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?

  • $\begingroup$ Could you specify in mathematical terms what you mean by $y_{(2)}[n]$? $\endgroup$ – Matt L. May 13 '18 at 12:43
  • $\begingroup$ 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. $\endgroup$ – 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 $Z(j\omega) = e^{-j\omega}Y_{(2)}(j\omega) = e^{-j\omega}2e^{-j4\omega}\frac{\sin(5\omega)}{\sin(\omega)} = 2e^{-j5\omega}\frac{\sin(5\omega)}{\sin(\omega)}$.

  • $\begingroup$ But isn't the new frequency for $\;y_{(2)}[n]\;$changed? $\endgroup$ – Tokugava May 13 '18 at 13:41
  • $\begingroup$ @Tokugava frequency is frequency, it does not change, only the function changes. 1Hz is 1Hz, 10 radians/second is always 10 radians/second. $\endgroup$ – Cath Maillon May 13 '18 at 14:21
  • $\begingroup$ 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. $\endgroup$ – Tokugava May 13 '18 at 14:25
  • $\begingroup$ @Tokugava is this not what I did? $\endgroup$ – Cath Maillon May 13 '18 at 14:44
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    $\begingroup$ why not? $Y_{(2)}(j\omega) = e^{-j4\omega}\frac{\sin(5\omega)}{\sin(\omega)}$ $\endgroup$ – Cath Maillon May 13 '18 at 14:52

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