More on Pure and Impure Sine Waves

Remember when we derived the cumulant as the solution to the pure nth-order sine-wave problem? It sounded good at the time, I hope. But here I describe a curious special case where the interpretation of the cumulant as the pure component of a nonlinearly generated sine wave seems to break down.

Let’s consider a very simple cyclostationary signal, the baseband binary pulse-amplitude modulated (PAM) signal with rectangular pulses. This signal is a building block for a slightly more complicated and slightly more realistic signal, the rectangular-pulse BPSK signal. The latter builds on the binary PAM signal by adding a non-zero carrier offset frequency, a symbol-clock phase, and carrier phase. But overall the binary PAM signal is pretty much the same thing as a BPSK signal.

Here is a graphical representation of our binary PAM signal:

rect_pam

We know this signal has non-conjugate cycle frequencies equal to harmonics of 1/T_0 and since it is real-valued, its conjugate cycle frequencies are identical to its non-conjugate cycle frequencies.

Let’s now take a look at the fourth-order parameters for the signal. First, consider the fourth-order temporal moment function with equal delays \tau_j = D,

\displaystyle R_x(t, \boldsymbol{\tau}; 4,m) = E \left[ \prod_{j=1}^4 s^{(*)_j} (t+D) \right]. \hfill (1)

The value of m doesn’t matter here because s(t) is real-valued, so we have

\displaystyle R_x(t, \boldsymbol{\tau}; 4,m) = E \left[ \prod_{j=1}^4 s(t+D) \right] = E \left[ s^4(t + D) \right] = E \left[ 1 \right] = 1. \hfill (2)

This moment function is trivially periodic, and so has a Fourier series expansion with a single coefficient. In other words, it is equivalent to a sine wave with zero frequency, unit amplitude, and zero phase. There is no symbol-rate fourth-order cyclic moment (no impure sine wave with non-zero frequency), nor any other non-trivial cyclic moment for the case of all delays being equal. Moreover, if we were to look at the second-order temporal moments with equal delays, they are also equal to one. So no non-trivial lower-order sine waves exist either.

What about when the delays are not all equal?

Let’s now consider the delay set [\tau_1\ \ \tau_2\ \ \tau_3\ \ \tau_4] = [0\ \ 0\ \ -T_0/2\ \ -T_0/2].

\displaystyle R_x(t, \boldsymbol{\tau}; 4,m) = E \left[ s(t)s(t) s(t-T_0/2)s(t-T_0/2) \right]. \hfill (3)

Now we know that s(t)s(t) = 1 and s(t+D)s(t+D) = 1 so that, once again,

\displaystyle R_x(t, \boldsymbol{\tau}; 4,m) = E \left[ 1 \right] = 1. \hfill (4)

Here, again, there are no impure sine waves with non-zero frequencies such as the symbol rate 1/T_0. What are the pure sine waves? Are they also absent?

Because the signal is real-valued, and we are considering only the fourth-order cumulant, our general moments-to-cumulants formula reduces to

\displaystyle C_s(t, [0\ 0\ -T_0/2\ -T_0/2]; 4,m) = R_s(t, [0\ 0\ -T_0/2\ -T_0/2]; 4,m)

\displaystyle -R_s(t, [0\ 0]; 2, \cdot)R_s(t, [-T_0/2\ -T_0/2]; 2, \cdot) - 2[R_s(t, [0\ -T_0/2]; 2, \cdot)]^2. \hfill (5)

From our previous analysis (for example (4)), we can replace several of these terms with unity:

\displaystyle R_s(t, [0\ 0\ -T_0/2\ -T_0/2]; 4,m) = 1, \hfill (6)

\displaystyle R_s(t, [0\ 0]; 2, \cdot) = 1,\hfill (7)

\displaystyle R_s(t, [-T_0/2\ -T_0/2]; 2, \cdot) = 1, \hfill (8)

so that our temporal cumulant becomes

\displaystyle C_s(t, [0\ 0\ -T_0/2\ -T_0/2]; 4,m) = 1 - (1)(1) - 2[R_s(t, [0\ -T_0/2]; 2, \cdot)]^2, \hfill (9)

or

\displaystyle C_s(t, [0\ 0\ -T_0/2\ -T_0/2]; 4,m) = - 2[R_s(t, [0\ -T_0/2]; 2, \cdot)]^2. \hfill (10)

The question then becomes: What is R_s(t, [0\ -T_0/2]; 2, \cdot)? The periodic component of the lag product s(t) s(t - T_0/2) is a square wave with period T_0 and duty cycle of 1/2. We can see that graphically:

rect_pam_second_order

So the second-order moment R_s(t, [0\ -T_0/2]; 2, \cdot) is equal to that square wave (green lines), which can be expressed as a Fourier series. Let’s do that:

\displaystyle c_j = \frac{1}{T_0} \int_{0}^{T_0} r_{T_0/2}(t-T_0/4) e^{-i 2 \pi j t/T_0}\, dt, \hfill (11)

\displaystyle c_j = \frac{1}{T_0} \int_{0}^{T_0/2} (1) e^{-i 2 \pi j t / T_0} \, dt. \hfill (12)

For j = 0, c_j = c_0 = 1/2. For all other j,

\displaystyle c_j = \frac{1}{T_0} \left. \frac{e^{-i 2 \pi j t/T_0}}{-i 2 \pi j/T_0} \right|_{0}^{T_0/2}. \hfill (13)

We end up with

\displaystyle c_j = \frac{1}{2} e^{-i \pi j/2} sinc (j/2), \ \ \ j \neq 0. \hfill (14)

In particular, |c_1| = 0.5 sinc(1/2) = 0.318. This means that the cyclic cumulant C_s^{1/T_0} ([0\ 0 \ -T_0/2\ -T_0/2]; 4, m) is not zero. That is, there is a pure fourth-order sine wave with frequency equal to the symbol rate 1/T_0 for the specified delay vector. But we already saw that there is no impure fourth-order sine wave with frequency equal to the symbol rate for that same delay vector.

Which is strange.

Let’s consider the moments and cumulants of a Gaussian signal for a moment. Recall that the nth-order cumulants of a Gaussian random variable, set of jointly Gaussian random variables, or a Gaussian process (signal) are zero for n \ge 3. So if we have a Gaussian signal x(t), we can write the following equation

\displaystyle C_x(t, \boldsymbol{\tau}; n,m) = \sum_{P = \{\nu_k\}_{k=1}^p} (-1)^{p-1}(p-1)! \prod_{j=1}^p R_{x_{\nu_j}}(t, \boldsymbol{\tau}_{\nu_j}; n_j, m_j), \hfill (15)

where the outer sum is the sum over all the distinct partitions of the index set \{1, 2, \ldots, n\}, each partition consists of p disjoint subsets of \{1, 2, \ldots, n\} denoted by \nu_k, the union of the \nu_k is the index set \{1, 2, \ldots, n\}, |\nu_k| = n_k, and m_k optional conjugations are included in \nu_k. (All that is review from this post.) Now consider n \ge 3 and for our Gaussian x(t) we have

\displaystyle 0 = \sum_{P = \{\nu_k\}_{k=1}^p} (-1)^{p-1}(p-1)! \prod_{j=1}^p R_{x_{\nu_j}}(t, \boldsymbol{\tau}_{\nu_j}; n_j, m_j), \hfill (16)

or

\displaystyle R_{x}(t, \boldsymbol{\tau}; n,m) = - \sum_{P = \{\nu_k\}_{k=1}^p, p\neq 1} (-1)^{p-1}(p-1)! \prod_{j=1}^p R_{x_{\nu_j}}(t, \boldsymbol{\tau}_{\nu_j}; n_j, m_j). \hfill (17)

If we look at a single Fourier component of this time-varying moment (that is, look at a specific cyclic moment), we obtain

\displaystyle R_x^\beta (\boldsymbol{\tau}; n,m) = -\sum_{P=\{\nu_k\}_{k=1}^p, p\neq 1} \left[ \sum_{\boldsymbol{\alpha} \boldsymbol{1}^\dagger = \beta}  (-1)^{p-1}(p-1)! \prod_{j=1}^p R_{x_{\nu_j}}^{\alpha_j} (\boldsymbol{\tau}_{\nu_j}; n_j, m_j) \right]. \hfill (18)

In other words, if you know all the lower-order cyclic moments with orders less than n, you know the nth-order cyclic moment.

Now consider some even-order temporal moment function for our binary PAM signal s(t), but also constrain the delay vector to be of the form

\displaystyle \boldsymbol{\tau} = [D_1\ D_1\ D_2\ D_2\ \ldots D_q\ D_q] \triangleq \boldsymbol{\lambda}. \hfill (19)

The moment function can be expressed in terms of the cumulant functions

\displaystyle R_s(t, \boldsymbol{\lambda}; n, m) = 1 = \sum_{P=\{\nu_k\}_{k=1}^p} \prod_{j=1}^p C_{s_{\nu_j}}(t, \boldsymbol{\lambda}_{\nu_j}; n_j, m_j). \hfill (19)

Rearranging this equation yields an expression for the nth-order cumulant

\displaystyle C_s(t, \boldsymbol{\lambda}; n,m) = 1 - \sum_{P=\{\nu_k\}_{k=1}^p, p\neq 1} \prod_{j=1}^p C_{s_{\nu_j}}(t, \boldsymbol{\lambda}_{\nu_j}; n_j, m_j). \hfill (20)

If we look at a Fourier component of this time-varying cumulant with \beta \neq 0, we get the cyclic cumulant,

\displaystyle C_s^\beta (\boldsymbol{\lambda}; n,m) = - \sum_{P=\{\nu_k\}_{k=1}^p, p\neq 1} \left[ \sum_{\boldsymbol{\alpha} \boldsymbol{1}^\dagger = \beta} \prod_{j=1}^p C_{s_{\nu_j}}^{\alpha_j}(\boldsymbol{\lambda}_{\nu_j}; n_j, m_j) \right]. \hfill (21)

In other words, if you know all the lower-order cyclic cumulants with orders less than n, you know the nth-order cyclic cumulant.

In some sense, then, the binary rectangular-pulse PAM signal is the dual of the Gaussian signal. They play special roles in the interplay between cyclic moments and cyclic cumulants. Gaussian signals have minimum cyclic cumulants and cyclic moments with a sort of redundancy, whereas the binary PAM signals have minimum cyclic moments (under the \boldsymbol{\lambda} constraint) and cyclic cumulants with a sort of redundancy.

I get the feeling I’m missing something fundamental here. If you see anything, or have corrections or issues with the post, I encourage you to leave a comment below.

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