I am not familiar with Fourier transform and signal processing. Is this part required for understanding the subsequent content of this course? Thanks.

abiagioli

@anonymous You do not need to be deeply familiar with the Fourier transform for this course. However you should know that functions can generally be decomposed into an orthonormal basis of sinusoids.

yongchi1

Does it mean that these sub-functions (sin/cos functions) from Fourier transform are "orthonormal" by the definition ||f(x)|| = sqrt(Integrate 0-1 f(x)^2) and <f,g> = Integrate 0-1 f(x)*g(x)?

abiagioli

@yongchi1 Almost. Here we are talking about periodic functions over 2pi, so we calculate the L2 inner product over the bounds [0, 2pi]. Indeed:

$$\int_{0}^{2 \pi} cos(mx)cos(nx)\,dx = 0\ \ \ \forall m, n \in \mathbb{N} s.t. m \neq n$$

$$\int_{0}^{2 \pi} sin(mx)sin(nx)\,dx = 0\ \ \ \forall m, n \in \mathbb{N} s.t. m \neq n$$

$$\int_{0}^{2 \pi} cos(mx)sin(nx)\,dx = 0\ \ \ \forall m, n \in \mathbb{N} s.t. m \neq n$$

Try and prove this yourself! Also see this page for more info on Fourier Series (the formal name for this set of orthonormal basis vectors)

jkalapos

Fourier transform always manages to appear again and again in the classes I take. Before one class I took I did not really consider functions as vectors but then my professor drew out a derivation of fourier transforms using hilbert spaces and the whole jazz and I can't say my life has been quite the same after that. All in all, I'm glad to see that I'm going to learn another interesting application of fourier transforms

dstiffler

Agreed. FFT seems to pop up in the most unexpected places, because the ability to multiply signals instead of convolving them is extremely efficient.

I am not familiar with Fourier transform and signal processing. Is this part required for understanding the subsequent content of this course? Thanks.

@anonymous You do not need to be deeply familiar with the Fourier transform for this course. However you should know that functions can generally be decomposed into an orthonormal basis of sinusoids.

Does it mean that these sub-functions (sin/cos functions) from Fourier transform are "orthonormal" by the definition ||f(x)|| = sqrt(Integrate 0-1 f(x)^2) and <f,g> = Integrate 0-1 f(x)*g(x)?

@yongchi1 Almost. Here we are talking about periodic functions over 2pi, so we calculate the L2 inner product over the bounds [0, 2pi]. Indeed:

$$\int_{0}^{2 \pi} cos(mx)cos(nx)\,dx = 0\ \ \ \forall m, n \in \mathbb{N} s.t. m \neq n$$

$$\int_{0}^{2 \pi} sin(mx)sin(nx)\,dx = 0\ \ \ \forall m, n \in \mathbb{N} s.t. m \neq n$$

$$\int_{0}^{2 \pi} cos(mx)sin(nx)\,dx = 0\ \ \ \forall m, n \in \mathbb{N} s.t. m \neq n$$

Try and prove this yourself! Also see this page for more info on Fourier Series (the formal name for this set of orthonormal basis vectors)

Fourier transform always manages to appear again and again in the classes I take. Before one class I took I did not really consider functions as vectors but then my professor drew out a derivation of fourier transforms using hilbert spaces and the whole jazz and I can't say my life has been quite the same after that. All in all, I'm glad to see that I'm going to learn another interesting application of fourier transforms

Agreed. FFT seems to pop up in the most unexpected places, because the ability to multiply signals instead of convolving them is extremely efficient.