Modulation Systems EMS310 - Ch 3 + 4

Modulation Systems EMS310 - Ch 3 + 4

Table of Contents

• Some self-study material (ELI revision)

• Signal transmission through a linear system

• Ideal vs practical filters

• Signal distortion over a channel

• Energy spectral density

• Power spectral density

• The Discrete Fourier Transform (DFT)

• Chapter 4

Self Study Material (ELI Revision)

• Sections 3.1-3.3:

  • 3.1 Aperiodic signal representation by Fourier integral

  • 3.2 Transforms of some useful functions

  • 3.3 Properties of the Fourier transform

Signal Transmission Through a Linear System

• LTI Systems:

  • Output in the time domain: y(t) = h(t) * x(t)

  • Output in the frequency domain: Y(f) = H(f)X(f)

Effect of Channel Distortion

• A channel can be modeled as an LTI system.

  • X(f) - Fourier transform of channel input x(t)

  • Y(f) - Fourier transform of channel output y(t)

  • H(f) - Fourier transform of channel impulse response h(t)

• Output of LTI (polar form): |Y(f)|e^{jθy(f)} = |H(f)||X(f)|e^{j[θx(f) + θ_h(f)]}

• Distortionless channel:

  • |H(f)| = k

  • θh(f) = -2πftd

    • Where t_d is the channel time delay (same for all f).

• Difference between an all-pass and distortionless channel.

Ideal vs Practical Filters

• Ideal causal filter:

  • y(t) = g(t - t_d)

  • h(t) = F^{-1}{H(f)} = 2Bsinc[2B(t - t_d)]

• Is impulse response realisable?

• How can we make a realisable filter?

Paley-Wiener Criterion

• A necessary and sufficient condition for |H(f)| to be a causal filter:

  • \int_{-\infty}^{\infty} \frac{|\ln |H(f)||}{1 + (2\pi f)^2} df < \infty

• |H(f)| = 0 in a band?

• |H(f)| = 0 at a single frequency?

Signal Distortion Over a Channel

• Linear Channel Distortion:

  • Amplitude response distortion: |H(f)| ≠ k

  • Phase response distortion: |θh(f)| ≠ -2πftd

  • Results in a non-ideal channel.

  • Results in dispersion (analog) or intersymbol interference (digital).

Linear Channel Distortion

• Pulse is dispersed when it passes through a system that is not distortionless.

Example 3.14

• A low-pass filter transfer function H(f) is given by:

H(f) = \begin{cases} (1 + k \cos 2\pi fT)e^{-j2\pi ft_d} & |f| < B \ 0 & |f| > B \end{cases}

• A pulse g(t) band-limited to B Hz is applied at the input of this filter. Find the output y(t).

• This filter has ideal phase and nonideal magnitude characteristics.

• Because g(t) \Leftrightarrow G(f), y(t) \Leftrightarrow Y(f) and

  • Y(f) = G(f)H(f) = G(f)\Pi(\frac{f}{2B})(1 + k \cos 2\pi fT)e^{-j2\pi ft_d}

• Note that in the derivation of Eq. (3.62) because g(t) is band-limited to B Hz, we have

  • G(f) \cdot \Pi(\frac{f}{2B}) = G(f). Then, by using the time-shifting property and Eq. (3.32a), we have:

  • y(t) = g(t - td) + \frac{k}{2}[g(t - td - T) + g(t - t_d + T)].

• The output is actually g(t) + (k/2)[g(t - T) + g(t + T)] delayed by td. It consists of g(t) and its echoes shifted by \pm td.

• The dispersion of the pulse caused by its echoes is evident from Fig. 3.29c. Ideal amplitude but nonideal phase response of H(f) has a similar effect.

Nonlinear Channel Distortion

• Consider a channel effect: y = f(g)

• Where y(t) = a0 + a1g(t) + a2g^2(t) + a3g^3(t) + … + a_kg^k(t).

• Note for g(t) with bandwidth B, g^k(t) has bandwidth kB.

• Causes frequency dispersion

• Interference between frequency channels

• What does equation (12) represent?

• Effect of using power-efficient amplifiers?

Multipath Distortion

• At specific time deltas and frequencies, and where two paths have similar gains

  • Constructive interference - same phases

  • Destructive interference - opposite phases

• This effect is frequency-selective fading.

Fading Channels

• In a mobile comms environment:

  • Multipath changes with time

  • Causes time variations (semi-periodic and random) and attenuation of signal known as fading

• Possible to reduce fading using automatic gain control (AGC)

• What will be the effect of noise in AGC?

Energy Spectral Density

• Parseval’s theorem in the frequency domain

• Signal energy can now be determined from g(t) in the time domain or from G(f) in the frequency domain

• Energy of signal composed of energies for all frequency components of g(t)

Energy Spectral Density

• Energy spectral density defined as \Psi_g(f) = |G(f)|^2

• And from the second equation in previous slide we have

• Eg = \int{-\infty}^{\infty} \Psi_g(f) df.

Essential Bandwidth of a Signal

• Spectra of most signals → ∞

• Practical signals have finite energy - hence spectrum → 0 as f → ∞

• There exists some bandwidth B for which the energy content at f > B becomes negligible

  • EXAMPLE: Select a B such that the energy inside B is 95% of the signal energy (other percentages possible)

• Suppression of components beyond b results in approximate signal \hat{g}(t)

• Homework: Ex 3.17 and 3.18

Energy of Modulated Signals

• Let g(t) be a baseband signal band-limited to B Hz. The amplitude-modulated signal \phi(t) is

  • \phi(t) = g(t) \cos 2\pi f_0t

• and the spectrum (Fourier transform) of \phi(t) is

  • \Phi(f) = \frac{1}{2}[G(f + f0) + G(f - f0)]

• The ESD of the modulated signal \phi(t) is |\Phi(f)|^2, that is,

\Psi{\phi}(f) = \frac{1}{4}|G(f + f0) + G(f - f_0)|^2

• If f0 \geq B, then G(f + f0) and G(f - f_0) are nonoverlapping (see Fig. 3.35), and

\Psi{\phi}(f) = \frac{1}{4}[|G(f + f0)|^2 + |G(f - f0)|^2] = \frac{1}{4} \Psig(f+f0) + \frac{1}{4} \Psig(f-f_0)

Energy of Modulated Signals

• The energy of a signal is equal to the area under the ESD, i.e.

• E{\phi} = \frac{1}{2}Eg , f_0 \geq B

Autocorrelation and ESD

• Autocorrelation given by

• \psig(\tau) = \int{-\infty}^{\infty} g(t)g^*(t + \tau) dt

• Note that \psig(\tau) = \psig(-\tau)

• The autocorrelation of g(t) and the ESD are Fourier transform pairs:

• \psig(\tau) \Leftrightarrow \Psig(f)

• Therefore:

• \Psig(f) = F{\psig(\tau)} = \int{-\infty}^{\infty} \psig(\tau)e^{-j2\pi f\tau} d\tau

• \psig(\tau) = F^{-1}{\Psig(f)} = \int{-\infty}^{\infty} \Psig(f)e^{j2\pi f\tau} df

• NOTE: Fourier transform performed in respect of \tau instead of t

ESD of an LTI Input and Output

• If x(t) and y(t) are the input and the corresponding output of a linear time-invariant (LTI) system, then

• Y(f) = H(f)X(f)

• Therefore,

  • |Y(f)|^2 = |H(f)|^2 |X(f)|^2

• \Psiy(f) = |H(f)|^2\Psix(f)

• Thus, the output signal ESD is |H(f)|^2 times the input signal ESD.

Power Spectral Density

• NOTE: Signal power is time-averaged signal energy

  • Pg = \lim{T \to \infty} \frac{E{gt}}{T} = \lim{T \to \infty} [\int{-\infty}^{\infty} \frac{|GT(f)|^2}{T} df]

  • = \int{-\infty}^{\infty} \lim{T \to \infty} \frac{|G_T(f)|^2}{T} df

• As T increases, the duration of g(t) increases, and so its energy E{GT} - therefore GT(f) increases with T

• Convergence implies |G_T(f)|^2 must grow at the same rate as T, which means that we can change the order of integral and limit in last step.

• Power spectral density and power:

  • Sg(f) = \lim{T \to \infty} \frac{|G_T(f)|^2}{T}

  • Pg = \int{-\infty}^{\infty} S_g(f) df

Time Autocorrelation Function

• The time autocorrelation function of real signal g(t) is defined as

• Rg(\tau) = \lim{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} g(t)g(t - \tau) dt

• As with autocorrelation, time autocorrelation is an even function of \tau, i.e. Rg(\tau) = Rg(-\tau)

• Fourier transform pairs:

• Rg(\tau) \Leftrightarrow \lim{T \to \infty} \frac{|GT(f)|^2}{T} = Sg(f)

ESD and PSD Relationships

Why Use Autocorrelation and PSD?

• Why not just use the Fourier transform of the signal?

  • Fourier transform of power signal generally does not exist

  • Fourier transform not available for random signals (random noise)

• Autocorrelation function and PSD exists for power signals and random signals

PSD of an LTI Input and Output

• If g(t) and y(t) are the input and output of an LTI system with a transfer function H(f), then

• Sy(f) = |H(f)|^2Sg(f)

Example 3.20

• A noise signal ni(t) with PSD S{n_i}(f) = K is applied at the input of an ideal differentiator

• Determine the PSD and the power of the output noise signal n_o(t).

• The transfer function of an ideal differentiator is H(f) = j2\pi f.

• If the noise at the demodulator output is n_o(t), then from Eq. (3.91),

• S{no}(f) = |H(f)|^2 S{ni}(f) = |j2\pi f|^2 K

• The output PSD S{no}(f) is parabolic.

• The output noise power N_o is the area under the output PSD. Therefore,

  • No = \int{-B}^{B} K(2\pi f)^2 df = 2K \int_{0}^{B} (2\pi f)^2 df = \frac{8\pi^2 B^3 K}{3}

PSD of a Modulated Signal

• Consider cosine amplitude modulated by signal g(t): \phi(t) = g(t) \cos 2\pi f_0t,

• then the PSD S_{\phi}(f) of the modulated signal is:

• S{\phi}(f) = \frac{1}{4}[Sg(f + f0) + Sg(f - f_0)],

• See detailed derivation in Sec 7.8 of textbook

• Modulation shifts PSD to \pm f_0

• Power of \phi(t) is half the power of g(t), i.e. P{\phi} = \frac{1}{2}Pg f_0 \geq B

The Discrete Fourier Transform (DFT)

Self Study

• NOTE: Section 3.9 is self-study / revision

Chapter 4

Double Sideband Suppressed Carrier (DSB-SC)

DSB-SC Modulation

• In amplitude-based modulation, the carrier phase remains constant, i.e. let \theta_c = 0 without loss of generality

• Consider a message signal m(t) with bandwidth B, where m(t) \Leftrightarrow M(f)

• then

• m(t)\cos(2\pi fct) \Leftrightarrow \frac{1}{2} [M(f + fc) + M(f - f_c)]

• Lower sidebands with bandwidth B where -fc < f < fc

• Upper sidebands with bandwidth B where f > fc and f < -fc

• fc > B, else the spectra will overlap, however practically fc >> B

DSB-SC Demodulation

• Consider a DSB-SC signal, multiplied by a cosine with the same frequency and phase as carrier:

  • e(t) = m(t)\cos^2(\omega_ct)

  • = \frac{1}{2} [m(t) + m(t)\cos(2\omega_ct)].

• The Fourier transform of e(t) is given by:

• E(f) = \frac{1}{2}M(f) + \frac{1}{4}[M(f + 2fc) + M(f - 2fc)]

DSB-SC Demodulation Example

• For a baseband signal m(t) = \cos \omegamt = \cos 2\pi fmt,

• find the DSB-SC signal and sketch its spectrum. Identify the USB and LSB. Verify that the
  DSB-SC modulated signal can be demodulated by the demodulator in Fig. 4.1e.

• The case in this example is referred to as tone modulation because the modulating signal
  is a pure sinusoid, or tone, \cos \omega_mt

• To clarify the basic concepts of DSB-SC modulation, we shall work this problem in
  the frequency domain as well as the time domain. In
  the frequency domain approach, we work with the signal spectra. The spectrum of the
  baseband signal m(t) = \cos \omega_mt is given by

\frac{1}{2} [\delta (f - fm) + \delta(f + fm)]

= \pi [\delta(\omega - \omegam) + \delta(\omega + \omegam)]

• In the time domain approach, we work directly with signals in the time domain. For the baseband signal m(t) = \cos \omegamt, the DSB-SC signal \phi{DSB-SC}(t) is

\phi{DSB-SC}(t) = m(t) \cos \omegact

= \cos \omegamt \cos \omegact

= \frac{1}{2} [\cos(\omegac + \omegam)t + \cos(\omegac - \omegam)t]

• We now verify that the modulated signal \phi{DSB-Sc}(t) = \cos \omegamt \cos \omegact, when applied to the input of the demodulator in Fig. 4.1e, yields the output proportional to the desired baseband signal \cos \omegamt.

• The signal e(t) in Fig. 4.1e is given by
  e(t) = \cos \omegamt \cos^2 \omegact

= \frac{1}{2} \cos \omegamt (1 + \cos 2\omegaet)

Amplitude Modulator Types

• Multiplier modulators

• Nonlinear modulators

• Switching modulators

Multiplier Modulators

• Obtained from variable gain amp where the gain parameter is controlled by m(t)

• Initially, multiplication of two signals over wide dynamic range was challenging - not such a problem anymore

• However, some other efficient amplitude modulation techniques are now presented

Nonlinear Modulators

• Nonlinear element:

• y(t) = ax(t) + bx^2(t)

• z(t) as in figure:

• z(t) = y1(t) - y2(t) = [ax1(t) + bx1^2(t)] - [ax2(t) + bx2^2(t]]

• Let x1(t) = \cos \omegact + m(t) and x2(t) = \cos \omegact - m(t), then

• z(t) = 2am(t) + 4bm(t) \cos \omega_ct

Switching Modulators

• Modulation can be achieved by multiplying m(t) with any periodic signal

• Such a periodic signal \phi(t) can be expressed by a trigonometric Fourier series:

  • \phi(t) = \sum{n=0}^{\infty} Cn \cos n\omegact + \thetan

• hence

• m(t)\phi(t) = \sum{n=0}^{\infty} Cnm(t) \cos n\omegact + \thetan

• Spectrum M(\omega) shifted to \pm \omegac, \pm 2\omegac, …, \pm n\omega_c

Switching Modulators (2)

• Consider the Fourier series expansion for a square wave:

• w(t) = \frac{1}{2} + \frac{2}{\pi} (\cos \omegact - \frac{1}{3} \cos 3\omegact + \frac{1}{5} \cos 5\omega_ct - … )

• The signal m(t)w(t) is then given by:

• m(t)w(t) = \frac{1}{2}m(t) + \frac{2}{\pi} (m(t) \cos \omegact - \frac{1}{3}m(t) \cos 3\omegact + \frac{1}{5} m(t) \cos 5\omega_ct - … )

• Spectrum M(\omega) shifted to \pm \omegac, \pm 3\omegac, \pm 5\omega_c, …

Diode Bridge Modulator

• Difference between this and chopper modulator?

Ring Modulator

• Consider case where m(t) is multiplied by a bi-polar rectangular pulse train

• The Fourier series expansion of the rectangular pulse train:

• w0(t) = \frac{4}{\pi} (\cos \omegact - \frac{1}{3} \cos 3\omegact + \frac{1}{5} \cos 5\omegact - … )

• resulting in a modulated signal:

• m(t)w(t) = \frac{4}{\pi} (m(t) \cos \omegact - \frac{1}{3} m(t) \cos 3\omegact + \frac{1}{5} m(t) \cos 5\omega_ct - … )

Ring Modulator

• How is this modulator related to the chopper modulator?

Examples 4.2 and 4.3

• Examples 4.2 and 4.3 for self-study

Amplitude Modulation

Amplitude Modulation (AM)

• DSB-SC is simple to analyse, but not so simple to implement

• DSB-SC needs coherent demodulation (carrier at receiver with same frequency and phase)

• Unknown transmission path and motion of TX or RX can induce Doppler effect, introducing frequency and phase shifts

• Local oscillator needs to reproduce cosine with the same frequency and phase, using only the received signal

• This requires the implementation of a phase-locked loop in the receiver, which is more expensive

Amplitude Modulation (AM)

• Amplitude modulation (AM) is a modulation method which allows for simplified receiver implementation

• This is at the expense of a higher power (more expensive transmitter)

• Makes sense in the case of a single transmitter and multiple receivers (radio)

• Consider the AM signal:

  • \Phi{AM}(t) = A \cos \omegact + m(t) \cos \omega_ct

  • = [A + m(t)] \cos \omega_ct

• This results in the Fourier transform pair:

\Phi{AM}(t) \Leftrightarrow \frac{1}{2} [M(f + fc) + M(f - fc)] + \frac{A}{2} [\delta(f + fc) + \delta(f - f_c)]

• How does this compare to DSB-SC?

• What does the spectrum look like?

AM - Two Cases

• AM signal: \Phi{AM}(t) = [A + m(t)] \cos \omegact

• Two cases of interest:

  1. [A + m(t)] \geq 0 \forall t, or A \geq |m(t)|

  2. [A + m(t)] < 0 \forall t, or A < |m(t)|

• If A \geq m(t), then we can recover m(t) by detecting the “envelope” of \Phi_{AM}(t)

AM - Envelope

• Definition: envelope

• If E(t) varies slowly in comparison with the sinusoidal carrier \cos \omegact, then the envelope of E(t) \cos \omegact is |E(t)|

• Therefore, envelope detection is only possible if and only if |A + m(t)| = A + m(t)

• To summarise,

  1. f_c >> the bandwidth of m(t)

  2. A + m(t) \geq 0, \forall t

Modulation Index

• m(t) with zero offset

• Let \pm mp the max and min values of m(t), then the condition for envelope detection implies A \geq mp

• Modulation index defined as

\mu = \frac{m_p}{A}

\mu = \frac{b}{A}

• For distortionless envelope detection:

• 0 \leq \mu \leq 1

Modulation Index

• m(t) with nonzero offset

• In some occasions m(t) can have a nonzero offset, i.e. m{min} ≠ m{max}

• Still, in this case, for distortionless envelope detection:

  • 0 \leq \mu \leq 1,

  • but here

  \mu = \frac{m{max} - m{min}}{2A + m{max} + m{min}}

Example 4.4

• Sketch \phi{AM}(t) for modulation indices of \mu = 0.5 and \mu = 1 when m(t) = b \cos \omegamt.

• This case is referred to as tone modulation because the modulating signal is a pure sinusoid (or tone).

• In this case, m{max} = b and m{min} = -b.

• Hence the modulation index according to Eq. M = \frac{b}{A}

therefore,
m(t) = b \cos \omegamt = \mu A \cos \omegamt

• \phi{AM} (t) = [A + m(t)] \cos \omegaet = A[1 + \mu \cos \omegamt] \cos \omegaet

Sideband and Carrier Power

• The advantage of AM envelope detection comes at a cost

• The carrier term does not carry information, and the carrier power is wasteful:

  • \Phi{AM}(t) = A \cos \omegact \overset{\text{carrier}}{} + m(t) \cos \omega_ct \overset{\text{sidebands}}{}

• The power of A \cos \omegact and m(t) \cos \omegact are [see textbook eq 3.93]:

  • P_c = \frac{A^2}{2}

  • Ps = \frac{1}{2}Pm

• The power efficiency is given by

\eta = \frac{\text{useful power}}{\text{total power}} = \frac{Ps}{Ps + Pc} = \frac{Pm}{P_m + \frac{A^2}{2}}

• For the special case of tone modulation:

  • \eta_{max} = 33\%

Example 4.5

Determine \eta and the percentage of the total power carried by the sidebands of the AM wave for tone modulation when \mu = 0.5 and when \mu = 0.3.

For \mu = 0.5

\eta = \frac{\mu^2}{2 + \mu^2} 100\% = \frac{(0.5)^2}{2 + (0.5)^2} 100\% = 11.11\%

Hence, only about 11% of the total power is in the sidebands. For \mu = 0.3

\eta = \frac{\mu^2}{2 + \mu^2} 100\% = \frac{(0.3)^2}{2 + (0.3)^2} 100\% = 4.3\%

Hence, only 4.3% of the total power is in the sidebands that contain the message signal.

Modulation and Demodulation of AM

• Modulation

  • Same as DSB-SC, but add a carrier component A \cos \omega_ct

• Demodulation

  • Rectifier detector

  • Envelope detector

Rectifier Detector

• In essence, the signal is multiplied by a periodic signal w(t) (approximated by a rectangular pulse train), resulting in the rectifier output:

• vR(t) = { [A + m(t)] \cos \omegact } w(t)

• = [A + m(t)] \cos \omegact [ \frac{1}{2} + \frac{2}{\pi} (\cos \omegact - \frac{1}{3} \cos 3\omegact + \frac{1}{5} \cos 5\omegact - … )]

• = \frac{1}{\pi} [A + m(t)] + \text{other higher frequency terms}

• After the low pass filter with cutoff B Hz, the output is [A + m(t)]/\pi

Envelope Detector

• Important design constraint of RC circuit

\frac{1}{\omega_c} << RC < \frac{1}{2\pi B}

\Rightarrow 2\pi B < \frac{1}{RC