# Question about time shifting property

I know that time shifting property of discrete-time Fourier Transform states that

$x[n]\leftrightarrow X(e^{j\omega})\implies x[n-n_0]\leftrightarrow e^{-j\omega n_0}X(e^{j\omega})$

So given a signal $\;y[n]=u[n]-u[n-5]\;,\;$What is $\;\mathfrak{F}\{ y_{(2)}[n-1]\}\:?$

$y_{(2)}[n]\;$ denotes the slowed-down version of $\:y[n].\;$And $\:y[n] \leftrightarrow e^{-j2\omega}\frac{\sin(5\omega/2)}{\sin(\omega/2)}$

From time expansion property $\:y_{(2)}[n]\leftrightarrow e^{-j4\omega}\frac{\sin(5\omega)}{sin(\omega)}\:$then follows the time shifting property

$\mathfrak{F}\{ y_{(2)}[n-1]\}=e^{-j6w}\frac{\sin(5\omega)}{\sin(\omega)}$

But the answer should be $\:e^{-j5\omega}\frac{\sin(5w)}{\sin(\omega)}$

What is wrong with my understanding?

• Could you specify in mathematical terms what you mean by $y_{(2)}[n]$? – Matt L. May 13 '18 at 12:43
• I am using the notation in the Oppenheim's book. It is $\; y_{(2)}[n]=y[n/2]\;$if n is a multiple of 2 and zero otherwise. – Tokugava May 13 '18 at 13:39

$y_{(2)}[n] = y[n/2]$ then $Y_{(2)}(j\omega) = 2 Y(j2\omega) = 2e^{-j4\omega}\frac{\sin(5\omega)}{\sin(\omega)}$.
$z[n] = y_{(2)}[n - 1]$ then $Z(j\omega) = e^{-j\omega}Y_{(2)}(j\omega) = e^{-j\omega}2e^{-j4\omega}\frac{\sin(5\omega)}{\sin(\omega)} = 2e^{-j5\omega}\frac{\sin(5\omega)}{\sin(\omega)}$.
• But isn't the new frequency for $\;y_{(2)}[n]\;$changed? – Tokugava May 13 '18 at 13:41
• What I mean is, the frequency of $\;y_{(2)}[n]\;$ isn't the same as the frequency of $\;y[n]\;$ But shifting property states we should use the frequency of the shifted function. – Tokugava May 13 '18 at 14:25
• why not? $Y_{(2)}(j\omega) = e^{-j4\omega}\frac{\sin(5\omega)}{\sin(\omega)}$ – Cath Maillon May 13 '18 at 14:52