Consider the signal x(n) with DTFT X(ejω). Let the signal y(n) be defined as;
\(y\left( n \right) = \left\{ {\begin{array}{*{20}{c}} {x\left( n \right)\;for\;n = 2k}\\ {0\;for\;n = 2k + 1} \end{array}} \right.\)
The DTFT of y(n) is
1
\(\frac{1}{2}X\left( {{e^{j\omega }}} \right) + \frac{1}{2}X\left( {{e^{j\left( {\omega + \pi } \right)}}} \right)\)
2
X(ej(ω +π))
3
\(\left( {{e^{j\frac{\omega }{2}}}} \right)\)
4
X(ej2ω)