Principles of Communications ECS 332 Asst. Prof. Dr. Prapun Suksompong prapun@siit.tu.ac.th 5. Angle Modulation Office Hours: BKD, 6th floor of Sirindhralai building Wednesday 4:3-5:3 Friday 4:3-5:3
Example 5.5: FM vs. PM Figure 32 m t Carrier: Acos 2πf t φ x PM t x FM t 2
x PM t Acos 2πf t φ k m t Phase Modulation Figure 32 m t =, t t, t t After time t, the message jumps to the value. same Acos 2πf t φ skipped Acos 2πf t φ 9 Acos 2πf t φ Here the phase modulator uses k 9 x PM t = Acos 2πf t φ, Acos 2πf t φ 9, t t t t After time t, the phase is skipped ahead (advanced) by 9. x FM t 3
f t f k m t Frequency Modulation Figure 32 Higher value m t Acos 2πf t φ When the message m t is, the x FM t has the same frequency as the carrier signal. When the message m t is, the frequency of x FM t is higher than the frequency of the carrier signal. x PM t x FM t 4 Higher frequency
Example 5.6: FM vs. PM Figure 33 m t Acos 2πf t φ x PM t x FM t 5
f t f k m t Frequency Modulation Figure 33 m t Acos 2πf t φ x PM t It should be evident that the frequency is changing. x FM t 6
f t f k m t Frequency Modulation Figure 33 m t The time at which m t is at its maximum value Acos 2πf t φ corresponds to the time at which x FM t has maximum x PM t frequency. x FM t 7
f t f k m t Frequency Modulation Figure 33 m t The time at which m t is at its minimum value Acos 2πf t φ corresponds to the time at which x FM t has minimum x PM t frequency. x FM t 8
f t f k m t Frequency Modulation Figure 33 m t is increasing Acos 2πf t φ x PM t The time interval during which m t corresponds to the time interval during which x FM t has increasing frequency. x FM t 9
f t f k m t Frequency Modulation Figure 33 m t The time interval during which m t is decreasing Acos 2πf t φ corresponds to the time interval during which x FM t has decreasing x PM t frequency. x FM t
x PM t Acos 2πf t φ k m t Phase Modulation Figure 33 m t In, the phase varies in proportion with. When and hence the phase of Acos change 2πf t φ continuously, it is difficult to see the connection with the actual plot of. x PM t x FM t
x PM t Acos 2πf t φ k m t Phase Modulation Figure 33 m t One may notice here that, in this example, is similar to Except that the graph is shifted. However, it Acos is still 2πf not t φ clear (visually) how the graph of is related to. x PM t x FM t 2
AM, FM, and PM Figure 34 m t x AM t x FM t x PM t 3
x AM t A m t cos 2πf t φ Amplitude Modulation Figure 34 m t x AM t In, the envelope varies in proportion to. x FM t 4
f t f k m t Frequency Modulation Figure 34 m t m t is increasing in this region x AM t f t is increasing in this region x FM t In, the frequency varies in proportion to. x PM t 5
x PM t Acos 2πf t φ k m t Phase Modulation Figure 34 Sudden drop in the value of m t m t m t is increasing in this region x AM t x FM t In, the phase varies in proportion to. Higher but constant frequency. Why? x PM t 6 Sudden change in the phase
Instantaneous Frequency Sinusoidal signal: g t Acos 2πf t φ Frequency f Generalized sinusoidal signal: Frequency? g t Acos t Observation: Frequency value may vary as a function of time. instantaneous frequency Why do we need to find the instantaneous frequency? Analyze Doppler effect (or Doppler shift) Implement frequency modulation (FM) where the instantaneous frequency will follow the message m t. 7
Instantaneous Frequency 2 t t x t cos 2 x() t.5 Figure 36.5 2 3 4 t At t = 2, frequency =? 4 8
Instantaneous Frequency 2 t t x t cos 2 x() t.5 Figure 36 9.5 By matching the terms with cos 2πf t, you may guess that f t t. 2 3 4 t At t = 2, f = t 2 = 4 Hz? Correct? 4
Instantaneous Frequency 2 t t x t cos 2 Figure 36 x() t cos( 2 4 t).5.5 2 By matching the terms with cos 2πf t, you may guess that f t t. 2 3 4 t At t = 2, f = t 2 = 4 Hz? Correct?
Instantaneous Frequency 2 t t x t cos 2 x() t cos( 2 4 t).5.5 Figure 36 2 3 4 t 4 Hz is too low!!! x() t cos( 2 4 t).5.5.9 2 2. t 2
Instantaneous Frequency 2 t t x t cos 2 x() t cos( 2 2 t).5.5 Figure 36 2 3 4 zoom in t 2 Hz fits better. x() t cos( 2 2 t).5.5 22.9 2 2. t 2 Hz?
First-order (straight-line) approximation/linearization How does the formula f t φ t work? Technique from Calculus: first-order (tangent-line) approximation/linearization 5 4 3 2-23 -2.5.5 2 2.5 3 3.5 4
First-order (straight-line) approximation/linearization How does the formula f t t work? Technique from Calculus: first-order (tangent-line) approximation/linearization When we consider a function t near a particular time, say, t t, the value of the function is approximately t t t t t t t t t t Therefore, near t t, cos slope slope constant t cos t t t t t Now, we can directly compare the terms with cos 2πf t φ. 5 4 3 2 - -2.5.5 2 2.5 3 3.5 4 24
First-order (straight-line) approximation/linearization For example, for t near t = 2, 5 2 3 2 3 t 2 3t t 2 2 t 2 2 t 2 6 t 2 t 2 4 3 2πt 2 2π 2 t 2π 6-25 -2.5.5 2 2.5 3 3.5 4
First-order (straight-line) approximation/linearization For example, for t near t = 2, 75 2 3 2 3 t 2 3t t 2 2 t 2 2 t 2 6 t 2 t 2 7 65 6 2πt 2π 2 t 2π 6 55 5 45 4 35 26 3.75.8.85.9.95 2 2.5 2. 2.5 2.2 2.25
First-order (straight-line) approximation/linearization For example, for t near t = 2, 54 2 3 2 3 t 2 3t t 2 2 t 2 2 t 2 6 t 2 t 2 53 52 2πt 2π 2 t 2π 6 5 5 49 48 27 47.96.97.98.99 2 2. 2.2 2.3 2.4
Same idea Suppose we want to find. Let. Note that Approximation: 5.9 is near 6.. 28 MATLAB:. >> sqrt(5.9) ans = 3.9874847475377
Phase Modulation Figure 33 m t When and hence the phase of change continuously, it is difficult to see the connection Acos 2πf t φ with the actual plot of. x PM t 29 New Fact: In varies in proportion with the slope of., the instantaneous x FM frequency t
Phase Modulation Figure 37 m t d m t dt Acos 2πf t φ x PM t f t d f k m t dt 3 New Fact: In proportion with the slope of., the instantaneous x FM frequency t varies in
Phase Modulation Figure 37 m t The time at which d the slope of m t m t dt is at its maximum Acos 2πf t φ value corresponds to the time at which x PM t has maximum x PM t frequency. 3 New Fact: In varies in proportion with the slope of., the instantaneous x FM frequency t
Phase Modulation Figure 37 m t The time at which d the slope of m t m t dt is at its minimum Acos 2πf t φ value corresponds to the time at which x PM t has minimum x PM t frequency. 32 New Fact: In varies in proportion with the slope of., the instantaneous x FM frequency t
Phase Modulation Figure 37 m t The time interval d during which the m t dt slope of m t is Acos 2πf t φ increasing corresponds to the time interval during which x PM t has x PM t increasing frequency. 33 New Fact: In varies in proportion with the slope of., the instantaneous x FM frequency t
Elements of digital commu. sys. Message Transmitter Information Source Recovered Message Destination Source Encoder Remove redundancy (compression) Source Decoder Channel Encoder Add systematic redundancy to combat errors introduced by the channel Receiver Channel Decoder Digital Modulator Map digital sequence to analog signal Digital Demodulator Channel Transmitted Signal Received Signal Noise & Interference 34
Digital Modulation/Demodulation Message Transmitter Information Source Recovered Message Destination Source Encoder Remove redundancy (compression) Source Decoder Channel Encoder Add systematic redundancy to combat errors introduced by the channel Receiver Channel Decoder Digital Modulator Map digital sequence to analog signal Digital Demodulator Channel Transmitted Signal Received Signal Noise & Interference 35
Digital Version Use digital signal to modulate the amplitude, frequency, or phase of a sinusoidal carrier wave. Think of m(t) as a train of scaled (rectangular) pulses. The modulated parameter will be switched or keyed from one discrete value to another. Three basic forms: amplitude-shift keying (ASK) frequency-shift keying (FSK) phase-shift keying (PSK) 36
Keying? To key = to enter or operate on (data) by means of a computer keyboard or telephone keypad she keyed in a series of commands A telegraph key is a switching device used primarily to send Morse code. 37
Binary ASK, FSK, and PSK m m t 38 [Carlson & Crilly, 29, Fig 4.- p 649]
Simple ASK: ON-OFF Keying (OOK) cos 2πf t Digital Modulator?.8.6 f c = 4 Hz.4.2 Bit rate = bps t -.2 -.4 -.6 -.8 T 2T 3T 4T 5T - 2 3 4 5 6 Seconds t 39 [ASK_playTones_Demo.m]
Simple ASK : ON-OFF Keying Smoke signal 4
Simple ASK: ON-OFF Keying (OOK) cos 2πf t Bit stream Digital Modulator?.8.6 f c = 4 Hz Bit rate = bps.4.2 -.2 -.4 t -.6 -.8-2 3 4 5 6 Seconds T 2T 3T 4T 5T f c = Hz.5 Bit rate = 2 bps t -.5 4 -.5.5 2 2.5 3 3.5 4 4.5 5 Seconds [ASK_playTones_Demo.m]
ASK: Higher Order Modulation 4 3 Digital Modulator? 2 32222233233333333223223333332322222223332233-2 42-4.5.5 2 2.5 3 3.5 4 4.5 5 Seconds f c = Hz Symbol rate = 2 symbols per second Bit rate = 4 bps
FSK M = 4 f f f f f c,,, 3,6,9,2 [Hz] 2 3 4 s t cos 2πf t s t cos 2πf t s t cos 2πf t s t cos 2πf t.5 -.5-43 [FSK_playTones_Demo.m]
FSK Digital Modulator? M = 4 f f f f f c,,, 3,6,9,2 [Hz] 2 3 4 [ ] f [2 2 3 2 9 3 6 9 2 2] Hz.8.6.4.2 -.2 -.4 -.6 -.8 44-2 3 4 5 6 7 8 9 Seconds [FSK_playTones_Demo.m]
FSK Digital Modulator? M = 4 f f, f, f, f, 2,3, 4 [Hz] c 2 3 4 [ ] f [4 4 4 3 2 3 4 4] Hz.8.6.4.2 -.2 -.4 -.6 -.8 45-2 3 4 5 6 7 8 9 Seconds [FSK_playTones_Demo.m]
FSK 2 5 = (random) bits Digital Modulator?.5 -.5 46 R s = 5 Bit rate = bps Magnitude - 2 3 4 5 6 7 8 9 Seconds 2.5.5-8 -6-4 -2 2 4 6 8 Frequency [Hz]
47 Digits of
48 Digits of
Digits of 49 https://youtu.be/wm-x3pucdeo
Figure 42 Five Frequencies cos 2πf t cos 2πf t cos 2πf t cos 2πf t cos 2πf t.5 -.5 -.5..5.2.25 Seconds Each tone lasts /R s sec. Rate = R s frequency-changes per second 5
Spectrum of Five Frequencies (/5).5 Hz 2 Hz 3 Hz 4 Hz 5 Hz cos 2πf t cos 2πf t cos 2πf t cos 2πf t cos 2πf t -.5 R s =.5-2 3 4 5 6 7 8 9 Seconds.8 Magnitude.6.4.2 5 - -8-6 -4-2 2 4 6 8 Frequency [Hz]
Spectrum of Five Frequencies (2/5).5 Hz 2 Hz 3 Hz 4 Hz 5 Hz cos 2πf t cos 2πf t cos 2πf t cos 2πf t cos 2πf t -.5 R s = 5 -..2.3.4.5.6.7.8.9 Seconds..8 Magnitude.6.4.2 52 - -8-6 -4-2 2 4 6 8 Frequency [Hz]
Cos vs. Cos Pulse x t cos 2π t x t cos 2π t,.4 t.6,, otherwise..5 -.5 -..2.3.4.5.6.7.8.9 Seconds.5.4.5 -.5 -..2.3.4.5.6.7.8.9 Seconds..8 Magnitude.3.2. Magnitude.6.4.2-2 -5 - -5 5 5 2 Frequency [Hz] -2-5 - -5 5 5 2 Frequency [Hz] 53
Figure 43 Cosine Pulse cos 2π t,.5 t.6, x t, otherwise..5 x(t) -.5 -..2.3.4.5.6.7.8.9 t [s].5.4 X(f).3.2. 54-2 -5 - -5 5 5 2 f [Hz]
Figure 44 Spectrum of Five Frequencies (3/5).5 Hz 2 Hz 3 Hz 4 Hz 5 Hz cos 2πf t cos 2πf t cos 2πf t cos 2πf t cos 2πf t -.5 R s = 2 -.5..5.2.25 Seconds.3 Magnitude.2. 55 - -8-6 -4-2 2 4 6 8 Frequency [Hz]
Fourier Transform Pairs (2) Time Domain Frequency Domain j 2 ft j2 ft g t G f e df G f g t e dt.284 -.272 -.93 56 sinc( x) sin x x
Spectrum of Five Frequencies (4/5) Hz 2 Hz 3 Hz 4 Hz 5 Hz.5 -.5 R s = 5.5 -.2.4.6.8..2 Seconds Magnitude..5 57 - -8-6 -4-2 2 4 6 8 Frequency [Hz]
Spectrum of Five Frequencies (5/5).5 Hz 2 Hz 3 Hz 4 Hz 5 Hz cos 2πf t cos 2πf t cos 2πf t cos 2πf t cos 2πf t -.5 R s =.5-2 3 4 5 6 7 8 9 Seconds 95.8 Magnitude.6.4.2 58 - -8-6 -4-2 2 4 6 8 Frequency [Hz]