I am not sure about intuition in general, but regarding the step-function FT being a sync function:
Note that the shape will remain the same, but the frequencies over which the FT of a particular step-function resides is a function of the pulse-width of the original signal. Namely, expanding a function in the time-domain actually shrinks the corresponding frequency-domain function (think slowing down voice recordings, the sound gets very low i.e. lower frequency).
That being said, as you decrease the pulse-width of a particular step-function the frequency components of that signal increase because now there is more change happening (to use a loose descriptor) in a shorter amount of time.
In contrast, if we expand the step-function in the time-domain to have a longer pulse-width then there is less change and the corresponding frequency components must be much lower.
In general, I look at a function and try and get a feel for how quickly it might be changing to get a rough idea. But as I said, I don't know of any general rule of thumb here.
This is a root locus analysis with variable K.
The poles are:
$$
p_{1,2}=\frac{1}{2}(1 \pm (1 - 4c)^{1/2}), |p_2|<|p_1|
$$
For \$c<=1/4\$, the poles are real. \$p_{1}=1\$ when \$c = 0\$.
For \$c>1/4\$, the poles are complex, and their magnitude is:
$$
|p_{1,2}|=\sqrt{\frac{1}{4}(1 + (4c-1))}=\sqrt{c}
$$
which is equal to 1 when \$c=1\$.
Hence, for \$c\$, \$p_1\$ and \$p_2\$:
\$c=0: p_1=1\$ (critical), \$p_2=0\$ (stable)
\$c=1: p_{1,2}=1/2 \pm i\sqrt(3)/2\$ (underdamped stable,\$<1\$)
\$c=1/4: p_1=1/2, p_2=1/2\$ (overdamped stable,\$<1\$)
Because the time response is given, with negative delays, the system is casual, independently of \$c\$.
If only \$H(z)\$ were given and no time response, we obtain the explicit division, for inspect each converging term involving infinite sums:
$$
H(Z) = \frac{1}{p_1-p_2}(\frac{p_1}{1 - p_1 z^{-1}}-\frac{p_2}{1 - p_2 z^{-1}})
$$
Hence:
For a Causal System (no diverging components on the sum, for any \$c\$):
$$
|p_1 z^{-1}|<1 and |p_2 z^{-1}|<1 \rightarrow max(|p_1|,|p_2|)<|z|
$$
if \$c<=1/4\$: \$\frac{1}{2}(1 + (1 - 4c)^{1/2})<|z|\$
if \$c>1/4\$: \$\sqrt{c}<|z|\$
For a Anti-Causal System (both components diverge):
$$
|p_1 z^{-1}|>1 and |p_2 z^{-1}|>1 \rightarrow min(|p_1|,|p_2|)>|z|
$$
if \$c<=1/4\$: \$\frac{1}{2}(1 - (1 - 4c)^{1/2})>|z|\$
if \$c>1/4\$: \$\sqrt{c}>|z|\$
For a Non-Causal System (one component diverge and the other converge):
$$
|p_1 z^{-1}|>1 and |p_2 z^{-1}|<1 \rightarrow |p_2|<|z|<|p_1|
$$
if \$c<=1/4\$: \$\frac{1}{2}(1 - (1 - 4c)^{1/2})<|z|<\frac{1}{2}(1 + (1 - 4c)^{1/2})\$
if \$c>1/4\$: \$\sqrt{c}=|z|\$
in MATLAB, making:
h = tf([1],[1 -1 0],1); rlocus(h); axis equal;
you obtain the 'variable locus', though the correct interpretation of the c values is artificial here.
Q.E.P.D.
Best Answer
We need to play a bit with the layout of the expression. We have:
$$ F=\frac{(1+a)(1-a)}{(1-ae^{\ jw})(1-ae^{\ -jw})} =\frac{(1+a)}{(1-ae^{\ -jw})}\frac{(1-a)}{(1-ae^{\ jw})} $$
We can rewrite it as:
$$ =\left( \frac{1}{(1-ae^{\ -jw})}+\frac{a}{(1-ae^{\ -jw})}\right) \left( \frac{1}{(1-ae^{\ jw})}-\frac{a}{(1-ae^{\ jw})}\right) \\ $$
We factor out a \$-ae^{\ jw}\$ from the right-most terms and do the inverse transform:
$$ = \left( \frac{1}{(1-ae^{\ -jw})}+a\frac{1}{(1-ae^{\ -jw})}\right) \left( -\frac{1}{a}\frac{e^{\ -jw}}{(1-\frac{1}{a}e^{\ -jw})}+\frac{e^{\ -jw}}{(1-\frac{1}{a}e^{\ -jw})}\right) \\ \implies \left( a^nu[n]+a(a^nu[n]) \right) * \left( -\tfrac{1}{a}({\tfrac{1}{a}}^nu[n-1])+({\tfrac{1}{a}}^nu[n-1]) \right) $$
Finally, cleaning up:
$$ = \left( a^nu[n](1+a) \right) * \left (\tfrac{1}{a}^{n-1}u[n-1](1-\tfrac{1}{a}) \right) $$
Sorry if it is too messy. Fourier tends to be a lot of writing. Tell me where you think I may have made an error or isn't clear! If someone finds a mistake, please let me know.