# Is this proof valid?

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:

$$y[n] = \cdots,0, x,0, x,0,x,,0,x,0,\cdots$$

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:

$$z[n] = \cdots, x, x , x, x, \cdots$$

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 $$(a)$$.

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