Modulation Basics

Frequency Division Multiplexing

Different TV channels occupy different frequency bands

But how do we shift the frequency of a signal from baseband to radio frequency band?
- through modulation

Modulation Revisited

Amplitude Modulation (AM)

multiplying with a sinusoid signal

carrier frequency $\omega_c$ is actually $2\pi f_c$
Therefore $y (t) = x(t)cos(\omega_c t) = x(t)cos(2\pi f_c t)$

Basic Equalities

Note:

$e^{j 2 \pi_{c} t} = cos(2\pi f_c t) + jsin(2\pi f_c t)$

$e^{-j 2 \pi_{c} t} = cos(2\pi f_c t) - jsin(2\pi f_c t)$

$x(t) e^{j 2 \pi_{c} t} \leftrightarrow X\left(f-f_{c}\right)$

$x(t) e^{-j 2 \pi_{c} t} \leftrightarrow X\left(f+f_{c}\right)$

$x(t) \cos \left(2 \pi f_{c} t\right) \leftrightarrow \frac{1}{2}\left(X\left(f-f_{c}\right)+X\left(f+f_{c}\right)\right)$

Frequency Domain Interpretation of Modulation

Demodulation

How to get back to the baseband?

To get back to the baseband:
- Multiplying with the same sinusoid + low pass filtering

Summary

Modulation:

$y(t) = x(t)cos(2\pi f_c t)$

Demodulation:

$w(t) = y(t)cos(2\pi f_c t) = x(t)cos^2(2\pi f_c t)$

Using the equality $cos^2(\theta) = \frac{1}{2}(1 + cos(2\theta))$

$w(t)=\frac{1}{2}\left(1+\cos \left(4 \pi f_{c} t\right)\right) x(t)=\frac{1}{2} x(t)+\frac{1}{2} x(t) \cos \left(4 \pi f_{c} t\right)$

After applying Lowpass filter, only the first term is retained and the second term is removed.

$\frac{1}{2}x(t)$

Real-life example

To transmit the three signals (programme), each signal is shifted to a different carrier frequency

How about demodulation?
- since we only select one program to watch, we use a demultiplexer to filter other programs out.
- Then apply the demodulation

Transmission Layer – Digital Modulation Techniques

Transmission Layer – Digital Modulation Techniques for Digital TV

Modulation for Digital TV

Recall: for each bit entered into the convolutional encoder, 2 bits are generated
For transmission of digital bits over analog channels
- Convert group of digital bits into analog waveforms (symbols)
- The analog waveforms are designed according to the desired carrier frequency

Modulation schemes used in DVB/DTMB

QPSK (Quadrature Phase-shift Keying) or 8-PSK for satellite
- the signal is rather weak (far away)
  - the carrier-to-noise ratio C/N can be very low (10dB or less)
- The channel bandwidth is large, generally between 27 and 36 MHz.
- QPSK: require wide channel bandwidth and low SNR.
16-QAM (Quadrature Amplitude Modulation), 32-QAM, 128-QAM or 256-QAM for cable
- the signal is strong (general more than 30dB) and signal amplifiers can be added
- different channel bandwidths are adopted
  - (6 MHz in the U.S., 7 MHz in Australia and some parts of Europe, 8 MHz in China, Russia, and some parts of Europe.)
- QAM: require higher SNR transmission threshold and lower bandwidth than QPSK.
16-QAM or 64-QAM COFDM (Coded Orthogonal Frequency Division Multiplex) for terrestrial transmission, 4G or 5G transmission

Digital Modulation Formats

Modulation may be viewed as changing certain characteristics of a carrier (usually a sinusoidal signal) in accordance with a modulating wave (input data).
In digital modulation, the modulating wave is either binary or M-ary coded data.
- Mapping a sequence of binary digits into a corresponding set of discrete amplitudes, phases, or frequencies of the carrier to distinguish one signal from another.

Phase-shift Keying (PSK)

PSK change the phase

Binary PSK or PSK

A digital signal has only two states, 1 and 0, and when it is used to modulate a carrier, only two states of the carrier phase are necessary to convey the digital information.

Input data can be either 1 or 0
From 1 to 0 or 0 to 1, a 180 degree phase difference is introduced

Quadrature PSK (QPSK), aka 4-PSK

M-ary Coded Data
- The modulation produces one of an available set of $M=2^m$ distinct signal in response to $m$ bits of source data at a time.
Example: $m=2$ $m = 2$ and $M=4$ $M = 4$
- Four possible patterns:
- 00, 01, 10, 11

For each of these combination, the carrier is set to a particular phase angle

8-PSK

For each carrier, it can transmit 3 bits
- more coding efficiency
Error are more easy to occur (due to smaller phase angle difference)

Amplitude-shift Keying (ASK)

ASK change the Amplitude

Binary ASK or ASK

Amplitude-shift keying (ASK) is a kind of Amplitude Modulation which represents the binary data in form of variations in the amplitudes of the signal.

Input data can be either 1 or 0
0 value output for 0 input
carrier output for 1 input

M-ary Coded Data in ASK

The modulation produces one of an available set of $M=2^m$ distinct signal in response to $m$ bits of source data at a time.

0 value output for 00 input
Different scale of carrier output for other inputs

Amplitude Modulation by Cosine Function

Fourier transform property for modulation by a cosine

$Y(\omega)=\frac{1}{2} F\left(\omega+\omega_{c}\right)+\frac{1}{2} F\left(\omega-\omega_{c}\right)$

Demodulation is modulation then lowpass filtering
Similar derivation for modulation with $sin(\omega_c t)$

Amplitude Modulation by Sine Function

$y(t) = f(t) sin(\omega_c t)$

Fourier transform property for modulation by a sine

$Y(\omega)=\frac{j}{2} F\left(\omega+\omega_{c}\right)-\frac{j}{2} F\left(\omega-\omega_{c}\right)$

Demodulation is modulation then lowpass filtering

Quadrature Amplitude Modulation(QAM)

Two signals are used to modulate a single carrier using two balanced modulators operating in phase quadrature:
- Utilize both sine wave and cosine wave as carriers.
- The modulated signals are sent over a single channel simultaneously
- The amplitude (including sign and magnitude) of each wave conveys the information (bits) being sent.
  - Orthogonality or Quadrature

Recall:

QAM method used to multiplex two signal components together onto a single carrier frequency

Transmit 4 input bits, divided into $I$ and $Q$
Single signal, occupying the whole available bandwidth
The symbol rate is the bandwidth of the signal being centered on carrier frequency

QAM Constellation

A two-dimensional representation of the QAM signal.
Bit-to-symbol mapping: from data bits (say, m bits) to $M$ ( $M=2^m$ ) possible symbols (or constellation points).
For $m=4$ , 16 points allow a unique point for any combination of bits. (16-QAM)

QAM example

If too much noise is present at the receiver $(e > \frac{d_{16}}{2})$ $(e > \frac{d _{16}}{2})$ ,
- the point projected on the constellation may be closer to the wrong known point than the accurate known point,
- resulting in an error estimating the recovered symbol
4-QAM is less sensitive to noise (because error occur when $e > \frac{d_{4}}{2}$

256-QAM can transfer 8 bits. ( $2^8 = 256$ )

So we pick the QAM depends on the bit error rate on the channels.

QAM Demodulation

Logical view of a QAM demodulator

Amplitude gives the estimation of $I$ and $Q$

Signal-Processing view of a QAM Demodulation

Orthogonality between a sine and a cosine wave allows them to transmit data simultaneously over a channel.
- Discrete sine and cosine of same frequency are orthogonal to each other, i.e., their correlation at zero lag (dot product) is zero.
- https://www.youtube.com/watch?v=3pU0e1r5JDE
Consider a single period of each wave, the orthogonality principle can be illustrated by

$\int_{0}^{T} \sin \left(\frac{2 \pi t}{T}\right) \cos \left(\frac{2 \pi t}{T}\right) d t=0$

Where $T$ is the period of the sine and cosine wave

detailed explaination:

$\int_{0}^{T} \sin \left(\frac{2 \pi t}{T}\right) \cos \left(\frac{2 \pi t}{T}\right) d t = \frac{1}{2}\int^T_0 \sin(\frac{4\pi t}{T})dt = \frac{-1}{2}\cos\frac{4\pi t}{T}|^T_0$

$= \frac{-1}{2}[\cos(4\pi\frac{T}{T})-cos(0)] = \frac{-1}{2}[1-1]= 0$

Sine and cosine functions are called basis functions

Note the $I$ and $Q$ values are from the Constellation map

Why $\int_{0}^{T} I_{i} \cos ^{2}\left(\omega_{c} t\right) d t = \frac{TI_i}{2}+0$ ?

$\int_{0}^{T} I_{i} \cos ^{2}\left(\omega_{c} t\right) d t$

$= \frac{1}{2}\int^T_0 (I_i + I_icos(2\omega_ct))dt$

$= \frac{1}{2}\int^T_0 I_i dt + \frac{1}{2}\int^T_0 cos(2\omega_ct)dt$

$= \frac{TI_i}{2} + \frac{1}{2}\sin(2\omega_c t)|^T_0$

Now we only consider $\frac{1}{2}\sin(2\omega_c t)|^T_0$

$\frac{1}{2}\sin(2\omega_c t)|^T_0 = \frac{1}{2}\sin(4 \pi f_c t)|^T_0$

$= \frac{1}{2}(\sin(4\pi f_cT) - sin(0) ) = \frac{1}{2}(1-1) = 0$

Therefore $\int_{0}^{T} I_{i} \cos ^{2}\left(\omega_{c} t\right) d t = \frac{TI_i}{2}+0$

The idea is same of the $Q$ term.

The terms going to zero in equations (2) and (3) because of the orthogonality property of sines and cosines.
The values at the output of the demodulator then contain a scaled estimate of the magnitudes of the received sine and cosine waves to project onto the receiver’s constellation map.
The scale factor can be removed before integrating by scaling the multiplication cosine and sine waves.

Intersymbol Interference (ISI)

Multiple Path Interference (Multipath)

In wireless or terrestrial communication, Multipath is the propagation phenomenon that result in signal reaching the reciever antenna by 2 or more paths.
- Signal from transmitter is the desired signal
- Reflected paths are undesirable
different paths that a signal takes in reaching an aerial from the transmitter, and the signal level varies greatly according to the distance.
interference and variable echoes due to multipath exist.
- worse channel conditions

Reflected signals take a longer time to reach the receiving aerial than direct signals
- Ghosting occur

Intersymbol Interference (ISI)

In digital TV transmission, delayed signal can cause intersymbol interference (ISI) and fading
The Recieve side will just add all the signals together, yielding a resulting signal like the yellow one in below picture.
- Completely cannot determine the recieved signal

Intersymbol interference (ISI) is a form of distortion of a signal in which one symbol interferes with subsequent symbols. This is an unwanted phenomenon as the previous symbols have similar effect as noise, thus making the communication less reliable. ISI is usually caused by multipath propagation of a channel.

It can be solved by COFDM.

ISI on SHORT symbol duration

Consider: data is sent serially over the channel by modulating one single carrier at a baud rate of R symbols per second.
- The data symbol period $T_s$ is then 1/R.

Recall:

What is a Symbol? What is a symbol rate? What is the different between bits and symbols?

If $T_s$ $T_{s}$ is very small, in a multi-path fading channel
- time dispersion can be significant compared to the symbol period
- results in serious inter symbol interference (ISI).

If the delay is in the region of 180°, the reflected signal cancels out the direct signal and complete picture failure occurs.

How about same delay, in a long symbol duration?

ISI on LONG symbol duration

In this region, the reflected signal is actively supporting the direct path signal.
With long symbol duration, a reflected signal will only be out of phase with the direct signal for a part of total duration

to avoid the grey shaded area, a guard interval (guard period) is added at the start of the symbol during which time the receiver pauses before starting the evaluation of the carriers.

Guard Period

To eliminate ISI almost completely, a guard period is added for each symbol.
- is chosen larger than the expected delay spread, such that multipath components from one symbol cannot interfere with the next symbol.
Instead of the delayed signals interfering with the direct-path symbol, since the same signals make the symbol, they simply increase the power in the received symbol before it is processed.

Reduce the transmission efficiency

Terrestrial TV reception using COFDM

Consider a bitstream consisting of 500 bits, with a bit duration of 0.1µs.

Case 1: if the 500 bits are used to modulate a single carrier, the duration over which each bit remains “active” i.e. the symbol duration = 0.1µs

Case 2: if the 500 bits are used to modulate 500 different carriers to form a COFDM symbol, the symbol duration = 0.1µs x 500 = 50 µs

The Coded Orthogonal-Frequency-Division-Modulation (COFDM) technology is employed.
- Using COFDM, instead of just a single carrier signal, multiple orthogonal (equally spaced) carriers are used, each of which is independently modulated by one or more bits from the encoded bitstream to be transmitted
- During each symbol period, the all modulated signals are added together to form the signal/symbol that is transmitted.

Conceptual view of an COFDM modulator

Involves the distribution of a high-rate serial bitstream over a large number of parallel carriers.
For each carrier, the bit rate is far below that of the original modulating bitstream
Due to the large number of carriers, the duration of the COFDM symbol is considerably larger than the duration of one bit of the modulating bitstream

Multicarrier Modulation

The total signal frequency band is divided into $N_c$ non-overlapping frequency subchannels.

$T_s$ $T_{s}$ becomes relatively large
- the symbol duration is made $N_c$ times larger
time dispersion is insignificant compared to the symbol period in a multi-path fading channel
then inter symbol interference (ISI) is not so serious.
A guard period is then needed for the each symbol

It seems good to avoid spectral overlap of channels to eliminate interchannel interference/intercarriers interference (ICI).
- inefficient use of the available spectrum
To cope with the inefficiency, the ideas is to use parallel data and FDM with overlapping subchannels.

But Overlapping subchannels will produce ICI (intercarrier interference).
- Affect each other
The selection of multiple carriers is to combat the multipath distortion, as well as to fully use the available bandwidth
We need to look for Orthogonality between the different modulated carriers, so they can cancel out
- How to select the multiple carriers that are orthogonal?

Principle of COFDM

The available bandwidth W is divided into a number of $N_c$ of subband

in single carrier modulation, data is sent serially over the channel by modulating one single carrier at a baud rate of $R$ $R$ symbols per second.
- Data symbol period $= \frac{1}{R}$
Instead of transmitting the data symbols in a serial way, at a band rate $R$ , a multicarrier transmitter partitions the data stream into blocks of $N_c$ data symbols that are transmitted in parallel by modulating the $N_c$ carriers.
Now , the symbol duration for a multicarrier scheme:
- $T_s = \frac{N_c}{R}$

Selection of subcarriers

To assure a high spectral efficiency, the subchannel waverforms must have overlapping transmit spectra.
They need to be orthogonal for enabling simple separation of these overlapping subchannels at the receiver.
Multicarrier modulations that fulfill these conditions are called coded orthogonal frequency division multiplex (COFDM) systems.

$f_k = \omega_k / 2\pi$ (subcarrier frequency)

$f_0 = w_0/2\pi$ ((the lowest frequency)

The spacing between the adjacent subcarriers is:

$\Delta f = \frac{1}{T_s} = \frac{R}{N_c}$

A general set of orthogonal waveforms, is given by:

$\begin{array}{l} \psi_{k}(t)=\left\{\begin{array}{cc} \frac{1}{\sqrt{T_{s}}} e^{j \omega_{k} t} & t \in\left[0, T_{s}\right] \\ 0 & \text { otherwise } \end{array}\right. \\ \text { with } \omega_{k}=\omega_{0}+k \omega_{s} ; \quad k=0,1, \ldots, N_{c}-1 \end{array}$

Recall:

the Sinc function

Multiplication by a complex exponential

$x(t) e^{j \omega_{c} t} \leftrightarrow X\left(\omega-\omega_{c}\right)$

These two signals are not affecting each other

When one is on maximum value, another one is on zero.

Expand the idea of “When one is on maximum value, another one is on zero”,

The Spectrum of a COFDM Signal will be like this:

To be able to determine which subset of the carrier-ensemble is present in a given symbol, carriers need to be ORTHOGONAL.
At carrier frequency, All other carriers are ZERO.
Strict fixed spacing of $1/T_s$ between adjacent carriers.

Example:

lets say we want to transmit 5 bits.

Receiver determines which of the five carrier signals are present in the symbol

All the signals are not affecting each other

Therefore we can determine the original bitstream that was transmitted

Successfully avoid ISI and ICI

Design Considerations

Although adding a guard interval ( $T_g$ ) avoids ISI and ICI, it also influences the max. encoded bit rate can be supported.
Longer $T_g$ $T_{g}$ => the lower the max. bit rate.
- Typically: $T_g < \text{max. value of } \frac{T_s}{4}$

Example: 2K specification for DVB-T

$T_g = 56$ us

$T_s = 56 \times 4 = 224$ us

=> a carrier spacing and fundamental frequency $f_s = \frac{1}{T_s} = 4.464$ kHz

Assuming a usable channel bandwidth is 7.607 MHz

Number of carriers $N_c = \frac{7.607M}{4.464K}+ 1 = 1704.07706 + 1 = 1705$

=> the carriers ( $N_c = 1705$ ) would be spaced at:

$0, 4.464, 8.928, 13.392, …, 7602.192, 7606.656$ kHz

Implementation of Set-top/Video Decoder Box

Also called integrated receiver decoder.
Three main parts
- Channel decoder
  - Demodulation, Forward Error control, output data are 188 byte transport packets
- Processor with transport stream processing
- Source decoder

The China Standard: DTMB

The DTMB system adopts multi-carrier transmission technique referred to as COFDM.
In the COFDM module, 64QAM (2^6 => 6bits) symbols as well as Transmission Parameter Signaling (TPS) are arranged properly to produce COFDM frames.
An COFDM frame defined in DTMB has 3780 sub-carriers and the guard interval length is of total data duration with time-domain synchronization and channel estimation.

DTMB(Digital Terrestrial Multimedia Broadcast)

TV standard for mobile and fixed terminals used in the People’s Republic of China, Hong Kong, and Macau. Although at first this standard was called DMB-T/H (Digital Multimedia Broadcast-Terrestrial/Handheld), the official name is DTMB.

Standard for terrestrial digital television in China — both fixed and mobile

time-domain synchronous orthogonal frequency-division multiplexing (TDS-OFDM) with mQAM/QPSK

Advantage of using COFDM

SFN problem is solved

Recall:

A single-frequency network or SFN is a broadcast network where several transmitters simultaneously send the same signal over the same frequency channel.

Advantage:

Only one frequency channel is required to cover the service area

Disadvantage:

If tranmitters’ covering area overlapped, ghosting (shadow of video frames) will happen due to echoes of the same signal (multi-path)

But with COFDM, we can adopt the SFN to save the bandwidth spectrum.

By using COFDM, the high tolerance to fading and multipath results in improving the quality of reception.
It allows broadcasting authorities to use a SFN throughout the country.
A signal from an adjacent transmitter broadcasting an identical signal will cause ghosting in analogue TV transmission.
- This problem can be solved by using DTMB (with the help of COFDM).

Practice Questions

A digital satellite channel interface uses the MPEG-2 transport stream multiplex, an RS(204,188) block code, a rate 3/4 convolutional encoder and QPSK modulation. Determine the number of overhead bits associated with the interface and hence the useful bit rate that is obtained with a channel that operates at (i) 20 Mbaud and (ii) 27.5 Mbaud.

What is baud? When we talks about baud, we need to convert them to bitrate.

baud means Symbol per second.

Therefore 20 Mbaud = 20 Symbol per second.

For QPSK, it use 2bits (4 phase!). Therefore each symbol can carry 2 bits.

Therefore 20 Mbaud = 20M x 2 = 40M bps.

What if we are using QAM? 16QAM uses 4 bits. Therefore each symbol can carry 4 bits.

Therefore 20 Mbaud = 20M x 4 = 80M bps.

Thus:

With QPSK, each symbol = 2 bits and hence raw bit rates are

(i) 20M x 2 = 40 Mbps

(ii) 27.5M x 2 = 55 Mbps.

MPEG-2 transport stream multiplex comprises a stream of 188-byte packets each of which has a 4-byte header. (The payload is only 184 byte, revisit Ch5)

The input: 188 bytes

The output of RS(204,188) is 204 bytes

Overhead introduced by RS = 204 - 188 = 16

after going into byte interleaver, the output is still 204 bytes. (no overhead)

Now the TCM convolutional encoder is at 3/4 rate.

The output of TCM = $204 \div \frac{3}{4} = 272$

Overhead introduced by TCM = 272 - 204 = 68

Finally the Modulation, no overhead

Therefore total overhead = 4 + 16 + 68 = 88 bytes.

The useful bit rate with a raw bit rate of 40Mbps: $40M \times \frac{184}{272} = 27.1Mbps$

The useful bit rate with a raw bit rate of 55Mbps: $55M \times \frac{184}{272} = 37.2Mbps$

Explain the principle of operation of COFDM transmission. Include the role of:

(a) the symbol mapper

In single carrier modulation, data is sent serially over the channel by modulating one single carrier at a baud rate of R symbols per second. In COFDM, instead of transmitting the data symbols in a serial way, at a baud rate R, the symbol mapper partitions the data stream into blocks of Nc data symbols that are transmitted in parallel by modulating Nc carriers.

(b) the generation of each transmitted symbol

Using COFDM, instead of just a single carrier signal, multiple orthogonal (equally-spaced) carrier are used, each of which is independently modulated by one or more bits from the encoded bitstream to be transmitted. The first carrier is a DC level, while the remaining four carriers are all sinusoidal signals of frequencies fs, 2fs, 3fs, and 4fs respectively. In the example shown in Fig. Q1, five carriers are used each of which is modulated by a single bit from the transmitted bitstream using on-off keying. The period Ts is called the symbol period and, since on-off keying is being used, is equal to the time to transmit 5 bits from the encoded bitstream being transmitted. During each symbol period, the five modulated signals are added together to form the signal/symbol that is transmitted.

The guard interval is used to ensure that all delayed versions of the direct-path signals that make up the symbol have been received. In this way, instead of the delayed signals interfering with the direct-path symbol, since the same signals make up the symbol, they simply increase the power in the received symbol before it is processed. The processing involves determining which of the five carrier signals are present in the symbol and, based on this, the receiver can determine the original bitstream that are transmitted.

In digital video broadcasting – terrestrial, assuming a guard interval of 50 us and a usable channel bandwidth of 7 MHz, derive (i) the symbol period and (ii) the number of carriers that would be used.

Although adding a guard interval ( $T_g$ ) avoids ISI and ICI, it also influences the max. encoded bit rate can be supported.

Longer $T_g$ => the lower the max. bit rate.

Typically: $T_g < \text{max. value of } \frac{T_s}{4}$

Usable channel bandwidth of 7 MHz

Guard interval $T_g = 50$ us

$T_s = 50 \times 4 = 200$ us

(i) the symbol period = 200us

=> a carrier spacing and fundamental frequency $f_s = \frac{1}{T_s} = 5$ kHz

Number of carriers $N_c = \frac{7000K}{5K} + 1 = 1400 + 1 = 1401$

=> the carriers ( $N_c = 1401$ ) would be spaced at:

$0, 5, 10, 15, …, 6995, 7000$ kHz