As stated in the previous blogs, the main problem we are facing is how to calculate the frequency of the signal. This is very important because to calculate the discharge power, the equation used includes the value of frequency and the sum of the products of the instantaneous voltage across the reactor and the change of the accumulated charge on the capacitor.
It used to be a problem as how to calculate the frequency using fast fourier transform before because each time we tried to generate the plot of the signal in the frequency domain using matlab, it turns out to be something like what the following figure shows.
In the figure there are two apparent frequency peaks, one is at around 0Hz and the other is about 2*10^9Hz. It is normal that there are two frequency peaks with the same amplitude in the frequency domain for one signal because it is such a characteristic for the fourier transform. Usually we would take the first half of the plot as our results. However, this time the frequency of the first peak is so close to 0Hz and it influences the results seriously. Thinking about the problem for a long time, I find there is a possible reason that the frequency of the first peak is so close to 0Hz might be that the range of the x-axis is too large. Therefore to solve the problem, I bound the border of the x-axis to a rather smaller range and suddenly the plot turns out to be successful and it can be seen in the following figure.
With this step successful, we are able to march on to the next step in the next week which is about to find the x-axis value of the first peak in the frequency domain which should be the frequency we are searching for and also the frequency of the target signal.
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