Peak to Average Power Ratio of an exponentially increasing signal
This feels a lot like there's quite a bit of homework you should do about this, but let's give you a start:

Write down your signal explicitly as 

$$s(t) = p(t) \cdot e^{\lambda t}\text,$$

with $p$ being your periodic oscillation. You sample that, and divide it into vectors. Each vector has $N=128$ samples. Let's write it this:

\begin{align}
s_k[n] &amp;= s\left(\frac{n+N\cdot k}{f_\text{sample}}\right),&amp; n = 0,\ldots, N-1\\
&amp;= p\left(\frac{n+N\cdot k}{f_\text{sample}}\right) \cdot e^{\lambda\frac{n+N\cdot k}{f_\text{sample}}}
\end{align}

where $k$ gives us the frame number.

Your question is how things develop when you increase the frame number!

So, consider the quotient of the next frame and the current frame:

\begin{align}
\frac{s_{k+1}[n]}{s_k[n]}
&amp;= \frac{p\left(\frac{n+N\cdot (k+1)}{f_\text{sample}}\right)}{p\left(\frac{n+N\cdot k}{f_\text{sample}}\right)} \cdot \frac{e^{\lambda\frac{n+N\cdot (k+1)}{f_\text{sample}}}}{e^{\lambda\frac{n+N\cdot k}{f_\text{sample}}}}\\
&amp;= \frac{p\left(\frac{n+N\cdot (k+1)}{f_\text{sample}}\right)}{p\left(\frac{n+N\cdot k}{f_\text{sample}}\right)} \cdot e^{\lambda\frac{n+N\cdot (k+1)}{f_\text{sample}}-\lambda\frac{n+N\cdot k}{f_\text{sample}}}\\
&amp;= \frac{p\left(\frac{n+N\cdot (k+1)}{f_\text{sample}}\right)}{p\left(\frac{n+N\cdot k}{f_\text{sample}}\right)} \cdot e^{\lambda\frac{N}{f_\text{sample}}}\\
\end{align}

Now, it'd be a nice simplification if we can just restrict the oscillations' period to something that is an divisor of $N$. That especially means that $p()$ becomes $N$-periodic (under the given sampling rate).

\begin{align}
&amp;= 1 \cdot e^{\lambda\frac{N}{f_\text{sample}}}\\
&amp;=\text{const.}
\end{align}

In other words, the next frame is identical to the previous frame, aside from a constant factor.

Since scaling with a constant factor doesn't change the PAPR at all, this was easy to answer for oscillations that have a frequency that divides the sampling rate, only. The rest isn't really unusual (it's leakage, essentially), but the math is slightly more annoying, so I'll leave it to you.