An FSK-signal with a frequency-shift $\Delta F$, a symbol-stream $x$, $x(n) \in {\{-1,1}\} $ and symbol duration $T$ has the complex envelope:

$$f[x](t)=\exp\left(2 \pi i \Delta F T \left(\sum_{l=0}^{j-1} x(l)\right)+\left(\frac{t}{T}-j\right)x(j)\right)\quad\text{for}\quad jT\leq t\leq (j+1)T$$

I am examining periodical symbol-streams for FSK-signals and I found a peculiar dependency between the strength of the periodicity and the modulation index,which I am very unsure of:

The phase of the samples at the start of a symbol is

$$2\pi i \Delta F T \left(\sum_{l=0}^{j-1} x(l)\right)=:\varphi(j,T).$$

So the phase deviation between the samples at the beginning of two consecutive symbols is given by

$$|(\varphi(j,T) -\varphi((j-1),T)|=\pi \eta, \tag{1}$$

where $\eta=2 \Delta F T$ is the modulation index of the FSK signal.

Due to $(1)$ for integer modulation indices $\eta$ the phase at the beginning of a symbol stays constant. Accordingly, a complex autocorrelation of a periodic signal shows a peak at right at the length of a period of the signal.

But, if the modulation index is not integer, the starting phase of the symbols shifts throughout the signal, i.e. $2\pi \nmid \pi \eta $.
A complex autocorrelation of same period signals, shifted in phase, does not show the same peaks and therefore does not exhibit periodicity in the same magnitude.

  1. Is my suspicion really sound for continuous phase frequency shift keying signals?
  2. Does this hold for MSK-signals too ?

1 Answer 1


After thinking for a while i came up with my own proof. I hope it is helpful to you and correct!

Definition: Frame-Structure

The frame structure used in the following is one of one-periodic signals, with a constant header and a uniform distributed payload. The ratio between payload and total frame is given as $$ p_{\%} = \frac{|payload|}{|frame|} $$ the ratio of samples in the header vs samples per frame.

Example: $$ |HHH|012|HHH|223|HHH|978|...$$ ,here: $p_{\%}=0.5$ .


The complex envelope of an FSK-signal is

\begin{array}{cc} x\left(t\right)=e^{i\left[2\pi\Delta FT\left(\varphi\left(j,T\right)+\left(\frac{t}{T}-j\right)d\left(j\right)\right)\right]} & ,\text{}jT\leq t\leq\left(j+1\right)T\end{array}


$$\varphi\left(j,T\right):=2\pi\Delta FT\sum_{l=0}^{j-1}d\left(l\right)+\varphi_{0}$$

and the modulation index $$\eta:=2\Delta FT.$$

We introduce the model by having a look at the ACF of truely periodic, i.e. $p_{\%}\left(x\right)=1 $, discrete signals. In a second step we generalize to partly periodic signals with $p_{\%}\left(x\right)<1$. Let $x$ be a digital FSK signal of length N+1, that is, $x$ is a sampled version of the complex envelope above. Then the ACF of x is defined by

\begin{array}{ccc} \rho\left[x\right]\left(\tau\right) & = & \sum_{k=0}^{N-\tau}x\left(k\right)x^{*}\left(k+\tau\right)\\ & = & \sum_{k=0}^{N-\tau}e^{i\left[\pi\eta\left(\left(\frac{k}{T}-j_{k}\right)d\left(j_{k}\right)+\varphi\left(j_{k},T\right)\right)\right]}e^{-i\left[\pi\eta\left(\left(\frac{k+\tau}{T}-\left(j_{k+\tau}\right)\right)d\left(j_{k+\tau}\right)+\varphi\left(j_{k+\tau},T\right)\right)\right]} \end{array}

Let $x$ be $\tau_{0}$-periodic, consequently $\tau_{0}$ is an integer-multiple of the symbol duration T measured in samples. Given our data model, we can expect to see a distinct peak in the ACF feature of such a signal. Let $j_{k}:=\left\lfloor \frac{k}{T}\right\rfloor$ be the index of the symbol being modulated at point of time $k\in\mathbb{N}$. While the signal $x$ is $\tau_{0}$-periodic, the series of symbols is $\frac{\tau_{0}}{T}$-periodic, i.e. $j_{k+\tau_{0}}=\left\lfloor \frac{k+\tau_{0}}{T}\right\rfloor =j_{k}+\frac{\tau_{0}}{T}$. Therefore the symbols of a signal are repeated periodically, i.e. $d\left(j_{k+\tau_{0}}\right)=d\left(j_{k}+\frac{\tau_{0}}{T}\right)=d\left(j_{k}\right)$. Hence we can simplify the ACF of x at lag $\tau_{0}$ to

\begin{array}{cc} \rho\left[x\right]\left(\tau_{0}\right)\\ = & \sum_{k=0}^{N-\tau_{0}}e^{i\left[\pi\eta\left(\left(\frac{k}{T}-j_{k}-\frac{k+\tau_{0}}{T}+j_{k}+\frac{\tau_{0}}{T}\right)d\left(j_{k}\right)\right)\right]}e^{i\left[\pi\eta\left(\sum_{l=0}^{j_{k}-1}d\left(l\right)-\sum_{l=0}^{j_{k}+\frac{\tau_{0}}{T}-1}d\left(l\right)\right)\right]}\\ = & \sum_{k=0}^{N-\tau_{0}}e^{i\left[\pi\eta\left(\left(-\frac{\tau_{0}}{T}+\frac{\tau_{0}}{T}\right)d\left(j_{k}\right)\right)\right]}e^{\pi\eta i\left(-\sum_{l=j_{k}}^{j_{k}+\frac{\tau_{0}}{T}-1}d\left(l\right)\right)}\\ = & \sum_{k=0}^{N-\tau_{0}}e^{\pi\eta i\left(-\sum_{l=j_{k}}^{j_{k}+\frac{\tau_{0}}{T}-1}d\left(l\right)\right)}. \end{array}

The value $\rho\left[x\right]\left(\tau_{0}\right)$ is decisively influenced by the term $p_{m}:=-\sum_{l=j_{k}}^{j_{k}+\frac{\tau_{0}}{T}-1}d\left(l\right)$. It contains exactly one period of the cyclic symbol stream and it shifts cyclically with time and, thus $p_{m}$ is constant. Accordingly we can simplify the ACF feature for periodical signals to

$$\left|\rho\left[x\right]\left(\tau_{0}\right)\right|=\left|\sum_{k=0}^{N-\tau_{0}}e^{\pi\eta i\left(-\sum_{l=j_{k}}^{j_{k}+\frac{\tau_{0}}{T}-1}d\left(l\right)\right)}\right|=\left(N-\tau_{0}+1\right).$$

The complex value of the ACF of a truely periodic signal is independent of modulation index and symbol stream.


In the second step we assume a partly-periodic signal x of ratio $p_{\%}$ with lag $\tau_{0}$. The ACF of a uniformly distributed series of samples converges to zero at all non-zero lags. Thus, according to our model the autocorrelation of the data part of the frames all but vanishes. To simplify the analysis, we introduce an indicator function b; indicating wether a sample is part of the periodic share of a signal: $$\begin{array}{ccc} b\left(k\right) & := & \begin{cases} 1 & k\in\text{header-part of x}\\ 0 & \text{else} \end{cases}\end{array}$$

We get

\begin{array}{cc} \rho\left[x\right]\left(\tau_{0}\right)\\ = & \sum_{k=0}^{N-\tau_{0}}b\left(k\right)e^{i\left[\pi\eta\left(\left(\frac{k}{T}-j_{k}-\frac{k+\tau_{0}}{T}+j_{k}+\frac{\tau_{0}}{T}\right)d\left(j_{k}\right)\right)\right]}e^{i\left[\pi\eta\left(\sum_{l=0}^{j_{k}-1}d\left(l\right)-\sum_{l=0}^{j_{k}+\frac{\tau_{0}}{T}-1}d\left(l\right)\right)\right]}\\ = & \sum_{k=0}^{N-\tau_{0}}b\left(k\right)e^{i\left[\pi\eta\left(\left(-\frac{\tau_{0}}{T}+\frac{\tau_{0}}{T}\right)d\left(j_{k}\right)\right)\right]}e^{\pi\eta i\left(-\sum_{l=j_{k}}^{j_{k}+\frac{\tau_{0}}{T}-1}d\left(l\right)\right)}\\ = & \sum_{k=0}^{N-\tau_{0}}b\left(k\right)e^{\pi\eta i\left(-\sum_{l=j_{k}}^{j_{k}+\frac{\tau_{0}}{T}-1}d\left(l\right)\right)}. \end{array}

The term $p_{m}$ (see [eq:ACF_FSK_PhaseMemory]) is not constant like it was above. As the frames are not truely periodic, the data-part of the frames introduce symbols randomly choosen from a uniform distribution. We split the results by modulation index:

If the modulation index $\eta$ is integer, $\eta p_{m}\in\mathbb{Z}$ and we get $e^{\pi\eta i\left(-\sum_{l=j_{k}}^{j_{k}+\frac{\tau_{0}}{T}-1}d\left(l\right)\right)}=1.$

Thus, $$ \left|\rho\left[x\right]\left(\tau_{0}\right)\right|=p_{\%}\left(N-\tau_{0}+1\right).$$

If $\eta$ is not integer,

$$\left|e^{\pi\eta i\left(-\sum_{l=j_{k}}^{j_{k}+\frac{\tau_{0}}{T}-1}d\left(l\right)\right)}\right|=1$$

but the argument of the summands in ${arg}\left(e^{\pi\eta i\left(-\sum_{l=j_{k}}^{j_{k}+\frac{\tau_{0}}{T}-1}d\left(l\right)\right)}\right)$

is not constant, as it depends on the random data frame, and thus consists of a sum of complex exponentials of differing phase. The sum decreases compared to the magnitude of the sum of an integer modulation index signal. I conclude that a random walk is in order to determine the effects of the cancellation.


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