# Peak to Average Power Ratio of an exponentially increasing signal

I am having an exponentially increasing sinusoidal signal.

After sampling, I am dividing the signal into frames of length 128 and doing windowing operation with a hanning window of length 128. Then find the FFT of this windowed sequence frame by frame.

The average power and peak power of the windowed signal in each frame is found out and Peak to average power ratio (PAPR) is calculated.

Will the PAPR increase with each frame?

• well, the quickest way to find out would be writing 20 lines of Python/matlab/R/octave/… simulation (which you'd want to do before dealing with the real signal, anyways). What's the result of that? Is it surprising? – Marcus Müller Nov 1 '19 at 12:37
• (hint: it's probably easiest if you start with a sinusoid signal that's has a period that's a divisor of 128 samples, so that you get an integer number of oscillations per FFT.) – Marcus Müller Nov 1 '19 at 12:42

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] &= s\left(\frac{n+N\cdot k}{f_\text{sample}}\right),& n = 0,\ldots, N-1\\ &= 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]} &= \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}}}}\\ &= \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}}}\\ &= \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} &= 1 \cdot e^{\lambda\frac{N}{f_\text{sample}}}\\ &=\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.