# Exercise 2.1: ACF and PSD 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  $\rm (ACF)$
$$\varphi_s(\tau) = \sum_{\lambda = -\infty}^{+\infty}{1}/{T} \cdot \varphi_a(\lambda)\cdot \varphi^{^{\bullet}}_{gs}(\tau - \lambda \cdot T)\hspace{0.05cm},$$
• the power-spectral density  $\rm (PSD)$
$${\it \Phi}_s(f) = {1}/{T} \cdot {\it \Phi}_a(f) \cdot {\it \Phi}^{^{\bullet}}_{gs}(f) \hspace{0.05cm}.$$

Here,  $\varphi_{a}(\lambda)$  denotes the discrete ACF 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 value 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,  use always  $s_{0}^{2} = 10 \ \rm mW$.

Notes:

• 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  ${\it \Phi}_{s}(f)$.

### Questions

1

What are the discrete ACF values  $\varphi_{a}(\lambda)$  of the amplitude coefficients? Enter the numerical values for  $\lambda = 0$,  $\lambda = 1$  and  $\lambda = 2$.

 $\varphi_{a}(\lambda = 0) \ = \$ $\varphi_{a}(\lambda = 1) \ = \$ $\varphi_{a}(\lambda = 2) \ = \$

2

What is the transmit power with the  NRZ basic transmission pulse?

 $P_{\rm S} \ = \$ $\ \rm mW$

3

What is the transmit power with  root-Nyquist characteristic  $(r = 0)$?

 $P_{\rm S} \ = \$ $\ \rm mW$

### Solution

#### 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 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 $r = 0$.