Monday, January 21, 2013

Communication Systems: SSB 2

Single Sideband (SSB) Modulation 2

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).

Consider these equations 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.

Consider these groups 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).

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 Fourier transform pair.

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.

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 the following Fourier transform pair. 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.

Also note a Fourier transform pair 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 here.

Now revisit the second two terms of the spectrum of a transmitted LSSB wave. Using the Fourier transform pair of a frequency shifted Hilbert transformed signal, the second two terms can be interpreted to be the following in the time domain. 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.

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 UMD's lecture slides on SSB. It deals with this stuff in yet another approach, using complex envelopes and the complex baseband representation of a signal.

Consider 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. The next step is to find the time domain representation. To do so, first consider the first term in the spectrum. 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.

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 this image (part 1) and this image (part 2). Note that the odd-symmetric double-exponential pulse is used to approximate the signum function.

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 first term. Here is the second term. This means that the USSB signal expressed succinctly from frequency to time domain is then as it appears in this image.

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.


  1. There is a sign error at

  2. You are right, I will update this and make a note of the mistake in the comments in case others were misled. Thanks!

    Fourier transform written as
    A_c/2 * m(t) * cos(2*pi*f_c*t) <=> A_c/4 * [M(f-f_c)+M(f-f_c)]

    Of course, it should be
    A_c/2 * m(t) * cos(2*pi*f_c*t) <=> A_c/4 * [M(f-f_c)+M(f+f_c)]
    because a cosine can be expressed as a summation of two complex sinusoids. If this is unclear, expand the Euler representation of exp(j*x) and exp(-j*x) and try to express cos(x) in terms of exp(j*x) and exp(-j*x).