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8

As you may know, orthogonality depends on the inner product of your vector space. In your question you state that: While sine and cosine are orthogonal functions... This means that you have probably heard of the "standard" inner product for function spaces: $$\langle f,g\rangle=\int\limits_{x_1}^{x_2} f(x)g(x) \ \mathrm{d}x$$ If you solve this integral ...


5

Orthogonality is indeed defined via an inner product, with an integral for a continuous ordinal time variable, with a sum for a discrete time variable. When you convert two (continuous) orthogonal signals into discrete ones (regular sampling, discrete amplitudes), possibly windowed (finite support), you can affect the orthogonality. In other words: two ...


5

No, that isn't possible. Think about it in the geometric sense, where each bit in the sequence is a dimension. In that context, you can recast your question as: In an $n$-dimensional space, are there any sets of $M$ vectors (where $M > n$) that are mutually orthogonal? For example, in three-dimensional space, can you construct a set of four vectors ...


5

Just use the formula for the geometric series (I use $l=h-k\neq mN$): $$\sum_{n=0}^{N-1}e^{-j\frac{2\pi}{N}nl}=\frac{1-e^{-j\frac{2\pi}{N}Nl}}{1-e^{-j\frac{2\pi}{N}l}}=\frac{1-e^{-j2\pi l}}{1-e^{-j\frac{2\pi}{N}l}}=\frac{1-1}{1-e^{-j\frac{2\pi}{N}l}}=0,\quad l\neq mN$$


4

Say you have two pulses, $p(t)$ and $q(t)$, such that the following three statments are true: \begin{align} \int_{-\infty}^\infty p(t) p(t) dt &= 1,\\ \int_{-\infty}^\infty q(t) q(t) dt &= 1,\\ \int_{-\infty}^\infty p(t) q(t) dt &= 0. \end{align} Then these pulses are said to be orthonormal. Now, let us say you want to transmit two information-...


3

I don't think you cannot state generally that $y_1$ and $y_2$ will be orthogonal. I'll try to sketch out my thinking: Since the Hamming code is a linear code, each parity bit can be represented as a linear combination of the information bits; that is, each bit can be represented as: $$ p_i = \mathbf{g_i^T}\mathbf{x} $$ where $\mathbf{x}$ is the ...


3

Sinusoids are power signals, and in the case of power signals, cross-correlation between two signals $x(t)$ and $y(t)$ is defined by $$R_{xy}(t)=\lim_{T\rightarrow\infty}\frac{1}{2T}\int_{-T}^{T}x(\tau)y(t+\tau)d\tau\tag{1}$$ If $x(t)=\cos(\omega_1 t)$ and $y(t)=\cos(\omega_2 t)$ then for $\omega_1\neq\omega_2$ the cross-correlation at $t=0$ is $$\begin{...


2

You first have to define an inner product for functions. You can't just multiply with each other. I am not sure about the properties of inner product myself, but according to this lecture an inner product has to be commutative, linear and the inner product of a function with itself should be positive definite. One option for an inner product for functions ...


2

I think I can answer the question after reading the article "The empirical mode decomposition and the Hilbert spectrum for nonlinear and non-stationary time series analysis" by Huang. In this paper (Page 927), Huang gave the defination of the orthogonality between two signals: And also, I'd like to share with you my MATLAB code: function OC=ort(x,y) x=x(:)'...


2

Signals that are plus or minus constants on unit intervals (here $\pm 2$) are great example, because they help you understand some of the concepts of orthogonality, like a combination of amplitude and support, and you can almost evaluate it visually, without much computations. Plus, you don't have to deal with complex conjugates (as Matt did). First, ...


2

I would draw a new function which is the multiplication of both functions at each point and afterwards compute the "area"(:= Integral) under the graph. This process describes the formula in an intuitive way.


2

The inner product of two continuous-time signals is defined by $$\langle\mathbf{s}_1,\mathbf{s}_2\rangle=\int_{-\infty}^{\infty}s_1(t)s^*_2(t)dt\tag{1}$$ where $*$ denotes complex conjugation. In your example the conjugation is irrelevant because both signals are real-valued. So you just need to multiply the signals and determine the area of the result. ...


2

The disadvantage is that the signal V21 = V1 + V2 = [2, 0, 2, 0] disappears every other chip interval which makes maintaining carrier and phase synchronization more difficult. Also, while the total energy in V21 is equals to the sum of the energies of V1 and V2, V21 uses 4 times the instantaneous power needed by either V1 or V2 and so the transmitter needs ...


2

If you have two sinusoidal length $N$ sequences $$x_1[n]=\sin\left(n\omega_1+\phi_1\right),\qquad n\in[0,N-1]\\ x_2[n]=\sin\left(n\omega_2+\phi_2\right),\qquad n\in[0,N-1]\tag{1}$$ then for orthogonality we require $$\sum_{n=0}^{N-1}x_1[n]x_2[n]=0\tag{2}$$ For the given sequences $(1)$ we obtain $$\begin{align}\sum_{n=0}^{N-1}x_1[n]x_2[n]=&\sum_{n=...


1

Bear with me, as this is really only tangential¹ to my area of expertise, but: Maybe the question is not how to extend wavelets to work on non-orientable surfaces, but to interpret said surface as something where wavelets natively work well. Enter differential geometry (differential topology? I don't know.): As you've notice: if you only consider a part ...


1

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 ...


1

In terms of matrix multiplication (such as for a DFT), the equivalent interval of integration for signals is determined by the size of the matrix (or the size of the input vector) and the sample rate. These are often chosen due to practical considerations (time or space of interest and/or of availability, etc.). Orthogonality is defined over that interval ...


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