4 Analog modulation 4.1 Modulation formats The message waveform is represented by a low-pass real signal mt) such that Mf) = 0 f W where W is the message bandwidth. mt) is called the modulating signal. Carrier modulation: Reversible transformation of mt) into a bandpass signal xt) centered around f c W f c is the carrier frequency). Demodulation is the inverse transformation of xt) into mt). xt) is the modulated signal. Two types of modulation schemes: xt) = x I t) cos πf c t x Q t) sin πf c t Linear modulation: linear relationship between the modulated signal and the message signal ex: AM, DSB-SC, SSB, VSB). Angle modulation: the angle of the carrier wave is varied according to the message signal ex: FM,PM). 4. Linear modulation a) Amplitude modulation AM xt) = A c 1 + a mt) ) cos πf c t = A c 1 + µmn t) ) cos πf c t x I t) = A c 1 + µmn t) ) x Q t) = 0 m n t) = mt) max mt) a is the amplitude sensitivity, µ = a max mt) is modulation index factor) 0 µ 1). Exercise: Plot a graph representing xt) and identify the message signal mt). 9
The Fourier transform of xt) is given by Xf) = A c [ δf fc ) + µm n f f c ) ] + A c [ δf + fc ) + µm n f + f c ) ] where M n f) = F {m n t)}. Exercise: Plot Xf) and find the condition on f c to avoid distortion. Let W be the bandwidth of Mf) = F {mt)} and M n f), then the bandwidth of the modulated signal xt) is B = W. Generation of AM: Ac1 + µmnt)) xt) cosπfct) Figure 4: Generation of an AM modulated signal Demodulation of AM: Synchronous coherent) detection Envelope detection Coherent detection: 30
xt) yt) ut) zt) cosπfct) W LPF Figure 5: Coherent detection of an AM modulated signal yt) = A c 1 + µmn t) ) cos πf c t) = A c 1 + µmn t) ) + A c 1 + µm nt)) cos4πf c t) ut) = A c 1 + µmn t) ) Output of low pass) zt) = A cµ m nt) = A c amt) Exercise: Assume that the local carrier is not synchronized to the signal, show that the output of the coherent demodulator is distorted. Envelope detection: The output of an envelope detector is the natural envelope xt). Envelope detection is feasible since 1 + µm n t) 0. Draw an example of a RC circuit implementing envelope detection. What conditions on the resistors and capacitor are necessary to ensure the envelope detector will function properly? 31
Envelope detection is feasible due to the inclusion of the carrier but the transmission of the carrier represents a waste of power contains no information). b) Double sideband-suppressed carrier: DSB-SC xt) = A c mt) cos πf c t x I t) = A c mt) x Q t) = 0 Exercise: Plot xt) and identify the message signal mt). What phenomenon characterizes DSB-SC? The Fourier transform of xt) is given by Xf) = A c Mf f c) + A c Mf + f c) where Mf) = F {mt)}. Exercise: Plot Xf) and find the condition on f c to avoid distortion. Let W be the bandwidth of Mf) = F {mt)}, then the bandwidth of the modulated signal xt) is B = W. Generation of DSB-SC: Draw a bloc diagram of a DSB-SC modulator. 3
Demodulation of DSB-SC: Only Synchronous coherent) detection Coherent detection: Using a local carrier synchronized to the received signal carrier, draw a bloc diagram of a coherent detector for DSB-SC. Envelope detection: The output of an envelope detector is the natural envelope xt) = A c mt) A c mt). AM and DSB-SC are wasting bandwidth, thus filtering of sidebands to reduce bandwidth results in a more bandwidth efficient scheme. c) Single sideband modulation: SSB xt) = A c mt) cos πf ct A c ˆmt) sin πf ct Upper sideband SSB) xt) = A c mt) cos πf ct + A c ˆmt) sin πf ct Lower sideband SSB) x I t) = A c mt) x Qt) = ± A c ˆmt) The Fourier transform of xt) is given by Xf) = A c where Mf) = F {mt)}. { 1 Mf f c) + 1 Mf + f c) 1 [ ˆMf fc ) j ˆMf + f c )] } = A c 4 Mf f c) 1 + sgnf f c ) ) + A c 4 Mf + f c) 1 sgnf + f c ) ) 33
Exercise: Plot Xf). Let W be the bandwidth of Mf) = F {mt)}, then the bandwidth of the modulated signal xt) is B = W. Generation of SSB: Using a Hilbert transformer but wideband π/ shifter difficult to implement) Using sideband filtering but demands a very sharp filter if Mf) contains very low frequencies components, hence Vestigial sideband modulation VSB) is also used) Demodulation of SSB: Only Synchronous coherent) detection Envelope detection by adding a strong carrier to the SSB signal but not a regular SSB anymore see VSB subsection) Coherent detection: Using a local carrier synchronized to the received signal carrier, draw a bloc diagram of a coherent detector for SSB. SSB is difficult to implement if the message signal mt) has a large bandwidth and it is rich in 34
low frequency components. In this case vestigial sideband modulation is used. d) Vestigial sideband modulation: VSB xt) = A c Kmt) cos πf ct A c m νt) sin πf c t x I t) = A c Kmt) x Qt) = A c m νt) = A c mt) h νt) Rationale generation of VSB): Assume that a DSB-SC signal is passed through a general bandpass filter to alter its sidebands ex. for SSB half of the sidebands filtered out). The Fourier transforms of the DSB-SC signal yt) and xt) are given by Y f) = A c { Mf fc ) + Mf + f c ) } DSB-SC Xf) = A c { Mf fc ) + Mf + f c ) } Hf) where Mf) = F {mt)}. The filter Hf) must have spectral characteristics such that the original message signal mt) can be recovered from xt) by coherent detection. Demodulation of VSB: Only Synchronous coherent) detection Envelope detection by adding a strong carrier to the VSB signal but not a regular VSB anymore Coherent detection: xt) vt) ut) cosπf c t) W LPF Figure 6: Coherent detection of VSB vt) = xt) cos πf c t 35
V f) = 1 [ Xf fc ) + Xf + f c ) ] = A { c [Mf fc ) + Mf) ] Hf f c ) + [ Mf) + Mf + f c ) ] Hf + f c ) } 4 = A c 4 Mf)[ Hf f c ) + Hf + f c ) ] + A c [ Mf fc )Hf f c ) + Mf + f c )Hf + f c ) ] 4 The output of the lowpass filter is given by For distortionless transmission Uf) = A c 4 Mf)[ Hf f c ) + Hf + f c ) ] Uf) = A c 4 KMf) i.e. ut) = A c 4 Kmt) where K is a constant. Thus the filter Hf) must satisfy the so-called vestigial symmetry condition: Hf f c ) + Hf + f c ) = K = const. f W Time domain representation of VSB signals: xt) = ht) yt) = xt) = 1 ht) ỹt) Since yt) = A c cosπf c t)mt), its complex envelope is given by ỹt) = A c mt) = Ỹ f) = A cmf) The Fourier transform of ht) is Hf) = Hf + f c ) f > f c hence Xf) = 1 Hf)Ỹ f) = A cmf)hf + f c ) f > f c 9) The inphase and quadrature components of xt) are x I t) = R { xt)} = 1 [ xt) + x t)] 36
x Q t) = I { xt)} = 1 j [ xt) x t)] their Fourier transform are then given by X I f) = 1 [ Xf) + X f)] = A c [Mf)Hf + f c) + M f)h f + f c )] from 9) = A c Mf) [Hf + f c) + Hf f c )] since mt) and ht) are real.) = A c X Q f) = 1 j Mf)K from the vestigial symmetry 10) [ Xf) f)] X = A c j Mf) [Hf + f c) Hf f c )] = A c Mf)H νf) 11) where the filter H ν f) is defined as H ν f) = 1 j [Hf + f c) Hf f c )] From 10) and 11), the inphase and quadrature components of a VSB signal are given by x I t) = A c Kmt) x Q t) = A c mt) h νt) = A c m νt) Bandwidth of mt): W Bandwidth of xt): W < B < W typically B VSB = 1.5B SSB ) If H ν f) = j sgnf), then we obtain the upper sideband SSB. Envelope detection of SSB and VSB: We add a strong carrier A c cosπf c t) with A c Ac to be demodulated is given by xt) = A c + A ) c Kmt) K mt), Ac m νt) ) such that the signal cosπf c t) A c m νt) sinπf c t) 37
= A c 1 + A ck mt) A c For simplicity assume K = 1, and let Ac A c xt) = A c 1 + a xt) = A c ) mt) ) 1 + a mt) ) cosπf c t) A c = a A c m A ν t) sinπf c t) c cosπf c t) A c am ν t) sinπf c t) 1) + ja a c m νt) complex envelope of xt)) The output of an envelope detector is the natural envelope of the input, hence the output of the envelope detector is given by xt) = [ = A c A c A c 1 + a mt) 1 + a ) + A ) [ mt) 1 + ) 1 + a mt) c a 4 m νt) ] 1/ a 4 m νt) 1 + a mt) ) since a mt) 1 and a m ν t) 1. This method is used in TV systems. Distortion due to the envelope detection of VSB is reduced by reducing a ensuring the conditions a mt) 1 and a m ν t) 1. ] 1/ 4.3 Multiplexing The purpose of multiplexing is to transmit several signals {m 1 t),..., m N t)} at the same time by the use of a single communication system. This can be achieved by combining the signals into one signal st) such that each of the signals m t) can be extracted from st). In this section, we present two types of multiplexing; quadrature carrier multiplexing and frequency division multiplexing FDM). A third multiplexing technique called Time Division Multiplexing TDM) will be considered in the context of signal sampling. a) Quadrature carrier multiplexing Since cosπf c t) and sinπf c t) are orthogonal functions, the principle of quadrature multiplexing of two signals is to transmit one signal using a carrier of the form cosπf c t) and to transmit the other signal using a carrier of the form sinπf c t). Let m 1 t) and m t) be two lowpass signals with bandwidth W. The schemes of multiplexing and demultiplexing follows: 38
m 1 t) cosπf c t) st) = m 1 t) cosπf c t) + m t) sinπf c t) m t) sinπf c t) Figure 7: Quadrature carrier multiplexing v 1 t) m 1 t) Complete the proof that the scheme of Fig. 8 is a quadrature carrier demultiplexer: v 1 t) = st) cosπf c t) st) cosπf c t) W LPF = + }{{} lowpass term }{{} bandpass term v t) m t) v t) = st) sinπf c t) sinπf c t) W LPF = + }{{} lowpass term }{{} bandpass term Since the bandpass terms are removed by the lowpass filter, the result follows. Figure 8: Quadrature carrier demultiplexing 39
b) Frequency division multiplexing FDM) The principle of frequency division multiplexing is to modulate each signal m 0 i t) using a different carrier frequency f ci such that the spectrum of the modulated signals x i t) do not overlap. Then a FDM signal is obtained by adding the modulated signal yielding a signal with a higher bandwidth. The multiplexed signal can be further modulated before transmission. Hence the modulation of the signals m i t) bandlimited version of m 0 i t)) to be multiplexed is called sub-modulation and the carrier {f ci } i=1,...,n are sub-carriers. In a FDM system, the sub-carriers are selected such that the spectrum of the sub-modulated signals do not overlap. Therefore the original message signals m 0 i t) have to be passed first through a lowpass filter that limit them to a predetermined bandwidth W. If the original message signals are already bandlimited to W, no lowpass filtering is required. Let B be the bandwidth of each of the sub-modulated signals x i t). To avoid overlapping of the spectrum of the sub-modulated signals and hence to ensure distortionless demultiplexing), we must have f ci f c B FDM multipler and demultiplexer follows: m 0 1t) LPF m 1t) Modulator f c1 x 1 t) st) = N i=1 x it) m 0 N t) LPF m N t) Modulator f cn x N t) Figure 9: Frequency Division multiplexing example of FDM: st) = st) = N 1 + ai m i t) ) cosπf ci t) AM sub-modulation BW of st): NW ) i=1 N i=1 A c [ mi t) cosπf ci t) ˆm i t) sinπf ci t) ] SSB sub-modulation BW of st): NW ) 40
BPF 1 x 1 t) Demodulator f c1 m 1 t) st) f c1 B BPF N x N t) Demodulator f cn m N t) f cn B Figure 10: Frequency Division demultiplexing Complete the proof that the scheme of Fig. 10 is a FDM demultiplexer: st) = + }{{}}{{} bandpass term around f ci sum of bandpass terms around f cj j i) The bandpass filter around f ci eeps the bandpass term around f ci namely x i t)) and removes all the other bandpass terms. It is seen that non-overlapping of the spectrums of x i t) is needed to avoid distortion. Then each x i t) can be demodulated. Note that for SSB sub-modulation, since the spectrum of x i t) contains only one sideband, the required bandpass filter should pass only one sideband as seen in Fig. 11. BPF n f cn f cn + W f Figure 11: Bandpass filter required for upper sideband SSB 41
4.4 Angle modulation With mt) the message signal, an angle modulated signal is defined as where ϕt) is given by xt) = A c cos θt) ) = A c cos πf c t + ϕt) ) 13) ϕt) = K t mτ)ht τ)dτ = πk f t mτ)ht τ)dτ where K = πk f is the phase sensitivity of the modulator) expressed in rd/volt, K f = K π is the frequency sensitivity of the modulator) expressed in Hz/Volt and ht) is assumed to be causal. ϕt) is the phase of xt) and θt) = πf c t + ϕt) is the angle of xt). The instantaneous frequency of xt) is defined as ft) = 1 dθt) π dt The maximum phase deviation of xt) is = f c + 1 dϕt) π dt ϕ = max ϕt) The maximum frequency deviation of xt) is f = max ft) f c = max 1 π The complex envelope of xt) is given by Phase modulation PM): xt) = A c e jϕt) dϕt) dt ht) = δt) Hf) = 1 = ϕt) = Kmt) The phase of a PM signal is proportional to the message signal. Frequency modulation FM): 1, t > 0 1 ht) = ut) =, t = 0 = ϕt) = K 0, t < 0 t mτ)dτ 4
Hf) = 1 jπf + 1 δf) The instantaneous frequency of an FM signal is proportional to the message signal up to a carrier frequency shift ft) = K π mt) + f c = K f mt) + f c Frequency modulation with pre-emphasis/de-emphasisfm): When noise analysis is done for FM, it can be shown that the power spectral density of noise at the FM receiver output is proportional to f in the frequency-band of the message, thus the noise power is higher at high frequencies. To increase the overall signal-to-noise ratio, practical systems use a pre-emphasis filter h pe t) before frequency modulation. The purpose of h pe t) is to artificially increase the high-frequency components of the message signal to compensate for the high noise level. After FM detection, the recovered emphasized message signal is passed through a de-emphasis filter h de t) which must be ideally the inverse filter of the pre-emphasis filter i.e. H de f) = 1 H pef) ). If pre-emphasis is used, the emphasized message signal is given by and the FM modulated signal is xt) = t m pe t) = mt) h pe t) m pe τ)dτ = m pe t) ut) = mt) h pe t) ut) Therefore 13) corresponds to an FM signal with pre-emphasis when ht) = h pe t) ut). Equivalently in the frequency domain 1 Hf) = H pe f) δf) + 1 ) = 1 jπf H pe0)δf) + H pef) jπf An example is H pe f) = 1 + jf c. a) Narrow-band angle modulation Narrow-band angle modulation: ϕt) 1 Hence If ϕt) 1, then xt) = A c e jϕt) A c 1 + jϕt)) xt) = R { xt)e jπfct} complete this line) A c cosπf c t) A c ϕt) sinπf c t) 43
Thus narrow-band angle modulation has similar features to AM. Bandwidth of narrow-band angle modulation: bandwidth of ϕt). Generation of narrow-band angle modulation: Exercise: Draw a bloc diagram that generates a narrow-band angle modulated signal. Detection of narrow-band angle modulation: Exercise: Draw a bloc diagram of a system with input xt) and output Ac ϕt): b) Wide-band angle modulation Wide-band angle modulation: ϕt) 1 xt) = A c e jϕt) Generation of wide-band angle modulation using indirect method Armstrong s method) : Generates a narrow-band angle modulated signal t ) x 0 t) = A c cos πf c t + K mτ)ht τ)dτ with K t mτ)ht τ)dτ 1. Pass x 0 t) through a Frequency multiplier by N whose bloc diagram is illustrated in Fig. 1. The output to the frequency multiplier by N is then t ) xt) = A c cos πnf c t + NK ht τ)mτ)dτ 44
x 0 t) ) N x N 0 t) BPF xt) Nf c Figure 1: frequency multiplier by N If N 1 we have a wide-band angle-modulated signal around a carrier at Nf c. Analysis of a frequency multiplier by N: yt) = x 0 t)) N = 1 t ) cos πnf N 1 c t + NK mτ)ht τ)dτ + other terms lie cosπnf c t +...) with n < N since [ cos α) n = 1 n 1 n cos α) n 1 = 1 n =0 [ n 1 n ) cos n )α ) + n 1 =0 ) ] n n ) cos n 1)α ) + ) ] n n The other terms are removed by the bandpass filter yielding xt) as output. Generation of any FM modulation using direct method; use of Voltage-Controlled-Oscillator VCO): V c t) VCO xt) Figure 13: Voltage-Controlled Oscillator VCO) If the input of a VCO is a voltage V c t) then the output of a VCO with unmodulated frequency of oscillation f c has an instantaneous frequency given by ft) = f c + K f V c t) 45
Thus its output is see for example Hartley oscillator. d) Tone modulation t ) xt) = A c cos πf c t + K V c τ)dτ Tone modulation corresponds to a sinusoidal message signal. Let mt) = A m cosπf c t) A m 0) applied at some time t 0 such that t 0 t. Let us calculate the steady state expression of ϕt) corresponding to a general angle modulated signal. ϕt) = K K = K = K t t 0 A m cosπf m τ)ht τ)dτ t = KA m A m cosπf m τ)ht τ)dτ A m cosπf m τ)ht τ)dτ steady state) ht) is causal) F τ {A m cosπf m τ)} F τ {ht τ)} df [δf f m ) + δf + f m )] [ H f)e jπft] df = KA m { H f m )e jπfmt + H f m )e jπfmt} = KA m R { Hf m )e jπfmt} ht) is real, thus H f) = Hf)) Generalization of Parseval theorem t fixed) = KA m Hf m ) cos πf m t + arg [Hf m )] ) steady state) 14) Show that for PM and FM ϕt) is given by { KA m cosπf m t) ϕt) = KA m πf m sinπf m t) PM FM The modulation index is defined as the maximum phase deviation, or equivalently the maximum deviation of the angle θt) from πf c t. β = max ϕt) = max θt) πf c t 46
= KA m Hf m ) assuming K, A m 0 Hence from 14) the steady state expression of ϕt) for tone modulation is also given by ϕt) = β cos πf m t + arg [Hf m )] ) where θ = arg [Hf m )] + π. Phase modulation PM): = β sin πf m t + θ) β PM = KA m = ϕ ϕ: maximum phase deviation Frequency modulation FM): β FM = KA m πf m = K fa m f m = f f m f = K f A m : maximum frequency deviation Transmission bandwidth of an angle modulated signal with tone modulation: For convenience, we define θ = arg [Hf m )] such that θ FM = 0 and θ PM = π and we use ϕt) = β sin πf m t + θ) The complex envelope of the angle-modulated signal is xt) = A c e jϕt) = A c exp {jβ sinπf m t + θ)} Unlie the original signal xt), the complex envelope xt) is periodic with period T m = 1 f m, therefore xt) admits a Fourier series representation xt) = where the Fourier coefficients are given by n= c n e jπnfmt Tm 1 c n = T m Tm xt)e jπnfmt dt = A c f m Tm = A cf m πf m Tm θ+π = A c e jnθ 1 π θ π θ+π xt)e j[β sinπfmt+θ) πnfmt] dt e j[β sinu) nu+nθ] du u = πf m t + θ) θ π e j[β sinu) nu] du 47
= A c e jnθ J n β) where J n ) is the n th Bessel function of the first ind defined as J n β) = 1 π = =0 θ+π θ π 1) β e j[β sinu) nu] du integral independent of θ 1 ) ) n+! + n)! Therefore the modulated signal xt) with tone modulation is given by { xt)e jπfct} xt) = R { = R n= A c e jnθ J n β)e j[πfct+πnfmt] } = A c n= J n β) cos πf c + nf m )t + nθ ) and has a Fourier transform given by Xf) = A c n= J n β) {[ δf f c nf m ) + δf + f c + nf m ) ] cosnθ) +j [ δf f c nf m ) δf + f c + nf m ) ] sinnθ) } Thus it is seen that angle-modulated signals have an infinite bandwidth. A c n= J nβ) [ δf f c nf m ) + δf + f c + nf m ) ] FM A Xf) = c { p= Jp β) [ δf f c pf m ) + δf + f c + pf m ) ] 1) p + jj p+1 β) [ δf f c p + 1)f m ) δf + f c + p + 1)f m ) ]} PM Using J n β) = 1) n J n β) and mathematical tables, the Fourier transform Xf) for FM is illustrated for example in Fig. 14 and Fig. 15. Note that usually only the magnitude of Xf) is drawn. It is obtained by reversing the negative peas to become positive peas. 1 Expand the integral as the sum of the three integrals π θ π + π π + θ+π and maes the change of variable v = π u + π in the first one. 48
Xf) β = ) f c - 3f m f - f c m f f - f c m f c f + f c m f + 3f c m positive frequencies only) Figure 14: Fourier transform of Xf) β = ) with FM modulation Xf) β = 8) f c - f m f c f + c 3f m f f c - 3f m f c+ f m positive frequencies only) Figure 15: Fourier transform of Xf) β = 8) with FM modulation Some properties of Bessel functions: J n β) = 1) n J n β) When β 1, J n β) lim J nβ) = β 0 ) n β 1 n! { 1 n = 0, 0 else. = J 0 β) 1 J 1 β) β J n β) 0, n 49
To obtain a definition of an effective or essential) bandwidth, that is a bandwidth that contains most of the total power usually 98% or 99%), let us consider the average power of xt). The average power of xt) is given by 1 P x = lim T T A c = lim T T = A c + = A c T T x t)dt n= p= n= p= T T T J n β)j p β) cos πf c + nf m )t + nθ ) cos πf c + pf m )t + pθ ) dt T [ T J n β)j p β) lim T 1 T T cos πn p)f m t + n p)θ ) ] dt [ 1 Tn,p J n β)j p β) cos T p= n,p T n,p T ) ] n,p πt cos + n p)θ dt n= + 1 T n,p T n,p T n,p cos πf c + n + p)f m )t + n + p)θ ) dt ) πt + n + p)θ dt T n,p where T n,p = [f c + n + p)f m ] 1 and T n,p = n p) 1 f 1 m. Since the two trigonometric integrals are zero unless n = p and are equal to 1 when n = p, the average power of xt) is given by P x = A c Jnβ) = P c + n= n= P n P c = A c J 0 β) P n = A c J nβ) n th side-band power P n = A c J nβ) = A c J nβ) = P n Note that from xt) = A c cos πf c t + ϕt)), the average power of xt) is also given by therefore we can deduce that n= P x = A c J nβ) = 1 50
An effective essential) bandwidth can be defined as B T = n max f m where n max is the largest n such that J n β) 0.01 and > n J β) < 0.01. It turns out that n max depends on β. In terms of power, it is equivalent to neglect side-bands that contribute to less than 0.01% of the total power. Another rule for the effective bandwidth is obtained by approximating n max by a linear curve, B T = f m β + c) 1 c When c = 1, the classical Carson s rule is obtained example: Show that Carson s rule for FM is B Cars T = f m β + 1) B FM T = f m + f) d) General modulating signal with bandwidth W If mt) is periodic, mt) should be expressed in terms of its Fourier series representation. If mt) is a general non-periodic deterministic signal, we can use e jϕt) = =0 Generalization of Carson s rule to a general modulating signal [ϕt)]!. Worst-case tone approach. Let mt) be a message signal with bandwidth W and maximum amplitude max mt). Assume we model mt) as an infinite number of tones of frequency f m and maximum amplitude A m, then its effective bandwidth would be B T = max {f m [β + 1]} = max max f m + max KA m f m Hf m ) f m + K max = W + K max mt) max A m max [f m Hf m ) ] 15) [f m Hf m ) ] where to get 15) we have applied the worse case tone approach. The worse case tone approach consists of evaluating the bandwidth obtained by considering a tone at the highest possible frequency and the highest possible amplitude, thus maximizing the product of the second term in B T by maximizing its two terms A m and f m Hf m ) separately. Let us consider now the term max [f m Hf m ) ] for the special case of FM and PM modulation. For FM modulation ] max [f m Hf m ) ] = max [ fm πf m = max 1 = 1 = W HW ) 51
consistent with the worse tone at W. Note that for FM maximizing A m is equivalent to maximize the frequency deviation f. For PM modulation Hence we obtain max [f m Hf m ) ] = max f m = W = W HW ) B T = W + K max mt) )W HW ) = W β + 1) where β called for general modulating signal the deviation ratio is defined as β = K max mt) ) HW ) = { f W = K f max mt) W ϕ = K max mt) FM PM Exercise: Calculate using Carson s rule the bandwidth of commercial FM broadcasting characterized by a maximum allowed maximal frequency deviation of 75Hz and a maximum audio signal of 15Hz. e) Detection of FM signals Basic structure: using a differentiator and an envelope detector Show that for large f c, if xt) is applied to a differentiator followed by an envelope detector, the resulting output is proportional to the message signal: t ) xt) = A c cos πf c t + K mτ)dτ dxt) = dt complete zt) = Output of envelope detector complete) Frequency domain differentiation slope demodulator) Draw the transfer function of a real filter that can implement the differentiation. 5
Note that the differentiation operation has to be implemented only in the bandwidth of the modulated signal. Time domain differentiation Based on the following approximation dx dt xt) xt t) t draw the bloc diagram of a time domain differentiator. Quadrature detector for an angle-modulated signal Using a differentiator and a π/ phase shifter, draw the bloc diagram of a demodulator for an angle modulated signal, that is based on similar principles as coherent detection for AM. 53
Show that the output of the quadrature detector is given by A c dϕt) dt. f) Form of Hf) for FM The phase of an FM modulated signal is given by ϕt) = K Its Fourier transform is given by Hence in theory ht τ)mτ)dτ = K F {ϕt)} = KHf)Mf) = K [ 1 jπf Hf) = 1 jπf + δf) t mτ)dτ ] M0) Mf) + δf) Let H f) = 1, then the corresponding phase jπf ϕ t) is given by ] [ ϕ t) = F [KH 1 f)mf) = F 1 K 1 ] jπf Mf) and the original phase is ϕt) = F 1 [KHf)Mf)] = F 1 [ K 1 [ = F 1 K 1 ] jπf Mf) + K M0) ] [ ] K jπf Mf) + F 1 δf)m0) Hence the difference between the two phases is only a constant phase shift which is equivalent to a change of the time of origin. Furthermore the instantaneous frequency given by ft) = f c + 1 dϕt) π dt 54
is independent of the constant KM0). Finally the two possible expressions of Hf) 1. Hf) = 1 jπf + 1 1 δf). Hf) = jπf yield the same result for the deviation ratio β. Hence in the derivation of results Hf) = 1 jπf can be considered instead of Hf) = 1 jπf + 1 δf). 55