Consider a discrete-time sequence, $y[n]$, defined as:

$$y[n] = \frac 12 x[n] + \frac 12 (-1)^n\: x[n]$$

where $x[n]$ is another discrete-time sequence.

The DTFT of $y[n]$ defined as $Y(e^{j\omega}) = \sum_{n=-\infty}^{\infty} y[n] e^{-j\omega n}$ can be easily written in terms of the DTFT of $x[n]$ as:

$$Y(e^{j\omega}) = \frac 12 X(e^{j\omega}) + \frac 12 X(e^{j(\omega + \pi)})$$

using the fact that $(-1)^n = e^{j\pi n}$.

Now let's consider a third sequence, $z[n] = y[2n]$. We can again write $Z(e^{j\omega})$ in terms of $X(e^{j\omega})$ by using the fact that:

$$Z(e^{j\omega}) = \frac 12 \left[ \underbrace{Y(e^{j\omega/2})}_{(i)} + \underbrace{Y(e^{j(\omega-2\pi)/2}}_{(ii)}\right]$$ which gives us:

$$Z(e^{j\omega}) = \frac 14 \underbrace{X(e^{j\omega/2}) + \frac 14 X(e^{j(\omega/2+\pi)})}_{\text{from } (i)} + \frac 14 \underbrace{X(e^{j(\omega-2\pi)/2}) + \frac 14 X(e^{j((\omega-2\pi)/2+\pi)})}_{\text{from } (ii)} \quad \tag{1}$$

However, let's look at what the sequence $y[n]$ actually is.

The sequence $y[n]$ is simply:

$$y[n] = \cdots,0, x[0],0, x[2],0,x[4],,0,x[6],0,\cdots $$

so it is essentially the even indices of $x[n]$. If I decimate $y[n]$ by $2$ I am merely removing the zeroes to get:

$$z[n] = \cdots, x[0], x[2] , x[4], x[6], \cdots$$

which can be expressed by this diagram:


which essentially means that $z[n]$ is just $x[n]$ downsampled by $2$, i.e. $z[n] = x[2n]$. Which means that $Z(e^{j\omega})$ should be:

$$Z(e^{j\omega}) = \frac 12 [X(e^{j\omega/2}) + X(e^{j(\omega -2\pi)/2}] \quad \tag{2}$$

I can verify this in the frequency domain by make $(1)$ equal to $(2)$ as follows:

$$(1) = \frac 14 X(e^{j\omega/2}) + \frac 14 \underbrace{X(e^{j(\omega/2+\pi)})}_{(a)} + \frac 14 X(e^{j(\omega-2\pi)/2}) + \frac 14 \underbrace{X(e^{j((\omega-2\pi)/2+\pi)})}_{(b)}$$

where $(a) = X(e^{j(\omega+2\pi)/2}) = X(e^{j(\omega-2\pi)/2})$ and $(b) = X(e^{j((\omega-2\pi)/2+2\pi/2)}) = X(e^{j\omega/2}) $ which makes $(1) = (2)$.

This means that $z[n] = x[2n] = y[2n]$ and $y[2n] = \frac 12 x[2n] + \frac 12 (-1)^{2n}x[2n] = x[2n]$. So all of the relations in both domains are satisfied but is the explanation that $(a) = X(e^{j(\omega+2\pi)/2}) = X(e^{j(\omega-2\pi)/2})$ true. I used the fact that the DTFT is always $2\pi$ periodic but I am not sure if that is actually satisfied in $(a)$.

  • $\begingroup$ Why are you suspicious that it may not be satisfied? $\endgroup$ Sep 10 at 15:34


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