ELG3175 Introduction to Communication Systems Lecture 12-13

Angle Introduction to Angle Modulation

• In angle modulation, the of the modulated signal remains fixed while the information is carried by the angle of the carrier. • The process that transforms a message signal into an angle modulated signal is a nonlinear one. • This makes analysis of these signals more difficult. • However, their modulation and demodulation are rather simple to implement. The angle of the carrier

• Let θi(t) represent the instantaneous angle of the carrier. • We express an angle modulated signal by:

s(t) = Ac cos(θi (t))

where Ac is the carrier amplitude.

Instantaneous frequency

• One cycle occurs when θi(t) changes by 2π radians, therefore the average frequency of s(t) on the interval t to t+Δt is: θ (t + Δt) −θ (t) f = i i Δt 2πΔt

• Therefore the instantaneous frequency is found in the limit as Δt tends towards 0. 1 dθ (t) f (t) = i i 2π dt

• There are two angle modulation techniques. – Phase modulation (PM) – (FM) • In PM, the phase of the carrier is a linear function of the message signal, m(t). Therefore sPM(t) is:

sPM (t) = Ac cos(2πf ct + k p m(t) + φc )

where kp is the phase sensitivity and φc is the phase of the unmodulated carrier.

• To simplify expressions, we will assume that φc = 0. Therefore the angle of a PM signal is given by θi(t) = 2πfct + kpm(t). FM

• For FM, the instantaneous frequency is a linear function of the message:

fi (t) = fc + k f m(t)

• where kf is the frequency sensitivity.

t t θi (t) = 2π ∫ fi (τ )dτ = 2πfct + 2πk f ∫ m(τ )dτ −∞ −∞ ⎡ t ⎤ s (t) A cos 2 f t 2 k m( )d FM = c ⎢ π c + π f ∫ τ τ ⎥ ⎣ −∞ ⎦ Instantaneous frequency of a PM signal / Instantaneous phase of an FM signal

• From sPM(t), we find k p dm(t) fi (t) PM = f c + 2π dt From s (t), we find FM t φi (t)FM = 2πk f ∫ m(τ )dτ 0

k /2 k Mod m(t) d/dt p π f s (t) FM PM

2 k k Mod m(t) ∫ π f/ p s (t) PM FM

Example

• Find sFM(t) and sPM(t) if m(t) = Acos(2πfmt).

– SOLUTION

! # sPM (t) = Ac cos"2! fct + Akp cos(2! fmt)$

t A Acos(2πfmτ )dτ = sin(2πfmt) ∫ 2 f −∞ π m ⎡ Ak f ⎤ sFM (t) = Ac cos⎢2πfct + sin(2πfmt)⎥ ⎣ fm ⎦ • The PM and FM of the example are shown here for Ac = 5, A = 1, fc = 1 kHz, fm = 100 Hz, kp = 2π rads/V and kf = 500 Hz/V. 6 4 s (t) PM 2 0 -2 -4 -6 0 0.005 0.01 0.015 0.02 0.025 t en secondes 6 4 sFM(t) 2 0 -2 -4 -6 0 0.005 0.01 0.015 0.02 0.025 t en secondes Characteristics of Angle Modulated Signals

PM Signal FM Signal

t Instantaneous phase k m(t) p 2πk f ∫ m(τ )dτ φi(t) 0

Instantaneous k p dm(t) f c + f c + k f m(t) frequency 2π dt Maximum phase 2 π k f | x ( t ) | max où k p | m(t) |max t deviation Δφmax x(t) = ∫ m(τ )dτ −∞ Maximum frequency k p où | x(t) | dm(t) deviation Δf max x(t) = k f | m(t) |max max 2π dt 2 2 Power Ac Ac 2 2 Modulation index

• Assume that m(t) = Amcos(2πfmt). The resulting FM signals is:

! Amk f $ sFM (t) = Ac cos#2! fct + sin(2! fmt)& " fm %

• For the FM signal

k f Am Δfmax βF = = fm fm FM Modulation index

• For any m(t) which has bandwidth Bm, we define the modulation index as :

k f | m(t) |max !fmax !F = = Bm Bm Example

• The signal m(t) = 5sinc2(10t).

Find the modulation index for FM modulation with

kf = 20 Hz/V. – SOLUTION

• Bm = 10Hz, therefore βF = 20×5/10 = 10. Narrowband FM

• Consider an FM signal : ⎡ t ⎤ s (t) A cos 2 f t 2 k m( )d FM = c ⎢ π c + π f ∫ τ τ ⎥ ⎣ −∞ ⎦ t 2 k m( )d 1 where π f ∫ τ τ << −∞

• We say that sFM(t) is a narrowband FM signal.

• For example, consider when m(t) = Amcos(2πfmt). ⎛ A k ⎞ s (t) A cos⎜2 f t m f sin(2 f t)⎟ FM = c ⎜ π c + π m ⎟ ⎝ fm ⎠ sFM (t) = Ac cos(2πfct + βF sin(2πfmt)) Narrowband FM

• When βF << 1, the FM signal is NBFM. • cos(A+B) = cos(A)cos(B)-sin(A)sin(B). Therefore

⎡ t ⎤ sFM (t) = Ac cos⎢2πfct + 2πk f ∫ m(τ )dτ ⎥ ⎣⎢ 0 ⎦⎥ ⎛ t ⎞ ⎛ t ⎞ = A cos 2πf t cos⎜2πk m(τ )dτ ⎟ − A sin 2πf t sin⎜2πk m(τ )dτ ⎟ c ( c ) ⎜ f ∫ ⎟ c ( c ) ⎜ f ∫ ⎟ ⎝ 0 ⎠ ⎝ 0 ⎠ ⎛ t ⎞ ≈ A cos 2πf t − A ⎜2πk m(τ )dτ ⎟sin 2πf t c ( c ) c ⎜ f ∫ ⎟ ( c ) ⎝ 0 ⎠ (if A << 1, cos(A) ≈ 1 and sin(A) ≈ A.) NBFM Modulator

Bandwidth of NBFM approx. = 2Bm

A cos(2 f t) c π c + + sNBFM(t)

Acsin(2πfct) Trans. - t Hilbert ∫ m(τ )dτ t 0 ( )d 2 k × m(t) ∫ • τ π f 0 Wideband FM - WBFM

• For an FM signal to be NBFM, βF << 1. • Any signal that is not narrowband is therefore wideband.

• However, typically βF > 1 for an FM signal to be considered wideband. • The bandwidth of WBFM signals is larger than NBFM

since Δfmax is increased. The Fourier series of the WBFM signal when m(t) = Amcos2πfmt.

• We can express the complex envelope of the WBFM signal using Bessel functions of first kind and order n as ∞ ~ j2πnfmt sFM (t) = ∑ Ac J n (β F )e n=−∞ • And the WBFM signal itself becomes:

~ j2πfct sFM (t) = Re{sFM (t)e } ∞ ⎧ j(2 f t 2 nf t) ⎫ Re A J ( )e π c + π m = ⎨ ∑ c n β F ⎬ ⎩n=−∞ ⎭ ∞ A J ( )cos(2 ( f nf )t) = ∑ c n β F π c + m n=−∞ Spectrum: Examples

1 1

β=0.3 β=1 0.5 0.5

0 0 10 0 15 0 20 0 25 0 . 30 0 10 0 15 0 20 0 25 0 . 30 0 0.4 0.4

0.3 β=10 β=5 0.2 0.2

0.1

0 0 0 10 0 20 0 30 0 40 0 0 10 0 20 0 30 0 . 40 0 . Spectrum of the WBFM signal when m(t)

= Amcos2πfmt.

• The spectrum of this signal is: A ∞ S ( f ) = c J (β ) δ ( f − f − nf ) +δ ( f + f + nf ) FM ∑ n F [ c m c m ] 2 n=−∞ • This expression shows that the FM signal’s spectrum is made up of an infinite number of impulses at frequencies f = fc+nfm. • Therefore, theoretically, this WBFM signal has infinite bandwidth. • However, the properties of the Bessel function show that most of these impulses contribute little to the overall power of the signal and are negligible. – We define the practical bandwidth as the range of frequencies which contains at least 99% of the total power of the WBFM signal. Lecture 6 Properties of Jn(β) 1) If n is an integer :

Jn(β) = J-n(β) for even n and

Jn(β) =-J-n(β) for odd n

2) 4) Im{Jn(β)}=0 when β << 1

J0(β) ≈ 1

J1(β) ≈ β/2 ∞ 2 and 3) ∑ J n (β ) = 1 n=−∞ Jn(β) ≈ 0, n > 1

Power of the FM signal

• The power of an FM signal is: A2 P = c FM 2

∞ sFM (t) = ∑ Ac J n (β F )cos(2π ( fc + nfm )t) n=−∞ • The power of the above expression is: 2 ∞ Ac 2 P = ∑ J n (β F ) 2 n=−∞ Filtering a WBFM signal to limit its bandwidth.

B sFM (t) = x(t) ∞ A J ( )cos(2 ( f nf )t) ∑ c n β F π c + m f n=−∞ -fc fc

We want to choose B so that the power of x(t) Is at least 0.99× the power of sFM(t). X x(t) = ∑ Ac J n (β F ) cos(2π ( f c + nf m )t) n=−X where X is the largest integer that satisfies : B B fc + Xfm ≤ fc + and f − Xf ≥ f − 2 c m c 2 • The power of x(t) is: A2 X P c J 2 ( ) x = ∑ n βF 2 n=− X • Therefore we must choose X so that: X J 2 ( ) 0.99 ∑ n β F ≥ n=− X

2 2 • We know that Jn (βF) = J-n (βF). Therefore

X J 2 ( ) 2 J 2 ( ) 0.99 0 β F + ∑ n β F ≥ n=1 Values of Jn(β)

n β=0.1 β=0.2 β=0.5 β=1 β=2 β=3 β=5 β=10 0 0.997 0.99 0.938 0.765 0.224 -0.2601 -0.178 -0.246 1 0.05 0.1 0.242 0.44 0.577 0.3391 -0.323 0.043 2 0.001 0.005 0.031 0.115 0.353 0.4861 0.047 0.255 3 2×10-5≈ 0 1.6×10-4 0.0026 0.02 0.129 0.3091 0.365 0.058 4 0.002 0.034 0.1320 0.391 -0.220 5 0.007 0.0430 0.261 -0.234 6 0.001 0.0114 0.131 -0.014 7 0.0025 0.053 0.217 8 0.018 0.318 9 0.006 0.292 10 0.001 0.207 11 0.123 12 0.063 13 0.029 Example

• The signal m(t) = Amcos(2πfmt) is to be transmitted using FM techniques. Find the practical bandwidth if

(a) Am = 5V, fm = 20 Hz and kf = 4 Hz/V

(b) Am = 10V, fm = 400 Hz and kf = 200 Hz/V. • SOLUTION

(a) IN this example, βF = (5)(4)/(20)X = 1. We need to J 2 ( ) 2 J 2 ( ) 0.99 find X so that S = 0 β F + ∑ n β F ≥ . n=1 • From the table, if X = 1, S = (0.7652+2×0.442) = 0.9648. If X = 2, S = 0.9648+2×0.1152 = 0.9912.

Therefore X = 2 and B = 4fm.

(b) Here, βF = (10)(200)/(400) = 5. We can show that X = 6 yields S = 0.994. Therefore B = 12fm. Carson’s Rule

• For m(t) = Amcos(2πfmt), When β is an integer, we always find that X = β+1. • Therefore we can estimate that the practical bandwidth of an FM signal is B = 2(βF+1)fm.

• For any random m(t) with maximum value Am and bandwidth Bm, the true bandwidth is difficult to find. • According to Carson, the worst case is when the

spectrum of m(t) is concentrated around f = Bm (such as a sinusoid). • Based on experiments by Carson, the bandwidth of a

WBFM signal, BFM, can be estimated by BFM = 2(β F +1)Bm (***) Generation of WBFM Signals • Direct method – Voltage Controlled Oscillator (VCO)

m(t) VCO sFM(t)

• Indirect method – Armstrong’s method

NBFM BPF m(t) nonlinearity @ nfc sWBFM(t) mod @ fc @ nfc Armstrong’s method

• Nonlinearity 2 3 – vo = a1vi+a2vi +a3vi +…

– vi(t) = sNBFM(t).

– Let sNBFM(t) = Accos(2πfct+2πkf∫m(t)dt) = Accos(θi(t)). 2 3 – vo(t) = a1sNBFM(t)+ a2s NBFM(t)+ a3s NBFM(t)… 2 2 3 3 – vo(t) = a1 Accos(θi(t))+a2 Ac cos (θi(t))+a3 Ac cos (θi(t)) … 2 2 – vo(t) = a1 Accos(θi(t))+a2 Ac /2+(a2 Ac /2)cos(2θi(t))+ 3 3 (3a3Ac /4)cos(θi(t))+(a3Ac /4)cos(3θi(t)) …

– nθi(t) = 2π(nfc)t+2π(nkf)∫m(t)dt (carrier frequency = nfc and kf’ = nkf therefore βF’ = nβF).

• BPF is used to pass the spectral component centred @ f = nfc. Demodulation of FM signals

• Differentiator plus envelope detection • Frequency discriminator. • Frequency counter. Differentiator and envelope

x(t) Envelope y(t) DC Block z(t)=Km(t) d/dt detector sFM(t) Differentiator and envelope detector ds (t) x(t) = FM dt d = (Ac cos(θi (t))) dt dθi (t) = − Ac sin(θi (t)) dt = 2πAc fi (t)sin(2πfct + 2πk f ∫ m(t)dt + π ) = 2πAc (fc + k f m(t))sin(2πfct + 2πk f ∫ m(t)dt +π )

fc >> |kfm(t)| then 2πAc(fc+kfm(t)) > 0. Example

• m(t) = cos2π10t, fc = 100, Ac = 2, kf = 40 Hz/V.

• sFM(t) = 2cos(2π200t+4sin2π10t) • x(t) = 4π(100+40cos2π10t)sin(2π100t+4sin2π10t+π) 1

0.5 m(t) 0

-0.5

-1 0 0.05 0.1 0.15 0.2 0.25

2

1 sFM(t) 0

-1

-2 0 0.05 0.1 0.15 0.2 0.25

2000 x(t) 1000 0

-1000

-2000 0 0.05 0.1 0.15 0.2 0.25

2πAc(fc+kfm(t)) Differentiator and envelope detector

• Output of envelope detector

– y(t) = 2πAc(fc+kfm(t)) = 2πAcfc + 2πAckfm(t) – Assuming that m(t) has no DC component (M(f) = 0 for f = 0), then • Output of DC block

– z(t) = 2πAckfm(t) = Km(t). Frequency discriminator

• Similar to differentiator • Input to envelope detector has lower amplitude.

x1(t) y1(t) H1(f) E.D + Km(t) sFM(t) + - H2(f) E.D y (t) x2(t) 2 FM Modulator

⎡⎤t variable xt( ) = Acccos⎢⎥ω t+∫ Ωττ( ) d +ϕ0 General resonant RF oscillator ⎣⎦⎢⎥0 principle: circuit

mt()

Practical implementation:

Difficulty: frequency stability. Suitable for narrowband FM only.

3-Jun-13 L.W. Couch II, Digital and Analog Communication Systems, Prentice Hall, 2001. Indirect Wideband Angle Modulator

Frequency multiplier:

1 BPF 1 cos2 (tt ) 1 cos 2 ( ) cos( 2ψ (t )) ψ=+⎣⎦⎡⎤( ψ) 2 2 2002 Hall, Prentice Systems Engineering, Communications J.Proakis, M.Salehi, Direct Wideband Angle Modulator

L.W. Couch II, Digital and Analog Communication Systems, Prentice Hall, 2001. how it operates? consider it without feedback first why is feedback required? why is frequency divider required? Balanced Discriminator: Block Diagram

L.W. Couch II, Digital and Analog Communication Systems, Prentice Hall, 2001. Balanced Discriminator: Circuit Diagram

L.W. Couch II, Digital and Analog Communication Systems, Prentice Hall, 2001. Phased Locked Loop (PLL) Detector

AA12 vt= Asinω t+ϕ ( t ) vt10( ) =sin[ϕ−ϕin ( t ) ( t )]+ (2ω c )term in( ) in[ c in ] 2

vt00( ) = Acosωc t+ϕ 0 ( t ) [ ] Informally, d ωVCO(tttvt) =(ω c+ϕ02()) =ω c +α () ϕ≈ϕ()tt () dt in 0 1 d vt2()≈ϕin () t α dt PLL Detector: Linear Model

ϕin ()t Δϕ()t vt1 ( )

ϕ0()t

AA AA12 12 vt20( ) =sinϕ−ϕ≈in ( t ) ( t ) vt10( ) =sin[ϕ−ϕin ( t ) ( t )]+ [ ] 2 2 +(2ω )term AA12 c ≈( ϕ−ϕin()ttKt0 ()) = d Δϕ () 2

d 11dd ()tvt () ϕ02=α vt2()=[ϕ−Δϕ≈ϕin () t () t] in () t dt ααdt dt Comparison of AM and FM/PM • AM is simple (envelope detector) but no noise/ interference immunity (low quality). • AM bandwidth is twice or the same as the modulating signal (no bandwidth expansion). • Power efficiency is low for conventional AM. • DSB-SC & SSB – good power efficiency, but complex circuitry. • FM/PM – spectrum expansion & noise immunity. Good quality. • More complex circuitry. However, ICs allow for cost-effective implementation. Important Properties of Angle- Modulated Signals: Summary

• FM/PM signal is a nonlinear function of the message. • The signal’s bandwidth increases with the modulation index. • The carrier spectral level varies with the modulation index, being 0 in some cases. • Narrowband FM/PM: the signal’s bandwidth is twice that of the message (the same as for AM). • The amplitude of the FM/PM signal is constant (hence, the power does not depend on the message).