Angle Modulation and Demodulation

Nonlinear Modulation

• In the 1920s, it was falsely believed that frequency modulation (FM) could eliminate sidebands inherent in amplitude modulation (AM), leading to zero-bandwidth signals.
• The idea was to modulate the carrier frequency ω(t) with m(t) such that ω(t) = ωc + km(t).
• By making k arbitrarily small, the maximum and minimum frequencies of ω(t) would approach ωc, theoretically resulting in zero bandwidth.
• This approach was unsuccessful.

Concept of Instantaneous Frequency

• FM signals vary the instantaneous frequency in proportion to m(t).
• Instantaneous frequency definition over less than one cycle of a sinusoid.
• Frequency definition using a sinusoidal signal with a generalized angle θ(t): \phi(t) = a \cos θ(t)
• For a generalized angle θ(t) as a linear function: \phi(t) = a \cos(ωct + θ0)

Relationship Between Phase and Frequency

• Instantaneous frequency and generalized angle are related:
  • ωi(t) = \frac{dθ(t)}{dt}
  • θ(t) = \int_{-∞}^{t} ωi(α) dα
• Phase Modulation (PM):
  • Modulating θ0(t) with m(t)
• Frequency Modulation (FM):
  • Modulating ωi(t) with m(t)
• Both PM and FM are types of angle modulation or exponential modulation.

Phase Modulation (PM)

• Angle varied linearly with m(t): θ(t) = ωct + θ0 + kpm(t), where kp is a constant and ωc is the carrier frequency.
• Without generality loss, let θ0 = 0, then θ(t) = ωct + kpm(t).
• The PM waveform: \phi(t) = A \cos(ωct + kpm(t))
• Instantaneous frequency ωi(t) = \frac{dθ}{dt} = ωc + kp\dot{m}(t)
• In PM, the frequency varies linearly with the derivative of m(t).

Frequency Modulation (FM)

• Frequency varied linearly with m(t): ωi(t) = ωc + kfm(t), where kf is a constant regulating the magnitude of frequency modulation.
• Angle θ(t) is given by: θ(t) = \int{-∞}^{t} [ωc + kfm(α)] dα = ωct + kf \int{-∞}^{t} m(α) dα
• Assuming constant term in θ(t) to be zero (without loss of generality).
• The FM waveform: \phi(t) = A \cos (ωct + kf \int_{-∞}^{t} m(α) dα)

FM and PM Relationship

• PM and FM are similar and interchangeable.
• Replacing m(t) with \int m(α) dα changes PM into FM.
• Replacing m(t) with \dot{m}(t) changes FM into PM.

General Angle Modulation

• Consider the angle θ(t) in PM (Fig 5.2b): H(s) = s
• Use any linear function H(s) to filter m(t), i.e., \phi{EM}(t) = \cos[ωct + ψ(t)] = \cos [ωct + \int{-∞}^{t} m(α)h(t − α) dα]
• H(s) must be reversible, i.e., one can obtain [H(s)]^{-1}.
• PM and FM are special cases with h(t) = kpδ(t) and h(t) = kpu(t) respectively.

Power of Angle Modulated Wave

• Always \frac{A^2}{2}

PM and FM Example - Indirect Method

• Example 5.1: Sketch FM and PM waves for the modulating signal m(t)
• Constants: kf = 2 × 10^5 and kp = 10π, carrier frequency fc = 100 MHz.
• The equation in terms of the variable f (frequency in hertz):
  • fi = fc + \frac{k_f}{2π}m(t) = 10^8 + 10^5m(t)
  • (fi){min} = 10^8 + 10^5 [m(t)]_{min} = 99.9 \text{ MHz}
  • (fi){max} = 10^8 + 10^5 [m(t)]_{max} = 100.1 \text{ MHz}
• Because m(t) increases and decreases linearly with time, the instantaneous frequency increases linearly from 99.9 to 100.1 MHz over a half-cycle and decreases linearly from 100.1 to 99.9 MHz over the remaining half-cycle of the modulating signal.
• PM for \dot{m}(t) is FM for m(t).
• For PM:
  • fi = fc + \frac{k_p}{2π} \dot{m}(t) = 10^8 + 5 \dot{m}(t)
  • (fi){min} = 10^8 + 5 [\dot{m}(t)]_{min} = 10^8 – 10^5 = 99.9 \text{ MHz}
  • (fi){max} = 10^8 + 5 [\dot{m}(t)]_{max} = 100.1 \text{ MHz}
• Because \dot{m}(t) switches back and forth from a value of -20,000 to 20,000, the carrier frequency switches back and forth from 99.9 to 100.1 MHz every half-cycle of \dot{m}(t).

PM and FM Example - Direct Method

• Example 5.2: Sketch FM and PM waves for the digital modulating signal m(t)

• Constants: kf = 2π × 10^5 and kp = π/2, and fc = 100 MHz.

• For FM:

  • fi = fc + \frac{k_f}{2π}m(t) = 10^8 + 10^5m(t)

• Because m(t) switches from 1 to -1 and vice versa, the FM wave frequency switches back and forth between 99.9 and 100.1 MHz.

• Frequency shift keying (FSK) is a scheme of carrier frequency modulation by a digital signal, where information digits are transmitted by keying different frequencies.

• For PM:

• fi = fc + \frac{1}{2π} k_p \dot{m}(t) = 10^8 + \frac{1}{4} \dot{m} (t)

• The derivative \dot{m}(t) is zero except at points of discontinuity of m(t) where impulses of strength ±2 are present.

• The frequency of the PM signal stays the same except at these isolated points of time.

  • \phi{PM} (t) = A \cos [wct + k_pm(t)]
  • \phi{PM} (t) = A \begin{cases} \cos [wct + \frac{\pi}{2} ] = -A \sin wct & \text{when } m(t) = 1 \ \cos [wct - \frac{\pi}{2} ] = A \sin w_ct & \text{when } m(t) = -1 \end{cases}

• Phase shift keying (PSK) is a scheme of carrier PM by a digital signal, because information digits are transmitted by shifting the carrier phase. Note that PSK may also be viewed as a DSB-SC modulation by m(t).

• Phase discontinuities in 4_{PM} (t) occur at instants where impulses of \dot{m}(t) are located. At these instants, the carrier phase shifts by π instantaneously. A finite phase shift in zero time implies infinite instantaneous frequency at these instants.

• The amount of phase discontinuity in PM (t) at the instant where \dot{m}(t) is discontinuous is kpmd, where md is the amount of discontinuity in \dot{m}(t) at that instant. In the present example, the amplitude of \dot{m}(t) changes by 2 (from -1 to 1) at the discontinuity. Hence, the phase discontinuity in PM (t) is kpm_d = (π/2) × 2 = π \text{ rad}

• When \dot{m}(t) is a digital signal, 4{pm} (t) shows a phase discontinuity where \dot{m}(t) has a jump discontinuity. To avoid ambiguity in demodulation, in such a case, the phase deviation kpm(t) must be restricted to a range (-π, π). For example, if k_p were 3π/2, then

Bandwidth of Angle Modulated Waves

• Angle modulation is non-linear, and standard Fourier analysis can’t be used to determine bandwidth.
• Calculate the bandwidth of an FM wave.
  • a(t) = \int_{-∞}^{t} m(α) dα
  • \hat{\phi}{FM} = A e^{j[ωct+kfa(t)]} = A e^{jkfa(t)} e^{jω_ct}
  • \phi{FM} = \Re[\hat{\phi}{FM}]
• Expand the exponential e^{jk_fa(t)}:
  • \hat{\phi}{FM} = A [1 + jkfa(t) − \frac{k^2f}{2!} a^2(t) + · · · + j^n \frac{k^nf}{n!} a^n(t) + · · · ] e^{jω_ct}
• The real part is given by:
  • \phi{FM} = Re[\hat{\phi}{FM}] = A (\cos ωct − kfa(t) \sin ωct − \frac{k^2f}{2!} a^2(t) \cos ωct + \frac{k^3f}{3!} a^3(t) \sin ω_ct + · · ·)
• The signal a(t) is the integral of m(t)
• If M(f) is band limited to B, then A(f) is band limited to B
• The spectrum of a^2(t) is A(f) ∗ A(f) and is band limited to 2B
• Similarly a^n(t) is band limited to nB
• Bandwidth of FM theoretically infinite, but practically finite, since n! increases much faster than |k_fa(t)|^n

Narrowband Angle Modulation

• Angle modulation non-linear, non-linear relationship between ϕ_{FM} and a(t) evident from a^n(t) terms
• Consider the case when kf is very small, i.e. |k_fa(t)| << 1,
• Then all of the higher order terms are zero except:
  • \phi{FM} = A[\cos ωct − kfa(t) \sin ωct]
• Linear modulation with a(t)
• If m(t) has bandwidth B, ϕ_{FM} will have bandwidth 2B
• A narrowband PM signal can be approximated by
  • \phi{PM} = A[\cos ωct − kpm(t) \sin ωct]

Wideband FM Analysis

• In practical FM signals, the constant kf is large enough that |k_fa(t)| << 1 does not hold
• Higher order terms cannot be ignored
• Consider a staircase signal approximation \hat{m} (t) of m(t), then
  • \text{rect}(\frac{t}{2B}) \cos ([ωc + kf m(tk)]t) ⇐⇒ \frac{1}{2} \text{sinc} [\frac{ω + ωc + kf m(tk)}{4B} ] + \frac{1}{2} \text{sinc} [\frac{ω − ωc − kf m(t_k)}{4B} ]

Approximate BW of FM

• The FM spectrum bandwidth is approximately:
  • B{FM} = \frac{1}{2π} (2kfmp + 8πB) = 2 (\frac{kfm_p}{2π} + 2B) \text{ [Hz]}
• Original fallacy - max and min frequencies would be in [ωc − kf mp, ωc + kf mp]
• Assumption: sinusoid of frequency ω has spectral density at ω
• True for sinusoid of infinite duration, but for finite duration (rect) has a sinc spectrum around ω_c
• Hence original optimism was based on false assumption

Approximate BW of FM (2)

• Peak frequency deviation \Delta f
  • \Delta f = \frac{kf}{2 · 2π} (m{max} − m{min}) = \frac{kf m_p}{2π}
• Alternative expression for FM bandwidth:
  • B_{FM} ≈ 2(\Delta f + 2B)
• This is based on a rather pessimistic staircase approximation of m(t). In reality m(t) is much smoother.
• A more realistic FM bandwidth approximation is:
  • B_{FM} ∈ [2\Delta f, 2\Delta f + 4B]

Approximate BW of NBFM

• A more realistic FM bandwidth approximation is:
  • B_{FM} ∈ [2\Delta f, 2\Delta f + 4B]
• Going back to the case of NBFM, kf and consequently ∆f is very small, i.e:
  • B_{NBFM} ≈ 4B
• But earlier we showed that B_{NBFM} = 2B \text{ Hz}
• A better estimate would be Carlson’s rule:
  • B{FM} = 2\Delta f + 2B = 2(\Delta f + B) = 2 (\frac{kf m_p}{2π} + B)
• Carlson’s rule i.t.o. the deviation ratio β:
  • B_{FM} = 2B(β + 1)
  • β = \frac{\Delta f}{B}
• For the special case of tone modulation, the deviation ratio IS the modulation index

Phase Modulation Results

• For PM the instantaneous frequency:
  • ωi = ωc + k_p\dot{m}(t)
• Peak frequency deviation:
  • \Delta f = \frac{kp}{2 · 2π} ([\dot{m}(t)]{max} − [\dot{m}(t)]_{min})
• Let \dot{m}p = [\dot{m}(t)]{max} = −[\dot{m}(t)]_{min}, then
  • \Delta f = \frac{kp \dot{m}p}{2π}
• Therefore
  • B{PM} = 2(\Delta f + B) = 2 (\frac{kp \dot{m}_p}{2π} + B)
• NOTE: In FM \Delta f = \frac{kf mp}{2π} depends on the peak value of m(t), whereas in PM \Delta f = \frac{kp \dot{m}p}{2π} depends on the peak value of \dot{m}(t)

Tone FM Analysis

• Considered the case of a staircase approximation of any m(t)
• Do a precise bandwidth analysis in the case of tone modulation and verify previous results
  • m(t) = α \cos ω_mt
  • a(t) = \frac{α}{ωm} \sin ωmt
  • \hat{\phi}{FM} = A e^{j[ωct+\frac{kfα}{ωm} \sin ω_mt]}
  • Δω = kf mp = αk_f

Tone FM Analysis (2)

• The bandwith of m(t) is 2πB = ω_m radians per second.
• The deviation ratio (and also modulation index) is: β = \frac{\Delta f}{B} = \frac{\Delta ω}{2πB} = \frac{αkf}{ωm}
• The expression for complex FM becomes:
  • \hat{\phi}{FM} = A e^{j[ωct+β \sin ωmt]} = A e^{jωct}e^{jβ \sin ω_mt}

Tone FM Analysis (3)

• Recall
  • \hat{\phi}{FM} = A e^{jωct}e^{jβ \sin ω_mt}
• The signal e^{jβ \sin ωmt} is a periodic signal with period 2π/ωm, and can be expressed as an exponential Fourier series:
  • e^{jβ \sin ωmt} = \sum{-∞}^{∞} Dn e^{jωmt}
  • Dn = \frac{ωm}{2π} \int{-π/ωm}^{π/ωm} e^{jβ \sin ωm t} e^{-jnωm t} dt = \frac{1}{2π} \int{-π}^{π} e^{j(β \sin x−nx)}dx
    *No analytic solution for the integral or the RHS, but can be approximated by an infinite series, the nth order Bessel function of the first kind J_n(β), i.e. eq (45) becomes
  • e^{jβ \sin ωmt} = \sum{-∞}^{∞} Jn(β) e^{jnωmt}

Tone FM Analysis (4)

• Substituting into (44):
  • \hat{\phi}{FM} = A\sum{-∞}^{∞} Jn(β) e^{j(ωct+nω_mt)}
• \phi{FM} = A\sum{-∞}^{∞} Jn(β) \cos (ωc + nω_m)t
• FM signal has carrier component and also infinite number of sidebands at ωc ± ωm, ωc ± 2ωm , . . ., ωc ± nωm
• The strength of the nth sideband is J_n(β)
• J_n(β) decreases with n, resulting in a finite number of significant spectral lines.

Single Tone FM Bandwidth

• J_n(β) negligible when n > β + 1, hence the number of significant sideband impulses is β + 1
• The bandwidth of single tone FM is given by
  • B{FM} = 2(β + 1)fm = 2(\Delta f + B)
• This corresponds to the previous result.
• When β << 1 (NBFM) then there is only one significant sideband and the bandwidth is 2B

Example 5.3

• Estimate B{FM} and B{PM} for the modulating signal m(t) in Fig. 5.4a for kf = 2 × 10^5 and kp = 5. Assume the essential bandwidth of the periodic m(t) as the frequency of its third harmonic.
• Repeat the problem if the amplitude of m(t) is doubled.
• The peak amplitude of m(t) is unity. Hence, m_p = 1. Determine the essential bandwidth B of m(t).
• The Fourier series for this periodic signal is given by
  • m(t) = \sum Cn \cos nω0t
  • ω_0 = \frac{2π}{2 × 10^{-4}} = 10^4π
  • C_n = \frac{\begin{cases} \frac{8}{πn^2} & n \text{ odd} \ 0 & n \text{ even} \end{cases}
• The harmonic amplitudes decrease rapidly with n. The third harmonic is only 11% of the fundamental, and the fifth harmonic is only 4% of the fundamental. This means the third and fifth harmonic powers are 1.21 and 0.16%, respectively, of the fundamental component power. Justified in assuming the essential bandwidth of m(t) as the frequency of its third harmonic, that is,
  • B = 3 × \frac{10^4}{2π} = 15 kHz
• For FM:
  • \Delta f = \frac{1}{2π} kf mp = \frac{1}{2π} (2π × 10^5)(1) = 100 kHz
  • B_{FM} = 2(\Delta f + B) = 2(100 + 15) = 230 kHz
• Alternatively, the deviation ratio β is given by
  • β = \frac{\Delta f}{B} = \frac{100}{15} = 6.66
  • B_{FM} = 2B(β + 1) = 30(6.66 + 1) = 230 kHz
• For PM: The peak amplitude of \dot{m}(t) is 20,000 and
  • \Delta f = \frac{1}{2π} kp \dot{m}p = 50 kHz
  • B_{PM} = 2(\Delta f + B) = 130 kHz

Example 5.4

• Repeat Example 5.1 if m(t) is time-expanded by a factor of 2: that is, if the period of m(t) is 4 × 10^{-4}.
• Time expansion of a signal by a factor of 2 reduces the signal spectral width (bandwidth) by a factor of 2. This is verified by observing that the fundamental frequency is now 2.5 kHz, and its third harmonic is 7.5 kHz. Hence, B = 7.5 kHz, which is half the previous bandwidth. Moreover, time expansion does not affect the peak amplitude and thus mp = 1. However, \dot{m}p is halved, that is, \dot{m}_p = 10,000.
• For FM:
  • \Delta f = \frac{1}{2π} kf mp = 100 kHz
  • B_{FM} = 2(\Delta f + B) = 2(100+ 7.5) = 215 kHz
• For PM:
  • \Delta f = \frac{1}{2π} kp \dot{m}p = 25 kHz
  • B_{PM} = 2(\Delta f + B) = 65 kHz
• Note that time expansion of m(t) has very little effect on the FM bandwidth, but it halves the PM bandwidth. This verifies our observation that the PM spectrum is strongly dependent on the spectrum of m(t).

Example 5.5

• An angle-modulated signal with carrier frequency w_c = 2 × 10^5π is described by the equation

  • \phi{EM} (t) = 10 \cos (ωct +5 \sin 3000πt+ 10 \sin 2000πt)

• Find:

  • the power of the modulated signal

• the frequency deviation Af

  • the deviation ratio B

• the phase deviation Aø.

  • Estimate the bandwidth

• The signal bandwidth is the highest frequency in m(t) (or its derivative). In this case

  • B = \frac{2000π}{2π} = 1000 Hz

• The carrier amplitude is 10, and the power is

  • P = \frac{\frac{10^2}{2}} = 50

• To find the frequency deviation Af, find the instantaneous frequency @i, given by

• The carrier deviation is 15,000 \cos 3000t+20,000π \cos 2000πt.

• The two sinusoids will add in phase at some point, and the maximum value of this expression is 15,000 + 20,000π. This is the maximum carrier deviation Δw. Hence,

  • Δw /= 15,000 + 20,000π
  • \Delta f = \frac{\Delta ω}{2π} = 12, 387.32 \text{ Hz}

• (c)

  • B = \frac{\Delta f}{B} = \frac{12,387.32}{1000} = 12.387

• The angle 0 (t) = wt + (5 \sin 3000t + 10 \sin 2000πt). The phase deviation is the maximum value of the angle inside the parentheses, and is given by A = 15 \text{ rad}.

• B_{em} = 2(\Delta f + B) = 26, 774.65 Hz

FM and PM Generation

• Two methods for generating FM:
• Direct and Indirect
• First consider indirect narrowband FM generation

NBFM Generation

• NBFM and NBPM signals:
  • \phi{FM} =A[\cos ωct − kfa(t) \sin ωct]
  • \phi{PM} =A[\cos ωct − kpm(t) \sin ωct]
• Both expressions are linear and similar to AM
• Use DSB-SC modulators to perform NB-FM/PM modulation
• NBFM has distortion and some amplitude modulation owing to approximation/omission of higher order terms
• A nonlinear device called a bandpass limiter can remove most of the distortion

Bandpass Limiter

• Consider a variable amplitude angle modulated signal:

  • v_i(t) = A(t) \cos θ(t)
  • where
  • θ(t) = ωct + kf \int_{-∞}^{t} m(α) dα

• The output of the hard limiter is

  • v_o(θ) = \begin{cases} +1 & \text{cos } θ ≥ 0 \ −1 & \text{cos } θ < 0 \end{cases}

• which is a rectangular wave with Fourier series expansion:
  v_o(θ) = \frac{4}{π} (\cos θ − \frac{1}{3} \cos 3θ + \frac{1}{5} \cos 5θ − . . .)

• The output of the hard limiter is angle modulated with

  • θ(t) = ωct + kf \int_{-∞}^{t} m(α)dα

• vo(θ(t)) = vo (ωct + kf \int{-∞}^{t} m(α)dα) = \frac{4}{π} (\cos (ωct + kf \int{-∞}^{t} m(α)dα) − \frac{1}{3} \cos 3 (ωct + kf \int{-∞}^{t} m(α)dα) + \frac{1}{5} \cos 5 (ωct + kf \int{-∞}^{t} m(α)dα) − . . .)

• After bandpass filter with center frequency ωc:

  • eo(t) = \frac{4}{π} \cos (ωct + kf \int{-∞}^{t} m(α)dα)

• This result also applies to PM.

Armstrong's Indirect Method

• Concept of non-linear devices and frequency multipliers.
  • y(t) = a2 \cos² [wet + kf ∫ m(α) dα] = 0.5a2 + 0.5a2 \cos [2wet + 2k_f ∫ m(α) dα]

Example 5.6

• Discuss the nature of distortion inherent in the Armstrong indirect FM generator.

• Two kinds of distortion arise in this scheme: amplitude distortion and frequency distortion.

• The NBFM wave is given by:

  • \phi{FM} (t) =A[\cos wct – kfa(t) \sin wct]
  • = AE(t) \cos [w_ct +\theta(t)]
  • where
    • E(t) = \sqrt{1+ k² a² (t)}
    • and
    • \theta (t) = tan^{-1}[k_fa(t)]

• Amplitude distortion occurs because the amplitude AE(t) of the modulated waveform is not constant. This is not a serious problem because amplitude variations can be eliminated by a bandpass limiter, as discussed earlier in the section (see also Fig. 5.9).

• Ideally, \theta (t) should be k_fa(t). Instead, the phase \theta (t) in the preceding equation is

  • \theta(t) = tan^{-1} [k_fa(t)]

• and the instantaneous frequency w_i(t) is

  • wi(t) = \dot{\theta} (t) = \frac{kf\dot{a}(t)}{1+k²f a² (t)} = \frac{kfm(t)}{1+k²_f a² (t)}
  • = kfm(t)[1 – k²fa²(t) + k^4_fa^4 (t) – … . . ]
    Ideally, the instantaneous frequency should be kfm(t). The remaining terms in this equation are the distortion.

• Let us investigate the effect of this distortion in tone modulation where

  • m(t) = a \cos w_mt
  • a(t)=a(\sin wmt)/wm
  • \beta =(a kf)/wm

• Immunity from non-linearity: primary reason for use of FM in microwave relay systems (use of efficient class C high power non-linear amplifiers)

• Constant amplitude FM provide immunity from rapid fading through AGC

• FM was also used in 1G cellular, owing to immunity to fading

Effect of Interference on FM/PM

*FM/PM less vulnerable than AM to small signal interference from adjacent channels

• Consider interference of unmodulated carrier A cos wct by another sinusoid I cos(wc + w)t.
• The received signal is given by
• r(t) = A cos wct+I cos(wc + w)t = A cos wct+I cos wt cos wct–I sin wt sin w_ct
• =(A + I cos wt) cos wct–I sin wt sin wct = Er(t) cos[wct + \Psi_d(t)]
• Where
  • \Psi_d(t) = tan^{-1} \frac{I sin wt}{ A + I cos wt}
• For a small interfering signal (I << A), we have
  • \Psi_d(t) ≈ \frac{I}{A} sin wt
• Recall for a small interfering signal (I << A), we have
  • $$\Psi_d(t) ≈ \frac{I}{A