# Exercise 5.2: Band Spreading and Narrowband Interferer

Considered model

A spread spectrum system is considered according to the given diagram in the equivalent low-pass range:

• Let the digital signal  $q(t)$  possess the power-spectral density  ${\it \Phi}_q(f)$,  which is to be approximated as rectangular with bandwidth  $B = 1/T = 100\ \rm kHz$   (a rather unrealistic assumption):
$${\it \Phi}_{q}(f) = \left\{ \begin{array}{c} {\it \Phi}_{0} \\ 0 \\ \end{array} \right. \begin{array}{*{10}c} {\rm{for}} \\ {\rm{otherwise}} \hspace{0.05cm}. \\ \end{array}\begin{array}{*{20}c} |f| <B/2 \hspace{0.05cm}, \\ \\ \end{array}$$
• Thus,  in the low-pass range,  the bandwidth  (only the components at positive frequencies)  is equal to  $B/2$  and the bandwidth in the band-pass range is  $B$.
• The band spreading is done by multiplication with the PN sequence  $c(t)$  of the chip duration  $T_c = T/100$
("PN" stands for "pseudo-noise").
• To simplify matters,  the following applies to the auto-correlation function:
$${\it \varphi}_{c}(\tau) = \left\{ \begin{array}{c}1 - |\tau|/T_c \\ 0 \\ \end{array} \right. \begin{array}{*{10}c} {\rm{for}} \\ {\rm{otherwise}} \hspace{0.05cm}. \\ \end{array}\begin{array}{*{20}c} -T_c \le \tau \le T_c \hspace{0.05cm}, \\ \\ \end{array}$$
• At the receiver,  the same spreading sequence  $c(t)$  is again added phase-synchronously.
• The interference signal  $i(t)$  is to be neglected for the time being.
• In subtask  (4)   $i(t)$  denotes a narrowband interferer at carrier frequency  $f_{\rm T} = 30 \ \rm MHz = f_{\rm I}$  with power  $P_{\rm I}$.
• The influence of the  (always present)  AWGN noise  $n(t)$  is not considered in this exercise.

Note:

### Questions

1

What is the power-spectral density  ${\it \Phi}_c(f )$  of the spreading signal  $c(t)$?  What value results at the frequency  $f = 0$?

 ${\it \Phi}_c(f = 0) \ = \$ $\ \cdot 10^{-6} \ \rm 1/Hz$

2

Calculate the equivalent bandwidth  $B_c$  of the spread signal as the width of the equal-area  $\rm PDS$  rectangle.

 $B_c \ = \$ $\ \rm MHz$

3

Which statements are true for the bandwidths of the signals  $s(t)$   ⇒   $B_s$ and  $b(t)$   ⇒   $B_b$?  The (two-sided) bandwidth of  $q(t)$  is  $B$.

 $B_s$  is exactly equal to  $B_c$. $B_s$  is approximately equal to  $B_c + B$. $B_b$  is exactly equal to  $B_s$. $B_b$  is equal to  $B_s + B_c = 2B_c + B$. $B_b$  is exactly equal to  $B$.

4

What is the effect of band spreading on a  "narrowband interferer"  at the carrier frequency?  Let  $f_{\rm I} = f_{\rm T}$.

 The interfering influence is weakened by band spreading. The interfering power is only half as large. The interfering power is not changed by band spreading.

### Solution

#### Solution

(1)  The power-spectral density  $\rm (PDS)$  ${\it \Phi}_c(f)$  is the Fourier transform of the triangular ACF,  which can be represented with rectangles of width  $T_c$  as follows:

$${\it \varphi}_{c}(\tau) = \frac{1}{T_c} \cdot {\rm rect} \big(\frac{\tau}{T_c} \big ) \star {\rm rect} \big(\frac{\tau}{T_c} \big ) \hspace{0.05cm}.$$
• From this follows  ${\it \Phi}_{c}(f) = {1}/{T_c} \cdot \big[ T_c \cdot {\rm sinc} \left(f T_c \right ) \big ] \cdot \big[ T_c \cdot {\rm sinc} \left(f T_c \right ) \big ] = T_c \cdot {\rm sinc}^2 \left(f T_c \right ) \hspace{0.05cm}$  with maximum value
$${\it \Phi}_{c}(f = 0) = T_c = \frac{T}{100}= \frac{1}{100 \cdot B} = \frac{1}{100 \cdot 10^5\,{\rm 1/s}} = 10^{-7}\,{\rm 1/Hz} \hspace{0.15cm}\underline {= 0.1 \cdot 10^{-6}\,{\rm 1/Hz}}\hspace{0.05cm}.$$

(2)  By definition,  with  $T_c = T/100 = 0.1\ \rm µ s$:

Power density spectrum of the pseudo-noise spread signal
$$B_c= \frac{1}{T_c} \cdot \hspace{-0.03cm} \int_{-\infty }^{+\infty} \hspace{-0.03cm} {\it \Phi}_{c}(f)\hspace{0.1cm} {\rm d}f = \hspace{-0.03cm} \int_{-\infty }^{+\infty} \hspace{-0.03cm} {\rm sinc}^2 \left(f T_c \right )\hspace{0.1cm} {\rm d}f$$
$$\Rightarrow \hspace{0.3cm} B_c= \frac{1}{T_c}\hspace{0.15cm}\underline {= 10\,{\rm MHz}} \hspace{0.05cm}$$

The graph illustrates,

• that  $B_c$  is given by the first zero of the  $\rm sinc^2$ function in the equivalent low-pass range,
• but at the same time also gives the equivalent  (equal area)  bandwidth in the band-pass region.

(3)  Solutions 2 and 5  are correct:

• The PDS  ${\it \Phi}_s(f)$  results from the convolution of  ${\it \Phi}_q(f)$  and  ${\it \Phi}_c(f)$.  This actually gives  $B_s = B_c + B$  for the bandwidth of the transmitted signal.
• Since the spreading signal  $c(t) ∈ \{+1, –1\}$  multiplied by itself always gives the value  $1$,  naturally  $b(t) ≡ q(t)$  and consequently  $B_b = B$.
• Obviously, the bandwidth  $B_b$  of the band compressed signal is not equal to  $2B_c + B$,  although the convolution  ${\it \Phi}_s(f) ∗ {\it \Phi}_c(f)$  suggests this.
• This is due to the fact that the power density spectra must not be convolved, but the spectral functions  (amplitude spectra)  $S(f)$  and  $C(f)$  must be assumed, taking into account the phase relations.
• Only then can the PDS  $B(f)$  be determined from  ${\it \Phi}_b(f)$.  Clearly,  the following is also true:   $C(f) ∗ C(f) = δ(f)$.

(4)  Only the  first solution  is correct.  The solution shall be clarified by the diagram at the end of the page:

• In the upper diagram the PDS  ${\it \Phi}_i(f)$  of the narrowband interferer is approximated by two Dirac delta functions at  $±f_{\rm T}$  with weights  $P_{\rm I}/2$.   Also plotted is the bandwidth  $B = 0.1 \ \rm MHz$  (not quite true to scale).
• The receiver-side multiplication with  $c(t)$  – actually with the function of the band compression,  at least with respect to the useful part of  $r(t)$ –  causes a band spreading with respect to the interference signal  $i(t)$.  Without considering the useful signal,  $b(t) = n(t) = i(t) · c(t)$.  It follows:
$${\it \Phi}_{n}(f) = {\it \Phi}_{i}(f) \star {\it \Phi}_{c}(f) = \frac{P_{\rm I}\cdot T_c}{2}\cdot {\rm sinc}^2 \left( (f - f_{\rm T}) \cdot T_c \right )+ \frac{P_{\rm I}\cdot T_c}{2}\cdot {\rm sinc}^2 \left( (f + f_{\rm T}) \cdot T_c \right ) \hspace{0.05cm}.$$
Power density spectra before and after band spreading
• Note that  $n(t)$  is used here only as an abbreviation and does not denote AWGN noise.
• In a narrow range around the carrier frequency  $f_{\rm T} = 30 \ \rm MHz$,  the PDS  ${\it \Phi}_n(f)$  is almost constant.  Thus,  the interference power after band spreading is:
$$P_{n} = P_{\rm I} \cdot T_c \cdot B = P_{\rm I}\cdot \frac{B}{B_c} = \frac{P_{\rm I}}{J}\hspace{0.05cm}.$$
• This means:   The interference power is reduced by the factor  $J = T/T_c$  by band spreading,  which is why  $J$  is often called  "spreading gain".
• However,  such a  "spreading gain"  is only given for a narrowband interferer.