How was this sum simplified to a ratio of Sinh functions?

Question

I came across this simplification in a paper and I do not understand how it was done.

$$\sum_{m=0}^{K-1} \frac{e^{-mT/Kt}(1-e^{-T/Kt})e^{jnw(m+0.5)T/K}} {(1-e^{-T/t})} = \frac{sinh( \frac{T}{2Kt})}{sinh(\frac{T(1-jnwt)}{2Kt})}$$

where $j$ is the imaginary unit, and $w=2\pi/T$

My attempts so far have been to expand the right hand side and get it to match the left hand side using the identity $$sinh(x) = \frac{1-e^{-2x}}{2e^{-x}}$$

However, I cannot find a way to get rid of the summation over $m$. Ultimately, the LHS simplified to something of the form: $$A \sum_{m=0}^{K-1} e^{-mT/Kt}e^{jn(m+0.5)2\pi/K}$$

Is it possible to simplify out the summation?

Answer 1

The sum in your final line is a geometric progression: $$\sum_{m=0}^{K-1} e^{-mT/Kt}e^{jn(m+0.5)2\pi/K} =\sum_{m=0}^{K-1} e^{jn\pi/K}\bigl(e^{-T/Kt+jn2\pi/K}\bigr)^m$$ and the sum is $$e^{jn\pi/K}\frac{1-e^{-T/t+jn2\pi}}{1-e^{-T/Kt+jn2\pi/K}}\ .$$ If $n$ is an integer (you didn't actually say so) then this is $$\eqalign{e^{jn\pi/K}\frac{1-e^{-T/t}}{1-e^{-T/Kt+jn2\pi/K}} &=e^{T/2Kt}\frac{1-e^{-T/t}}{e^{T/2Kt-jn\pi/K}-e^{-T/2Kt+jn\pi/K}}\cr &=e^{T/2Kt}\frac{1-e^{-T/t}}{2\sinh(T/2Kt-jn\pi/K)}\cr &=e^{T/2Kt}\frac{1-e^{-T/t}}{2\sinh(T(1-jnwt)/2Kt)}\ .\cr}$$ This is fairly close to what you want, hopefully your constant $A$ will do the rest. Good luck!

Answer 2

You can break the indexed exponentials in the numerator into three, two of which involve the summation index. Factor out everything not involving the summation index. Then you have something of the form $$A(T,K,t,n)\sum_{m=0}^{K-1}{e^{\alpha m}},$$ where $\alpha=-\frac{T}{Kt}+\frac{2πnj}{K}.$

Now the sum is just a geometric series with $K$ terms, first term being $1$ and ratio being $e^{\alpha}.$ Thus this sum works out to $$\frac{1-e^{\alpha K}}{1-e^{\alpha}}.$$

You can make the appropriate simplifications and substitutions.