oscillator
According to the solution, it says it's because the phase shift is wrong. The gain network is non-inverting, so I assume that means that the LC feedback network was not 0 degrees for that particular frequency.
So I tried to work it out by hand but I am stuck here. Is my knowledge that the LC feedback not being 0 degrees for f = 1/root(2LC) the reason why it doesn't work? If so, what is the shift exactly, and how can I calculate the phase shift from the equations below
C1 to ground in series with the crystal and then another C2 to ground.
It's not obvious exactly what your circuit looks like, but to clarify my answer, I'll show you what imagine it looks like based on your description:
simulate this circuit – Schematic created using CircuitLab
Incidentally, this circuit, with C1 in series with the crystal, is not what I think of as an especially common oscillator circuit.
If the RF signal is AC why doesnt the AC signal just go straight to ground?
The very simple description of a capacitor is "at low frequencies it's an open circuit and at high frequencies it's a short circuit." And this is a reasonable model for a lot of cases. But saying you have an AC signal doesn't mean you have a low or a high frequency, it just means the frequency is not exactly 0 Hz. And what's happening here is that the frequency you're operating at is neither a "low" frequency or a "high" frequency, it's somewhere in between.
Between the two extremes you need to look at the impedance model of the capacitor:
\$Z = \dfrac{1}{j2\pi{}fC}\$
This also tells you what is meant by "low" and "high" frequencies. When the frequency is low enough that Z is so large it doesn't affect your circuit differently than an open circuit would, that's a "low" frequency. When the frequency is high enough that it doesn't affect your circuit differently than a short circuit would, that's a "high" frequency.
I'm not answering this in the classical "colpitts explanation way" because I think certain things need to be said about the driving source to pin 1 and the off resonance of L and C2 (ignoring C1's effect initially)....
If you apply an oscillating voltage at pin 1, L and C2 form a resonant low pass filter with pin 3 being the output. See below: -
Depending on the Q of the circuit there may be voltage amplification but there will be round about 90 degrees of phase shift between pin 1 and pin 3.
However, this isn't quite how the colpitts works. C1 plays a role and pin 1 isn't a driven by hard voltage source but by a weak voltage source. This adds up to an overall phase shift that can become 180 degrees; the output resistance of the driving source produces a phase shift with C1 that may be (say) 70 degrees and at the right frequency (slightly higher than resonance of L and C2), L and C2 produce another 110 degrees. All this adds up to 180 degrees and the oscillator then oscillates at this frequency.
Best Answer
Consider the Wien bridge oscillator circuit: -
It is very similar to the oscillator in the question except that the lower capacitor is is an inductor in the question. At this point alarm bells should possibly start ringing to inform you that replacing the cap with an inductor will not work.
However, lets look at the frequency and phase response of the RC Wien bridge network. This is the filter section of the Wien bridge oscillator circuit (C2 is replaced by L in the question): -
Here's the phase response: -
As can be seen, at resonance, the phase shift is zero as one expects - this produces oscillation at Fr. With C2 changing to an inductor, the phase shift will be 90 degrees and will therefore not oscillate. This last point presumes you understand what happens when L and C are combined at resonance. If not then look on this page for a simulation.