converting complex sum of exponential signal to real signal

Question

I have the following Signal model, that generates a discrete-time complex signal, where $a_k$ - amplitudes, $\phi_k$ - phases, $\alpha_k$ - damping factors and $f_k$ - frequencies are the parameters of the model and $n$ is the time index and $\Delta t$ is the time interval.

\begin{equation} x_n = \sum_{k=1}^{K} (a_k e^{j\phi_k})(e^{\{(- \alpha_k + j2\pi f_k )\Delta t\}n}), \quad n = 0,1,...,N-1 \end{equation}

For what values of the model parameters, does it generate a real signal? I worked out to write the phase $\phi$ as a function of $n$ and $f$ as: $$ \phi_k = n\pi (1-2f_k \Delta t)$$ or simply, $$ \phi_k = -2n\pi f_k \Delta t$$

But can phase be a function of time index? When I set this value, then the frequency has no role to play. Is there any other values of the parameter, that makes this signal sequence real?

Answer 1

For what values of the model parameters, does it generate a real signal?

That's probably not possible. Roughly speaking, your frequencies are all positive and a real signal must have a conjugate symmetric spectrum, i.e. equal amount of positive and negative frequencies.

If you make the sum run from $-K$ to $+K$ and set $a_k = a_{-k}, \alpha_k = \alpha_{-k}, \phi_k = -\phi_{-k}$, the result will indeed be real. But that's more or less equivalent by replacing the complex exponential with a cosine (phase shifted by $\phi_k$).

With the damping you may be able to enforce real values for some samples simply by solving for $\Im\{x[n]\} = 0 $, but you only have $3K$ model parameters, so you can't do it for all samples of the infinitely long signal