# Complex damped exponential signal with repetitive poles and the significance of falling factorial

I was modelling a complex damped exponential signal (discrete) with unique poles as below: $$\begin{equation} x = \sum_{k=1}^{K} (a_k e^{(j\phi_k)})(e^{\{(j2\pi f_k - \alpha_k)\Delta t\}t}), \quad t = 0,1,...,N-1 \end{equation}$$ where $$K \in \mathbb{N}$$, $$a_k$$ amplitudes, $$\phi_k$$ phases, $$\alpha_k$$ damping factors, $$f_k$$ - frequencies in Hz and $$\Delta t$$ - time interval. The above equation is simplified as: $$\begin{equation} x = \sum_{k=1}^{K} (c_k)(z_k^t), \quad t = 0,1,...,N-1 \end{equation}$$

This signal satisfies below homogeneous linear recursion form: $$\begin{equation} x(t) + p_1x(t-1) +...+ p_Kx(t-K) = 0 \end{equation}$$ where $$p_1,...,p_K$$ are coefficients of polynomial $$\begin{equation} P(z) = \prod_{k=1}^K (z-z_k) = \sum_{k=0}^K p_{K-k}z^k \end{equation}$$ where $$p_0=1$$ and $$p_K \neq 0$$

Now, I want to convert the model for repetitive poles case (i.e. my prediction polynomial $$P(z)$$ can have multiple roots). $$\begin{equation} P(z) = \prod_{k=1}^K (z-z_k)^{M_k} = \sum_{k=0}^K p_{K-k}z^k \end{equation}$$ where $$M_k \in \mathbb{N}$$ and can be greater than 1.

I came across this paper High resolution spectral analysis of mixtures of complex exponentials modulated by polynomials, that talks about falling factorial ($$F_m$$) in the modified formulation of signal for multiple poles case where the original signal $$x(t)$$ is modified as below: $$\begin{equation} x = \sum_{k=1}^{K} \sum_{m=0}^{M_k-1} (c_k)F_m(t)(z_k^{t-m}), \quad t = 0,1,...,N-1 \end{equation}$$ where the falling factorial $$F_m$$ is defined as: $$\begin{equation} F_m(t) = \begin{cases} 0, \quad & m < 0,\\ 1, \quad & m = 0,\\ \frac{1}{m!} \prod_{j=0}^{m-1} (t-j), \quad & m > 0 \end{cases} \end{equation}$$

Could someone please explain the significance of $$F_m$$ and how does a signal with multiple poles vary from the same signal with unique poles? only in terms of Amplitude?

• Could you please expand your question or specifically the final paragraph to show two examples of what you are referring to (links may expire over time and also will invite help from others with less time to read the paper). If you could show an example of what you mean as a signal with multiple poles and then what the equivalent is for a signal with unique poles that would likely clarify things better. Jul 1 at 13:29
• Thank you for your feedback. I hope my correction is clear now. Please feel free to comment for more clarity, if required. Jul 1 at 15:32
• if $t$ is time and not the sample index, then there should be no $f_\text{sample}$ in your expressions. Jul 1 at 16:25
• Thank you, i corrected it. Jul 1 at 16:47
• nope. not intending to do that. Jul 1 at 19:17

A complex-valued analytic signal:

\begin{align} x_\mathrm{a}(t) &= \sum\limits_{k=1}^{K} g_k \ e^{-\alpha_k t} \ e^{j(2\pi f_k t + \phi_k)} \ u(t) \\ &= \sum\limits_{k=1}^{K} \underbrace{g_k e^{j\phi_k}}_{c_k} \ e^{(-\alpha_k+j2\pi f_k)t} \ u(t) \\ &= \sum\limits_{k=1}^{K} c_k \ e^{(-\alpha_k+j2\pi f_k)t} \ u(t) \\ &= x(t) + j\hat{x}(t) \\ \end{align}

where $$\quad K\in\mathbb{Z}>0, \quad g_k,\alpha_k,f_k\in\mathbb{R}>0, \quad \phi_k\in\mathbb{R} \quad$$ and

$$\hat{x}(t) = \mathscr{H}\Big\{ x(t) \Big\}$$

is the Hilbert Transform.

$$u(t)$$ is the unit step function:

$$u(t) \triangleq \begin{cases} 0 \quad & t < 0, \\ 1 \quad & t \ge 0, \\ \end{cases}$$

It is not too hard to see that

$$x(t) = \sum\limits_{k=1}^{K} g_k \ e^{-\alpha_k t} \ \cos(2\pi f_k t + \phi_k) \ u(t)$$

and

$$\hat{x}(t) = \sum\limits_{k=1}^{K} g_k \ e^{-\alpha_k t} \ \sin(2\pi f_k t + \phi_k) \ u(t)$$

If this complex analytic signal is uniformly sampled at sample rate $$f_\mathrm{s}=\frac{1}{T} \quad T>0$$, then

\begin{align} x_\mathrm{a}[n] &\triangleq x_\mathrm{a}(t) \Bigg|_{t=nT} \\ \\ &\triangleq x_\mathrm{a}(nT) \\ &= \sum\limits_{k=1}^{K} g_k \ e^{-\alpha_k nT} \ e^{j(2\pi f_k nT + \phi_k)} \ u(nT) \\ &= \sum\limits_{k=1}^{K} \underbrace{g_k e^{j\phi_k}}_{c_k} \ e^{(-\alpha_k+j2\pi f_k)nT} \ u(nT) \\ &= \sum\limits_{k=1}^{K} c_k \ (\underbrace{e^{(-\alpha_k+j2\pi f_k)T}}_{p_k})^n \ u(nT) \\ &= \sum\limits_{k=1}^{K} c_k \ (p_k)^n \ u[n] \\ \end{align}

where $$n\in\mathbb{Z}$$ is the sample index and

\begin{align} c_k &\triangleq g_k e^{j\phi_k} \\ \\ p_k &\triangleq e^{-\alpha_k T}\ e^{j2\pi f_k T} \\ \end{align}

So applying the Z Transform

\begin{align} X_\mathrm{a}(z) &= \mathcal{Z}\Big\{ x_\mathrm{a}[n] \Big\} \\ &= \mathcal{Z}\left\{ \sum\limits_{k=1}^{K} c_k \ p_k^n \ u[n] \right\} \\ &= \sum\limits_{k=1}^{K} c_k \ \mathcal{Z}\Big\{ p_k^n \ u[n] \Big\} \\ &= \sum\limits_{k=1}^{K} \frac{c_k}{1-p_k z^{-1}} \\ &= \sum\limits_{k=1}^{K} \frac{c_k \, z}{z-p_k} \\ \end{align}

So this signal looks like the impulse response of a complex-coefficient LTI system having $$p_k$$ as poles of the transfer function.

Now to confirm that this signal satisfies the below homogeneous linear recursion form:

$$x_\mathrm{a}[n] + a_1 x_\mathrm{a}[n-1] +...+ a_K x_\mathrm{a}[n-K] = 0$$

where $$a_1,...,a_K$$ are coefficients of polynomial

$$P(z) = \prod\limits_{k=1}^K (z-p_k) = \sum\limits_{m=0}^K a_{K-m}\ z^m$$

where $$\ a_0=1 \$$ and $$\ a_K \neq 0 \$$.

The Z Transform is:

$$X_\mathrm{a}(z) + a_1 X_\mathrm{a}(z) z^{-1} +...+ a_K X_\mathrm{a}(z) z^{-K} = 0$$

which, after factoring out $$X_\mathrm{a}(z)$$ is

$$\sum\limits_{m=0}^K a_m z^{-m} = 0$$

But this doesn't say anything about $$X_\mathrm{a}(z)$$. So let's try this:

\begin{align} 0 &\stackrel{?}{=} \sum\limits_{m=0}^K a_m x_\mathrm{a}[n-m] \\ &= \sum\limits_{m=0}^K a_m \sum\limits_{k=1}^{K} c_k \ (p_k)^{n-m} \ u[n-m] \\ \end{align}

Now, to make our lives simpler, let's say for the moment that we're looking at times where the sample index $$n \ge K$$ so that the unit step function $$u[n-m]=1$$ goes away. So far, the right-hand limit $$N-1$$ suggested in the OP hasn't appeared to us yet.

\begin{align} 0 &\stackrel{?}{=} \sum\limits_{m=0}^K a_m x_\mathrm{a}[n-m] \\ &= \sum\limits_{m=0}^K a_m \sum\limits_{k=1}^{K} c_k \ (p_k)^{n-m} \\ &= \sum\limits_{m=0}^K \sum\limits_{k=1}^{K} a_m c_k \ (p_k)^{n-m} \\ &= \sum\limits_{k=1}^K \sum\limits_{m=0}^{K} a_m c_k \ (p_k)^{n-m} \\ &= \sum\limits_{k=1}^K \sum\limits_{m=0}^{K} a_m c_k \ p_k^{n} \ p_k^{-m} \\ &= \sum\limits_{k=1}^K c_k \ p_k^{n} \sum\limits_{m=0}^{K} a_m \ p_k^{-m} \\ \end{align}

So, if it can be made true that

$$\sum\limits_{m=0}^{K} a_m \ p_k^{-m} = 0$$

for every $$\ p_k, \quad 1 \le k \le K \quad$$ then the "homogeneous linear recursion form" above will be satisfied. That means that $$p_k$$ are roots to the polynomial

$$P(z) = \prod\limits_{k=1}^K (z-p_k) = \sum\limits_{m=0}^K a_{K-m}z^m$$

Okay, so this means that for the All-pole model (which is a small misnomer because there can be zeros in the $$z$$-plane, but all zeros are located at the origin $$z=0$$), then

$$\sum\limits_{m=0}^K a_m x_\mathrm{a}[n-m] = 0$$

which means that this recursion formula works:

\begin{align} x_\mathrm{a}[n] &= -\sum\limits_{m=1}^K a_m x_\mathrm{a}[n-m] \\ &= - \big( a_1 x_\mathrm{a}[n-1] + a_2 x_\mathrm{a}[n-2] + ... + a_K x_\mathrm{a}[n-K] \big) \\ \end{align}

So the current complex sample of $$x_\mathrm{a}[n]$$ can be defined recursively, solely in terms of the $$K$$ most recent samples in the past, $$x_\mathrm{a}[n-m]$$ for $$1 \le m \le K$$, with fixed feedback coefficients: $$-a_m$$.

Now to get this complex function to appear exactly as the originally defined

$$x_\mathrm{a}[n] = \sum\limits_{k=1}^{K} c_k \ p_k^n \ u[n]$$

that will require setting the initial states of $$x_\mathrm{a}[n]$$ for $$-K \le n \le -1$$ to specific values. I'm pretty sure that those initial states must fit the given exponential form for $$-K \le n \le -1$$ even though the unit step function explicitly would set those samples to zero. So the initial states must be:

$$x_\mathrm{a}[n] = \sum\limits_{k=1}^{K} c_k \ p_k^n \qquad \text{for} \ -K \le n \le -1$$

So far, whether there are multiple poles or not, does not seem to be an issue. And the right-hand limit of $$n \le N-1$$ also does not seem to be an issue as long as it's larger than $$K$$.

Now consider a single set of coincident multiple poles. The Binomial Theorem provides:

\begin{align} (z-p)^K &= \sum\limits_{k=0}^{K} \frac{K!}{(K-k)! \, k!} z^k (-p)^{K-k} \\ \\ \prod\limits_{m=1}^K (z-p) &= \sum\limits_{k=0}^K a_{K-k} z^k \\ \end{align}

That says that the polynomial coefficients are:

$$a_{K-k} = \frac{K!}{(K-k)! \, k!} (-p)^{K-k}$$

or

$$a_k = \frac{K!}{(K-k)! \, k!} (-p)^k$$

and the recursion is

\begin{align} x_\mathrm{a}[n] &= -\sum\limits_{k=1}^K a_k \ x_\mathrm{a}[n-k] \\ &= -\sum\limits_{k=1}^K \frac{K!}{(K-k)! \, k!} (-p)^k \ x_\mathrm{a}[n-k] \\ \end{align}

Now I am not sure what that tells us, but your feedback coefficients $$a_k$$ have the form as shown above.

Okay, i don't wanna do different multiple poles, just a set of identical coincident poles. With multiple sets of poles you can split them into parallel sections but that's partial fraction expansion and that's hard.

With a set of identical multiple poles, the recursion equation comes out as above with the factorials in the denominator.

• i will return to this later this evening. Jul 2 at 21:28
• to whomever upvoted: i ain't done yet. just trying to set up the problem using math conventions i think is most common in signal processing done from the electrical engineering perspective. Jul 3 at 14:32