Assume I have a matrix $D$ whose its entries are as below :
Where $A$ and $B$ can be written using using the trigonometric functions for (1) as:
My question, Is it possible to simplify (1) more? how can we do that ?
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4 Answers
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Observe:
$$ \cos( \alpha ) e^{j \beta} = \frac{e^{j \alpha}+e^{-j\alpha}}{2} \cdot e^{j \beta} =\frac{1}{2} \left( e^{j (\beta + \alpha)}+e^{j(\beta - \alpha)} \right) $$
Where:
$$ \alpha = (k-1)(2n-1) \cdot \frac{\pi}{2N} = \frac{ 2nk - 2n - k + 1}{4} \cdot \frac{2\pi}{N} $$
$$ \beta = nk \cdot \frac{2\pi}{N} $$
Plug this in and you can transform jbondu's equation into the sum of two pure complex exponentials. That might be considered a simplification. This also allows for the definition of your matrix as the sum of two matrices.
answered May 25, 2020 at 21:51
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If you use the Euler's formula, you can simplify like this: $$ [d]_{k,n} = \frac{\sqrt{2}}{N}\left( \cos{\left[ \frac{(k-1)(2n-1)\pi}{2N} \right]} e^{j\frac{2 \pi nk}{N}} \right) $$
I think we can't simplify more.
PS: If you use your expressions of $A$ and $B$ and again the Euler's formula, you will get the same result.
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The cosine and sine terms can be combined to a single exponential function using:
$$ \cos(\theta) + j\sin(\theta) = e^{j\theta}$$
answered May 25, 2020 at 13:32
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$\begingroup$ I see .. but it won't be simplified, just writing it in exponential term. $\endgroup$– GzeMay 25, 2020 at 14:05
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$\begingroup$ You have two distinct frequencies so to that extent they are in simplest form other than doing this $\endgroup$ May 25, 2020 at 14:05
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You can use, from Euler's formula, the fact that $$ \exp\left(j\theta\right) = \cos\theta + j\sin\theta \qquad \text{with}\qquad j = \sqrt{-1} $$ and combine the sine and cosine into exponential terms. You basically have the first exponential inside the square bracket equal to $$ \exp\big[j\left(6\pi nk + \pi -k\pi - 2\pi n \right)\big] = \exp\bigg\{j\big[k\pi \left(6n - 1\right) - \pi\left(2n - 1\right) \big]\bigg\} $$ Which can be simplified further as follows: \begin{align} \exp\bigg\{j\big[k\pi \left(6n - 1\right) - \pi\left(2n - 1\right) \big]\bigg\} &= \exp\bigg\{j\big[k\pi \left(6n - 1\right)\big]\bigg\}\cdot\exp\bigg\{-j\big[\pi\left(2n - 1\right) \big]\bigg\} \\ &= \left(-1\right)^k\cdot \left(-1\right)\ \qquad \forall \ k\in \mathbb Z\ \text{and}\ \forall n\in \mathbb Z\\ & = \left(-1\right)^{k+1} \end{align} You can simplify the second exponential and proceed in the same way.
N.B. The factor $\frac 12$ has been taken out of the square brackets in (1) to get the factor $\frac 1{N\sqrt{2}}$ outside.
EDIT: I overlooked the $N$ factor, not really the general case. so I assumed $N = 1$, my bad. But I leave this here anyway.
