If the Arduino is anything like a PIC µC then you have no hope of sampling at 44KHz. Most simple µC have quite a slow sampling rate (like 100's of samples per second).
If you want faster then you'd be looking at using something like a dsPIC which has an actual audio grade ADC in it, or use an audio ADC externally that can send I²S data to a µC that is fast enough to respond to it.
I have done some similar work recently while designing a digitally controlled amp.
I had the output of the first stage of the amp going into an analog input on the controlling PIC to then control a bargraph of LEDs for a simple VU meter.
For an output from a PC soundcard you're probably looking at around 1 to 2 volts voltage swing. For my system I wasn't too fussed about frequency and such - just pure peak amplitude - so I passed the signal through a small shottky diode first to trim off the negative voltages. This simplified my design a whole lot.
I am also designing a small frequency analyzer at the moment, and am looking at having selectable op-amp based band-pass filters based around this design: http://www.wa4dsy.net/robot/bandpass-filter-calc which so far has given quite good results. I am varying some of the resistor values by a combination of digital pots and analog multiplexers.
I would certainly recommend at least protecting your analog input(s) with op-amps to limit the maximum voltage they get - just in case. You don't want a voltage spike blowing up your Arduino now do you? Easier to replace a blown op-amp.
And as for a signal for testing? There are many free signal generators for the PC available for download if you do a little google for them. They will let you select waveform, frequency, amplitude, phase, etc. Even allow summing of waveforms to give new waveforms if you're lucky.
You can even use your PC soundcard as a rudimentary scope as well with the right software and a small home-made probe. There is software and designs around for this too on the net.
Oh, and remember to isolate different stages / voltage levels with capacitors in the audio signal. As a rule of thumb, if I am changing PSU voltage levels, I always introduce a capacitor to isolate the stages. So, I had one on the input signal, one on the stage 1 -> stage 2 (+/-5V to +/-12V power supply), one on the stage 1 -> analog input, and one again on the output. It pays to take no chances with stray DC offsets wandering into the wrong part of the circuit.
The cap has to go. The diode needs to be reverse biased, and with the cap in place, its not. Do you have the Anode going to ground, or the cathode? It should be the anode. Also, R1 should probably be removed. Your photodiode is functioning as a current source which will generate a voltage through the feedback resistor alone. R1 does nothing but generate another voltage drop which will lower the back biasing of your photodiode.
Best Answer
You are asking an xy question. The question is not, "How do I charge a powerbank with a microphone?", but rather "Can I charge a powerbank with a microphone?"
The first thing you should have done before you got yourself into this mess was to look at the basics of the question.
Let's assume that your sound emitter is producing 1 watt of acoustic power, broadcast with a hemispherical radiation pattern. A microphone with an collecting area 1 cm x 1 cm is located 3 meters from the sound source. How much power can the microphone collect?
Well, that's easy. The microphone has an area of 10^-4 square meters. The watt from the emitter is spread over an area 2 pi r^2, so the available power P is $$P = \frac{10^{-4}}{2 \pi r^2} = \frac{10^{-4}}{56.5} = 1.7 \mu W $$ Of course, your numbers may be different, so you should do your own calculations - but frankly, they're not going to get all that much better. 1 watt of power is pretty loud, especially compared to normal conversation levels.
If you compare this to the power needed to charge a powerbank, you will see that it's going to take a really, really long time to do the job. To make matters worse, using the numbers given and 5 volts as the battery voltage, the charge current will be on the order of 0.3 microamps. Now you need to find a fascinating number for your powerbank - the self-discharge current. I suspect that, for a commercially-available battery that number will be more than your charging current, which means that your charger will not actually be able to increase the amount of capacity lost during non-operation, merely slow the process down.
I am, frankly, astonished that a professor would be willing to commit to supervising your thesis - it is almost certainly a waste of time, except as a lesson in asking the wrong question. If you can, I suggest that you attempt to find a new thesis subject.