tag:blogger.com,1999:blog-41482950767299510752017-03-04T15:08:57.549-08:00Electronics, finance, politics, programming, science, researchwellhereiamhttp://www.blogger.com/profile/17067740844274290145noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-4148295076729951075.post-74703596203142857462013-01-21T20:05:00.002-08:002013-01-21T20:05:14.271-08:00Communication Systems: SSB 3<h2>Single Sideband (SSB) Modulation 3</h2>To tie the previous two posts together, LSSB and USSB are given by <a href="http://adf.ly/3115744/ssb-summary-frequency-and-time-domain">the following spectrum and time domain expressions</a>. So what does this expression give? It provides the basis for the construction of the block diagram level SSB transmitter.<br /><br />This means that the message signal m(t) gets sent to two separate paths. The top path multiplies m(t) by (1/2)*A_c*cos(2*pi*f_c*t). The lower branch takes the message and passes it through a Hilbert transformer to generate ^m(t). ^m(t) is then multiplied by (1/2)*A_c*sin(2*pi*f_c*t). The resulting outputs of each of the multipliers (also known as frequency mixers) is fed to a summing junction. If LSSB is desired, then the sine term is added to the cosine term. If USSB is desired, then the sine term is subtracted from the cosine term.<br /><br />Why do all of this when the most intuitive thing to do would have been to put a low-pass filter in front of a DSB-SC transmitter (for LSSB) or a high-pass filter in front of a DSB-SC transmitter (for USSB)? The reason is that the filter requirements are not practical and trying to build something that approaches the ideal low-pass/high-pass filter means a lot of components will be necessary.<br /><br />Of course, as with most things in communication systems, there is a trade-off for using the first technique. Though the filters have to be basically ideal in the latter case, in the former case, a wideband phase shifter is necessary in the phasing method of SSB transmission. The Hilbert transformer has to phase shift every frequency in the bandwidth of the message signal by -90 degrees. The Hilbert transformer, luckily, operates near the baseband and so it's not impractical to build such a Hilbert transformer.<br /><br />According to the IIT Delhi lecture previously linked, in practice both are used and both have their pros and cons. Realizing a -90 phase shifter at a single frequency is easy theoretically. An ideal capacitor has an impedance that is such that the capacitor output is precisely -90 degrees out of phase with the input.<br /><br />Since SSB is essentially DSB-SC with one sideband chopped off, coherent demodulation can be done in the same way that it was done with DSB-SC: a product modulator followed by a low-pass filter recovers the message -- exploiting the fact that cosine squared can equivalently be written in terms of a constant and a cosine of twice the frequency (just filter out the high-frequency term and recover the message). This assumes that there is phase coherence between the transmitter and the receiver of course. This is enough theory for now. I will post more later. wellhereiamhttp://www.blogger.com/profile/17067740844274290145noreply@blogger.com0tag:blogger.com,1999:blog-4148295076729951075.post-80634718215099885702013-01-21T17:54:00.000-08:002013-01-21T19:35:15.362-08:00Communication Systems: SSB 2<h2>Single Sideband (SSB) Modulation 2</h2>To continue from last time, consider the mathematical representation of a lower sideband (LSSB) transmitted wave's spectrum in terms of a DSB-SC wave's spectrum multiplied by an ideal low-pass filter's spectrum (where the spectrum is given as a rectangular function specified in terms of signum functions).<br /><br />Consider <a href="http://adf.ly/3115744/lssb-in-terms-of-dsb-sc-and-lpf">these equations</a> to understand what that collection of words above meant mathematically. Now to understand what each of these four terms in the final line really mean. It is possible to collect these terms into two groups. The first group will explicitly list the spectral components that made up the DSB-SC spectrum. The second group will be the ones that "lop off" the upper sideband.<br /><br />Consider <a href="http://adf.ly/3115744/lssb-in-terms-of-dsb-sc-and-mystery-part">these groups</a> mathematically. Note that the LSSB transmitted signal can be written in terms of a DSB-SC transmitted signal and a mystery term. The reason that some of the signum function terms drop between the second equation and the third equation is that each of the signum functions attached to each of the shifted copies of the message spectrum relevant to representing DSB-SC act as a kind of an all-pass filter (which begins at f=-infinity and ends at f=f_c or which begins at f=-f_c and ends at f=+infinity).<br /><br />So what is the time-domain expression of the first two terms represented in the frequency domain by A_c/4 * [M(f-f_c) + M(f+f_c)]? Well, it has to be a DSB-SC representation so it ought to be A_c/2 * m(t) * cos(2*pi*f_c*t). This can be seen as <a href="http://adf.ly/3115744/fourier-transform-pair-for-dsb-sc-wave">a Fourier transform pair</a>.<br /><br />So what is the time-domain expression of the second two terms represented in the frequency domain by A_c/4 * [M(f+f_c)*sgn(f+f_c) - M(f-f_c)*sgn(f-f_c)]? This is difficult to say directly and a little more mathematical background is necessary. By now, the Fourier transform ought to be intuitive, but another transform that is useful will be reviewed -- the Hilbert transform.<br /><br />Remember that if a signal m(t) is given, the Hilbert transform of m(t) is given by \hat{m}(t) (pronounced em hat of tee and alternately written here as ^m(t)). The Hilbert transform of m(t) is given by <a href="http://adf.ly/3115744/fourier-transform-of-a-hilbert-transform">the following Fourier transform pair</a>. If this isn't clear, remember that the point of the Hilbert transform is to phase shift all components of a signal by -pi/2 (i.e. -90 degrees). This Fourier transform pair looks strange until it is noticed that exp(-j*pi/2) = cos(-pi/2)+j*sin(-pi/2)=0-j*sin(pi/2)=-j*1=-j by Euler's formula. The amplitude of a complex exponential is 1. Therefore, by multiplying by -j, only the phase of each frequency component is changed -- not the amplitude.<br /><br />Also note <a href="http://adf.ly/3115744/frequency-shifting-property-of-ft">a Fourier transform pair</a> which highlights the frequency translation property or frequency shifting property of the Fourier transform. Now combine the two -- what would be the Fourier transform pair given a time domain Hilbert transformed signal that is also frequency shifted (i.e. what is the Fourier transform of ^m(t) * exp(j*2*pi*f_c*t) and what is the Fourier transform of ^m(t) * exp(-j*2*pi*f_c*t)?) The answer is that it will be the same spectrum of ^m(t) shifted to plus or minus f_c as seen <a href="http://adf.ly/3115744/ft-of-frequency-shifted-ht-signal">here</a>.<br /><br />Now revisit the second two terms of the <a href="http://adf.ly/3115744/lssb-in-terms-of-dsb-sc-and-mystery-part">spectrum of a transmitted LSSB wave</a>. Using <a href="http://adf.ly/3115744/ft-of-frequency-shifted-ht-signal">the Fourier transform pair of a frequency shifted Hilbert transformed signal</a>, the second two terms can be interpreted to be <a href="http://adf.ly/3115744/lssb-in-terms-of-dsb-sc-and-ht-of-mt">the following in the time domain</a>. This concludes the derivation of LSSB. All that was involved was really just the properties of the Fourier transform and properties of the Hilbert transform. However, it's not easy to visualize this waveform in the time-domain.<br /><br />But to understand SSB fully, USSB should also be derived. Again, start with DSB-SC and figure out what kind of a filter needs to be placed at the end of a DSB-SC modulator to "lop off" the unwanted sideband. Since the lower sideband of a bandpass signal can be seen as those closest to f=0 and therefore can be low-pass filtered, the upper sideband of a bandpass signal can be seen as those farthest away from f=0 and therefore need to be high-pass filtered. This is easier to express with the unit step function in time. Previously, signum functions were used, though unit step functions could have worked just as well. This alternate strategy will strengthen your signal processing skills and force you to become more familiar with different Fourier transforms. An excellent resource for this is <a href="http://adf.ly/3115744/umd-lecture-on-ssb">UMD's lecture slides on SSB</a>. It deals with this stuff in yet another approach, using complex envelopes and the complex baseband representation of a signal.<br /><br />Consider <a href="http://adf.ly/3115744/ussb-derivation-before-time-domain-rep">a USSB signal's spectrum written in terms of a DSB-SC signal's spectrum and two shifted unit step functions in the frequency domain</a>. The next step is to find the time domain representation. To do so, first consider the <a href="http://adf.ly/3115744/ussb-spectrum-first-term-up-to-ift">first term in the spectrum</a>. The inverse Fourier transform of M(f) is m(t), but the inverse Fourier transform of u(f) is not as straightforward. First, u(f) is not an energy signal, so technically it does not pass the Dirichlet conditions.<br /><br /><br />However, the delta function allows a larger class of functions to be represented by the Fourier transform. To exploit this useful generalized function known as the delta function, rewrite u(t) in terms of sgn(t) and find the Fourier transform of the signum function in terms of a piecewise function consisting of two decaying exponentials. The point is to approximate the signum function with two functions that definitely have Fourier transforms (because the decaying exponentials are finite energy signals). In the limit that the exponentials are allowed to very slowly decay, they approximate the signum function with the error going to 0. Once the Fourier transform of the sgn(t) is found, then the Fourier transform of the unit step function u(t) can be found in terms of the Fourier transform of the signum function. Once that is found, then the duality property can be used to find the Fourier transform of u(f). This is all done in <a href="http://adf.ly/3115744/inverse-fourier-transform-of-uf-part-1">this image (part 1)</a> and <a href="http://adf.ly/3115744/inverse-fourier-transform-of-uf-part-2">this image (part 2)</a>. Note that the odd-symmetric double-exponential pulse is used to approximate the signum function.<br /><br />All this work was just to get the first term of the USSB signal's spectrum translated back into the time domain (where it isn't as intuitive to visualize why it is bandwidth efficient)? The first term then reduces to a convolution integral. However, the convolution integral is difficult to compute. So instead, consider a simpler derivation that replaces the step function with a signum. Here is the <a href="http://adf.ly/3115744/first-term-of-ussb-spectrum-in-time">first term</a>. Here is the <a href="http://adf.ly/3115744/second-term-of-ussb-spectrum-in-time">second term</a>. This means that the USSB signal expressed succinctly from frequency to time domain is then as it appears in <a href="http://adf.ly/3115744/ussb-frequency-domain-and-time-domain">this image</a>.<br /><br />So that's how to derive SSB's pecular time domain waveforms. The key is to think first in frequency domain intuitively about what SSB is. For LSSB, a low-pass filter is used to transmit the lower sideband. For USSB, a high-pass filter is used to transmit the upper sideband. Then do some algebra in the frequency domain. Then in the process of trying to inverse Fourier transform the resulting terms, make sure to find a way to crowbar the signum function. This way, the Hilbert transform can be introduced, and a much quicker and simpler route to the time domain expression can be had.wellhereiamhttp://www.blogger.com/profile/17067740844274290145noreply@blogger.com2tag:blogger.com,1999:blog-4148295076729951075.post-17171950216439108372013-01-19T22:04:00.003-08:002013-01-20T17:13:02.706-08:00Communication Systems: SSB 1<h2><span style="font-weight: normal;">Single Sideband (SSB) Modulation 1</span></h2><br />I found that Haykin and Moher's Introduction to Analog & Digital Communications gave a quick overview of the development of SSB for a general signal. I decided to expand upon it and give my own explanation for how it all works.<br /><br />First, consider the motivation. Wikipedia shows <a href="http://adf.ly/3115744/dsb-sc-wikipedia-article">Double Sideband Suppressed Carrier (DSB-SC)</a> is simply the result of multiplying a carrier signal by a message signal. In this way, the amplitude of the carrier is varied in proportion to the message.<br /><br />From the perspective of a transmitter, it is relatively simple to construct. All that is needed is something that can multiply two signals together (e.g. a mixer), a local oscillator, and an information source. However, the transmitter wastes a lot of power and bandwidth transmitting both sidebands when only one is necessary to properly transmit information. The reason that both sidebands go out is that real signals have equal amounts of positive and negative frequencies. When a cosine carrier, which can be expressed as a summation of two complex exponentials, multiplies a message signal, the result is that the positive and negative frequencies are copied and shifted to the right as well as copied and shifted to the left.<br /><br />This is where SSB comes in. SSB only transmits one sideband as the name suggests. Wikipedia makes a good point in the <a href="http://adf.ly/3115744/ssb-wikipedia-article">Single Sideband (SSB) article</a> that SSB is really a special case of Quadrature Amplitude Modulation (QAM) where only one signal is sent out instead of the usual two. But other than that, the Wikipedia article requires a lot of clicking on links to figure out what all the words mean.<br /><br />Intuitively, all SSB has to do is take a DSB-SC wave and chop off one sideband that's translated up near a carrier frequency f_c. Theoretically, this would require an infinitely sharp transition band, meaning the filter can suddenly stop passing frequencies. This is difficult and limits the use of SSB in practice to situations where there is an energy gap in the message signal.<br /><br />I wanted one resource where all this information could be found. Lectures, in general, are easier to follow than concise textbooks, but they take time and it's difficult to pinpoint information unless someone has already done it for you or you got around to identifying key points yourself. However, I find that the Indian Institute of Technology video lectures on YouTube to be of very good quality. Some of the insights here are due to the <a href="http://adf.ly/3115744/ssb-iit-delhi-lecture-10-youtube">lecture on SSB from IIT Delhi</a>. Note that some of their videos, particularly for the communication systems lectures have horrible quality (very distorted, difficult to hear even with the volume up, but it is audible).<br /><br />With the motivation explained and some of the resources used pointed out, there's not much else left to do but to get technical and do some math. SSB is very easy to understand in the frequency domain, but it's not as easy to understand in the time domain.<br /><br />Consider a <a href="http://adf.ly/3115744/dsb-sc-spectrum-for-arbitrary-message">DSB-SC signal spectrum</a> first. The goal is to lop off one sideband (in this case, first, the upper sideband) represented by two triangles: the right angle triangle with side length W (from f_c up to f_c+W) and the right angle triangle with side length W (from -f_c down to -f_c-W). This is LSSB (lower sideband).<br /><br /><a href="http://adf.ly/3115744/dsb-sc-spectrum-with-low-pass-filter">See how it looks when an ideal low-pass filter does just that</a>. How can this be written mathematically then? The low-pass filter is lopping off the upper sideband leaving only the lower sideband behind. So LSSB can be represented, in the frequency domain, as the product of the DSB-SC wave's spectrum and the ideal low-pass filter's spectrum. <a href="http://adf.ly/3115744/rectangular-function-in-terms-of-signum">A rectangular function can also be written in terms of two sgn (signum) functions</a>.<br /><br />Wikipedia defines the <a href="http://adf.ly/3115744/signum-wikipedia-article">signum function</a> as being -1 below x=0, as being +1 above x=0, and as being 0 at x=0. This alternate representation of the rectangular function works because one signum is 1 for a portion of the frequency axis while the other signum is -1. The net result is a sort of "destructive interference" and they cancel out. This explains the stopband region (i.e. outside of -f_c to f_c). How about f=-f_c and f=f_c? At those points, one signum is 1 and the other is 0, so the result is 0.5. Within -f_c to f_c, both signums are 1 and so the result of the expression is 1. More next time.wellhereiamhttp://www.blogger.com/profile/17067740844274290145noreply@blogger.com6