EE4512 Analog and Digital Communications Chapter 6. Chapter 6 Analog Modulation and Demodulation

Transcription

1 Chapter 6 Analog Modulation and Demodulation

2 Chapter 6 Analog Modulation and Demodulation Amplitude Modulation Pages

3 The analytical signal for double sideband, large carrier amplitude modulation (DSB-LC AM) is: s DSB-LC AM AM(t) ) = A C (c + s(t)) cos (2π f C t) where c is the DC bias or offset and A C is the carrier amplitude. The continuous analog signal s(t) is a baseband signal with the information content (voice or music) to be transmitted.

4 The baseband power spectral density (PSD) spectrum of the information signal s(t) ) or S(f) ) for voice has significant components below 500 Hz and a bandwidth of < 8 khz: S(f) ) = F(s(t)) The single-sided sided spectrum of the modulated signal is: F(A C (c + s(t)) cos (2π f C t)) = S(f f C ) db Power Spectral Density of s(t) 500 Hz 8 khz

5 The single-sided sided (positive frequency axis) spectrum of the modulated signal replicates the baseband spectrum as a double-sided spectrum about the carrier frequency. Double-sided spectrum Carrier 25 khz Baseband spectrum

6 The double-sided modulated spectrum about the carrier frequency has an lower (LSB)) and upper (USB)) sideband.

7 The modulated DSB-LC AM signal shows an outer envelope that follows the polar baseband signal s(t).

8 The analytical signal for double sideband, suppressed carrier amplitude modulation (DSB-SC SC AM) is: s DSB-SC SC AM AM(t) ) = A C s(t) cos (2π f C t) where A C is the carrier amplitude. The single-sided sided spectrum of the modulated signal replicates the baseband spectrum as a double-sided spectrum about the carrier frequency but without a carrier component.

9 The analytical signal for double sideband, suppressed carrier amplitude modulation (DSB-SC SC AM) is: s DSB-SC SC AM AM(t) ) = A C s(t) cos (2π f C t) where A C is the carrier amplitude. The modulated signal s DSB-SC SC AM(t (t)) looks similar to s(t) but has a temporal but not spectral carrier component.

10 The DSB-LC AM and the DSB-SC SC AM modulated signals have the same sidebands. DSB-LC AM Carrier 25 khz DSB-SC AM No carrier

11 The modulated DSB-LC AM and the DSC-SC AM signals are different.

12 The modulated DSB-SC SC AM signal has an envelope that follows the polar baseband signal s(t) but not an outer envelope.

13 Chapter 6 Analog Modulation and Demodulation Coherent Demodulation of AM Signals Pages

14 The DSB-SC SC AM coherent receiver has a bandpass filter centered at f C and with a bandwidth of twice the bandwidth of s(t) because of the LSB and USB.. The output of the multiplier is lowpass filtered with a bandwidth equal to the bandwidth of s(t). r(t) ) = γ s DSB-SC SC(t) ) + n(t) The DSB-SC SC AM received signal is r(t) = γ s DSB-SC SC(t) ) + n(t). The bandpass filter passes the modulated signal but filters the noise: z(t) ) = γ s DSB DSB-SC SC(t) ) + n o (t) S&M Eq. 6.3 n o (t) has a Gaussian distribution. The bandpass filter has a center frequency of f C = 25 khz and a -33 db bandwidth of 8 khz (25 ± 4 khz).

15 EE4512 Analog and Digital Communications Chapter 5 The filter noise n o (t) has a flat power spectral density within the bandwidth of the bandpass filter: n o (t) PSD 21 khz 29 khz f C = 25 khz

16 The filter noise n o (t) can be described as a quadrature representation: n o (t) = W(t) cos (2π f C t) ) + Z(t) sin (2π( f C t) ) S&M Eq. 5.62R In the coherent receiver the noise is processed: n o (t) cos (2π f C t) = W(t) cos 2 (2π f C t) ) + S&M Eq. 6.5 Z(t) cos (2π f C t) sin (2π f C t) PSD 21 khz 29 khz f C = 25 khz

17 Applying the trignometric identity the filter noise n o (t) is: n o (t) cos (2π f C t) = ½ W(t) + ½ W(t) cos (4π f C t) ) + ½ Z(t) sin (4π f C t) ) S&M Eq. 6.5 After the lowpass filter in the receiver the demodulated signal is: s demod (t) = ½ γ A C s(t) + ½ W(t) S&M Eq. 6.7 PSD 21 khz 29 khz f C = 25 khz

18 The transmitted DSB-SC SC AM signal is: s DSB-SC SC AM AM(t) ) = A C s(t) cos (2π f C t) The average normalized bi-sided power of s DSB-SC SC(t) ) is found in the spectral domain with S(f) ) = F (s(t)): 2 1 P trans =A [ S(f f C) + S(f+f C) ] df 2 2 S&M Eq. 6.8

19 The dual-sided spectral do not overlap (at zero frequency) and the cross terms are zero so that: P trans =A [ S DSB-SC(f f C) + S DSB-SC(f+f C )] df 2 2 A P trans = P 2 s where P s is the average normalized power of s(t). S&M Eq. 6.9

20 The average normalized power of s(t) is found in the spectral domain: 2 2 P s = S(f) df = S(f + f C) df S&M Eq In a noiseless channel the power in the demodulated DSB-SC SC AM signal is: γ P = γ A P= P 4 2 demod, noiseless s trans S&M Eq. 6.11

21 The average normalized power of the processed noise is: 1 P processed noise = N o(2 B) 4 The signal-to to-noise power ratio then is: 2 γ P trans 2 γ Ptrans SNRcoherent DSB-SC = 2 = 1 N(2 B) N o B 4 o 2 B S&M Eq. 6.12

22 The DSB-SC SC AM coherent receiver requires a phase and frequency synchronous reference signal. If the reference signal has a phase error φ then: SNR coherent DSB-SC phase error 2 2 γ cos ϕ P N B o trans = S&M Eq. 6.17

23 The DSB-SC SC AM coherent receiver requires a phase and frequency synchronous reference signal. If the reference signal has a frequency error f then: S demod frequency error (t (t)) = ½ γ A C s(t) cos (2π f f t) + ½ X(t) ) cos (2π f f t) + ½ Y(t) ) sin (2π f f t) S&M Eq Although the noise component remains the same, the amplitude of the demodulated signal varies with f:

24 Chapter 6 Analog Modulation and Demodulation Non-coherent Demodulation of AM Signals Pages

25 The non-coherent AM (DSB-LC) receiver uses an envelope detector implemented as a semiconductor diode and a low- pass filter: The DSB-LC AM analytical signal is: s DSB-LC AM where c is the DC bias (offset). AM(t) ) = A C (c + s(t)) cos (2π f C t)

26 EE4512 Analog and Digital Communications Chapter 5 The envelope detector is a half-wave rectifier and provides a DC bias (c)) to the processed DSB-LC AM signal : c = DC bias

27 EE4512 Analog and Digital Communications Chapter 5 The output of the half-wave diode rectifier is low-pass filtered to remove the carrier frequency and outputs the envelope which is the information:

28 The DSB-LC AM signal can be decomposed as: s DSB-LC AM AM(t) ) = s(t) cos (2π f C t) + A C c cos (2π f C t) S&M Eq. 6.20R The average normalized power of the information term: 2 AC P info term = P 2 S S&M Eq. 6.23

29 The average normalized transmitted power is: P P T 1 = carrier term T 0 C C 2 2 AC c carrier term = 2 [ ] A c cos(2πf t) dt Since s(t) ) + c must be >= 0 to avoid distortion in the DSB-LC AM signal: c min [s(t[ s(t)] or c 2 s 2 (t) for all t. 2 S&M Eq. 6.24

30 Therefore c 2 P s and for DSB-LC AM: P P S&M Eq carrier term info term The power efficiency η of a DSB-LC AM signal is: Pinfo term Pinfo term η = = 0.5 P + P P carrier term info term trans DSB-LC AM term S&M Eq. 6.29

31 The DSB-LC AM signal wastes at least half the transmitted power because the power in the carrier term has no information: P P η 0.5 carrier term info term The modulation index m is defined as: [ ] min[ s(t) + c] [ ] [ ] max s(t) + c m = S&M Eq max s(t) + c + min s(t) + c

32 The modulation index m defines the power efficiency but m must be less than 1. If m > 1 then min [s(t[ s(t) ) + c] < 0 and distortion occurs. [ ] min[ s(t) + c] [ ] [ ] max s(t) + c m = S&M Eq max s(t) + c + min s(t) + c

33 The average normalized power of the demodulation noiseless DSB-LC AM signal is: P demod, noiseless = 2 2 γ Pinfo term Then the signal-to to-noise power ratio for the DSB-LC AM signal is: S&M Eq γ Pinfo term γ Ptrans DSB-LC SNRnoncoherent DSB-LC = = η N (2 B) N B o 2 B o S&M Eq. 6.39

34 The non-coherent AM (DSB-LC) receiver is the crystal radio which needs no batteries! Power for the high- impedance ceramic earphone is obtained directly from the transmitted signal.. For simplicity, the RF BPF is omitted and the audio frequency filter is a simple RC network.

35 Chapter 6 Analog Modulation and Demodulation Frequency Modulation and Phase Modulation Pages

36 The analytical signal for an analog phase modulated (PM) signal is: (t)) = A C cos [2π f C t + α s(t)] S&M Eq s PM (t where α is the phase modulation constant rad/v and A C is the carrier amplitude. The continuous analog signal s(t) is a baseband signal with the information content (voice or music) to be transmitted.

37 The analytical signal for an analog frequency modulated (FM) signal is: s FM (t (t)) = A C cos{ { 2π 2 [f C + k s(t)] t + φ] ] S&M Eq where k is the frequency modulation constant Hz / V, A C is the carrier amplitude and φ is the initial phase angle at t = 0. The continuous analog signal s(t) is a baseband signal with the information content.

38 The instantaneous phase of the PM signal is: Ψ PM (t (t)) = 2π2 f C t + α s(t) ) S&M Eq The instantaneous phase of the FM signal is: Ψ FM (t (t)) = 2π [f C + k s(t)] t + φ] ] S&M Eq The instantaneous phase is also call the angle of the signal. The instantaneous frequency is the time rate of change of the angle: f(t) ) = (1/2π) dψ(t) / dt S&M Eq. 6.58

39 The instantaneous frequency of the unmodulated carrier signal is: f carrier (t (t)) = dψd carrier (t) ) / dt = d/dt {2π f C t + φ} } S&M Eq The instantaneous phase is also: t t Ψ(t) = f (λ)( ) dλ d = f (λ)( ) dλ d + φ S&M Eq There are practical limits on instantaneous frequency and instantaneous phase. To avoid ambiguity and distortion in FM signals due to phase wrapping: k s(t) f C for all t S&M Eq. 6.61

40 To avoid ambiguity and distortion in PM signals due to phase wrapping: -π < α s(t) π radians for all t S&M Eq Since FM and PM are both change the angle of the carrier signal as a function of the analog information signal s(t), FM and PM are called angle modulation. For example, is this signal FM, PM or neither: t x(t) ) = A C cos { 2π2 f C t + k s(λ) ) dλ d + φ} S&M Eq

41 The instantaneous phase of the signal is: t Ψ x (t) ) = 2π2 f C t + k s(λ) ) dλ d + φ S&M Eq. p which is not a linear function of s(t) ) so the signal is not PM. The instantaneous frequency of the signal is: f x (t) ) = (1/2π) dψ x (t) ) / dt = f C + k s(t) ) / 2π2 and the frequency difference f x f C is a linear function of s(t) so the signal is FM. The maximum phase deviation of a PM signal is max αs(t) ). The maximum frequency deviation of a FM signal is f f = max k s(t) ).

42 The spectrum of a PM or FM signal can be developed as follows: S&M Eqs through 6.71 v(t) = A C sin(2π ft C + β sin 2π ft) m v(t) = Re { exp(j 2π ft C +j β sin 2π ft) m } now exp(j 2π ft C +j β sin 2π ft) m = cos (2π ft C + β sin 2π ft) m + j sin (2π ft C + β sin 2π ft) m v(t) = Im { A exp(2π ft + jβ sin 2π ft) } now C C exp(j β sin 2π f t) = c exp(j 2π n f t) after further development m n m n = - exp(j β sin 2π ft) = J( β) exp(j 2π n f t) m n m n = - m Bessel function of the first kind

43 Bessel functions of the first kind J n (β) are tabulated for FM with single tone f m angle modulation (S&M Table 6.1): β n

44 For single tone f m angle modulation the spectrum is periodic and infinite in extent: C n m C n = - v(t) = A J ( β) sin[2π (n f + f ) t] S&M Eq n β

45 The complexity of the Bessel function solution for the spectrum of a single tone angle modulation can be simplified by the Carson s s Rule approximation for the bandwidth B.. Since β = f f / f m : B = 2 (β( + 1) f m = 2 ( f( f + f m ) Hz S&M Eq n β

46 Carson s s Rule for the approximate bandwidth of an angle modulated signal was developed by John R. Carson in 1922 while he worked at AT&T. Prior to this in 1915 he presaged the concept of bandwidth efficiency in AM by proposing the suppression of a sideband (see S&M p ) 333): B = 2 (β( + 1) f m = 2 ( f( f + f m ) Hz

47 The normalized power within the Carson s s Rule bandwidth for a single tone angle modulated signals is: P in-band, sinusoid 2 β+1 AC = 2 n = -(β+1) 2 J ( β ) Note that J -n (β)) = ± J n (β)) so that J -n2 (β)) = J n2 (β)) and for the normalized power calculation the sign of J(β) is not used. n S&M Eq Spectrum of single tone FM modulation

48 The analog FM power spectral density PSD of the voice signal has a bandwidth predicted only by Carson s s Rule since it is not a single tone. Voice PSD

49 Here f max = 4 khz, k = 25 Hz/V and f max = 40(25) = 1 khz. The Carson s s Rule approximate maximum bandwidth B = 2 ( f( f + f m ) = 10 khz or ± 5 khz (but seems wrong!) Voice 40 f C PSD Bandwidth

50 A 200 Hz single tone FM signal has a PSD with periodic terms at f C ± n f m = 25 ± 0.2 n khz. PSD f C 200 Hz

51 Here f m = 200 Hz, k = 25 Hz/V and f max = 40(25) = 1 khz. The Carson s s Rule approximate maximum bandwidth B = 2 ( f( f + f m ) = 2.4 khz or ± 1.2 khz: PSD f C 200 Hz Bandwidth

52 Since β = f f / f m = 1 khz / 0.2 khz = 5 and the Bessel function predicts a bandwidth of 2 n f m = 2(12)(200) = 4.8 khz (since n = 12 for β = 5 from Table 6.1): PSD f C 200 Hz Bandwidth

53 Chapter 6 Analog Modulation and Demodulation Noise in FM and PM Systems Pages

54 A general angle modulated transmitted signal, where Ψ(t) is the instantaneous phase, is: s angle-modulated modulated(t The received signals is: r angle-modulated modulated(t (t)) = A C cos [Ψ(t)] S&M Eq (t)) = γ A C cos [Ψ(t)] + n(t) ) S&M Eq. 6.87

55 The analytical signal for PM is: s PM (t) ) = A C cos [Ψ(t)] = A C cos [2π f C t + α s(t)] S&M Eq After development the SNR for demodulated PM is: SNR PM = (αγ( A C ) 2 P S / (2 N o f max ) S&M Eq where π < α s(t) π for all t.

56 The analytical signal for FM is: s FM (t) ) = A C cos [Ψ(t)] = A C cos [2π f C t + k s(λ) ) dλ] d S&M Eq After development the SNR for demodulated FM is: SNR FM = 1.5 (k γ A C /(2π) ) 2 P S / (N o f max3 ) S&M Eq where k s(t) f C for all t.

57 End of Chapter 6 Analog Modulation and Demodulation