# Orthogonal basis of signal space and the projection of white noise

I am trying to make the relation between linear modulation and nonlinear one by using orthonormal expansion. The purpose is to understand what is orthonormal set in each case and to understand the operation of projection.

Linear modulation

In digital linear modulation, the baseband signal can be written as

$$x(t) = \sum_n a_n p(t-nT)$$ where $a_n$ is data, $\lbrace p_n(t) = p(t-nT)\textrm{, }n\in \mathbb{Z}\rbrace$ is orthonormal set, i.e. $<p_n(t),p_m(t)> = \delta_{m-n}$ with $<x(t),y(t)> = \int x(t)y^*(t)dt$ is inner product defined for the signal space we are working on. To get $a_n$ back, we need project $x(t)$ to $p_n(t)$.

Use matched filter $q(t) = p^*(-t)$ thus $q(t-\tau) = p^*(\tau - t)$.

by defining $g(t) = p(t) \star p^*(-t)$ where $\star$ is convolution. $g(t=0) = 1$ and $g(t=nT) = 0$ because $\delta_{m-n} = <p_n(t),p_m(t)> = g((m-n)T)$.

\begin{align} y(t) &= x(t) \star p^*(-t)\\ &= \sum_n a_n p(t-nT)\star p^*(-t)\\ &= \sum_n a_ng(t-nT)\\ \implies y(t=mT) &= \sum_n a_ng((m-n)T) = a_m \end{align}

If we have white noise $n(t)$ : $y(t) = x(t) + n(t)$. The noise is processed as

\begin{align} z(t) &= n(t) \star q(t)\\ &= \int n(\tau)q(t-\tau) d\tau\\ &= \int n(\tau) p^*(\tau -t) d\tau\\ \implies z(t=mT) &= \int n(\tau) p^*(\tau - mT) d\tau\\ &= <n(t), p(t-mT)> \end{align}

Then we can say that for a given $y(t)$, each processing $m$ at the receiver which ends with sampling at $t=mT$, the noise process is projected to the vector $p_m(t)$ creating $z_m$ and the data part is projected to the same vector creating $a_m$. $\lbrace p_m(t) = p(t-mT)\textrm{, }m\in \mathbb{Z}\rbrace$ is orthonormal set, thus $z_m$ are uncorrelated, with Gaussian assumption, $z_m$ are independent.

This is famous process in classical technical books. Two conclusions I get

1. The set of orthonormal vectors $\lbrace p_m(t) = p(t-mT)\textrm{, }m\in \mathbb{Z}\rbrace$ has infinite number of elements as $m \in \mathbb{Z}$.

2. The orthonormal vectors $p_m(t)$ can be no-time-limited and the orthogonal property holds on the entier time support.

Non-linear modulation

Let's take a binary FSK modulation as example. $M$ basis functions, $k \in \mathbb{S} = \lbrace 1, ..., M \rbrace$:

$$\phi_k(t) = \begin{cases}\sqrt{\frac 2T}\sin\left(\frac{k\pi t}{T}\right)& \text{for}\quad 0 \leq t \leq T\\0 &\text{otherwise}\end{cases}$$

I write the waveform as $x(t) = \sum_n \gamma_n(t-nT)$ where $\gamma_n(t) = \phi_k(t) \textrm{ if } a_n = k$.

Questions

• Question 1 : At the receiver, for the $n^\rm{th}$ symbol, we try projecting to $M$ functions $\phi_k(t-nT)$. If $k=k_0$ gives the largest power, we decode $a_n = k_0$. Is $\lbrace \phi_k(t-nT) \textrm{, } n \in \mathbb{Z}\textrm{, } k \in \mathbb{S}\rbrace$ the orthonormal set as $<\phi_k(t-nT),\phi_l(t-mT)> = \delta_{(m-n)\times(k-l)}$ ?

I mean I expect something equivalent to the two properties (infinite number of element and basis functions are not time-limited) of orthonormal set in the linear modulation case.

• Question 2 : is it correct if I say the noise samples are uncorrelated because the projection are $<n(t),\phi_k(t-nT)>$ and $<n(t),\phi_l(t-mT)>$ and $<\phi_k(t-nT),\phi_l(t-mT)> = \delta_{(m-n)\times(k-l)}$ ?

• Question 3 : The $M$ basis functions are interpreted as constellation of $M$ points. Is there any analogy with the linear modulation case ? If we consider M-QAM, we have also a constellation of $M$ complex points but if I write $a_n = a_{rn} + ja_{in}$, the two functions $p(t-nT)$ and $jp(t-nT) = e^{j\pi/2}p(t-nT)$ are not orthogonal.

• I dont understand why you conclude that the basis functions are not time-limited? Why wouldn't it work with time-limited functions? (Hint: Rectangular functions forms an orthogonal set). Apr 25, 2017 at 11:58
• @MaximilianMatthé you may refer to my question 1 ? I used the wrong words, I meant "basis functions can be not time-limited" as in the conclusion 2 in the linear modulation part. I meant if they are time limited, their time-shifted versions are orthogonal by the time-limited property. I just want to understand in the non linear modulation, what is the orthonormal set (question 2) ? Because in all books, they mention only the function defined orthogonal from 0 to $T$. Apr 25, 2017 at 12:29
• @MaximilianMatthé If we get back to your hint, rectangular functions forms an orthogonal set, but it is not the rectangular but the time-shifted versions of the rectangular function, and the orthogonal property must hold on the entier time domain, i.e. the rectangular functions that are able to form an orthonormal sets are non-zero in a duration of $T$ and zero outside and they are not time-limited. Do I understand it correctly ? Apr 25, 2017 at 12:29
• Let me rewrite it properly : a rectangular functions are time-limited in the sense of energy, but when they are used to form an orthogonal set, they are not time-limited in the sense the functions must be defined $\phi_n(t) = 1$ in $nT \leq t \leq (n+1)T$ and zero otherwise so that the orthogonality holds on the entier time axis. Apr 25, 2017 at 12:56
• Karhunan Loeve Expansion
– user28715
Jun 18, 2017 at 22:33

Leave time dimension aside for a moment and deal with just one symbol at a time. In a PAM modulation the generated functions are not ortho and not normal. Actually, in a 4-PAM modulation you have four basis pulses: $p_0(t)=a_0 \cdot p(t)$, $p_1(t)=a_1 \cdot p(t)$, $p_2(t)=a_2 \cdot p(t)$ and $p_3(t)=a_3 \cdot p(t)$. The pulse $p(t)$ is explicitly defined just because it makes the math easier and coherent: it is a function that generates a 1-dimensional Hilbert subspace.

The task is about mapping data symbols to signals in the Hilbert space, in such a way that at the receiver side the mapping can be reverted, thus, such a mapping must be bijective. On the receiver side, we must compute some kind of metric over the symbol period to come back to the discrete domain. For a Hilbert space, such metric is the inner product. So, in both cases (FSK and PAM) you end up performing, one way or another, the same computation:

$$a = \int_T f(t) g(t) dt$$

And this inner product is performed against each orthogonal basis function (one or more). By choosing more than one orthogonal basis functions, you get more parallel channels (more dimensions) to transmit onto if you wish. Then, choose how to encode the data:

1. BFSK encodes data by signaling one of those channels or the other. So for one of them, $a \neq 0$ (gets mapped to symbol $s_1$) and for the other (assuming orthogonality) $a = 0$ (mapped to symbol $s_0$). If both channels get signaled, you cannot reverse the mapping.

2. 2-PAM encodes data by signaling, within the same channel, different levels. So that if $a \gt threshold$, symbol $s_1$ is decided. If two levels are sent, you cannot reverse the mapping.

3. M-QAM uses both methods: different levels and different channels within each symbol period.

In short, data can be encoded in the value of $a$ within the same channel (same dimension, like a scaled basis function) so you further have to define decision boundaries for mapping $a$ to discrete symbols, or can be encoded by which channel (or dimension the orthogonal basis function generates) you signal. This resembles vector-space math because Hilbert space is the space of continuous vectors.

So, for the questions:

Q1. Yes. Harmonic sines within the fundamental period are orthogonal under the inner product. Orthonormal just means the norm for each basis function equals 1.

Q2. No. When it is said that noise is uncorrelated it refers to the fact that AWGN has no memory (time dimension), the noise is already uncorrelated before projection onto any basis. Within a symbol period the noise is correlated, its value is its power over that period.

Q3. When the constellation is said to have M points, it means M points get mapped to M different functions in the chosen Hilbert sub-space. The way you choose those functions to be is not defined. You can choose a M-dimensional space (so you only need to find the orthogonal basis function which fits better) or a N-dimensional space with $M \gt N$ and define different regions in that space for symbol mapping. Also, in QAM, $p_r(t) = cos(\omega t) \ne sin(\omega t) = p_i(t)$, and the inner product is zero within one period.

I avoided to talk about the time dimension (when possible) because IMHO it would not help. Unless you are trying to exploit temporal correlation (viterbi, equalizers, noise shaping, etc...) it does not help and only introduces noise in the explanation. Usually, this kind of analysis is performed on a symbol based period. However, according to Fourier analysis, every function in the Hilbert space can be expressed in terms of amplitude, frequency and phase, and a shift in time is a shift in phase, thus the same logic applies. In real implementations you do not find neither infinite signal levels ($a$'s), since power is constrained, nor frequencies (basis function's shape) which are constrained by bandwidth. Time shift (phase) is not infinite neither since you need to start and stop sometime.