Exercise 2.1: ACF and PSD with Coding

Power-spectral density with coding

We consider the digital signal $s(t)$, using the following descriptive quantities:

$a_{\nu}$ are the amplitude coefficients,
$g_{s}(t)$ indicates the basic transmission pulse,
$T$ is the symbol duration (spacing of the pulses).

Then holds:

$$s(t) = \sum_{\nu = -\infty}^{+\infty} a_\nu \cdot g_s ( t - \nu \cdot T) \hspace{0.05cm}.$$

To characterize the spectral properties resulting from the coding and pulse shaping, one uses, among other things

the auto-correlation function (ACF)

$$\varphi_s(\tau) = \sum_{\lambda = -\infty}^{+\infty}{1}/{T} \cdot \varphi_a(\lambda)\cdot \varphi^{^{\bullet}}_{gs}(\tau - \lambda \cdot T)\hspace{0.05cm},$$

the power-spectral density (PSD)

$${\it \Phi}_s(f) = {1}/{T} \cdot {\it \Phi}_a(f) \cdot {\it \Phi}^{^{\bullet}}_{gs}(f) \hspace{0.05cm}.$$

Here, $\varphi_{a}(\lambda)$ denotes the discrete auto-correlation function of the amplitude coefficients related to the power-spectral density ${\it \Phi}_{a}(f)$ via the Fourier transform. Thus, for this holds:

$${\it \Phi}_a(f) = \sum_{\lambda = -\infty}^{+\infty}\varphi_a(\lambda)\cdot {\rm e}^{- {\rm j}\hspace{0.05cm} 2 \pi f \hspace{0.02cm} \lambda T} \hspace{0.05cm}.$$

Furthermore, the energy ACF and energy spectrum are used in above equations:

$$\varphi^{^{\bullet}}_{gs}(\tau) = \int_{-\infty}^{+\infty} g_s ( t ) \cdot g_s ( t + \tau)\,{\rm d} t \hspace{0.4cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet \hspace{0.4cm} {\it \Phi}^{^{\bullet}}_{gs}(f) = |G_s(f)|^2 \hspace{0.05cm}.$$

In the present exercise, the following function is to be assumed for the power-spectral density of the amplitude coefficients (see graph):

$${\it \Phi}_a(f) = {1}/{2} - {1}/{2} \cdot \cos (4 \pi f \hspace{0.02cm} T)\hspace{0.05cm}.$$

The following assumptions are made for the basic transmission pulse:

In question (2), let $g_{s}(t)$ be an NRZ rectangular pulse, so that there is a triangular energy ACF confined to the range $|\tau| ≤ T$. The maximum here is

$$\varphi^{^{\bullet}}_{gs}(\tau = 0) = s_0^2 \cdot T \hspace{0.05cm}.$$

For question (3), assume a root-Nyquist characteristic with rolloff factor $r = 0$. In this case holds:

$$|G_s(f)|^2 = \left\{ \begin{array}{c} s_0^2 \cdot T^2 \\ 0 \\ \end{array} \right.\quad \begin{array}{*{1}c} {\rm{for}} \\ {\rm{for}} \\ \end{array} \begin{array}{*{20}c} |f| < {1}/({2T}) \hspace{0.05cm}, \\ |f| > {1}/({2T}) \hspace{0.05cm}.\\ \end{array}$$

For numerical calculations, always use $s_{0}^{2} = 10 \ \rm mW$.

Notes:

The exercise belongs to the chapter Basics of Coded Transmission.

Consider that the transmit power $P_{\rm S}$ is equal to the ACF $\varphi_{s}(\tau)$ at the point $\tau = 0$, but can also be calculated as an integral over the PSD $\Phi_{s}(f)$.

Questions

Solution

(1) Since ${\it \Phi}_{a}(f)$ as a power-spectral density is always real (plus even and positive, but that does not matter here) and the ACF values $\varphi_{a}(\lambda)$ are symmetric about $\lambda = 0$, the given equation can be transformed as follows:

$${\it \Phi}_a(f) = \sum_{\lambda = -\infty}^{+\infty}\varphi_a(\lambda)\cdot {\rm e}^{- {\rm j}\hspace{0.05cm} 2 \pi f \hspace{0.02cm} \lambda T} = \varphi_a(0) + \sum_{\lambda = 1}^{\infty}2 \cdot \varphi_a(\lambda)\cdot\cos ( 2 \pi f \hspace{0.02cm} \lambda T) \hspace{0.05cm}.$$

By comparison with the sketched function

$${\it \Phi}_a(f) = {1}/{2} - {1}/{2} \cdot \cos (4 \pi f \hspace{0.02cm} T)\hspace{0.05cm}.$$

one obtains:

$${\it \varphi}_a(\lambda = 0)\hspace{0.15cm}\underline { = 0.5}, \hspace{0.2cm} {\it \varphi}_a(\lambda = 2) = {\it \varphi}_a(\lambda = -2) \hspace{0.15cm}\underline {= -0.25} \hspace{0.05cm}.$$

All other ACF values result to zero, so also $\varphi_{a}(\lambda = ±1)\hspace{0.15cm}\underline {=0}$.

(2) For the rectangular NRZ basic pulse, due to the limitation of the energy ACF to the range $|\tau| ≤ T$, we obtain:

$$P_{\rm S} = \varphi_s(\tau = 0) = \frac{1}{T} \cdot \varphi_a(\lambda = 0)\cdot \varphi^{^{\bullet}}_{gs}(\tau = 0)= \frac{1}{T} \cdot \frac{1}{2} \cdot s_0^2 \cdot T = \frac{s_0^2}{2} \hspace{0.15cm}\underline {= 5\,\,{\rm mW}}\hspace{0.05cm}.$$

(3) For rectangular spectral function, it is more convenient to calculate the transmit power by integration over the power-spectral density:

$$P_{\rm S} = \ \int_{-1/(2T)}^{+1/(2T)} {\it \Phi}_s(f) \,{\rm d} f = \frac{1}{T} \cdot \int_{-1/(2T)}^{+1/(2T)} {\it \Phi}_a(f) \cdot {\it \Phi}^{^{\bullet}}_{gs}(f) \,{\rm d} f$$

$$\Rightarrow\hspace{0.3cm}P_{\rm S} = \ \frac{1}{T} \cdot \left [ s_0^2 \cdot T^2 \right ] \cdot \int_{-1/(2T)}^{+1/(2T)} \left( {1}/{2} - {1}/{2} \cdot \cos (4 \pi f \hspace{0.02cm} T)\right ) \,{\rm d} f\hspace{0.05cm} = {s_0^2}/{2}\hspace{0.15cm}\underline { = 5\,\,{\rm mW}} .$$

Here it is considered that for this exercise the energy PSD $|G_{s}(f)|^{2}$ is given as constant (within the integration interval) and thus can be drawn in front of the integral.
In spite of a completely different signal form $s(t)$, the same transmit power results here, since the integral yields the value $1/(2T)$.
It should be noted that this simple calculation is only possible for the rolloff factor $r = 0$.