#### Transcript Ch5 - Department of Engineering and Physics

ENGR 4323/5323 Digital and Analog Communication Ch 5 Angle Modulations and Demodulations Engineering and Physics University of Central Oklahoma Dr. Mohamed Bingabr Chapter Outline • Nonlinear Modulation • Bandwidth of Angle Modulation • Generating of FM Waves • Demodulation of FM Signals • Effects of Nonlinear Distortion and Interference • Superheterodyne Analog AM/FM Receivers • FM Broadcasting System 2 Baseband Vs. Carrier Communications Angle Modulation: The generalized angle θ(t) of a sinusoidal signal is varied in proportion the message signal m(t). 𝜑 𝑡 = 𝐴 𝑐𝑜𝑠𝜃(𝑡) 𝜑 𝑡 = 𝐴 cos 𝜔𝐶 𝑡 + 𝜃0 𝑡1 < 𝑡 < 𝑡2 Instantaneous Frequency 𝑑𝜃 𝜔𝑖 (𝑡) = 𝑑𝑡 Two types of Angle Modulation • Frequency Modulation: The frequency of the carrier signal is varied in proportion to the message signal. • Phase Modulation: The phase of the carrier signal is varied in proportion to the message signal. 3 Frequency and Phase Modulation 𝜑 𝑡 = 𝐴 𝑐𝑜𝑠𝜃(𝑡) Phase Modulation: 𝜃 𝑡 = 𝜔𝐶 𝑡 + 𝑘𝑝 𝑚(𝑡) 𝜑𝑃𝑀 𝑡 = 𝐴 cos 𝜔𝐶 𝑡 + 𝑘𝑝 𝑚(𝑡) Frequency Modulation: 𝜔𝑖 𝑡 = 𝜔𝑐 + 𝑘𝑓 𝑚(𝑡) 𝑡 𝜃 𝑡 = 𝑡 𝜔𝑖 𝑡 𝑑𝑡 = 𝜔𝑐 𝑡 + 𝑘𝑓 −∞ 𝑡 𝜑𝐹𝑀 𝑡 = 𝐴 cos 𝜔𝐶 𝑡 + 𝑘𝑓 𝑚 𝛼 𝑑𝛼 −∞ 𝑚 𝛼 𝑑𝛼 −∞ Power of an Angle-Modulated wave is constant and equal A2/2. 4 Example of FM and PM Modulation Sketch FM and PM waves for the modulating signal m(t). The constants kf and kp are 2π x105 and 10π, respectively, and the carrier frequency fc is 100 MHz. 5 Example of FM and PM Modulation Sketch FM and PM waves for the digital modulating signal m(t). The constants kf and kp are 2π x105 and π/2, respectively, and the carrier frequency fc is 100 MHz. Frequency Shift Keying FSK Phase Shift Keying PSK Note: for discontinuous signal kp should be small to restrict the 6 phase change kpm(t) to the range (-π,π). PSK and DPSK 7 Bandwidth of Angle Modulated Waves 𝑡 𝜑𝐹𝑀 𝑡 = 𝐴 cos 𝜔𝐶 𝑡 + 𝑘𝑓 𝑎(𝑡) where 𝑎(𝑡) = 𝜑𝐹𝑀 𝑡 = 𝐴 Re 𝑒 𝑗 [𝜔𝐶 𝑡+𝑘𝑓 𝑎(𝑡)] = 𝐴 Re 𝑚 𝛼 𝑑𝛼 −∞ 𝑒 𝑗𝑘𝑓𝑎(𝑡) 𝑒 𝑗 𝜔𝐶 𝑡 Expand 𝑒 𝑗𝑘𝑓 𝑎(𝑡) using power series expansion 𝜑𝐹𝑀 𝑡 = 𝐴 Re 𝜑𝐹𝑀 𝑛 𝑘𝑓2 2 𝑘 𝑓 𝑛 1 + 𝑗𝑘𝑓 𝑎 𝑡 − 𝑎 𝑡 + ⋯ + 𝑗 𝑛 𝑎 𝑡 + ⋯ 𝑒 𝑗 𝜔𝐶 𝑡 2! 𝑛! 𝑘𝑓2 2 𝑘𝑓3 3 𝑡 = 𝐴 𝑐𝑜𝑠𝜔𝑐 𝑡 − 𝑘𝑓 𝑎 𝑡 𝑠𝑖𝑛𝜔𝑐 𝑡 − 𝑎 𝑡 𝑐𝑜𝑠𝜔𝑐 𝑡 + 𝑎 𝑡 𝑠𝑖𝑛𝜔𝑐 𝑡 + ⋯ 2! 3! The bandwidth of a(t), a2(t), an(t) are B, 2B, and nB Hz, respectively. From the above equation it seems the bandwidth of angle modulation is infinite but for practical reason most of the power reside at B Hz since higher terms have small power because of n!. 8 Narrowband PM and FM 𝜑𝐹𝑀 𝑘𝑓2 2 𝑘𝑓3 3 𝑡 = 𝐴 𝑐𝑜𝑠𝜔𝑐 𝑡 − 𝑘𝑓 𝑎 𝑡 𝑠𝑖𝑛𝜔𝑐 𝑡 − 𝑎 𝑡 𝑐𝑜𝑠𝜔𝑐 𝑡 + 𝑎 𝑡 𝑠𝑖𝑛𝜔𝑐 𝑡 + ⋯ 2! 3! When kf a(t) << 1 𝜑𝐹𝑀 𝑡 ≈ 𝐴 𝑐𝑜𝑠𝜔𝑐 𝑡 − 𝑘𝑓 𝑎 𝑡 𝑠𝑖𝑛𝜔𝑐 𝑡 The above signal is Narrowband FM and its bandwidth is 2B Hz. Same steps can be carried out to find the Narrowband PM. 𝜑𝑃𝑀 𝑡 ≈ 𝐴 𝑐𝑜𝑠𝜔𝑐 𝑡 − 𝑘𝑝 𝑚 𝑡 𝑠𝑖𝑛𝜔𝑐 𝑡 9 Wideband FM (WBFM) In many application FM signal is meaningful only if its frequency deviation is large enough, so kf a(t) << 1 is not satisfied, and narrowband analysis is not valid. 𝑟𝑒𝑐𝑡 2𝐵𝑡 𝑐𝑜𝑠 𝜔𝑐 𝑡 + 𝑘𝑓 𝑚 𝑡𝑘 𝑡 Fourier Transform 𝜔+𝜔𝑐 +𝑘𝑓 𝑚(𝑡𝑘 ) 1 𝑠𝑖𝑛𝑐 2 4𝐵 𝐵𝐹𝑀 = 1 2𝜋 + 𝜔−𝜔𝑐 −𝑘𝑓 𝑚(𝑡𝑘 ) 1 𝑠𝑖𝑛𝑐 2 4𝐵 2𝑘𝑓 𝑚𝑝 + 8𝜋𝐵 Hz 10 Wideband FM (WBFM) 𝐵𝐹𝑀 = 1 2𝜋 2𝑘𝑓 𝑚𝑝 + 8𝜋𝐵 =2 𝑘𝑓 𝑚𝑝 2𝜋 + 2𝐵 = 2 ∆𝑓 + 2𝐵 Hz 𝑘𝑓 𝑚𝑝 Peak frequency deviation in hertz ∆𝑓 = 2𝜋 A better estimate: Carson’s rule 𝐵𝐹𝑀 = 2 ∆𝑓 + 𝐵 Hz ∆𝑓 is the deviation ratio = 2𝐵 𝛽 + 1 Where 𝛽 = 𝐵 When Δf >> B the modulation is WBFM and the bandwidth is BFM = 2 Δf When Δf << B the modulation is NBFM and the bandwidth is BFM = 2B 11 Wideband PM (WBPM) The instantaneous frequency 𝜔𝑖 = 𝜔𝑐 + 𝑘𝑝 𝑚(𝑡) 𝑚𝑝 ∆𝑓 = 𝑘𝑝 2𝜋 𝐵𝑃𝑀 = 2 ∆𝑓 + 𝐵 Hz 12 Example a) Estimate BFM and BPM for the modulating signal m(t) for kf =2π x105 and kp = 5π. Assume the essential bandwidth of the periodic m(t) as the frequency of its third harmonic. b) Repeat the problem if the amplitude of m(t) is doubled. c) Repeat the problem if the time expanded by a factor of 2: that is, if the period of m(t) is 0.4 msec. 13 Example An angle-modulated signal with carrier frequency ωc = 2 105 is described by the equation 𝜑𝐸𝑀 𝑡 = 10𝑐𝑜𝑠 𝜔𝑐 𝑡 + 5 𝑠𝑖𝑛3000𝑡 + 10 𝑠𝑖𝑛2000𝜋𝑡 a) b) c) d) e) Find the power of the modulated signal. Find the frequency deviation Δf. Find the deviation ration . Find the phase deviation Δø. Estimate the bandwidth of 𝜑𝐸𝑀 𝑡 . 14 Indirect NPFM Generation NBFM generation 𝜑𝐹𝑀 𝑡 ≈ 𝐴 𝑐𝑜𝑠𝜔𝑐 𝑡 − 𝑘𝑓 𝑎 𝑡 𝑠𝑖𝑛𝜔𝑐 𝑡 𝜑𝑃𝑀 𝑡 ≈ 𝐴 𝑐𝑜𝑠𝜔𝑐 𝑡 − 𝑘𝑝 𝑚 𝑡 𝑠𝑖𝑛𝜔𝑐 𝑡 With this method of generation, the amplitude of the NBFM modulator will have some amplitude variation due to approximation. 15 Generating FM Waves Bandpass Limiter 𝑣𝑖 𝑡 = 𝐴 𝑡 𝑐𝑜𝑠𝜃(𝑡) 𝑡 𝜃 𝑡 = 𝜔𝑐 𝑡 + 𝑘𝑓 𝑚 𝛼 𝑑𝛼 −∞ +1 𝑣𝑜 (𝜃) = −1 𝑐𝑜𝑠𝜃 > 0 𝑐𝑜𝑠𝜃 < 0 4 1 1 𝑣𝑜 𝜃 = cos 𝜃 − cos 3𝜃 + cos 5𝜃 + ⋯ 𝜋 3 5 𝑡 𝑣𝑜 𝜃(𝑡) = 𝑣𝑜 𝜔𝑐 𝑡 + 𝑘𝑓 𝑚 𝛼 𝑑𝛼 −∞ 16 Generating FM Waves 𝑡 𝜃 𝑡 = 𝜔𝑐 𝑡 + 𝑘𝑓 4 1 1 𝑣𝑜 𝜃 = cos 𝜃 − cos 3𝜃 + cos 5𝜃 + ⋯ 𝜋 3 5 𝑚 𝛼 𝑑𝛼 −∞ 4 𝑣𝑜 𝜃 = cos 𝜔𝑐 𝑡 + 𝑘𝑓 𝜋 𝑡 −∞ 1 𝑚 𝛼 𝑑𝛼 − cos 3 𝜔𝑐 𝑡 + 𝑘𝑓 3 𝑡 𝑚 𝛼 𝑑𝛼 −∞ 𝑡 1 + cos 5 𝜔𝑐 𝑡 + 𝑘𝑓 𝑚 𝛼 𝑑𝛼 + ⋯ 5 −∞ The output of the bandpass filter 4 𝑒𝑜 𝑡 = 𝑐𝑜𝑠 𝜔𝑐 𝑡 + 𝑘𝑓 𝜋 𝑡 𝑚 𝛼 𝑑𝛼 −∞ 17 Indirect Method of Armstrong NBFM is generated first and then converted to WBFM by using additional frequency multipliers. For NBFM << 1. For speech fmin = 50Hz, so if Δf = 25 then 25/50 = 0.5, 𝜑𝐹𝑀 𝑡 ≈ 𝐴 𝑐𝑜𝑠𝜔𝑐 𝑡 − 𝑘𝑓 𝑎 𝑡 𝑠𝑖𝑛𝜔𝑐 𝑡 18 Example 5.6 Discuss the nature of distortion inherent in the Armstrong indirect FM generator. Amplitude and frequency distortions. 𝜑𝐹𝑀 𝑡 = 𝐴 𝑐𝑜𝑠𝜔𝑐 𝑡 − 𝑘𝑓 𝑎 𝑡 𝑠𝑖𝑛𝜔𝑐 𝑡 𝜑𝐹𝑀 𝑡 = 𝐴𝐸 𝑡 𝑐𝑜𝑠 𝜔𝑐 𝑡 + 𝜃(𝑡) 𝐸 𝑡 = 1 + 𝑘𝑓2 𝑎2 (𝑡) 𝜃 𝑡 = 𝑡𝑎𝑛−1 𝑘𝑓 𝑎(𝑡) 𝑘𝑓 𝑎(𝑡) 𝑘𝑓 𝑚 (𝑡) 𝑑𝜃 𝜔𝑖 𝑡 = 𝜔𝑐 + = 𝜔𝑐 + = 𝜔𝑐 + 2 2 𝑑𝑡 1 + 𝑘𝑓 𝑎 (𝑡) 1 + 𝑘𝑓2 𝑎2 (𝑡) 𝜔𝑖 𝑡 = 𝜔𝑐 + 𝑘𝑓 𝑚 𝑡 1 − 𝑘𝑓2 𝑎2 𝑡 + 𝑘𝑓4 𝑎4 𝑡 − ⋯ 19 Direct Generation 1. The frequency of a voltage-controlled oscillator (VCO) is controlled by the voltage m(t). 𝜔𝑖 (𝑡) = 𝜔𝑐 + 𝑘𝑓 𝑚(𝑡) 2. Use an operational amplifier to build an oscillator with variable resonance frequency ωo. The resonance frequency can be varied by variable capacitor or inductor. The variable capacitor is controlled by m(t). 𝜔𝑜 = 𝜔𝑜 = 1 𝐿𝐶 = 1 𝐿 𝐶0 − 𝑘𝑚(𝑡) 1 𝑘𝑚 𝑡 𝐿𝐶0 1 − 𝐶0 1/2 ≈ 1 = 𝐿𝐶0 1 − 1 𝐿𝐶0 1+ 𝑘𝑚 𝑡 𝐶0 𝑘𝑚 𝑡 2𝐶0 𝑘𝑚 𝑡 ≪1 𝐶0 20 Direct Generation 𝑘𝑚 𝑡 𝜔𝑜 = 𝜔𝑐 1 + 2𝐶0 𝜔𝑜 = 𝜔𝑐 + 𝑘𝑓 𝑚(𝑡) 𝜔𝑐 = 1 𝐿𝐶0 𝑘𝜔𝑐 𝑘𝑓 = 2𝐶0 The maximum capacitance deviation is 𝐶 = 𝐶0 − 𝑘𝑚(𝑡) 2𝑘𝑓 𝐶0 𝑚𝑝 ∆𝐶 = 𝑘𝑚𝑝 = 𝜔𝑐 ∆𝐶 2𝑘𝑓 𝑚𝑝 2∆𝑓 = = 𝐶0 𝜔𝑐 𝑓𝑐 In practice Δf << fc 21 Example Design an Armstrong indirect FM modulator to generate an FM signal with carrier frequency 97.3 MHz and Δf = 10.24 kHz. A NBFM generator of fc1 = 20 kHz and Δf = 5 Hz is available. Only frequency doublers can be used as multipliers. Additionally, a local oscillator (LO) with adjustable frequency between 400 and 500 kHz is readily available for frequency mixing. 22 Demodulation of FM Signals A frequency-selective network with a transfer function of the form |H(f)|=2af + b over the FM band would yield an output proportional to the instantaneous frequency. 𝑡 𝜑𝐹𝑀 𝑡 = 𝐴 cos 𝜔𝐶 𝑡 + 𝑘𝑓 𝑚 𝛼 𝑑𝛼 −∞ 𝜑𝐹𝑀 𝑑 𝑡 = 𝐴 cos 𝜔𝐶 𝑡 + 𝑘𝑓 𝑑𝑡 𝑡 𝑚 𝛼 𝑑𝛼 −∞ 23 Demodulation of FM Signals 𝜑𝐹𝑀 𝑑 𝑡 = 𝐴 cos 𝜔𝐶 𝑡 + 𝑘𝑓 𝑑𝑡 𝑡 𝑚 𝛼 𝑑𝛼 −∞ 𝑡 𝜑𝐹𝑀 𝑡 = 𝐴 𝜔𝑐 + 𝑘𝑓 𝑚(𝑡) sin 𝜔𝐶 𝑡 + 𝑘𝑓 𝑚 𝛼 𝑑𝛼 −∞ 24 Practical Frequency Modulators 𝑗2𝜋𝑓𝑅𝐶 𝐻 𝑓 = ≈ 𝑗2𝜋𝑓𝑅𝐶 1 + 𝑗2𝜋𝑓𝑅𝐶 𝜔𝑐 < 𝜔0 = if 2𝜋𝑓𝑅𝐶 ≪ 1 1 𝑅𝐶 The slope is linear over small band, so distortion occurs if the signal band is larger than the linear band. Zero-crossing detectors: First step is to use amplitude limiter and then the zero-crossing detector. Instantaneous frequency = the rate of zero crossing 25 Effect of Nonlinear Distortion and Interference Immunity of Angle Modulation to Nonlinearities 𝑦 𝑡 = 𝑎0 + 𝑎1 𝑥 𝑡 + 𝑎2 𝑥 2 𝑡 + ⋯ + 𝑎𝑛 𝑥 𝑛 (𝑡) 𝑥 𝑡 = 𝐴 cos 𝜔𝐶 𝑡 + 𝜓(𝑡) 𝑦 𝑡 = 𝑐0 + 𝑐1 cos 𝜔𝐶 𝑡 + 𝜓(𝑡) + 𝑐2 cos 2𝜔𝐶 𝑡 + 2𝜓(𝑡) + ⋯ + 𝑐𝑛 cos 𝑛𝜔𝐶 𝑡 + 𝑛𝜓(𝑡) Vulnerability of Amplitude Modulation to Nonlinearities 𝑦 𝑡 = 𝑎 𝑚 𝑡 cos(𝜔𝐶 𝑡) + 𝑏𝑚3 (𝑡)𝑐𝑜𝑠 3 (𝜔𝐶 𝑡) 3𝑏 3 𝑏 3 𝑦 𝑡 = 𝑎𝑚 𝑡 + 𝑚 (𝑡) cos 𝜔𝐶 𝑡 + 𝑚 𝑡 𝑐𝑜𝑠(3𝜔𝐶 𝑡) 4 4 26 Interference Effect Angle Modulation is less vulnerable than AM to small-signal interference from adjacent channels. 𝑟 𝑡 = 𝐴 cos 𝜔𝑐 𝑡 + 𝐼𝑐𝑜𝑠( 𝜔𝐶 + 𝜔)𝑡 𝑟 𝑡 = (𝐴 + 𝐼 𝑐𝑜𝑠 𝜔𝑡) cos 𝜔𝑐 𝑡 − 𝐼 𝑠𝑖𝑛 𝜔𝑡 sin 𝜔𝑐 𝑡 𝑟 𝑡 = (𝐴 + 𝐼 𝑐𝑜𝑠 𝜔𝑡) cos 𝜔𝑐 𝑡 − 𝐼 𝑠𝑖𝑛 𝜔𝑡 sin 𝜔𝑐 𝑡 𝑟 𝑡 = 𝐸𝑟 (𝑡) cos [𝜔𝑐 𝑡 + 𝜓𝑑 (𝑡)] Instantaneous frequency is 𝜔𝑐 + 𝜓𝑑 (𝑡) 𝐼 𝑠𝑖𝑛 𝜔𝑡 𝐼 𝜓 𝑡 ≈ 𝑠𝑖𝑛 𝜔𝑡 𝜓𝑑 𝑡 = for I << A 𝑑 𝐴 𝐴 + 𝐼 𝑐𝑜𝑠 𝜔𝑡 The output of ideal phase and frequency demodulators are 𝑡𝑎𝑛−1 𝑦𝑑 𝑡 = 𝐼 𝑠𝑖𝑛 𝐴 𝜔𝑡 for PM 𝑦𝑑 𝑡 = 𝐼𝜔 𝑐𝑜𝑠 𝐴 𝜔𝑡 for FM 27 Preemphasis and Deemphasis in FM With white noise, the amplitude interference is constant for PM but increase with ω for FM. For audio signal the PSD is concentrated at low frequency below 2.1 kHz, so interference at high frequency will greatly deteriorate the quality of audio signal. Preemphasis Filter Deemphasis Filter Noise Preemphasis and Deemphasis (PDE) in FM Preemphasis Filter Deemphasis Filter Preemphasis and Deemphasis (PDE) in FM 𝑗2𝜋𝑓 + 𝜔1 𝐻𝑝 𝑓 = 𝐾 𝑗2𝜋𝑓 + 𝜔2 Where K is the gain and = ω2/ ω1 For 2f << ω1 𝐻𝑝 𝑓 ≅ 1 For ω1 << 2f << ω2 𝑗2𝜋𝑓 𝐻𝑝 𝑓 = 𝜔1 𝜔1 𝐻𝑑 𝑓 = 𝐾 𝑗2𝜋𝑓 + 𝜔1 PDE is used in many applications such as recording of audiotape and photograph recording, where PDE depends on the band. Superheterodyne Analog AM/FM Receivers fIF = 455KHz (AM radio); 10.7 MHz (FM); 38 MHz (TV) AM stations that are 2 fIF apart are called image stations and both would appear simultaneously at the IF output. RF section filter out undesired image station, while IF section filter out undesired stations. FM Broadcasting System FCC specifications for FM communication - Frequency range = 88 to 108 MHz - Channel separations = 200 kHz, - Peak frequency deviation = 75 kHz - Transmitted signal should be received by monophonic and stereophonic receivers. FM Broadcasting System 𝑚 𝑡 = 𝐿 + 𝑅 ′ + 𝐿 − 𝑅 ′ cos 𝜔𝑐 𝑡 + 𝛼𝑐𝑜𝑠 𝜔𝑐 𝑡 2 % fs=400; % sampling rate % Window = 4; % length of window in second % t=0:1/fs:Window; % w =-fs/2:1/Window:fs/2; % f0= 10; % fundamental frequency % x=2*sin(2*pi*f0*t) + sin(2*pi*3*f0*t); % plot(t, x) [x fs]=wavread('Fill.wav'); Window=length(x)/fs; w =-fs/2:1/Window:fs/2; FFTx= fftshift(fft(x)); figure; plot(abs(FFTx)) FFTx(1197:1202)=0; FFTx(900:905)=0; xM=ifft(ifftshift(FFTx)); sound(xM,fs) 34