A Pseudo-random Noise (PN) sequence is a sequence of binary numbers, e.g. ±1, which appears to be random; but is in fact perfectly deterministic. The sequence appears to be random in the sense that the binary values and groups or runs of the same binary value occur in the sequence in the same proportion they would if the sequence were being generated based on a fair "coin tossing" experiment. In the experiment, each head could result in one binary value and a tail the other value.

Consider a model, $$ x(t) = h^T u(t) +w(t) \tag{1}$$ where $u$ is the input and $w$ is Additive White Gaussian noise.

In Book : Fundamentals of Statistical Signal Processing by Steven Kay, Chapter 4 : Eq(4.21) finds the variance of the estimated coefficients when the input is PRN as

$$var(\hat{h_i} )= \sigma_w^2/Nr_{uu}[0] \tag{2}$$ under the assumption of the input to be a PN . where $\sigma_w^2$ is the variance of the measurement noise; $\sigma_u^2$ is the variance of the input The MVU estimator using a PRN sequence is mentioned as $\hat{\theta }= (H^TH)^{-1} H^T x$

It is stated that for the deterministic sequence $u$ the $H^TH$ becomes a symmetric Toeplitz autocorrelation matrix. For $H^TH$ to be diagonal we should use a PRN sequence as input which yileds $H^TH = Nr_{uu}[o]I$. These expressions can be found in document Pg 7 Section 4.1.3

My problem is that I am confused if PRN means binary or $\pm1$ sequence or only $iid$ real valued White Gaussian noise. If PRN is different from PN or binary, then what will be the estimates and the CRLB?

  • $\begingroup$ Can you include the definition of $\hat{h}(I)$ for the benefit of those who don;t have access to Eq.4.2.1 of Kay's book? $\endgroup$ Jul 18, 2015 at 3:09
  • $\begingroup$ @DilipSarwate: I have updated the Question and added the link to the reference of the Example as well. Thank you for your help. $\endgroup$
    – SKM
    Jul 20, 2015 at 21:32
  • $\begingroup$ PRN sequence is the PN sequence. $\endgroup$
    – Batman
    Jul 21, 2015 at 2:28

1 Answer 1


If the output $x[\cdot]$ of a discrete-time linear system is related to the input via $$x[n] = \sum_{i=0}^m h[i]u[n-i] \tag{1}$$ where $u[\cdot]$ is a periodic sequence with period $N \gg m$, then $x[\cdot]$ is also a sequence with period $N$ and the periodic cross-correlation function $R_{x,u}[\cdot]$ between $x[\cdot]$ and $u[\cdot]$ is the periodic convolution between the sequence $h[\cdot]$ and the periodic autocorrelation function $R_{u,u}[\cdot]$. That is, $$R_{x,u} = x \star u = (h * u) \star u = h * (u \star u) = h* R_{u,u} \tag{2}$$ If $u[\cdot]$ is a binary pseudonoise (PN) a.k.a. pseudorandom noise (PRN) sequence that takes on values $\pm 1$, then $$R_{u,u}[n] = \begin{cases}N, & n \equiv 0 \bmod N,\\ -1, &\text{otherwise}\end{cases}\tag{3}$$ which is essentially a periodic digital impulse function, and by ignoring that $-1$ as inconsequential compared to $N$, we get from $(2)$ that $$N h[n] = R_{x,u} \implies h[n] = N^{-1} R_{x,u}[n].\tag{4}$$ If we include additive white Gaussian noise $w[n]$ of variance $\sigma^2$ on the right side of $(1)$, then $(2)$ becomes $$R_{x+w,u} = (x + w)\star u = (h * u + w) \star u = h * (u \star u) + w \star u= h* R_{u,u} + \hat{w} \tag{5}$$ where $\hat{w} \sim \mathcal N(0, N\sigma^2)$. Thus, the measured cross-correlation function is now a Gaussian random variable with mean $Nh[n]$ and variance $N\sigma^2$. The estimator for $h[n]$ is thus just $N^{-1}$ times the measured value of the cross-correlation function, and its variance is thus $\displaystyle \frac{\sigma^2}{N}$, This matches up perfectly with Kay's result $$\operatorname{var}(\hat{h}[n]) = \frac{\sigma_{w}^2}{Nr_{u,u}[0]}$$ since we have assumed that $u[n] \in \{+1, -1\}$ and hence $r_{u,u}[0] = 1$: all the $u[n]$, regarded as random instead of pseudorandom, have variance $1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.