EEM 306 Introduction to Communications
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1 EEM 306 Introduction to Communications Lecture 5 Department o Electrical and Electronics Engineering Anadolu University April 8, 2014 Lecture 5 1/20
2 Last Time Bandpass Systems Phase and Group Delay Introduction to Modulation Amplitude Modulation Double-Sideband Suppressed Carrier (DSB-SC) AM: bandwidth Lecture 5 2/20
3 Upper and Lower Sidebands Upper ( > c ) and Lower ( < c ) sidebands: U( ) A A c 2 c c c 0 c c c U ( ) u A A c 2 Upper sideband c c 0 c c Ul ( ) Ac A 2 Lower sideband c c Either one o the sidebands o U() contains all the requencies that are in M(). Lecture 5 3/20 0 c c
4 Double-Sideband Suppressed Carrier AM cont d U() contains both the upper and lower sidebands double-sideband (DSB) AM signal u(t) does not contain a carrier component. No impulse in U() at = c suppressed-carrier signal. Lecture 5 4/20
5 Example 1: Suppose that the modulating signal m(t) = a cos(2π m t), m c Determine a. The DSB-SC AM signal (in time and requency) b. Its upper and lower sidebands Lecture 5 5/20
6 Power Content o DSB-SC Signals Assume that the phase o the signal is set to zero. u(t) = A c m(t) cos(2π c t) The time-avg autocorrelation unction 1 R u (τ)= lim T T 1 = lim T T T/2 T/2 T/2 = A2 c 1 2 lim T T T/2 T/2 u(t)u(t τ)dt A 2 cm(t)m(t τ) cos(2π c t) cos(2π c (t τ))dt T/2 = A2 c 2 R m (τ) cos(2π c τ) where we have used lim T/2 T T/2 m(t)m(t τ)[cos(4π c t 2π c τ) + cos(2π c τ)]dt m(t)m(t τ) cos(4π c t 2π c τ)dt = 0 Lecture 5 6/20 Prove!
7 Power Content o DSB-SC Signals cont d R u (τ) = A2 c 2 R m(τ) cos(2π c τ) By taking the FT o both sides S u () = A2 c 4 [S m( c ) + S m ( + c )] To obtain the total power in the modulated signal Substitute τ = 0 in the time-avg autocorrelation unction P u = A2 c 2 R m(τ) cos(2π c τ) = A2 c τ=0 2 R m(0) = A2 c 2 P m or Integrate the power spectral density Lecture 5 7/20
8 Example 2: In Example 1, determine a. The power-spectral density o the modulated signal b. The power in the modulated signal c. The power in each o the sidebands. Lecture 5 8/20
9 Demodulation o DSB-SC AM Signals In the absence o noise, and with the assumption o an ideal channel, the received signal r(t) = u(t) = A c m(t) cos(2π c t + φ c ) rt () yl () t X LPF cos(2 t c ) ~ r(t) cos(2π c t + φ) = A c m(t) cos(2π c t + φ c ) cos(2π c t + φ) = 1 2 A cm(t) [cos(φ c φ) + cos(4π c t + φ + φ c )] Ater LPF, y l (t) = 1 2 A cm(t) cos(φ c φ) What happens i φ c φ = 90 0? The need or a phase-coherent or synchronous demodulator! Lecture 5 9/20
10 Conventional Amplitude Modulation The modulated signal: u(t) = A c [1 + am n (t)] cos(2π c t + φ c ) a: modulation index m n (t): normalized message signal m n (t) = m(t) max m(t) The spectrum o the modulated signal: U() = Ac 2 [am n( c )e jφc + am n ( + c )e jφc... + δ( c )e jφc + δ( + c )e jφc ] Occupies a bandwidth twice the bandwidth o the message signal. Lecture 5 10/20
11 Example 3: Suppose that the modulating signal m(t) = cos(2π m t), m c Assuming a modulation index o a, determine a. The DSB AM signal (in time and requency) b. Upper and lower sidebands c. Its spectrum. A c U( ) 2 2 Aa Aa c Aa c c Aa c c c c m c m m c c m A c Lecture 5 11/20
12 Power or Conventional AM Signal Conventional AM signal is similar to DSB-SC when m(t) is substituted with 1 + am n (t). Recall or DSB-SC, P u = A2 c 2 P m. For the conventional AM P m 1 = lim T T 1 = lim T T T/2 T/2 T/2 T/2 (1 + am n (t)) 2 dt (1 + a 2 m 2 n(t))dt where we have assumed that the average o m n (t) is zero. Thus, or conventional AM P m = 1 + a 2 P mn P u = A2 c 2 + A2 c 2 a2 P mn Less power eicient! Lecture 5 12/20
13 Demodulation o Conventional DSB AM Signals The major advantage o conventional AM signal transmission: No need or a synchronous demodulator As long as [1 + am n (t)] > 0, to demodulate: Use envelope detector (a combination o rectiier and LPF) The simplicity o the demodulator a practical choice or AM radio broadcasting! Lecture 5 13/20
14 Single-Sideband AM The transmission o either sideband is suicient to reconstruct the message signal m(t). 1 AM(0) 2 c DSB-SC c c c 0 c c c S( ) : upper sideband c c 0 c c S( ) : lower sideband c c 0 c c Lecture 5 14/20
15 Single-Sideband AM cont d Consider an upper SSB amplitude modulated signal. s(t) = s c (t) cos(2π c t) s s (t) sin(2π c t) In-phase component: { S( c ) + S( + S c () = c ), W W 0, else S( ) c 0 c S ( ) c S c () = 1 2 A cm(). Thus s c (t) = 1 2 A cm(t) Lecture 5 15/20 2 c
16 Single-Sideband AM cont d Quadrature component: { j[s( c ) S( + S s () = c )], W W 0, else S( ) c 0 c S ( ) s 2 c S s () = j 2 A c[ sgn()]m() = 1 2 A c ˆM(). Thus s s (t) = 1 2 A c ˆm(t) Lecture 5 16/20
17 Single-Sideband AM cont d s(t) = 1 2 A cm(t) cos(2π c t) 1 2 A c ˆm(t) sin(2π c t) ( ): upper SSB (+): lower SSB Generation o an SSB AM signal: 2 st ( ) (+) LSSB Lecture 5 17/20
18 Single-Sideband AM cont d Another method to generate an SSB AM signal: Generate a DSB-SC AM signal Employ a ilter which selects either the upper sideband or the lower sideband Lecture 5 18/20
19 Demodulation o SSB AM Signals Given the USSB signal, Multiply with the carrier s(t) cos(2π c t + φ) = [ 1 2 A cm(t) cos(2π c t) 1 2 A c ˆm(t) sin(2π c t) ] cos(2π c t + φ) = 1 4 A cm(t) cos(φ) A c ˆm(t) sin(φ) + double requency terms LPF y l (t) = 1 4 A cm(t) cos(φ) A c ˆm(t) sin(φ) The eect o the phase oset is not only to reduce the amplitude o the desired signal m(t) by cos(φ), but it also results in an udesirable sideband signal due to the presence o ˆm(t) in y l (t). Require a phase coherent or synchronous demodulator! Lecture 5 19/20
20 Demodulation o SSB AM Signals cont d The spectral eiciency o SSB AM attractive or use in voice communications over telephone channels (wire lines and cables)! Usually, a pilot tone is transmitted or synchronous demodulation and shared among several channels. Lecture 5 20/20
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