Continuous varying impedances are used all the time for impedance matching. If you have a very capacitive part of a trace (for example, where a large component pad might be), you can have a relatively inductive transition before or after it to "balance" it out.
What will end up happening is that the reflections will "stack up" but, instead of being at one point (a VSWR peak), it will be moderately spread out. You can still imagine it discretely, but in small steps.
And also remember, if you have a small reflection point, any backward reflection after THAT will be reflected slightly FORWARD, and so on.
Anyway, the good gents at http://www.microwaves101.com/encyclopedia/klopfenstein.cfm always have a nice, in depth explanation.
edit: I didn't completely answer your question. "How it would look" is dependent a bit on how you are describing it. In the frequency domain, what you'll probably get is a VSWR that is "de-Q'd". You'll go from a nice sharp peak at midband to a more gradual, broader band response.
In the time domain....well, I don't work with the time domain as much but I would imagine you would have a lower amplitude, longer pulsewidth "ringing" or reflection.
The admittance of this circuit can be written as :
Y = \$\dfrac{1}{sL} + \dfrac{1}{R + \dfrac{1}{sC}}\ = \dfrac{CLs^2 + CRs + 1}{Ls(CRs + 1)}\$.
Substituting \$s = j\omega\$, multiplying the denominator by its complex cojugated and simplifying into real and imaginary parts gives us:
\$\dfrac{R}{\frac{1}{C^2 \omega^2} + R^2} + j\left(\dfrac{1}{C\omega\left(\frac{1}{C^2\omega^2} + R^2 \right)} - \dfrac{1}{L\omega}\right)\$.
A complex admittance consists of a conductance (real part) and a susceptance (imaginary part).
Substituting the value of the resistance and frequency, we want
\$\dfrac{50}{\frac{1}{C^2 (2*\pi*10^9)^2} + 50^2} = 10^{-3}S.\$
Solving for C gives C \$\approx\$ 0.73 pF.
Plugging this value of C and R into
\$j\left(\dfrac{1}{C\omega\left(\frac{1}{C^2\omega^2} + R^2 \right)} - \dfrac{1}{L\omega}\right)\ = -j10^{-3}\$
and solving for L gives L \$\approx\$ 30 nH.
The admittance of this inductance is \$\approx 5.3*10^{-9}\$ S.
Looking here as a reference, the equation for the length of an open-circuited transmission line to act as an inductor is:
l = \${\frac {1}{\beta }}\left[\pi(n+1) -\operatorname{arccot} \left({\frac {\omega L}{Z_{0}}}\right)\right]\$, where \$L = 30*10^{-9}\$\$, \beta = \dfrac{2\pi f }{c_l}\$, \$f = 10^9\$, and \$c_{l} \approx 0.8*3.0*10^9 \frac{m}{s}\$.
Best Answer
Above about 1MHz, normal cable characteristic impedance is fairly contant with frequency as per the equation: -
\$Z_0 = \sqrt{\dfrac{L}{C}}\$ where L and C are inductance and capacitance per unit distance.
The theoretically correct equation for all frequencies is: -
\$Z_0 = \sqrt{\dfrac{R+j\omega L}{G + j\omega C}}\$ where R and G are series resistance and shunt conductance per unit distance.
This tends to reduce to a complex value in the sub-1MHz area of: -
\$Z_0 = \sqrt{\dfrac{R}{j\omega C}}\$
If both frequencies are well above 1MHz then you can take Z0 to be a constant.