( f )G( f )Copyright1999 by BG Mobasseri 67
  • 68. Find HT of a Sinusoid Q: what is the HT of cosine?Ans:sine g(t ) = cos(2πfc t ) we know ⎡ ⎛ ⎞⎤ Gˆ ( f ) = − j sgn( f )G( f ) = − j sgn( f )⎢1 ⎜δ ( f − fc ) + δ ( f + fc )⎟⎥ ⎢2 ⎜14 2 43 14 2 43 ⎟⎥ ⎣ ⎝ pos. freq neg. freq ⎠⎦ j j 1 = − δ ( f − fc ) + δ ( f + fc ) = [δ ( f − fc ) − δ ( f + fc )] 2 2 2j 1 4 4 4 4 2 4 4 4 43 Fourier transform of sineCopyright1999 by BG Mobasseri 68
  • 69. HT Properties Property 1 • g and HT(g) have the same amplitude spectrum Property 2 Property 3 HT ( g(t )) = −g(t ) ˆ • g and HT(g) are orthogonal, i.e. ∫ g(t )g(t )dt = 0 ˆCopyright1999 by BG Mobasseri 69
  • 70. Using HT: Pre-envelope From a real-valued signal, we can extract a complex-valued signal by adding its HT as follows g+ (t ) = g(t ) + jg(t ) ˆ g+(t) is called the pre-envelope of g(t)Copyright1999 by BG Mobasseri 70
  • 71. Question is Why? It turns out that it is easier to work with g+ (t) than g(t) in many comm. situations We can always go back to g(t) g(t ) = Re { g+ (t )} where Re stands for " real part of"Copyright1999 by BG Mobasseri 71
  • 72. Pre-envelope Example Find the pre-envelope of the RF pulse g(t ) = m (t )cos( 2πfc t + θ ) We can re-write g(t) as follows j (2πfc t +θ ) g(t ) = Re{ m (t )e } because j (2πfc t +θ ) e = cos( 2πfct + θ ) + j sin( 2πfc t + θ )Copyright1999 by BG Mobasseri 72
  • 73. Pre-envelope is... Compare the following two j (2πfc t +θ ) g(t ) = Re{ m (t )e } g(t ) = Re{ g+ ( t )} g+ ( t ) = m(t )e j ( 2πfc t +θ ) Pre-envelope ofis g(t ) = m (t )cos( 2πfc t + θ ) j ( 2πfc t +θ ) g+ ( t ) = m(t )eCopyright1999 by BG Mobasseri 73
  • 74. Pre-envelope in the Frequency Domain How does pre-envelope look in the frequency domain? We know g+ (t ) = g(t ) + jg( t ). Fourier transform is ˆ F{ g+ (t )} = G+ ( f ) = G( f ) + j[ − j sgn( f )]G( f )Copyright1999 by BG Mobasseri 74
  • 75. Pre-envelope in positive and negative frequencies Let’s evaluate G+(f) for f>0 G(f)f >0G+ ( f ) = G( f ) + G( f ) = 2G( f ) ff <0 G+(f)G+ ( f ) = G( f ) − G( f ) = 0 fCopyright1999 by BG Mobasseri 75
  • 76. Interpretation Fourier transform of Pre-envelope exists only for positive frequencies As such per-envelope is not a real signal. It is complex as shown by its definition g+ (t ) = g(t ) + jg(t ) ˆCopyright1999 by BG Mobasseri 76
  • 77. Corollary To find the pre-envelope in the frequency domain, take the original spectrum and chop off the negative partCopyright1999 by BG Mobasseri 77
  • 78. Example Find the pre-envelope of a modulated message g(t ) = m (t )cos( 2πfc t + θ ) G(f) G+(f) AM signalCopyright1999 by BG Mobasseri 78
  • 79. Another Definition for Pre-envelope Pre-envelope is such a quantity that if you take its real part, it will give you back your original signal g(t ) = m (t )cos( 2πfc t + θ ) original signal j ( 2πfc t +θ ) g(t ) = Re{ m (t )e j (2πft +θ ) g+ ( t ) = m(t )e } g(t ) = Re{ g+ ( t )}Copyright1999 by BG Mobasseri 79
  • 80. Bringing Signals Down to Earth Communication signals of interest are mostly high in frequency Simulation and handling of such signals are very difficult and expensive Solution: Work with their low-pass equivalentCopyright1999 by BG Mobasseri 80
  • 81. Tale of Two Pulses Consider the following two pulses Which one carries more “information”?Copyright1999 by BG Mobasseri 81
  • 82. Lowpass Equivalent Concept The RF pulse has no more information content than the square pulse. They are both sending one bit of information. Which one is easier to work with?Copyright1999 by BG Mobasseri 82
  • 83. Implementation issues It takes far more samples to simulate a bandpass signal 0.01 sec. Sampling rate=200Hz 4cycles/0.01 sec Sampling rate=800Hz -->fc=400HzCopyright1999 by BG Mobasseri 83
  • 84. Complex Envelope Every bandpass signal has a lowpass equivalent or complex envelope Take and re write as g(t ) = m (t )cos( 2πfc t + θ ) ⎧ ⎫ j (2πfc t +θ ) ⎪ jθ j 2πfc t ⎪ g(t ) = Re{ m (t )e } = Re ⎨ m (2)e3 1 t e ⎬ ⎪ lowpass ⎪ ⎩ complex envelop or ⎭Copyright1999 by BG Mobasseri 84
  • 85. Complex Envelope: The Quick Way Rewrite the signal per following model j (2πfc t +θ ) g(t ) = Re { m (t )e } = Re { m(t )e jθ e j 2πfc t } The term in front of j2πf t is the complex envelope shown by e c jθ g( t ) = m(t )e ˜ m and theta contain all the informationCopyright1999 by BG Mobasseri 85
  • 86. Signal Representation Summary Take a real-valued, baseband signal G(f) g(t)Copyright1999 by BG Mobasseri 86
  • 87. Pre-envelope Summary: baseband G(f)g+ (t ) = g(t) + jg(t) ˆg(t) = HT { g(t)}ˆ G+(f)g(t) = Re{ g+ (t)} Nothing for f<0Copyright1999 by BG Mobasseri 87
  • 88. Pre-envelope Summary: bandpass Baseband signal: g(t ) = m (t )cos( 2πfc t + θ ) G(f) G+(f)Copyright1999 by BG Mobasseri 88
  • 89. Complex Envelope Summary Complex/pre envelope are related − j 2πfc t G+(f)g( t ) = g+ (t)e˜or ˆ G( f ) j 2πfc tg+ (t) = g(t )e ˜Copyright1999 by BG Mobasseri 89
  • 90. RF Pulse: Complex Envelope Find the complex envelope of a T second long RF pulse at frequency fc t g(t ) = Arect ⎛ ⎞ cos(2πfc t ) ⎝T ⎠Copyright1999 by BG Mobasseri 90
  • 91. Writing as Re{ } Rewrite g(t) as follows ⎧Arect ⎛ t ⎞e j 2πfct ⎫ g(t ) = Re ⎨ ⎝T ⎠ ⎬ ⎩ ⎭ compare g(t ) = Re{ g (t)e j2πfct } ˜ Comp.Env=just a square pulse then t g (t) = complex _ envelope = Arect ⎛ ⎞ ˜ ⎝T ⎠Copyright1999 by BG Mobasseri 91
  • 92. RF Pulse Pre-envelope Recall − j 2πfc t Then g( t ) = g+ (t)e ˜ g+ (t ) = Arect ⎛ t ⎞e j 2πf ct ⎝T ⎠Copyright1999 by BG Mobasseri 92
  • 93. Story in the Freq. Domain RF PULSE SPECTRUM:two sincs at pulse freq RF PULSE SPECTRUM:two sincs at pulse freq -4 0 4 -4 0 4 FREQUENCY (HZ) FREQUENCY (HZ) Original RF pulse spectrum Pre-env. Spectrum (only f>0 portion)Copyright1999 by BG Mobasseri 93
  • 94. Complex Envelope Spectrum Complex envelope=low pass portion 1 0.8 0.6 0.4 0.2 0 -0.2 -0.4 -10 -8 -6 -4 -2 0 2 4 6 8 10Copyright1999 by BG Mobasseri 94
  • 95. CHANNELS AND SIGNAL DISTORTIONSome of the material not in the book
  • 96. Signal Transmission Modeling One of the most common tasks in communications is transmission of RF pulses through bandpass channels Instead of working at high RF frequencies at great computational cost, it is best to work with complex envelope representationsCopyright1999 by BG Mobasseri 96
  • 97. Channel I/O To determine channel output, we can work with complex envelopes ˜ 1 ˜ ˜ Y ( f ) = H ( f ) X( f ) 2 where ˜ X( f ) : C.E. of input(transmitted) signal ˜ H ( f ) : C.E. of channel transfer function ˜ Y ( f ) : C.E. of output(received) signalCopyright1999 by BG Mobasseri 97
  • 98. Passing an RF Pulse through a Bandpass Channel Here is the problem: what is the output of an ideal bandpass channel in response to an RF pulse? H(f) use ˜( f ) = 1 X ( f ) H( f ) Y ˜ ˜ 2Copyright1999 by BG Mobasseri 98
  • 99. What is the Complex envelope of H(f)? It is the lowpass equivalent of H(f) 2 ˜ ( f ) = 2rect ⎛ f ⎞ H 1 ⎝2B⎠ 2B B H(f)Copyright1999 by BG Mobasseri 99
  • 100. What is the Complex Envelope of the RF Pulse? We found this before ⎧Arect ⎛ t ⎞e j 2πfct ⎫ g(t ) = Re ⎨ ⎝T ⎠ ⎬ ⎩ ⎭ compare g(t ) = Re{ g (t)e j2πfct } ˜ then t g (t) = complex _ envelope = Arect ⎛ ⎞ ˜ ⎝T ⎠Copyright1999 by BG Mobasseri 100
  • 101. Channel Output Here is what we have f H ( f ) = 2rect ⎛ ⎞ ˜ • Channel complex envelope ⎝2B⎠ • Input complex envelope B=bandwidth t x (t ) = Arect ⎛ ⎞ ⇔ X ( f ) = AT sin c( fT ) ˜ ˜ • Output ⎝T ⎠ 1 f Y ( f ) = [ AT sin c( fT)] ⎡2rect( )⎤ ˜ 2 ⎣ 2B ⎦Copyright1999 by BG Mobasseri 101
  • 102. Interpretation ˜ f 1 Y ( f ) = AT sin c( fT)rect( ) 2B 0.8 0.6For the pulse to get through 0.4unscathed, channel bandwidth 0.2must be larger than pulse bw 0 -0.2B>=1/T=bit rate -0.4 -10 -8 -6 -4 -2 0 2 4 6 8 10 1/T BCopyright1999 by BG Mobasseri 102
  • 103. What Does Distortion Do? Channel Distortion creates pulse “dispersion” Channel interferenceCopyright1999 by BG Mobasseri 103
  • 104. Case of No Distortion There are two “distortions” we can live with • Scaling • Delay ToCopyright1999 by BG Mobasseri 104
  • 105. Modeling Distortion-free Channels The input-output relationship for a distortion-free channel is y(t)=Ax(t-Td) • x(t):input • y(t)=output • A:scale factor • Td: delayCopyright1999 by BG Mobasseri 105
  • 106. Response of a Distortion-free channel What is channel’s frequency response? Take FT of the I/O expression Y ( f ) = AX ( f )e − j2πfTd Then Y( f ) H( f ) = = Ae − j2πfTd X( f )Copyright1999 by BG Mobasseri 106
  • 107. Amplitude and Phase Response |H(f)| Const amplitude response f /_H(f) Linear phase response f −(2πTd ) fCopyright1999 by BG Mobasseri 107
  • 108. Complete Model The complete transfer function is H ( f ) = Ae − j 2πfTd Since this is a lowpass function, its complex envelope is the same as H(f) H ( f ) = Ae − j 2πfTd ˜Copyright1999 by BG Mobasseri 108
  • 109. Lowpass Channel Is a first order filter an appropriate model for a distortion-free channel? To answer this question we have to test the definition of the ideal channel R CCopyright1999 by BG Mobasseri 109
  • 110. Amplitude and Phase Responseamplitude _ response = aH( f ) = 2 a + (2πf ) 3-dB bandwidth=phase _ response = a/2pi=1/(2piRC) −1 ⎛ 2πf ⎞θh ( f ) = −tan ⎝ a ⎠ 1a= RCCopyright1999 by BG Mobasseri 110
  • 111. Response for RC=10^-3 AMPLITUDE RESPONSE PHASE RESPONSE 1.5 1 0.50.7 0 -0.5 -1 0 159 -1.5 0 159 FREQUENCY(HZ) FREQUENCY(HZ) bandwidth=159 Hz Copyright1999 by BG Mobasseri 111
  • 112. An “ideal” Channel? We must have constant amplitude response and linear phase response. Do we?. Deviation of H(f) from the ideal is tolerated up to .707form the peak. The frequency at which this occurs is the 3dB bandwidth No signal distortion if input frequencies are kept below 3dB bandwidth or 159 Hz hereCopyright1999 by BG Mobasseri 112
  • 113. Linear Distortion If any of the ideal channel conditions are violated but we are still dealing with a linear channel, we have linear distortion amplitude H ( f ) = { (1 + k cos (2πfT ))e− j2πftd f phaseCopyright1999 by BG Mobasseri 113
  • 114. Pulse Dispersion Putting a pulse g(t) through this filter produces 3 overlapping copies channel with T distortion >TCopyright1999 by BG Mobasseri 114
  • 115. Why? Let g(t) and r(t) be the transmitted and received signals. Then R( f ) = G( f )H( f ) = G( f )[1 + kG( f )cos(2πfT )]e− j2πftd = G( f )e− j2πftd + kG( f )cos(2πfT)e− j2πftd Taking the inverse FT k r(t) = g(t − td ) + [ g(t − td − T ) + g(t − td + T )] 2Copyright1999 by BG Mobasseri 115
  • 116. Nonlinear Distortion This is the most serious kind where input and output are related by a nonlinear equation Nonlinear g channel r r r=g^2 gCopyright1999 by BG Mobasseri 116
  • 117. Impact of Nonlinear Dist. Nonlinear channels generate new frequencies at the output that did not exist in the input signal. Why? G(f) if f W r(t ) = g 2 (t ) R(f) then R( f ) = G( f ) * G( f ) 2W fCopyright1999 by BG Mobasseri 117
  • 118. Practice Problems For pre-envelope: 2.23 For filtering using complex envelope: 2.32Copyright1999 by BG Mobasseri 118