Exercise 5.2Z: About PN Modulation

Models of PN modulation (top) and BPSK (bottom)

The diagram shows the equivalent circuit of $\rm PN$ modulation $($Direct Sequence Spread Spectrum, abbreviated $\rm DS–SS)$ in the equivalent low-pass range, based on AWGN noise $n(t)$. Shown below is the low-pass model of binary phase modulation $\rm (BPSK)$.

The low-pass transmitted signal $s(t)$ is set equal to the rectangular source signal $q(t) ∈ \{+1, –1\}$ with rectangular duration $T$ for reasons of uniformity.

The function of the integrator can be described as follows:

$$d (\nu T) = \frac{1}{T} \cdot \hspace{-0.1cm} \int_{(\nu -1 )T }^{\nu T} \hspace{-0.3cm} b (t )\hspace{0.1cm} {\rm d}t \hspace{0.05cm}.$$

The two models differ by multiplication with the $±1$ spread signal $c(t)$ at transmitter and receiver, where only the degree of spread $J$ is known from $c(t)$.

It has to be investigated whether the lower BPSK model can also be used for PN modulation and whether the BPSK error probability

$$p_{\rm B} = {\rm Q} \left( \hspace{-0.05cm} \sqrt { {2 \cdot E_{\rm B}}/{N_{\rm 0}} } \hspace{0.05cm} \right )$$

is also valid for PN modulation, or how the given equation should be modified.

Notes:

This exercise belongs to the chapter Direct-Sequence Spread Spectrum Modulation.
For the solution of this exercise, the specification of the specific spreading sequence $($M-sequence or Walsh function$)$ is not important.

Questions

Solution

(1) The last Lösungsvorschlag is correct:

We are dealing here with an optimal receiver.
Without noise, the signal $b(t)$ within each bit is constantly equal to $+1$ or $-1$.
From the given equation for the integrator

$$d (\nu T) = \frac{1}{T} \cdot \hspace{-0.1cm} \int_{(\nu -1 )T }^{\nu T} \hspace{-0.3cm} b (t )\hspace{0.1cm} {\rm d}t $$

it follows that $d(νT)$ can take only the values $+1$ and $-1$.

(2) Again the last solution is correct:

In the noise– and interference-free case ⇒ $n(t) = 0$, the twofold multiplication by $c(t) ∈ \{+1, –1\}$ can be omitted,
so that the upper model is identical to the lower model.

(3) Solution 1 is correct:

Since both models are identical in the noise-free case, only the noise signal has to be adjusted: $n'(t) = n(t) · c(t)$.
In contrast, the other two solutions are not applicable:
The integration must still be done over $T = J · T_c$ and the PN modulation does not reduce the AWGN noise.

(4) The last solution is correct:

Multiplying the AWGN noise by the high-frequency $±1$ signal $c(t)$, the product is also Gaussian and white.
Because of ${\rm E}\big[c^2(t)\big] = 1$, the noise variance is not changed either.
Thus, the equation $p_{\rm B} = {\rm Q} \left( \hspace{-0.05cm} \sqrt {{2 E_{\rm B}}/{N_{\rm 0}} } \hspace{0.05cm} \right )$ valid for BPSK is also applicable for PN modulation, independent of the spreading factor $J$ and the specific spreading sequence.
Ergo: For AWGN noise, band spreading neither increases nor decreases the error probability.

	$d(νT)$ can be Gaussian distributed.
	$d(νT)$ can take the values $+1$, $0$ and $-1$.
	Only the values $d(νT) = +1$ and $d(νT) = -1$ are possible.

	The noise $n(t)$ must be replaced by $n'(t) = n(t) · c(t)$.
	The integration must now be done over $J · T$.
	The noise power $σ_n^2$ must be reduced by a factor of $J$.

	The larger $J$ is chosen, the smaller $p_{\rm B}$ is.
	The larger $J$ is chosen, the larger $p_{\rm B}$ is.
	Independent of $J$, the value $p_{\rm B} ≈ 2.3 · 10^{–3}$ is always obtained.