Can we exactly calculate the sum $S=\sum\limits_{n=1}^{N} e^{2ib\cos n\theta} e^{-in\theta}$?

Question

Consider the sum $$S=\sum\limits_{n=1}^{N} e^{2ib\cos n\theta} e^{-in\theta}$$ where $\theta=\frac{2\pi}{N}$, $b$ is any real number (not small), $N$ is an integer. Here, $n$ is also an integer which takes the values $1,2,...,N$. By expanding cosine as the sum of two exponentials we are faced with exponentiation of sum of exponentials which look formidable. Is there a way to exactly calculate this sum?