Exercise 4.6Z: Locality Curve for Phase Modulation

A possible locality curve with phase modulation

We assume a source signal $q(t)$, which is considered normalised.

The maximum value of this signal is $q_{\rm max} = 1$ and the minimum signal value is $q_{\rm min} = -0.5$.
Otherwise nothing is known about $q(t)$.

The modulated signal with phase modulation ⇒ "transmission signal" is:

$$s(t) = s_0 \cdot {\cos} ( \omega_{\rm T}\hspace{0.05cm} t + \eta \cdot q(t)).$$

Here $\eta$ denotes the so-called "modulation index". Let the constant envelope $s_0$ also be a normalise quantity, which is set to $s_0 = 2$ in the following (see diagram).

If one replaces the cosine function with the complex exponential function, one arrives at the analytical signal

$$s_{\rm +}(t) = s_0\cdot {\rm e}^{{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}( \omega_{\rm T} \hspace{0.05cm}\cdot \hspace{0.05cm} t + \eta \hspace{0.05cm} \cdot \hspace{0.05cm} q(t)) }.$$

From this, one can calculate the equivalent low-pass signal sketched in the graph as follows:

$$s_{\rm TP}(t) = s_{\rm +}(t) \cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot\hspace{0.05cm} \omega_{\rm T} \hspace{0.05cm}\cdot\hspace{0.05cm} t } = s_0\cdot {\rm e}^{{\rm j}\hspace{0.05cm}\cdot\hspace{0.05cm} \eta \hspace{0.05cm} \cdot \hspace{0.05cm} q(t) }.$$

Hints:

This exercise belongs to the chapter Equivalent Low-Pass Signal and its Spectral Function.

You can check your solution with the interactive applet Physical Signal & Equivalent Low-Pass Signal ⇒ "Locality Curve".

Questions

$a(t = 0)\ = \ $

$\phi_{\rm min}\ = \ $

$\text{deg}$

$\phi_{\rm min}\ = \ $

$\text{deg}$

$\eta\ = \ $

	From $q(t) = -0.5 = \text{const.}$ follows $s(t) = s_0 \cdot \cos (\omega_T \cdot t)$.
	With a rectangular signal $($with only two possible signal values $q(t)=\pm 0.5)$ the locality curve degenerates to two points.
	With the signal values $\pm 1$ $(q_{\rm min} = -0.5$ is then no longer valid$)$ the locality curve degenerates to one point: $s_{\rm TP}(t) = -s_0$.

Solution

(1) The locus curve is a circular arc with radius $2$. Therefore, the magnitude function is constant $\underline{a(t) = 2}$.

(2) From the graph it can be seen that the following numerical values apply:

$\phi_{\rm min} =- \pi /2 \; \Rightarrow \; \underline{-90^\circ}$,
$\phi_{\rm max} = +\pi \; \Rightarrow \; \underline{+180^\circ}$.

(3) In general, the relation $s_{\rm TP}(t) = a(t) \cdot {\rm e}^{{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} \phi(t)}.$ applies here. A comparison with the given function yields:

$$\phi(t) = \eta \cdot q(t).$$

The maximum phase value $\phi_{\rm max} = +\pi \; \Rightarrow \; {180^\circ}$ is obtained for the signal amplitude $q_{\rm max} = 1$. From this follows directly ${\eta = \pi} \; \underline{\approx 3.14}$.
This modulation index is confirmed by the values $\phi_{\rm min} = -\pi /2$ and $q_{\rm min} = -0.5$ .

Locus curve (phase diagram) for a rectangular signal

(4) The second and third proposed solutions are correct:

If $q(t) = \text{const.} =-0.5$, the phase function is also constant:

$$\phi(t) = \eta \cdot q(t) = - {\pi}/{2}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} s_{\rm TP}(t) = - {\rm j} \cdot s_0 = - 2{\rm j}.$$

Thus, for the actual, physical signal:

$$s(t) = s_0 \cdot {\cos} ( \omega_{\rm T}\hspace{0.05cm} t - {\pi}/{2}) = 2 \cdot {\sin} ( \omega_{\rm T} \hspace{0.05cm} t ).$$

In contrast, $q(t) = +0.5$ leads to $\phi (t) = \pi /2$ and to $s_{\rm TP}(t) = 2{\rm j}$.
If $q(t)$ is a rectangular signal that alternates between $+0.5$ and $–0.5$ , then the locus curve consists of only two points on the imaginary axis, regardless of how long the intervals with $+0.5$ and $–0.5$ last.
If, on the other hand, $q(t) = \pm 1$, then the possible phase values $+\pi$ and $-\pi$ result purely formally, but they are identical.
The „locus curve” then consists of only one point: $s_{\rm TP}(t) = - s_0$
⇒ the signal $s(t)$ is „minus-cosine” for all times $t$ .