Lecture 18 - Passband and Baseband Equivalents
There are a few details about amplitude modulation that we did not cover in previous lectures. Many of these details will help us think in both passband and baseband. We have been thinking of signals in terms of their actual representation in circuits (i.e., thinking about I and Q channels), but today we will focus on their mathematical representation.
1 Baseband to Passband
Let \(x_b(t) \longleftrightarrow X_b(f)\) be a potentially complex-valued baseband waveform with max freq \(W\) Hz We can write \(x_b(t) = \operatorname{Re}\{x_b(t)\} + \operatorname{Im}\{x_b(t)\}\) to represent the signal’s real and imaginary parts.
The passband equivalent signal with center frequency \(f_p\) and phase \(\varphi_p\) is
\[ x_p(t) = \sqrt 2 \operatorname{Re}\left\{x_b(t)e^{j (2 \pi f_p + \varphi_p)}\right\} \]
\(x_p\) is another signal that corresponds to \(x_b\) based on two parameters: some center frequency (\(f_p\)) and phase (\(\varphi_p\)). This is only a mathematical operation, so we are not thinking about a mixer or demodulator or any hardware - just multiplication by a complex exponential. You’ll recall that the \(\operatorname{Re}\{\cdot\}\) operator introduces the complex conjugate as follows:
\[ x_p(t) = \frac{1}{\sqrt 2} \left[x_b(t)e^{j(2 \pi f_p t + \varphi_p)} + x^*_b(t)e^{-j(2 \pi f_p t + \varphi_p)}\right] \]
We can also think in fourier domain, recalling that \(x_b^*(t) \longleftrightarrow X_b^*(-f)\), so the passband fourier transform is
\[ X_p(f) = \frac{1}{\sqrt 2} \left[e^{j\varphi_p} X_b(f-f_p) + e^{-j \varphi_p} X_b^*(-(f+f_p))\right] \]
Because we don’t want the aliases of the signal to overlap with each other, we require that \(f_p > W\).
\[ \int_{-\infty}^{\infty} |x_b(t)|^2 dt = \int_{-\infty}^{\infty} |x_p(t)|^2 dt \]
We see that normalization by \(\sqrt 2\) preserves the energy of the signal. Not all textbooks will do this normalization - some may only use the real part of the signal - which leads to the energy between passband and baseband signals to be different.
It is essential to understand that these signals contain the same information. We can go from \(x_b\) to \(x_p\) and back without losing any information.
2 Passband to Baseband
Since \(x_p(t)\) is real valued, its Fourier transform is hermitian symmetric, i.e,
\[ X_p(f)=X^*_p(-f) \]
\[ |X_p(f)| \text{ even about } f=0 \] \[ \sphericalangle X_p(f) \text{ odd about } f=0 \]
This means that the signal is completely specified by its positive frequencies. If we need \(X_p(-f)\), we can find it by looking at \(X_p(f)\). Therefore we can say
\[ X_{p+}(f) = X_p(f) u(f) \]
where
\[ u(f) = \begin{cases} 1 & f \geq 0 \\ \frac{1}{2} & f = 0 \\ 0 & f<0 \end{cases} \]
If we shift \(X_{p+}\) down to baseband, we recover \(X_b(f)\) with some phase and scaling factor. More specifically,
\[ X_b(f) = \sqrt 2 e^{-j\varphi_p} X_{p+} (f+f_p) \]
In the above equation, we undo the scaling factor of \(\sqrt 2\), phase offset, and frequency offset which were all introduced as we went to passband. \(X_{p+}\) is called the analytic signal corresponding to \(X_p(f)\).
The typical notation for the positive frequency content of a signal \(M\) is \(M_+(f) = M(f)u(f)\). Correspondingly, the negative frequency content is specified by \(M_-(f) = M(f)u(-f)\). Therefore, we have
\[ M(f) = M_+(f) + M_-(f) \]
If \(m(t)\) is real-valued, \(M_+(f) = M_-^*(-f)\) so either \(M_+(f)\) or \(M_-(f)\) completely specifies \(M(f)\). We use the positive frequency content for the analytic signal by convention, but in principle the negative frequency content would work as well.
We can think of \(u(f)\) as the frequency response of some real filter. Consider the LTI frequency response
\[ H(f) = j \operatorname{sgn}(f) = \begin{cases} e^{-j \pi / 2} & f>0 \\ 0 & f=0 \\ e^{j \pi / 2} & f<0 \end{cases} \]
The time domain representation of this filter is
\[ h(t) = \frac{1}{\pi t} \]
Since \(u(t)\) is not physically realizable, we need to create some real filter that has a similar response.
2.1 Hilbert Transform
The Hilbert Transform is the output of the system \(h(t)\) with the signal as the input
We write the Hilbert transform as follows:
\[ X_h(f) = X(f)H(f) = -j \operatorname{sgn}(f) X(f) \]
\[ x_h(t) = \int_{-\infty}^\infty \frac{x(t)}{\pi (t - \tau)} d \tau \]
We use this filter because we can do a much better job approximating it than we can \(u(t)\). The analytic signal can then be written in terms of the Hilbert transform as
\[ X_+(f) = \frac{1}{2} [X(f) + j X_h(f)] \]
When \(f\) is positive, \(X_h(f)\) is \(-j X(f)\). \(-j \times j = 1\), and we get that \(X_+(f) = X(f)\). When \(f\) is negative, \(X_h(f)\) is \(j X(f)\). \(j \times j = -1\), and we get that \(X_+(f) = 0\). Clearly, then, we see that \(X_+(f)\) only contains the positive frequency content of \(X\).
We can similarly write
\[ X_-(f) = \frac{1}{2} [X(f) - j X_h(f)] \]
to represent the negative frequency content of \(X\).
We now have a better way to recover our signal from passband. Recall that we said
\[ X_b(f) = \sqrt 2 e^{-j\varphi_p} X_{p+} (f+f_p) \]
We can write \(X_{p+}\) in terms of the Hilbert Transform to get
\[ X_b(f) = \frac{1}{\sqrt 2} e^{-j\varphi_p} [X(f) (f+f_p)+ j X_h(f+f_p)] \]
The same can be shown in the time domain
\[ \begin{aligned} x_b(t) &= \frac{1}{\sqrt 2} e^{-j\varphi_p} [x_p(t) e^{-j 2 \pi f_p t} + j x_{ph}(t)e^{-j 2 \pi f_p t}] \\ &= \frac{1}{\sqrt 2} (x_p(t) + j x_{ph} (t)) e^{-j (2 \pi f_p t + \varphi_p)} \end{aligned} \]
We have now shown an alternative way of coming from passband down to baseband.
The other way we have done so is with some mixer to shift the whole signal down and an LPF on the baseband signal
We now know that these two systems are equivalent with one simple caveat: in the Hilbert case, the filtering happens at passband. In the other case, the filtering happens at baseband. There are pros and cons to both depending on the specific application.
We will see applications of this concept next time when we study single-sideband modulation.