0
192views
Find DTFT of $x(n)=\sin (n \theta) u(n)$
1 Answer
0
4views

Solution:

$ x(n)=\sin (\mathrm{n} \theta) \mathrm{u}(\mathrm{n})\\ $

$ X\left(e^{j \omega}\right)=\sum_{n=0}^{\infty} \sin (n \theta) e^{-\mathrm{j} \omega n}\\ $

$ =\sum_{n=0}^{\infty}\left(\frac{e^{j \theta n}-e^{-j \theta n}}{2 j}\right) e^{-\mathrm{j} \omega n}\\ $

$ =\frac{1}{2 j}\left(\sum_{n=0}^{\infty} \mathrm{e}^{\mathrm{j}(\theta-\omega) \mathrm{n}}-\sum_{n=0}^{\infty} \mathrm{e}^{-\mathrm{j}(\theta+\omega) \mathrm{n}}\right)\\ $

$ =\frac{1}{2 j}\left(\frac{1}{1-\mathrm{e}^{\mathrm{j}(\theta-\omega)}}-\frac{1}{1-\mathrm{e}^{-\mathrm{j}(\theta+\omega)}}\right)\\ $

$ =\frac{1}{2 j}\left(\frac{1-\mathrm{e}^{-\mathrm{j}(\theta+\omega)}-1+\mathrm{e}^{\mathrm{j}(\theta-\omega)}}{1-2 \mathrm{e}^{-\mathrm{j} \omega} \cos \theta+e^{-2 j \omega}}\right)\\ $

$ =\frac{1}{2 j}\left(\frac{2 \mathrm{je}^{-\mathrm{j} \omega} \sin \theta}{1-2 \mathrm{e}^{-\mathrm{j} \omega} \cos \theta+e^{-2 j \omega}}\right)\\ $

$ =\frac{\mathrm{e}^{-\mathrm{j} \omega} \sin \theta}{1-2 \mathrm{e}^{-\mathrm{j} \omega} \cos \theta+e^{-2 j \omega}}\\ $

Please log in to add an answer.