# Exercise 5.2: Band Spreading and Narrowband Interferer

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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

### 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 $T_c = T/100 = 0.1\ \rm µ s$:

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

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:

- $${\it \Phi}_{n}(f) = {\it \Phi}_{i}(f) \star {\it \Phi}_{c}(f) = \frac{P_{\rm I}\cdot T_c}{2}\cdot {\rm sinc}^2 \left( (f - f_{\rm T}) \cdot T_c \right )+ \frac{P_{\rm I}\cdot T_c}{2}\cdot {\rm sinc}^2 \left( (f + f_{\rm T}) \cdot T_c \right ) \hspace{0.05cm}.$$

- 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.