2
$\begingroup$

Let $n$ be even and consider the non-normalized discrete Sine transform of type 5 which is

$$S=\left(\sin(k+1)(l+1)\frac{\pi}{n+\frac12}\right)_{k,l=0}^{n-1}$$

Let us denote $s_{-,l}$ by the $l^{th}$-column of $S$. It can be considered as a $n$-tuple in $\mathbb{R}^n$.

Q. I am looking for $n$-tuples $v=(v_0,\cdots,v_{n-1})$ in $\mathbb{R}^n$ satisfying the following conditions:

1- $v_j=\sin2(j+1)\frac{\pi}{n+\frac12}$ if $j$ is even.

2- The following are valid concerning inner products: $$\langle v , s_{-,l} \rangle=\left\{ \begin{array}{cl} 1 & l=0 \\ 0 & l\neq 0~,~ l \operatorname{is even} \end{array} \right.$$

$\endgroup$
1
  • $\begingroup$ is this a spectral leakage question? $\endgroup$ Nov 27, 2022 at 10:28

2 Answers 2

3
$\begingroup$

Let's restate your problem as a linear equation

Let $U_{ee}$, $U_{eo}$, $U_{oe}$, $U_{oo}$ be the partition of the $S$ matrix in even-odd row-column index, similarly $v_e$ and $v_o$ are the partitions of $v$.

Restating your conditions in terms of matrices

1 - $v_e$ is fixed

2 - $v_e U_{ee} + v_o U_{oe} = I_1$, where $I_1$ denotes the first row of the identity matrix.

Placing all the constant terms on the right side of the condition 2, $v_o U_{oe} = (I_1 - v_e U_{ee})$, transposing we get a linear equation system in the standard form $U_{oe}^{T} v_o = (I_1 - v_e U_{ee})^T$

Proof that $U_{oe}$ has no pair of linearly dependent columns

Assuming

$s_{2i+1, 2j} = \sin\left( \frac{2\pi (2i+2)(2j+1)}{2n+1} \right)$

Two columns will be linearly dependent if, and only if, there are two integers $0 \le j_1, j_2 \le n/2-1$, for all integer $0 \le i \le n/2-1$

$$(2i+2)(2j_1 + 1) \equiv (2i+2)(2j_2 + 1) \operatorname{mod} (2n+1), $$

$$(2j_1 + 1) \equiv (2j_2 + 1) \operatorname{mod} (2n+1)$$

$$j_1 \equiv j_2 \operatorname{mod} (2n+1)$$

since both $j_1$ and $j_2$ are positive integers, smaller than $2n+1$, the condition can only be satisfied with $j_1 = j_2$, thus the matrix $U_{oe}$ is non-singular.

$\endgroup$
12
  • $\begingroup$ That is a nice approach, but seems there is only a gap! How can we sure that the entries in the diagonal of the upper triangular matrix $R$ are all non-zero? $\endgroup$
    – ABB
    Nov 23, 2022 at 8:38
  • $\begingroup$ If the rows of the input matrix are linearly independent, the diagonal elements are non-zero. I will add more details. $\endgroup$
    – Bob
    Nov 23, 2022 at 10:23
  • $\begingroup$ Yes, and so the proof linear independency will be equivalent to find the vector $q$. $\endgroup$
    – ABB
    Nov 23, 2022 at 10:25
  • $\begingroup$ To be honest I couldn't unambiguously parse your equation, do you mean $$s_{k,l} = \sin\left(\frac{(k+1)(l+1)\pi}{n+1}\right)$$ $\endgroup$
    – Bob
    Nov 23, 2022 at 10:49
  • $\begingroup$ I expressed in terms of a linear equation, the necessary and sufficient condition for a solution to exist is that $s_{k,l}$ for odd $k$ and even $l$ must be non-singular. Bear with me, and double check if I didn't mess with the indices ;) $\endgroup$
    – Bob
    Nov 23, 2022 at 11:20
0
$\begingroup$

1.- for any other reader not familiar with DST type 5 or DST-V, I'd like to include the definition of Discrete Sine Transform and types.

Following a list with DST types 1 to 5, from

https://planetmath.org/discretesinetransform

enter image description here

1D DST types 1st to 8th

2.- Bob's assumption is correct. The question expression

enter image description here

actually means

enter image description here

and I'd like to add that the question expression for DST-V

enter image description here

is this

enter image description here

3.- DST Discrete Sine Transform and the counterpart DCT Discrete Cosine Transform are just partials of the DFT Discrete Fourier Transform.

Like Wolfram concisely explains here

https://mathworld.wolfram.com/DiscreteFourierTransform.html

and here

https://mathworld.wolfram.com/Leakage.html

Unless the signal or signal fragment is periodic in the watched interval, there's always (spectral) leakage.

4.- So, if the unknown input signal were all ones, satisfying the 2 conditions in the question would go the Bob answered.

5.- Yet it is up to x(:)*sin((k+1)*(L+1)*pi/(N+1)) not just the term sin((k+1)*(L+1)*pi/(N+1) to meet the non-leakage condition mentioned in point 2.

And since x is not supplied in the question in any shape or manner. there's no way to tell without knowing x .

5.- Is this a MATLAB question anyway?

I politely suggest this question to be moved to a Stack Overflow Maths page, or to for instance signal analysis or sampling tags in this same Signal Processing page

Thanks for reading.

$\endgroup$

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.