List of Muddy Points in Comms IV

Introduction

If you found something confusing in a Comms IV lecture and thought it was a rather "muddy point" then ask a question on this page. Anyone can answer: either myself or any student or anyone in the department can answer. I will, of course, tweak everyone's answers to make them clear and check they are correct. So don't worry, just go for it and have a stab at it! You are allowed to correct my answers and make them better.

Section 0: Overview

Muddy Point 1: On slides 20 and 21 why is there a negative sign inside the function?

Answer 1: Consider a function [math]f(t)\,[/math] which takes the value [math]f(0)\,[/math] at [math]t = 0\,[/math]. If we shift this function to the right, so that it takes the value [math]f(0)\,[/math] when [math]t = T\,[/math], it is clear that the function must be [math]f(t-T)\,[/math], because putting [math]t = T\,[/math] in this gives [math]f(0)\,[/math].

Muddy Point 2: On slide 25, derive the result for completing the Square, how exactly is it done?

Answer 2: As mentioned in the lecture, there is a less general example here but it should give you the idea.

Section 1: Introduction

Muddy Point 1: What is a heterodyne receiver again?

Answer 1: It's basically what is used to detect frequency modulated (FM) signals. An antenna, photodetector, or any kind of detector only directly senses amplitude. So the trick for FM is to exploit interference effects between two signals. A heterodyne receiver is just a way of fooling an amplitude detector into sensing the phase difference between two signals. One of these signals is your received signal that carries the message, and the other is a reference signal called the local oscillator (LO). You mix the two signals and tune the LO so that you get interference beats. The principle is that the amplitude of the beats is proportional to the phase difference between the two signals. "Hetero" is Greek for "different" so this refers to the fact that we are mixing two different signals, and "dyne" is Greek for "force" that refers to the signal itself.

Muddy Point 2: So what is the difference between heterodyne and superheterodyne?

Answer 2: In heterodyne detection you convert from the received RF frequency directly down to baseband. In superheterodyne detection, you go from RF to an intermediate frequency (IF). This makes life easier as you only have to build an amplifier at IF, instead of a tunable amplifier over the RF range. This will become clearer as the lectures evolve, but that is the basic idea. Why conversion to IF gets to be called "super" I have no idea...if anybody finds out please insert here!

Comment: Superheterodyne is short for supersonic heterodyne. Supersonic because the heterodyne (beat) frequencies [math]f-f_{LO}\,[/math] and [math]f+f_{LO}\,[/math] (one of which is your IF) are higher in frequency than the baseband audio (sonic) signal. --Mattjb 16:31, 14 March 2008 (CST)

Reply: Excellent, you are correct.

Muddy Point 3: Why do we need to bother modulating signals?

Answer 3: This will become clearer when we cover this later in the lectures. But in a nutshell we can split this into three pertinent questions: what, which, and why? (A) What is modulation? Basically, in order to impress our message signal on a carrier frequency, we do two things: (i) code the information and (ii) modulate the signal. Imagine I communicate with you by flashing a light on and off. The light is the carrier in this case. For the coding, I might choose Morse code that uses two binary states. For the modulation, the light "on" could be a "1" and "off" could be a "0". Alternatively "on" for 2 seconds could be a "1" and "on" for 1 second could be a "0". Or I could rapidly switch the light 10 times for a "1" and 5 times for a "0" and so on. I could endlessly keep coming up with as many different silly modulation schemes, as the drugs I am on will allow. The goal for the engineer is to choose the modulation and coding scheme that gives the best trade-off between bandwidth and error rate, for a given scenario. (B) Which? Flashing the light on and off was a "toy" example of digital modulation. However, we could do it another way: using a dimmer switch we could continuously change the intensity of the light. This is what we would call analog modulation. So there are different modulation schemes to choose from. (C) Why? We modulate the carrier to allow for encoding of the message. But why do we need the carrier? The choice of a carrier could be light in the case of the above example or it could be any other frequency such as a radio signal. Firstly, our choice of carrier depends on our channel. If our channel is glass fibre, we go for optical light or infrared. If it is free-space we might go for a radio frequency for a long distance link or infrared if we only want a short distance (eg. your TV remote). In the case of radio signals, we may choose our carrier frequency to be much higher than our message signal in order to reduce antenna size. There are two basic reasons we need the carrier: (i) for efficient propagation through the channel, and (ii) so we can allocate non-overlapping bandwidths to allow many simultaneous communications.

Comment: I've heard that we also need to modulate because the frequency at which we are transmitting is related to the size of the antenna we need. So if I was to transmit voice at baseband I would need a really massive antenna. In comparison, if my carrier is transmitting something in the GHz range (like mobile phones) the antenna size is very small.

Reply: Yes, the selected modulation frequency will indeed determine your antenna size. In practice, however, one has to modulate for the above reasons, for free-space propagation, and if you are restricted to a certain antenna size you then have to choose the appropriate frequency.

Muddy Point 4: History of the telegraph was muddy.

Answer 4: Not sure where your difficulty was but you can get a nice summary of the history here. The take home message about the history that I wanted to put across is the principle that Morse code uses short codes for common letters and long codes for letters that are used less frequently.

Muddy Point 5: Solution to complete the square question is not available in exercise solutions which only start for exercises after first lecture

Muddy Point 6: The formula sheet doesn't have the definition for the Fourier transformations of sinx and cosx. What are they? using the definition of a Fourier Transform we are using for this course

Answer 6:

For sin we have [math] \cos(\omega_c t + \phi) \Leftrightarrow \frac{1}{2} \left[ e^{j\phi}\delta(f - f_c) + e^{-j\phi}\delta(f + f_c) \right] \quad [/math] and for cos we have [math] \sin(\omega_c t + \phi) \Leftrightarrow \frac{j}{2} \left[ e^{j\phi}\delta(f - f_c) - e^{-j\phi}\delta(f + f_c) \right] \quad [/math]

Section 2: Frequency Domain Analysis

Muddy Point 1: On slide 4, what does "dimensions V" mean again?

Answer 1: It just means that [math]X\,[/math] has the units or dimensions of Volts. It doesn't necessarily have to be Volts, but we picked on Volts purely by way of example, because in practice we most commonly look at signal amplitudes in the voltage domain.

Muddy Point 2: What exactly does it mean again if something is Hermitian?

Answer 2: A function [math]g(t)\,[/math] is Hermitian if [math]-g = g^*\,[/math]. It is named after the French mathematician Charles Hermite (1822-1901). Given a real signal in the time domain, [math]g(t)\,[/math], the Fourier transform results in the Hermitian property that [math]-G(f) = G(f)^*\,[/math]. It then follows that the real part of [math]G\,[/math] is and even function, i.e. [math]{\rm Re}[G(f)] = {\rm Re}[G(-f)]\,[/math], and that the imaginary part of [math]G\,[/math] is odd, i.e. [math]{\rm Im}[-G(f)] = {\rm Im}[G(-f)]\,[/math]. This means that the amplitude spectrum is even, and the phase spectrum is odd. Given the amplitude is an even function, it can sometimes simplify the maths to replace the double-sided Fourier integral by twice its single-sided form. It's a cool trick that comes in handy for some exercises. Also, as we will see in later lectures this property comes in handy when we define what is called the analytic signal. More about that later.

Muddy Point 3: I was just wondering if there was a difference between narrowband and baseband signals? Or can the two terms be used interchangeably?

Answer 3: For informal chatting amongst engineers, these terms can get slightly misused interchangeably without anyone misunderstanding what you really mean. However, for this course, we should stick strictly to the proper meanings. Strictly the baseband is the frequency band containing the information (before it has been modulated). We refer to this as the narrowband frequency (after it has been modulated), as it is generally very narrow compared to the carrier frequency, but it is upshifted from the baseband.

Muddy Point 4: With the solutions to Exercise on slide 2.16 (link), I was wondering how the equations on the left of line 3 equals that on the right of line 3? Isn't the small t' in the exponential on the right supposed to be t'+kT in order to be consistent?

Section 3: Analog Modulation Systems

Muddy Point 1: I'm unsure of instantaneous frequency and peak frequency deviation on Slide 35 of Section 3.

Answer 1: The mathematical description is on the slide and on Page 97 of Proakis. However, I think your question is asking, "what does it all mean?" Well, for an FM signal you are going to have a certain carrier frequency, but you are going to modulate that frequency to carry your message. So the frequency is constantly changing due to the modulation. The frequency at a given instant in time is the instantaneous frequency. And the maximum shift in frequency from the nominal carrier frequency is called the peak frequency deviation.

Muddy Point 2: The area of noise is muddy.

Answer 2: Well, please look at the noise section of the notes in the mid-semester break, whilst listening to the relevant podcast. Then get back to me with some specific questions about the noise that are troublesome. At the moment your question is a little too broad!

Muddy Point 3: Amplitude modulation is muddy.

Answer 3: Again, please review this section on the podcasts and get back to me with more specific questions. If your question is asking how to visualize the difference between AM and FM, see this animated gif.

Animated Gif

Muddy Point 4: For carrier recovery, why is the local carrier frequency [math]2f_o\,[/math]?

Answer 4: Well, it isn't. It is just [math]f_o\,[/math]. In Section 3, slide 19, [math]2f_o\,[/math] is mentioned, but that is not the recovered carrier frequency....it is mentioned as the frequency above which we want to filter to remove unwanted components.

Muddy Point 5: In the FM modulation notes, we take the integral of the message [math]m(t)\,[/math] and modulate the angle of the carrier with it. Why does that integral go from [math]-\infty\,[/math] to [math]t\,[/math]? (As opposed to [math]0\,[/math] to [math]t\,[/math]).

Answer 5: Because the instantaneous (radian/sec) frequency is [math]\omega(t)=\frac{\phi(t)}{dt}\,[/math], we obtain [math]\phi(t) = \int_{t_i}^t \omega(\lambda) d\lambda\,[/math]. What you choose for [math]{t_i}\,[/math] is somewhat arbitrary, but we use [math]-\infty\,[/math] assuming the system had been running for a long time. No harm in using [math]t_i = 0\,[/math], as this assumes the system starts operating at [math]t = 0\,[/math]. Using a different value of [math]{t_i}\,[/math] simply introduces a change in carrier phase (constant of integration), which is not of interest. The upper limit must be [math]t\,[/math] though.

Muddy Point 6: For carrier recovery, you start with a [math]\cos\left(2\pi f_o t\right)[/math] term and square it only to recover that same term in the end. So looking at it mathematically, why can't we grab the initial [math]\cos\left(2\pi f_o t\right)[/math] term in the first place?

Answer 6: The message [math]m(t)[/math] has no DC component and [math]m(t)\cos\left(2\pi f_o t\right)[/math] has no component at [math]f_o[/math], as we are talking about a suppressed carrier. However, as you see from the calculation, by squaring [math]m^2(t)[/math] it has a DC component and hence there is a component at [math]2 f_o[/math]. In theory one might be able to use some other non-linear operation instead of squaring, but squaring is the simplest option and easier for a real implementation.

Muddy Point 7: I calculated the clipping probability of [math]1.6\times10^{-3}[/math] for [math]\langle{m^2(t)}\rangle=0.1[/math], as it said to do on the lecture slides. However, I'm not sure how to interpret the probability and what it means.

Answer 7: If [math]\langle{m^2(t)}\rangle=0.1[/math], this means the RMS value is [math]\sqrt(0.1) = 0.3162[/math]. The signal [math]m(t)[/math] is constrained to be [math]-1 \leq m(t) \leq +1[/math], so the clipping level is 1/0.3162 = 3.162 times the RMS value. Hence the probability that the signal [math]m(t)[/math] exceeds the limits of +/-1 is [math]2 \times Q(3.162) = 1.565 \times 10^{-3}[/math]. For an ergodic signal, the meaning of the clipping probability is it the fraction of time that [math]m(t)[/math] exceeds the limits of +/-1.

Muddy Point 8: In the case of synchronous demodulation, does the phase difference between the two signals multiplied together matter?

Answer 8: Good question. It depends. For synchronous demodulation of AM or DSB, the phase of the local carrier must be the same as that of the received signal, so the usual procedure is to use a phase locked loop to extract the local carrier from the received signal. However, the phase locked loop will produce a carrier 90 degrees shifted from the received carrier, so this must be shifted by 90 degrees to do the modulation. With DSBSC it is a bit more tricky since there is no carrier component in the received signal, so the usual procedure is to apply the received signal to a square law (or non-linear) device and use a phase locked loop to extract the double frequency carrier component, divide the frequency by two, and use this as the local carrier (phase corrected if necessary). In SSB the phase can be different. For DSBSC is [math]x(t)\cos(wt)[/math] and squaring gives [math]x^2(t)\cos^2(wt)[/math] which contains a [math]\cos(2wt)[/math] component. A phase locked loop will generate [math]\sin(2wt)[/math] (as a square wave). Frequency divide by two on the negative edge gives [math]\cos(wt)[/math] (as a square wave), which can be used as the local carrier.

Section 4: Random Processes and Linear Systems

Muddy Point 1: I'm unsure of in phase and quadrature noise components on Slide 41 of Section 4.

Answer 1: Just think of noise as a phasor that can be resolved into two orthogonal components (just like you do with vectors). Now look at slide 14 of Section 5. The horizontal component is in phase with the signal, as they line up. But the vertical component is in quadrature with the signal because it is at right angles. [By Pythagoras, a right angled component can only be added in quadrature, which means adding sums of squares...that is the origin of the term quadrature, which is Latin for 'square'.]

Muddy Point 2: Crosscorrelation and autocorrelation are vague.

Answer 2: These are defined on slide 6 of Section 4. Autocorrelation, [math]R_{xx}\,[/math], is simply a signal [math]x(t)\,[/math] multiplied by a time shifted copy of itself [math]x(t-t')\,[/math] averaged over all [math]t'\,[/math]. So if [math]x(t)\,[/math] is totally random, then [math]R_{xx} = 0\,[/math]. The more repetitive structure the signal [math]x(t)\,[/math] has in it the bigger [math]R_{xx}\,[/math] will be. Now the crosscorrelation, [math]R_{xy}\,[/math], is exactly the same idea but applied to two separate signals [math]x(t)\,[/math] and [math]y(t-t')\,[/math]. So if [math]x\,[/math] and [math]y\,[/math] are totally uncorrelated then [math]R_{xy} = 0\,[/math], and if they have some structure in common then [math]R_{xy}\,[/math] will have some value.

Muddy Point 3: The relationship between [math]Rxx[/math] and [math]Sxx[/math].

Answer 3: The autocorrelation, [math]Rxx[/math], and the power spectral density (PSD), [math]Sxx[/math], are Fourier pairs. That is how they are related. This is on slide 10 and is called the Weiner-Khinchin theorem. It is easy to derive and you should check it out on page 179 of Proakis. The idea is that the PSD of [math]x(t)[/math] is related to [math]|X(f)|^2[/math]. So you then substitute in the Fourier integral of [math]x(t)[/math] into this expression. Because it is squared, you end up with two integrals that can easily be manipulated to give you [math]Rxx[/math].

Muddy Point 4: Energy and power spectral density are vague.

Answer 4: If I have a signal [math]x(t)[/math], the Fourier transform is [math]X(f)[/math] and the energy spectral density (ESD) is [math]|X|^2[/math] as defined on slide 8 of Section 4. Energy is proportional to voltage squared, so if [math]x(t)[/math] was in volts then the integral of [math]|X|^2[/math] is proportional to the energy content in the signal. The power spectral density (PSD) is like the ESD except it is ESD per unit time, and is defined on Slide 9. If we integrate PSD over all frequency we get the total power in the signal. Another important result is that the PSD is also the Fourier transform of the autocorrelation [math]Rxx[/math].

Muddy Point 5: Exercise 5.5: Where does [math]5\times10^{-7}[/math] come from in part (1) of the question?

Answer 5: See the solution here.

[math]SNR_\text(AM) = \frac{A^2 a^2 \langle m^2 \rangle}{2N_o W} = \frac{P_r}{N_oW} [/math]

[math]5\times10^{-7}[/math] comes from the [math]A^2/2[/math]

Muddy Point 6: Exercise 4.10, part 3: In the solutions [math]p(g) = 0.5 δ(g) + u(g) f_\text{cdf}(x) [/math]. Where does the delta function come from?

Answer 5:

Section 5: Effect of Noise on Analog Systems

Muddy Point 1: Why does pre-emphasis reduce noise? After all, if you are boosting the signal, you are amplifying the noise with it. And why does increasing signal reduce noise anyway?

Answer 1: Pre-emphasis in FM works because the high-frequency content of the message signal (speech or music) usually falls off rapidly at high frequencies. Hence it is possible to boost the high frequencies without significantly increasing the signal amplitude. At these boosted frequencies the signal then becomes larger relative to the background noise in the channel. The channel noise is not boosted because boosting is before the signal hits the channel. So when we restore the signal by attenuating it to its original unboosted state, using de-emphasis, the noise gets attenuated.

Muddy Point 2: For the theoretical SNR I'm using the equation on slide 43 of section 5 (and factoring in for the non-ideal filter by dividing it by the scaling factor before converting it to dB). It just seems a bit strange to me that my simulated SNR goes up and down and that it is so far below the theoretical value (assuming my theoretical value is correct, that is). Looking a bit closer the amplitude of my s_o(t) is around 2 x 10^-3, whilst n_o(t) is about 0.15V and v_o is about 2.5 volts so the signal that is demodulated is healthy as far as I can see. What is strange is that the value of lambda changes a fair bit. This is for increasing values of Pr_over_NoW.

Answer 2:

Muddy Point 3: In the notes on slide 17 of section 5.2, you've given the output noise power as 2*N_0*W, which i agree with...however on slide 21 where you are giving the predetection SNR, you are using a predetection noise power of N_0*B. I this correct? I thought it would have been 2*N_0*W as before (unless detecting the signal somehow doubles the noise power). This means the denominator of the predetection SNR would be 4*N_0*B, not 2*N_0*B as stated (slide 21 again).

Answer 3:

Muddy Point 4: Regarding, pre-Detection SNR for AM...when working out Pr, we ignore any channel attenuation right? So if we are given a specified transmitter power (with modulation) we just use this as the power received for any calculations?

Answer 4:

Muddy Point 5: I noticed that the transmitted power of DSB-SC according to the lecture notes is half that of SSB-SC. This intuitively doesn't make any sense. If I look at the diagram of the DSB-SC case the spectrum will be twice as big as for SSB-SC (since DSB-SC has the lower side-band as well). We're told that signal power is the same in both the time and frequency domains. So shouldn't the power in the DSB-SC signal be twice as big as SSB-SC since the spectrum is twice as big?

It appears as though in the calculation of transmitting power the power in the Hilbert transform has been included, i.e. added. Shouldn't it be discarded, though, since it cancels out half the message signal spectrum?

Answer 5: The power calculation is more easily understood in the time domain. For SSB there is a cosine modulated by the message, added to a sine modulated by the Hilbert transform of the message. As you've pointed out, both components have the same power and when added together are twice that of the DSBSC. The power is the energy per unit time that is physically required to transmit the SSB signal. Since the full SSB signal, including both cos and sin terms, is being transmitted the power must take both into account.

Now consider the frequency domain. For DSB, the PSD at both positive and negative frequencies is [math]0.5A[/math] times the power due to the PSD content of [math]m(t)[/math]. Converting this to power gives two terms each with power [math]0.25A^2{\langle}m(t)^2{\rangle}[/math], and total power [math]0.5A^2{\langle}m(t)^2{\rangle}[/math].

For SSB, the PSD at both positive and negative frequencies is [math]A[/math] times the power due to the PSD content of [math]m(t)[/math]. Converting this to power gives two terms each with power [math]A^2[/math] times [math]0.5{\langle}m(t)^2{\rangle}[/math], and total power [math]A^2{\langle}m(t)^2{\rangle}[/math].

So there are two differences to be aware of, and the [math]0.5{\langle}m(t)^2{\rangle}[/math] factor for SSB is likely the more difficult to understand:

(i) In DSB the [math]0.5A[/math] in the spectrum becomes [math]0.25A^2[/math] in power, which when doubled to take into account both positive and negative frequencies becomes [math]0.5A^2[/math]. In SSB the [math]A[/math] becomes [math]A^2[/math] in power, and when doubled for pos/neg freqs becomes [math]2A^2[/math].

(ii) the remaining difference takes into account the reduced bandwidth in SSB, but note that power is calculated from the PSD of [math]m(t)[/math] due to integration. By symmetry, there will be half the power in SSB compared to DSB. Since the power for DSB is [math]{\langle}m(t)^2{\rangle}[/math], it is [math]0.5{\langle}m(t)^2{\rangle}[/math] for SSB.

Section 6: Information Theory

Muddy point 1: In section 6 it says v_ref is the value of v_o with no added noise. Does that imply that v_ref = message_signal, because isn't the message signal simply the demodulated output signal without noise?

Answer 1:

Muddy point 2: When finding the probability of bit error after correction, [math]P_{cbe}[/math], will [math]n[/math] always equal 1 because Hamming blocks can only correct 1 error at time? I take it the formula for [math]P_{cbe}[/math] does not assume Hamming block coding.

Answer 2:

Muddy point 3: I am a little confused about this section. What does Hamming distance involve? What are the essentials that we need to know regarding this topic.

Answer 3:

Muddy point 4: What is the source coding theorem? I can't see how this sets the maximum efficiency or that efficiency = source entropy/channel capacity...

Answer 4: A consequence of the source coding theorem is that for a source with entropy [math]H[/math], it is possible to find a "good" source code of average length [math]R \gt= H[/math]. In practice, this means that if we know the entropy of a source, then it is worth searching for a code of average length [math]R = H[/math] or [math]R = H + \epsilon[/math], where [math]\epsilon[/math] is small. The closer the average code length, [math]R[/math], is to [math]H[/math], the more efficient is the code. Therefore, the efficiency is [math]H/R[/math]. If [math]H=R[/math], the efficiency is 1. If [math]R \gt H[/math], the efficiency is less than one. Now, if we actually had [math]R \lt H[/math], it would turn out that the code couldn't be uniquely decoded.

Muddy point 5: So information is a measure of surprisal. If I am receiving a voice, say from the radio, that has a high information content because I can't guess what is going to come next. I have a high level of surprisal. But what if my radio is tuned to noise? Noise is random, and hence must have a high information content, but noise is just... well, noise. It's not information, is it? Information theory says it should be full of information, but noise is not useful to me. This to me seems like a paradox. Am I understanding this right?

Answer 5: Listen to the podcast of that lecture again! I do indeed point out that apparent paradox. Furthermore, I explain that the solution is to totally scrub the everyday English meaning of "information" from your mind. The engineering definition is that information measures surprise and that's all. The more random something is the more surprising it is. In a comms scenario we don't usually think of noise as "useful," but its high information content is actually very rich and does tell us about the physical process that generated it...however, in the context of comms that is not information that is of interest to us. Another way to resolve the "paradox" is to imagine compressing the text on this page using a zip program. Before zipping, you are not very surprised when an "h" follows a "t", because this is English. Then when you look at the binaries of the zipped file you are very surprised by every bit you look at; as you cannot predict any bit from the preceding bits in a zip file. So the zip file appears totally random, and yet it contains all the "useful" stuff on this page. Bottom line: all noise is maximally packed with information, and this information only becomes "useful" when you have a context that can interpret it.

Section 7: Digital Modulation Systems

Muddy point 1: What is a "symbol"? Your lecture notes don't actually define it.

Answer 1: Think of a symbol as a means of representing data, and to think of this in the context of digital modulation using pulses. For our purposes, we are focussed on symbols for representing binary data. For example, PAM has two symbols, QAM has four symbols etc. Any given M-ary modulation scheme has M symbols, and each of these [math]M[/math] symbols represents [math]K = \log_2(M)[/math] bits per symbol.

Muddy point 2: What is a sub-carrier?

Answer 2: "Sub-carrier" has several meanings (see TV broadcasting in chapter 3). However in the context of the modem, a sub-carrier can be thought of as just like the carrier signal in AM or FM etc... just that it does not have a frequency much larger than the bandwidth. In the modem example, we have a channel that passes frequencies between 300-3300 Hz. Given that we want to transmit data using pulses that are baseband signals, we would lose some of our frequency content if we tried to transmit them through this channel. Therefore we modulate the pulses onto a carrier within the channel's bandwidth. Since we have the whole 3000 Hz channel, we may as well use its whole bandwidth, so we choose a carrier equal to 1800 Hz and use a pulse with baseband bandwidth 1500 Hz.

Muddy point 3: What is a root Nyquist Pulse?

Answer 3: If you check out the notes, you'll see a Nyquist pulse is simply one that is designed to take up as little bandwidth as we can get away with, and yet has smaller lobes than a sinc pulse. We hate the lobes on sinc pulses because they are stealing signal energy from the central part of the pulse. Signal energy in the lobes is bad, because any misalignment in pulse timing will turn that lobe energy into intersymbol interference (as shown in the lectures). Now a root Nyquist pulse is simply the pulse you must start with at the transmitter so that it ends up as a Nyquist pulse by the time it gets to the receiver. In the lecture, we showed in the simplest case that the relationship between the transmitter and receiver pulses is a square root.

Muddy point 4: For QAM I read that in order to transmit two signals simultaneously from one antenna the signals are modulated onto cosine and sine waves, with real and imaginary coefficients respectively. Does this mean that I can only ever transmit two signals at once (since the complex plane only has two axes and hence only two orthogonal basis functions)? So I couldn't transmit 5 or 11 signals at once?

Answer 4: The use of phasors (and hence the complex plane) is for a single frequency only. The functions [math]\cos\left(2\pi f t\right)[/math] and [math]\sin\left(2\pi f t\right)[/math] are orthogonal not only to each other, but to their counterparts at other frequencies. The result of this is that if the signals are bandlimited and sufficiently separated in frequency then everything works out if one modulates using different carrier frequencies.

Section 9: Channel Capacity and Coding

Muddy point 1: Do Hamming codes always have a minimum distance of three, ie. [math]d_{\rm min}=3[/math]? And if so, why?

Answer 1: Yes, basic Hamming codes always have [math]d_{\rm min}=3[/math]. There are more advanced extended Hamming codes with larger distances, but we only deal with the basic ones in this course. For this course always assume basic Hamming codes with a minimum Hamming distance of three.

To understand why the minimum distance is always three, consider that Hamming codes are constructed by having [math]q[/math] check digits added to [math]k[/math] message digits, giving [math]n = k + q[/math] transmitted digits, but are constructed with the constraint [math]n = 2^q - 1[/math] imposed. Now the syndromes are [math]1 \times q[/math] vectors that allow [math]2^q[/math] possible syndromes, which can then be used to identify [math] 2^q = n+1 [/math] possibilities that are 0 errors or 1 error in one of [math]n[/math] places. This is provided, of course, that the parity generation matrix [math]P[/math] is correctly designed. Hence a Hamming code can correct one error, which means that the minimum distance between codewords must be three. A way to visualise this to see if we move a distance 1 from an error free message, corresponding to one error, then this must not be distance 1 from another error-free message, otherwise, we do not know which message was in error. So this error must be at least distance 2 from every other error free message. For each of the [math]2^k[/math] transmitted codes there are [math]n + 1[/math] possible received codes (ie. 0 errors or 1 error in any one of [math]n[/math] places) so there [math](n+1) \times 2^k = 2^q \times 2^k = 2^n[/math] possible received codes. For example, with [math] q = 3 [/math] we have [math]n = 7[/math] and [math] k = n - q = 4[/math], so there are 16 transmitted codes and [math]16 \times 8 = 128[/math] possible received codes so that this exactly matches [math]2^n[/math].

See Also

• Communications IV

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