Q show that 1/(1-e^jx) = 1/2(1+jcot(x/2)) where x=theta,sorry bwt x
the first line of woking is to answer ( e^-jx/2 ) / ( (e^-jx/2 - e^jx/2 )
but how does 1 = e^-jx/2
could someone show me how to get ans pls???
Q show that 1/(1-e^jx) = 1/2(1+jcot(x/2)) where x=theta,sorry bwt x
the first line of woking is to answer ( e^-jx/2 ) / ( (e^-jx/2 - e^jx/2 )
but how does 1 = e^-jx/2
could someone show me how to get ans pls???
Two handy identites to know are:
$\displaystyle sin{\theta}=\frac{e^{i{\theta}}-e^{-i{\theta}}}{2i}$
and $\displaystyle cos{\theta}=\frac{e^{i{\theta}}+e^{-i{\theta}}}{2}$
$\displaystyle e^{i{\theta}}=cos{\theta}+isin{\theta}$
$\displaystyle e^{-i{\theta}}=cos{\theta}-isin{\theta}$
Another one handy to know is the identity $\displaystyle tan(\frac{\theta}{2})=\frac{1-cos{\theta}}{sin{\theta}}$
So, you have $\displaystyle \frac{1}{1-e^{-i{\theta}}}=\frac{1}{1-cos{\theta}+isin{\theta}}$