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: $θ(t) = \int<em>{-∞}^{t} [ωc + kfm(α)] dα = ωct + kf \int</em>{-∞}^{t} m(α) dα$
• 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., $\phi<em>{EM}(t) = \cos[ωct + ψ(t)] = \cos [ωct + \int</em>{-∞}^{t} m(α)h(t − α) dα]$
• $H(s)$ 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):
  • $f<em>i = f</em>c + \frac{k_f}{2π}m(t) = 10^8 + 10^5m(t)$
  • $(f<em>i)</em>{min} = 10^8 + 10^5 [m(t)]_{min} = 99.9 \text{ MHz}$
  • $(f<em>i)</em>{max} = 10^8 + 10^5 [m(t)]_{max} = 100.1 \text{ MHz}$
• 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:
  • $f<em>i = f</em>c + \frac{k_p}{2π} \dot{m}(t) = 10^8 + 5 \dot{m}(t)$
  • $(f<em>i)</em>{min} = 10^8 + 5 [\dot{m}(t)]_{min} = 10^8 – 10^5 = 99.9 \text{ MHz}$
  • $(f<em>i)</em>{max} = 10^8 + 5 [\dot{m}(t)]_{max} = 100.1 \text{ MHz}$
• 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:

  • $f<em>i = f</em>c + \frac{k_f}{2π}m(t) = 10^8 + 10^5m(t)$

• 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:

• $f<em>i = f</em>c + \frac{1}{2π} k_p \dot{m}(t) = 10^8 + \frac{1}{4} \dot{m} (t)$

• 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.

  • $\phi<em>{PM} (t) = A \cos [w</em>ct + k_pm(t)]$
  • $\phi<em>{PM} (t) = A \begin{cases} \cos [w</em>ct + \frac{\pi}{2} ] = -A \sin w<em>ct & \text{when } m(t) = 1 \ \cos [w</em>ct - \frac{\pi}{2} ] = A \sin w_ct &amp; \text{when } m(t) = -1 \end{cases}$

• 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 $k<em>pm</em>d$, where $m<em>d$ 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 $k</em>pm_d = (π/2) × 2 = π \text{ rad}$

• When $\dot{m}(t)$ is a digital signal, $4<em>{pm} (t)$ shows a phase discontinuity where $\dot{m}(t)$ has a jump discontinuity. To avoid ambiguity in demodulation, in such a case, the phase deviation $k</em>pm(t)$ 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.
  • $a(t) = \int_{-∞}^{t} m(α) dα$
  • $\hat{\phi}<em>{FM} = A e^{j[ω</em>ct+k<em>fa(t)]} = A e^{jk</em>fa(t)} e^{jω_ct}$
  • $\phi<em>{FM} = \Re[\hat{\phi}</em>{FM}]$
• Expand the exponential $e^{jk_fa(t)}$:
  • $\hat{\phi}<em>{FM} = A [1 + jk</em>fa(t) − \frac{k^2<em>f}{2!} a^2(t) + · · · + j^n \frac{k^n</em>f}{n!} a^n(t) + · · · ] e^{jω_ct}$
• The real part is given by:
  • $\phi<em>{FM} = Re[\hat{\phi}</em>{FM}] = A (\cos ω<em>ct − k</em>fa(t) \sin ω<em>ct − \frac{k^2</em>f}{2!} a^2(t) \cos ω<em>ct + \frac{k^3</em>f}{3!} a^3(t) \sin ω_ct + · · ·)$
• 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:
  • $\phi<em>{FM} = A[\cos ω</em>ct − k<em>fa(t) \sin ω</em>ct]$
• 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
  • $\phi<em>{PM} = A[\cos ω</em>ct − k<em>pm(t) \sin ω</em>ct]$

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
  • $\text{rect}(\frac{t}{2B}) \cos ([ω<em>c + k</em>f m(t<em>k)]t) ⇐⇒ \frac{1}{2} \text{sinc} [\frac{ω + ω</em>c + k<em>f m(t</em>k)}{4B} ] + \frac{1}{2} \text{sinc} [\frac{ω − ω<em>c − k</em>f m(t_k)}{4B} ]$

Approximate BW of FM

• The FM spectrum bandwidth is approximately:
  • $B<em>{FM} = \frac{1}{2π} (2k</em>fm<em>p + 8πB) = 2 (\frac{k</em>fm_p}{2π} + 2B) \text{ [Hz]}$
• Original fallacy - max and min frequencies would be in $[ω<em>c − k</em>f m<em>p, ω</em>c + k<em>f m</em>p]$
• 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 $\Delta f$
  • $\Delta f = \frac{k<em>f}{2 · 2π} (m</em>{max} − m<em>{min}) = \frac{k</em>f m_p}{2π}$
• 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:
  • $B<em>{FM} = 2\Delta f + 2B = 2(\Delta f + B) = 2 (\frac{k</em>f m_p}{2π} + B)$
• 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:
  • $ω<em>i = ω</em>c + k_p\dot{m}(t)$
• Peak frequency deviation:
  • $\Delta f = \frac{k<em>p}{2 · 2π} ([\dot{m}(t)]</em>{max} − [\dot{m}(t)]_{min})$
• Let $\dot{m}<em>p = [\dot{m}(t)]</em>{max} = −[\dot{m}(t)]_{min}$, then
  • $\Delta f = \frac{k<em>p \dot{m}</em>p}{2π}$
• Therefore
  • $B<em>{PM} = 2(\Delta f + B) = 2 (\frac{k</em>p \dot{m}_p}{2π} + B)$
• NOTE: In FM $\Delta f = \frac{k<em>f m</em>p}{2π}$ depends on the peak value of $m(t)$, whereas in PM $\Delta f = \frac{k<em>p \dot{m}</em>p}{2π}$ 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
  • $m(t) = α \cos ω_mt$
  • $a(t) = \frac{α}{ω<em>m} \sin ω</em>mt$
  • $\hat{\phi}<em>{FM} = A e^{j[ω</em>ct+\frac{k<em>fα}{ω</em>m} \sin ω_mt]}$
  • $Δω = k<em>f m</em>p = αk_f$

Tone FM Analysis (2)

• The bandwith of m(t) is $2πB = ω_m$ radians per second.
• The deviation ratio (and also modulation index) is: $β = \frac{\Delta f}{B} = \frac{\Delta ω}{2πB} = \frac{αk<em>f}{ω</em>m}$
• The expression for complex FM becomes:
  • $\hat{\phi}<em>{FM} = A e^{j[ω</em>ct+β \sin ω<em>mt]} = A e^{jω</em>ct}e^{jβ \sin ω_mt}$

Tone FM Analysis (3)

• Recall
  • $\hat{\phi}<em>{FM} = A e^{jω</em>ct}e^{jβ \sin ω_mt}$
• The signal $e^{jβ \sin ω<em>mt}$ is a periodic signal with period $2π/ω</em>m$, and can be expressed as an exponential Fourier series:
  • $e^{jβ \sin ω<em>mt} = \sum</em>{-∞}^{∞} D<em>n e^{jω</em>mt}$
  • $D<em>n = \frac{ω</em>m}{2π} \int<em>{-π/ω</em>m}^{π/ω<em>m} e^{jβ \sin ω</em>m t} e^{-jnω<em>m t} dt = \frac{1}{2π} \int</em>{-π}^{π} e^{j(β \sin x−nx)}dx$
    *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
  • $e^{jβ \sin ω<em>mt} = \sum</em>{-∞}^{∞} J<em>n(β) e^{jnω</em>mt}$

Tone FM Analysis (4)

• Substituting into (44):
  • $\hat{\phi}<em>{FM} = A\sum</em>{-∞}^{∞} J<em>n(β) e^{j(ω</em>ct+nω_mt)}$
• $\phi<em>{FM} = A\sum</em>{-∞}^{∞} J<em>n(β) \cos (ω</em>c + nω_m)t$
• FM signal has carrier component and also infinite number of sidebands at $ω<em>c ± ω</em>m, ω<em>c ± 2ω</em>m , . . ., ω<em>c ± nω</em>m$
• 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
  • $B<em>{FM} = 2(β + 1)f</em>m = 2(\Delta f + B)$
• 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 $B<em>{FM}$ and $B</em>{PM}$ 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
  • $m(t) = \sum C<em>n \cos nω</em>0t$
  • $ω_0 = \frac{2π}{2 × 10^{-4}} = 10^4π$
  • 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:
  • $\Delta f = \frac{1}{2π} k<em>f m</em>p = \frac{1}{2π} (2π × 10^5)(1) = 100 kHz$
  • $B_{FM} = 2(\Delta f + B) = 2(100 + 15) = 230 kHz$
• 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
  • $\Delta f = \frac{1}{2π} k<em>p \dot{m}</em>p = 50 kHz$
  • $B_{PM} = 2(\Delta f + B) = 130 kHz$

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 $m<em>p = 1$. However, $\dot{m}</em>p$ is halved, that is, $\dot{m}_p = 10,000$.
• For FM:
  • $\Delta f = \frac{1}{2π} k<em>f m</em>p = 100 kHz$
  • $B_{FM} = 2(\Delta f + B) = 2(100+ 7.5) = 215 kHz$
• For PM:
  • $\Delta f = \frac{1}{2π} k<em>p \dot{m}</em>p = 25 kHz$
  • $B_{PM} = 2(\Delta f + B) = 65 kHz$
• 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

  • $\phi<em>{EM} (t) = 10 \cos (ω</em>ct +5 \sin 3000πt+ 10 \sin 2000πt)$

• 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:
  • $\phi<em>{FM} =A[\cos ω</em>ct − k<em>fa(t) \sin ω</em>ct]$
  • $\phi<em>{PM} =A[\cos ω</em>ct − k<em>pm(t) \sin ω</em>ct]$
• 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
  • $θ(t) = ω<em>ct + k</em>f \int_{-∞}^{t} m(α) dα$

• The output of the hard limiter is

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

• 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

  • $θ(t) = ω<em>ct + k</em>f \int_{-∞}^{t} m(α)dα$

• $v<em>o(θ(t)) = v</em>o (ω<em>ct + k</em>f \int<em>{-∞}^{t} m(α)dα) = \frac{4}{π} (\cos (ω</em>ct + k<em>f \int</em>{-∞}^{t} m(α)dα) − \frac{1}{3} \cos 3 (ω<em>ct + k</em>f \int<em>{-∞}^{t} m(α)dα) + \frac{1}{5} \cos 5 (ω</em>ct + k<em>f \int</em>{-∞}^{t} m(α)dα) − . . .)$

• After bandpass filter with center frequency $ωc$:

  • $e<em>o(t) = \frac{4}{π} \cos (ω</em>ct + k<em>f \int</em>{-∞}^{t} m(α)dα)$

• This result also applies to PM.

Armstrong's Indirect Method

• Concept of non-linear devices and frequency multipliers.
  • $y(t) = a<em>2 \cos² [w</em>et + k<em>f ∫ m(α) dα] = 0.5a</em>2 + 0.5a<em>2 \cos [2w</em>et + 2k_f ∫ m(α) dα]$

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:

  • $\phi<em>{FM} (t) =A[\cos w</em>ct – k<em>fa(t) \sin w</em>ct]$
  • $= AE(t) \cos [w_ct +\theta(t)]$
  • 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

  • $w<em>i(t) = \dot{\theta} (t) = \frac{k</em>f\dot{a}(t)}{1+k²<em>f a² (t)} = \frac{k</em>fm(t)}{1+k²_f a² (t)}$
  • $= k<em>fm(t)[1 – k²</em>fa²(t) + k^4_fa^4 (t) – … . . ]$
    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

  • $m(t) = a \cos w_mt$
  • $a(t)=a(\sin w<em>mt)/w</em>m$
  • $\beta =(a k<em>f)/w</em>m$

• 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
• $r(t) = A cos w<em>ct+I cos(w</em>c + w)t = A cos w<em>ct+I cos wt cos w</em>ct–I sin wt sin w_ct$
• $=(A + I cos wt) cos w<em>ct–I sin wt sin w</em>ct = E<em>r(t) cos[w</em>ct + \Psi_d(t)]$
• 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