We do not know the values at the odd indices of $X'$. (for convenience of this and FFT, please assume that $N$ is a power of 2, and thus assume that $x'$ and $X'$ both have lengths $2 N$). Of course, this can be accomplished by computing $x$ from $X$ (via the inverse discrete Fourier transform), then padding it with zeros, and finally computing $X'$ however, the even indices of $X'$ are already given by $X_$$ Of course if you like you can change the definition (by adding a pi/2) but then it will not be in line with international definitions.Given $X$ (the discrete Fourier transform of some unknown vector $x$ of length $N$), is there any shortcut to computing $X'$ (the Fourier transform of $x$ after padding it with $N$ zeros)? And I cannot stress enough, the creators of these inverse trig, function decided as codomains for symmetrical intervals centered in origin. I am rusty myself and our system of education back in the day, in my old country was mainly based on the French/German style. By definition (and you need to check yourself in the books), the codomain of the arcsin(x) or asin(x) (same thing) is defined as (-pi/2,+pi/2) and not (0,pi). As I remember correctly from 7th grade a function can have an inverse only on a domain on which it is bijective.
The way we learned them was: injectivity, surjectivity and bijectivity. Go check in the middle school books about function properties. Grace, I will just give you some hints and I apologize if my language is not proper (might not in line with the US terminology). I imagined the FT would highlight this frequency?Īm i mistaken? Sorry for adding the detail, but i think this FT calculator is very useful, if i can get it working properly for this experiment. It appears from the raw data, that when the ultrasound is on and the bubble is at it minimum size, corresponds to pixel data at its minimum, but the bubble is oscillating to the frequency of the ultrasound, so the time difference between minimum is approximately every 16us(manually calculated from the graph), which refers to 62.5 KHz, so the bubble is oscillating at this frequency? Here is the raw data for a single video: 102 points, dt= 2us. I have tried using your algorithm, but for example if i put the frequency range from 0 to 2MHz, there is a peak at 1MHz and spectrum is mirrored both sides, if i change it to 1 MHz, there is a peak at 500 Khz. Which i then put into excel and plotted the change in pixel size over time.
Matlab was used to analyze the image and get an average pixel count for each frame. The camera is limited too 102 picture frames in each exposure, so i cannot create any more( i wish i could).