I know it's a bit late, but I've just got this sabe doubt. After looking around, I've come across this Microchip Doc that shows some examples.
First, we calculate \$\text{PR2}\$. From this formula,
$$ F_\text{PWM} = \dfrac{1}{(\text{PR2} + 1) \times 4 \times T_\text{OSC} \times \text{T2CKPS}} $$
we get
$$ \text{PR2} = \dfrac{1}{F_\text{PWM} \times 4 \times T_\text{OSC} \times \text{T2CKPS}} - 1 $$
where \$T_\text{OSC} = 1/F_\text{OSC}\$, and \$\text{T2CKPS}\$ is the Timer2 prescaler value (1, 4 or 16).
Therefore, if we want \$F_\text{PWM} = 20\text{kHz}\$, and choosing \$\text{T2CKPS} = 1\$, we get \$\text{PR2} = 249\$. We should choose higher values for \$\text{T2CKPS}\$ only if \$\text{PR2}\$ exceeds 8 bits (\$\text{PR2} \gt 255\$) for the given prescale.
Now we calculate the max PWM resolution for the given frequency:
$$ \text{max PWM resolution} = \log_2(\;\dfrac{F_\text{OSC}}{F_\text{PWM}}\;) $$
That gives us \$9.9658\$ bits (I know, it sounds weird, but we'll use it like that later).
Now, let's calculate the PWM duty cycle. It is specified by the 10-bit value \$\text{CCPRxL:DCxB1:DCxB0}\$, that is, \$\text{CCPRxL}\$ bits as the most significant part, and \$\text{DCxB1}\$ and \$\text{DCxB0}\$ (bits 5 and 4 of \$\text{CCPxCON}\$) the least significant bits. Let's call this value \$\text{DCxB9:DCxB0}\$, or simply \$\text{DCx}\$. (x is the CCP number)
In our case, since we have a max PWM resolution of \$9.9658\$ bits, the PWM duty cycle (that is, the value of \$\text{DCx}\$) must be a value between \$0\$ and \$2^{9.9658} - 1 = 999\$. So, if we want a duty cycle of 50%, \$\text{DCx} = 0.5 \times 999 = 499.5 \approx 500\$.
The formula given on the datasheet (also on the linked doc),
$$\text{duty cycle} = \text{DCx} \times T_\text{OSC} \times \text{T2CKPS}$$
gives us the pulse duration, in seconds. In our case, it's equal to \$25\text{ns}\$. Since \$T_\text{PWM} = 50\text{ns}\$, it's obvious that we have a 50% duty cycle.
That said, to calculate DCx in terms of duty cycle as \$r \in [0,1]\$, we do:
$$ \text{DCx} = \dfrac{r \times T_\text{PWM}}{T_\text{OSC} \times \text{T2CKPS}} = \dfrac{r \times F_\text{OSC}}{F_\text{PWM} \times \text{T2CKPS}} $$
Answering your other questions:
2) The resolution of your PWM pulse with period \$T_\text{PWM}\$ is
$$ \dfrac{T_\text{PWM}}{2^\text{max PWM res}} $$
3) Because CCPRxL, along with DCxB1 and DCxB0, determine the pulse duration. Setting CCPRxL with a higher value than \$2^\text{max PWM res} - 1\$ means a pulse duration higher than the PWM period, and therefore you'll get a flat \$V_{DD}\$ signal.
Best Answer
Definition:
I think that Oxford have it slightly wrong and that it should be "a curve showing the shape of its graph as a function of time". Wikipedia is better.
The waveform represents the variation in a value with respect to time. This could be a voltage (electrical), a height (wave on the sea), pressure (sound), etc. and simply shows the instantaneous value at any point on the time axis.
Not quite. We can deduce two things from the graph:
The diagram is comparing the relative phase of three different waves of the same frequency with a periodic time of T.
Typically t = 0 is just chosen for convenience to make the illustration clear.