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| Thursday, July 03, 2008 |
| Activity 5: Physical measurements from Discrete Fourier Transforms |
Using the code given by Dr. Soriano on generating a 1D sinusoid and obtaining its Fourier Transform and the shifted transformation:
//to generate a 1D sinusoid
T = 2;
N = 256;
dt = T/256;
t = [0:dt:(N-1)*dt];
f = 5;
y = sin(2*%pi*f*t);
f1 = scf(1); plot(t,y);
//obtaining Fourier Transform and Frequency scale
FY = fft(y);
F = 1/(2*dt);
df = 2*F/256;
f = [-(df*(N/2)):df:df*(N/2 -1)];
//displaying the FFT shifted output with the frequency axis
scf(2);
f2 = scf(2); plot(f, fftshift(abs(FY)));
As expected from a sine function, the resulting transformation is a delta function.
Applying Discrete Fourier Transforms in image processing involves decomposing image into sine and cosine components. The process then becomes similar with the process of getting the Fourier Transforms of temporal signals. But this time, in images, we don't use "time" instead we use "pixels" (so instead of temporal signals, we might call it pixel signals). Obviously then, instead of a 1D Fourier Transform, in the context of images, a 2D Fourier Transform must be done instead.
Answering Guide Questions:
a.) Light from a fluorescent lamp is known to flicker at 120Hz. what should be the threshold sampling interval for accurate Fourier Transform analysis?
Using the Nyquist Theorem which states that the maximum frequency is inversely proportional to twice the threshold sampling interval: Fmax = 1/2Δt; then Δt = 0.004167 seconds.
b.) What is the effect of increasing the number of samples (N) in the Fourier Transform?
Increasing the number of samples (N) would give rise to widening the domain of our Fourier Transform.
c.) What is the effect of decreasing the sampling interval (Δt) in the Fourier Transform?
Decreasing Δt, T inevitably decreases for the same number of samples (N). The maximum frequency will increase, N stays the same, so Δf increases also.
d.) What is the effect of fixing the time interval T but increasing the number of samples (N)?
If we increase (N) and set the total time to be fixed, we inevitably decrease dt. We then would discover that there is minimal difference on the peak frequency, and that the amplitude is half of N. This is caused by the addition of the signal's components as compared to lower samples.
I give myself 9/10 neutrinos since even if I know I gave the right answers I'm not particularly confident of my grasp of the subject matter. Nevertheless I acknowledge Jeric for his grasp of the subject which helped me substantially in answering the questions.
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posted by poy @ 7:11 PM  |
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