When the receiver gets the signal $m_t$ sent by the transmitter, the signal will have an additive interference $i_t$. Thereby signal received will be : $$r_t = m_t + i_t$$

I know that to get back the data from the signal, we have to multiply by the pseudo-noise sequence (PN).

That will take out the pseudo-noise from the signal $m_t$ allowing us to get the data.

What do we do with the noise $i_t$?

It has been multiplied by the PN but I supposed we cant take out the data from the $r_t$ until we eliminate the noise. I don't really see what kind of signal do we get multiplying the additive interference by the PN. I've read that we use an low pass filter but i dont really undestand why.

  • 1
    $\begingroup$ Welcome to signal.SE! Don't worry about your English, it's very good :) $\endgroup$ Nov 30, 2016 at 15:51
  • $\begingroup$ You can never "eliminate" the noise. You have to make the best decisions you can in the presence of the noise. The use of DSSS has no effect on the noise performance of your communication system in principle; it is a reversible operation, so it can't have an effect on the bit-error rate of the optimum receiver. So, you can take the DSSS modulation/demodulation out of the equation entirely and analyze the simpler system to get a feel for how it works. $\endgroup$
    – Jason R
    Nov 30, 2016 at 16:00

2 Answers 2


Multiplying $r_t$ by the pseudo random spreading code does different things to $m_t$ and $i_t$: Because $m_t$ is correlated to the spreading code, it is despread while keeping its power, so the power spectral density will be higher in a small bandwidth (due to remaining modulation). On the other hand, $i_t$ is not correlated and is translated to wideband noise again.

This gives you the opportunity to filter with a bandpass and just use the band, where the signal to noise ratio is above unity.

The noise is not eliminated, but its overall power is reduced, because your bandpass rejects some of the noise (that is still wideband) and passes all of the signal (that is now narrowband).


Suppose that the received signal in a direct-sequence spread-spectrum communications system is $r(t) = \pm As(t) + \mathcal N(t)$ where $s(t) = \sum_{n=0}^{N-1} x_n p(t-nT_c)$ is the PN signal of duration $T = NT_c$ that is used to spread the spectrum (the $x_n$ have value $\pm 1$, $N$ is the number of "chips" in the sequence, $p(t)$ is the unit-power baseband chip pulse shape of duration $T_c$) and $\mathcal N(t)$ is white Gaussian noise with two-sided power spectral density $\frac{N_0}{2}$. As @JasonR's cogent comment points out, this is just a canonical "antipodal signaling in white Gaussian noise" system using antipodal signals $\pm As(t)$ and we can ignore the fact that $s(t)$ is not a rectangular pulse or a raised cosine pulse or what-have-you: all the antipodal signaling analysis and methodology holds. The receiver doesn't just multiply $r(t)$ by $s(t)$, it multiplies and then integrates the product $r(t)\cdot s(t)$. That is, we correlate $r(t)$ with the locally-generated spreading signal $s(t)$:

\begin{align} Y &= \int_{0}^T r(t)\cdot s(t)\ \mathrm dt \tag{1}\\ &= \int_{0}^T \pm As(t)\cdot s(t)\ \mathrm dt + \int_{0}^T \mathcal N(t)\cdot s(t)\ \mathrm dt\\ &= \pm AT + \eta \tag{2} \end{align} where $\eta$ is a zero-mean Gaussian random variable with variance $\frac{N_0T}{2}$. The bit error probability is thus $$P_e = Q\left(\frac{AT}{\sqrt{\frac{N_0T}{2}}}\right) = Q\left(\sqrt{\frac{2A^2T}{N_0}}\right) = Q\left(\sqrt{\frac{2\mathscr E_b}{N_0}}\right)$$ as with the usual antipodal signaling.

With direct-sequence spread-spectrum signaling, an alternative implementation of the receiver uses a chip correlator followed by digital correlation. The receiver now computes

\begin{align} Y_n &= \int_{nT_c}^{(n+1)T_c} r(t)\cdot p(t-nT)\ \mathrm dt, \quad n = 0, 1, \ldots N-1 \tag{3}\\ &= \int_{nT_c}^{(n+1)T_c} \pm As(t)\cdot p(t-nT)\ \mathrm dt + \int_{nT_c}^{(n+1)T_c} \mathcal N(t)\cdot p(t-nT)\ \mathrm dt\\ &= \pm Ax_nT_c + \eta_n \tag{4} \end{align} where the $\eta_n$ are zero-mean Gaussian random variables. The receiver now does a digital correlation by computing $$\sum_{n=0}^{N-1} Y_n \cdot x_n = \sum_{n=0}^{N-1} \pm Ax_nT_c\cdot x_n + \eta_n\cdot x_n = Y + \eta \tag{5}$$ and makes a decision on $Y$ just as before.

It is a major mistake to make individual decisions $\hat{x}_n$ on the chip variables $Y_n$ and then sum $$\sum_{n=0}^{N-1} \hat{x}_n \cdot x_n.$$ The sum is easy to compute because the $\hat{x}n$ and the $x_n$ have value $\pm 1$ but the error probability of those decisions is $$P_{e, chip} = Q\left(\frac{AT_c}{\sqrt{\frac{N_0T_c}{2}}}\right) = Q\left(\sqrt{\frac{2A^2T_c}{N_0}}\right) = Q\left(\sqrt{\frac{2\mathscr E_c}{N_0}}\right)$$ where the chip energy $\mathscr E_c$ is smaller than $\mathscr E_b$ by a factor of $N$. Even when the bit SNR is reasonably large, the chip SNR is very small and the chip decisions are very little different from random guesses. Consequently, hard chip decisions are the wrong way to go, it is far far better to keep the soft decisions $Y_n$ and use them in a digital correlation to create the decision variable that is the optimum one to use.


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