Lecture 19 - Single Sideband AM
1 Motivation
Recall double sideband amplitude modulation from several lectures ago. We took some baseband signal \(s(t)\) and modulated it onto \(\cos (2 \pi f_c t)\) to produce a passband signal \(x(t)\). If the bandwidth of \(s(t)\) is \(W\), then we know \(x(t)\) will have bandwidth \(2W\).
We fixed this inefficiency in double sideband by modulating another signal onto the \(\sin (2 \pi f_c t)\) so that we transmit twice as much information in the passband spectrum.
We can look at this problem in another way: what if we remove the other half of the passband spectrum that is redundant? Our passband bandwidth would be the same as our baseband bandwidth (\(W\)), so there would be no unused spectrum.
2 Single Sideband Amplitude Modulation
We can draw on the analytic signal concept from last lecture to address the issue of passband redundancy. The spectrum of any signal modulated to passband using a real mixer contains 2 copies of both its positive frequency content and its negative frequency content. Our goal is to construct a signal that contains only one copy of each at passband.
There are two types of single sideband AM: upper sideband and lower sideband. One contains only the positive half of the spectrum while the other contains only the negative half. Mathematically we refer to the upper half of the frequency spectrum using the analytic signal (\(S_+(f)\)) and refer to the negative half using (\(S_-(f)\)).
We will need to use the Hilbert transform to generate a single sideband signal. Recall the Hilbert transform
and the definitions of the positive and negative components of signals
\[ X_+(f) = \frac{1}{2} [X(f) + j X_h(f)] \]
\[ X_-(f) = \frac{1}{2} [X(f) - j X_h(f)] \]
We can write the upper sideband signal in terms of the Hilbert transform as follows, noting that it should be the sum of the postive frequency content at \(+f_c\) and the negative frequency content at \(-f_c\).
\[ \begin{aligned} S_{USB}(f) &= \frac{1}{2}[S(f-f_c) + j S_h(f-f_c)] + \frac{1}{2}[S(f+f_c) - j S_h(f+f_c)] \\ &= \frac{1}{2}[S(f-f_c) + S(f+fc)] - \frac{1}{2j}[S_h(f-f_c) - S_h(f+f_c)] \end{aligned} \]
We can write the corresponding time domain equation:
\[ s_{USB}(t) = s(t) \cos(2 pi f_c t) - s_h(t) \sin(2 pi f_c t) \]
Similarly, we have
\[ s_{LSB}(t) = s(t) \cos(2 pi f_c t) + s_h(t) \sin(2 pi f_c t) \]
2.1 Demodulation
For simplicity of notation, let \(y(t) = s_{USB}(t)\). We know that we want to shift the positive frequencies down and the negative frequencies up, so we can write an expression like the following:
\[ Y_+(f) e^{-j 2 \pi f_c t} + Y_-(f)e^{j 2 \pi f_c t} \]
We can rewrite each of these terms using the Hilbert notation:
\[ \frac{1}{2} [Y(f) + j Y_h(f)] e^{-j 2 \pi f_c t} + \frac{1}{2} [Y(f) - j Y_h(f)] e^{j 2 \pi f_c t} \]
We can translate this into the time domain and get
\[ \frac{1}{2} [y(t) + j y_h(t)][\cos(2 \pi f_c t) - j \sin(2 \pi f_c t)] + \frac{1}{2} [y(t) + j y_h(t)][\cos(2 \pi f_c t) + j \sin(2 \pi f_c t)] \]
Terms cancel out until we end up at the final expression for our baseband signal \(s(t)\):
\[ s(t) = y(t) \cos(2 \pi f_c t) + y_h(t) \sin(2 \pi f_c t) \]
We can represent this demodulator with the following block diagram.
