Suppose: $$ s(t) = \sin(2\pi{f_0}t) $$
Suppose I'm sampling the signal with a sample frequency $f_s >2f_0$ .
However, every $M$ samples there is a dead-time of $L$ samples.
Traditionally, the (continuous) sampling function is just a delta comb function: $$ p(t) = \sum_{n=-\infty}^\infty \delta(t-nT_s) , \quad T_s=1/f_s $$ However, in my scenario the impulse train misses some impulses in repetitive way. Here's the mathematical description:
Start by defining: $$ v(t) = \sum_{n=0}^{M-1} \delta(t-nT_s) $$ This represents essentially 1 frame in which M samples are sampled.
Denote $K = (M+L)\cdot{T_s}$ , the period between subsequent frames. Then, I could define: $$ r(t)=\sum_{l=-\infty}^{\infty} v(t-lK) $$
Here's a graphical representation: 
So my question is, how can I obtain a good estimation of the FFT of s(t)?
Bonus question: what happens if K isn't an integer multiple of $T_s$ ?