Chebyshev pole calculation I am designing a Chebyshev filter and I am required to calculate the poles of it. It requires $1\textrm{ dB}$ pass-band ripple, a ripple factor of $0.5088$ and an order of $3$.

The equations for calculating the real and imaginary poles are shown below:

\begin{align}
s_k &amp;= \sigma_k + j\omega_k\\
\sigma_k &amp;= \sin\left(\frac{(2k-1)\pi}{2n}\right)\sinh(v)\\
\omega_k &amp;= \cos\left(\frac{(2k-1)\pi}{2n}\right)\cosh(v)
\end{align}

However, when I calculate the imaginary part I get the following values:

\begin{align}
\sigma_1 &amp;= -0.96\\
\sigma_2 &amp;= 0\\
\sigma_3 &amp;= +0.96
\end{align}

However, according to my solutions I should get $+0.96$ and $-0.96$. This checks out with the values from this link.

Is there some sort of rule to invert the values?
As I've mentioned in a comment, the poles are not "ordered" in any way. You can come up with different versions of the formula for the pole locations, which will give you the same poles in different orders.

But anyway, looking at the formulae in your question, the first thing to notice is that there should be a negative sign in the formula for the real parts $\sigma_k$. Second, the values that you got must be the imaginary parts $\omega_k$ (and not $\sigma_k$ as written in your question). If I evaluate the formula for $\omega_k$ given in your question for $k=1,2,3$ I get the following values (for $n=3$ and $1$ dB pass band ripple):
$$\begin{align}\omega_1&amp;=0.966i\\
\omega_2&amp;=0\\
\omega_3&amp;=-0.966i\end{align}$$

So I get a positive sign for $k=1$ and a negative sign for $k=3$.