Leture 4: Amplitude Modulation (Double Sideband Suppressed Carrier, DSB-SC) Dr. Mohammed Hawa Eletrial Engineering Department University of Jordan EE421: Communiations I Notation 2 1
Three Modulation Types A m(t); ω = onstant; θ o = onstant Amplitude Modulation (AM) Amplitude Shift Keying (ASK) A= onstant; ω m(t); θ o = onstant Frequeny Modulation (FM) Frequeny Shift Keying (FSK) A= onstant; ω = onstant; θ o m(t) Phase Modulation (PM) Phase Shift Keying (PSK) 3 DSB-SC Modulator: Mixer Modulator m(t) Mixer Output ϕ(t) Channel (t) = os(ω t) Osillator 4 2
m(t) M( ) V t 2 B 2 B (t) C( ) 1 V V 1 V t (t) 18 18 /2 ( ) /2 V t 5 2 B + 2 B Drawing a DSB-SC modulated signal 6 3
DSB-SC Demodulator: Mixer again! Demodulator Mixer Channel ϕ(t) x(t) LPF H(ω) Output y(t) (t) = os(ω t) 7 8 4
9 The LPF is the missing piee! x(t) Smoothed Output V t 1 5
Filter Speifiations! 11 1 18 Phase Shift -1.1.2.3.4.5.6.7.8.9.1 1.5-1.1.2.3.4.5.6.7 -.5.1.2.3.4.5.6.7.8.9.1 12 6
Example Assume we perform DSB-SC modulation for the baseband signal m(t) = αos(ω m t) [the ase of tone modulation], where ω >> ω m : Sketh the time-domain modulated signal ϕ(t). Sketh the Fourier transform of the modulated signal Φ(ω) [frequeny domain]. Find the bandwidthof m(t) and ϕ(t). Find the average power in both m(t) and ϕ(t). Show the demodulator hardware. Sketh and in the demodulator. 13 Solution 14 7
Homework For the following signals, sketh: The modulated signal at the modulator The Fourier transform Φ The signals and at the demodulator The Fourier transform and Find the average power and bandwidth for the signals and. Are there any phase shifts in? If so, where? Determine the DC value in, and. 15 Homework 16 8
Solution: Part(b) 2 1.5 m(t) M(ω) 2π α -1 2π α 1 1 3 5 t (se) 3ω ω ω 2ω 3ω.5 2π α -2 2π α -3 2π α 2 2π α 3 ω ϕ(t) Φ(ω) 2 18 18 18 2π α -1 /2 2π α 1 /2 2π α -1 /2 2π α 1 /2 1 1 3 5 t (se) ω ω 2ω ω + 2ω ω ω 2 17 Homework For the following iruit, sketh and, along with the Fourier transform and. 18 9
How to build a Mixer? Variable Gain Amplifier The basi design. Gilbert Cell (e.g., MC 1496) Popular (used in Integrated Ciruits). Uses variable gain differential amplifiers. Swithing Modulator Uses diodes. Cheaper design (was popular before ICs). 19 Variable Gain Amplifier m(t) G m(t) G Fixed Gain m(t) G k (t) m(t) Variable Gain G= k (t) 2 1
Gilbert Cell inside Phones 21 LTE Modem (CAT 4 up to 15Mbps) 22 11
23 RF Solid State Ciruits 24 12
25 Gilbert Cell (MC1496) Shemati 26 13
Differential Amplifier (Differential Pair) 27 Gilbert Cell Four Modes Gilbert Cell (MC1496) an operate in one of four possible modes based on balaning and/or saturating the upper differential amplifiers. Gilbert Cell Modes Balaned Mode (DSB-SC)... Linear Mode m(t) os(ω t)......... 28 14
Swithing Modulator 29 DSB-SC Swithing Modulator (Series-bridge diode modulator) Osillator D1 Cos( t) D2 Band-pass Filter Gain = 1 Bandwidth = 2BHz Center Frequeny = m(t) D3 D4 x(t) y(t) Channel 3 15
31 The BPF is not optional! V x(t) t -3-2 -1 X( ) 1 BPF 2 3 y(t) 2 1 m(t) os( t) Y( ) 18 18-1 1 V t 2 B + 2 B 32 16
DSB-SC Swithing Modulator (Shunt-bridge diode modulator) 33 34 17
The BPF is not optional! V x(t) t -3-2 -1 X( ) 1 BPF 2 3 y(t) 2 1 m(t) os( t 18 ) Y( ) 18 18-1 1 V t 2 B + 2 B 35 DSB-SC Swithing Modulator (Ring modulator) 36 18
37 The BPF is not optional! x(t) -3 X( ) BPF -1 1 3 V t y(t) 2 1 m(t) os( t) Y( ) 18 18-1 1 V t 2 B + 2 B 38 19
Gilbert Cell as Swithing Modulator It is reommended to use the Gilbert Cell (MC1496) in the swithing mode, rather than the linear mode. 39 Summary 4 2
Homework If we use a swithing demodulator (say a series-bridge diode demodulator), answer the following: What are the speifiations of the filter? What is the value of kin the output y(t) = k m(t) if the input is ϕ(t) = m(t) os(ω t) What is the value of kin the output y(t) = k m(t) if the input is ϕ(t) =2 α 1 m(t) os(ω t) Sketh the Fourier transform of the solution. 41 42 21
Synhronizing Osillator at the RX TX RX Problem m(t) ϕ(t) ϕ(t) x(t) y(t)=? LPF os(ω t) os(ω t + θ ) Loal Osillator Phase Error m(t) os(ω t) ϕ(t) ϕ(t) x(t) y(t)=? LPF os([ω + ω]t) Loal Osillator Frequeny Error 43 Frequeny Error at the Reeiver 44 22
To avoid problems due to phase and frequeny errors Solution #1: Use a PLL (Phase-Loked Loop) at the RX. A PLL an, by observing, reover the exat frequeny and phase of the arrier at the TX, and hene use these values at the RX. The PLL is alled a arrier-reovery iruit (omplex and expensive). The reeiver in this ase is known as a synhronous or oherent reeiver. Solution #2: Do not generate a arrier at the RX. Rather, let the TX send an extra opy of the arrier (e.g., DSB-LC) to help the RX demodulate. The RX is known as asynhronous or inoherent reeiver(heaper), but the TX is power ineffiient. 45 23