Aufgaben:Exercise 5.2: Band Spreading and Narrowband Interferer

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Considered model
of band spreading

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:

• This exercise belongs to the chapter  Direct-Sequence Spread Spectrum Modulation.

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.