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Consider an $N$-length vector $\mathbf{u} = [1, e^{j\theta}, e^{2j\theta}, \ldots, e^{(N-1)j\theta}]$. Let's say I represent this vector as the Kronecker product of two vectors $\mathbf{u}_1$ and $\mathbf{u}_2$. Here,

$$ \mathbf{u}_1 = [1, e^{Qj\theta}, e^{2Qj\theta}, \ldots, e^{\left\lfloor \frac{N-1}{Q} \right\rfloor Qj\theta}] $$

and

$$ \mathbf{u}_2 = [1, e^{j\theta}, \ldots, e^{(Q-1)j\theta}] $$

It can be easily noted that $\mathbf{u} = \mathbf{u}_1 \otimes \mathbf{u}_2$. My aim is to compute the DFT of $\mathbf{u}$ correctly using the DFT on the Kronecker product of $\mathbf{u}_1$ and $\mathbf{u}_2$. In the first single $\theta$ case, I can perform this without any error:

$$ \mathbf{u}(n) = (\mathbf{u}_1 \otimes \mathbf{u}_2)(n) = \mathbf{u}_1 \left( \left\lfloor \frac{n}{Q} \right\rfloor \right) \times \mathbf{u}_2 (n \% Q) $$

Now, I am working on an $N$-length vector with two$\theta$ values : $\mathbf{v} = [1, \alpha_1 e^{j\theta_1} + \alpha_2 e^{j\theta_2}, \alpha_1 e^{2j\theta_1} + \alpha_2 e^{2j\theta_2}, \ldots, \alpha_1 e^{(N-1)j\theta_1} + \alpha_2 e^{(N-1)j\theta_2}]$.

If I consider

$$ \mathbf{v}_1 = [1, \alpha_1 e^{Qj\theta_1} + \alpha_2 e^{Qj\theta_2}, \alpha_1 e^{2Qj\theta_1} + \alpha_2 e^{2Qj\theta_2}, \ldots, \alpha_1 e^{\left\lfloor \frac{N-1}{Q} \right\rfloor Qj\theta_1} + \alpha_2 e^{\left\lfloor \frac{N-1}{Q} \right\rfloor Qj\theta_2}] $$

and

$$ \mathbf{v}_2 = [1, \alpha_1 e^{j\theta_1} + \alpha_2 e^{j\theta_2}, \ldots, \alpha_1 e^{(Q-1)j\theta_1} + \alpha_2 e^{(Q-1)j\theta_2}] $$

Here, the Kronecker product seems to be

$$ \tilde{\mathbf{v}}(n) = \mathbf{v}_1 \otimes \mathbf{v}_2 = \mathbf{v}_1 \left( \left\lfloor \frac{n}{Q} \right\rfloor \right) \times \mathbf{v}_2 (n \% Q) = \left( \alpha_1 e^{\left( \left\lfloor \frac{n}{Q} \right\rfloor - 1 \right) Qj\theta_1} + \alpha_2 e^{\left( \left\lfloor \frac{n}{Q} \right\rfloor - 1 \right) Qj\theta_2} \right) \times \left( \alpha_1 e^{\left( n \% Q - 1 \right) j\theta_1} + \alpha_2 e^{\left( n \% Q - 1 \right) j\theta_2} \right) $$

But here it seems to have cross terms. I want to estimate the $N$-point DFT of the original $\mathbf{v}$ vector using the DFT on the Kronecker product of $\mathbf{v}_1$ and $\mathbf{v}_2$. I know that if I plot the DFT spectrum there will be 2 peaks corresponding to $\theta_1$ and $\theta_2$. These are as per the original spectral components in $\mathbf{v}$. But there will be additional interference peak components due to the cross terms.

  1. How can I identify the locations (DFT bin indices) of these interference terms so that I can remove them?

Note: I tried to plot the spectrum and I saw 2 interference peaks and 2 expected peaks. The interference peaks seams to be the same distance away from the actual expected peaks. Hence, I am thinking there might be a way to characterize their exact location

Any help is appreciated.

Notes:

$a \% b$ is the remainder when a is divided by b. For example $5 \% 2 =1$. Similarly, $\lfloor{a}\rfloor $ is the nearest integer less than or equal to $a$.

wanderer
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  • I find your notation difficult to follow. I am probably not the only one. Please try to be more clear. – AnonSubmitter85 Aug 06 '24 at 19:39
  • What does the % symbol mean? – Andreas Aug 08 '24 at 20:32
  • @Andreas % symbol means remainder operator and $\lfloor{x} \rfloor$ means nearest integer less than or equal to x. – wanderer Aug 09 '24 at 03:39
  • I'm not sure I can follow you even in the first case. You state $\mathbf{u} = \mathbf{u}_1 \otimes \mathbf{u}_2$. However, following the usual understanding of Kronecker products of vectors (e.g. https://math.stackexchange.com/questions/3981122/), we have $\mathbf{u}_1 \otimes \mathbf{u}_2 = [1, e^{j\theta}, e^{2 j\theta}, \ldots, e^{[ \left\lfloor \frac{N-1}{Q} \right\rfloor Q + (Q-1) ] j\theta}]$. But for many cases of $Q,N$ we have $\left\lfloor \frac{N-1}{Q} \right\rfloor Q + (Q-1) \ne N-1$, so the result is not $\mathbf{u}$. Can you explain? – Andreas Aug 12 '24 at 14:10

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