In response to one student’s question, I suggested that

you also plot the $f_k$ over $k$ (helps visualize the input function)

If you plot the real and imaginary parts separately and not only the magnitude, please use a fixed y-axis scale (for example from -3 to +3). Unless you do this, very small values that should be zero, but aren’t due to rounding errors, may appear hugely magnified.

Could you please clarify question 3 what you mean by “in the case of the sine wave?” Are we comparing the magnitudes of r=1 when fk is found using sine, or are we using r=1 of imaginary values?

In class, I introduced the idea of a transform with a simple “rule”, that is, to transform a time-domain function $f(t) = A \cdot sin(\omega t)$ into $A$ over $\omega$ space, where it can be represented by a single dot. This dot would be placed at $A=1$, which makes intuitive sense. However, if you use the DFT, you find two points with a magnitude of 0.5 each.

My question is, can you show that this is correct (even if not intuitive)? A simple way to do this: Set up the inverse DFT. With only two coefficient $F_r \neq 0$, the summation reduces to a sum of two terms, and it can be shown that the two amplitudes of 0.5 add up to 1.

To do this right, however, you need to consider the complex coefficients rather than their magnitude. The transform coefficients of $f(t) = \sin (2 \pi t)$ are imaginary and conjugate, meaning, $F_1 = -j/2$ and $F_{19} = -F_{-1} = +j/2$.

Just for those who want to dig deeper: There are some symmetries that we did not discuss. We always get the Hermitian symmetry for negative frequencies, that is,

$$
F(-r) = F^*(r)
$$

However, there is a lot more when it comes to even and odd functions. You may recall that I briefly mentioned that even functions are mirror-symmetric about the vertical axis, meaning $f(t) = f(-t)$, while odd functions are point-symmetric around the origin, or $f(-t) = -f(t)$. Examples are the cosine and sine functions, respectively. Moreover, cosine and sine are orthogonal (their inner product is zero). This means that even functions would cancel out the sine component, while odd functions cancel out the cosine component. For the 1D Fourier transform, this yields the following symmetries, always assuming that the input function $f(t)$ is real-valued:

If $f(t)$ is even, its Fourier transform is real-valued and even

If $f(t)$ is odd, its Fourier transform is imaginary and odd.

One additional and important property is the linearity of the Fourier transform, meaning, it obeys the scaling and superposition principle. If a real-valued constant $a$ is given, and two functions $f(t)$ and $g(t)$ have the Fourier transforms of $F(\omega)$ and $G(\omega)$, then

A real-valued multiple of $a \cdot f(t)$ results in an equally scaled Fourier transform $a \cdot F(\omega)$

The Fourier transform of $f(t)+g(t)$ is $F(\omega)+G(\omega)$.

In response to one student’s question, I suggested that

Could you please clarify question 3 what you mean by “in the case of the sine wave?” Are we comparing the magnitudes of r=1 when fk is found using sine, or are we using r=1 of imaginary values?

In class, I introduced the idea of a transform with a simple “rule”, that is, to transform a time-domain function $f(t) = A \cdot sin(\omega t)$ into $A$ over $\omega$ space, where it can be represented by a single dot. This dot would be placed at $A=1$, which makes intuitive sense. However, if you use the DFT, you find

twopoints with a magnitude of 0.5 each.My question is, can you show that this is correct (even if not intuitive)? A simple way to do this: Set up the inverse DFT. With only two coefficient $F_r \neq 0$, the summation reduces to a sum of two terms, and it can be shown that the two amplitudes of 0.5 add up to 1.

To do this right, however, you need to consider the complex coefficients rather than their magnitude. The transform coefficients of $f(t) = \sin (2 \pi t)$ are imaginary and conjugate, meaning, $F_1 = -j/2$ and $F_{19} = -F_{-1} = +j/2$.

It seems to me that the signs for F[r=1] and F[r=-1] you gave should be opposite…Shouldn’t F[r=-1] = +j/2?

Agreed… and corrected.

Thanks for pointing out my mistake.

Just for those who want to dig deeper: There are some symmetries that we did not discuss. We always get the Hermitian symmetry for negative frequencies, that is,

$$

F(-r) = F^*(r)

$$

However, there is a lot more when it comes to even and odd functions. You may recall that I briefly mentioned that even functions are mirror-symmetric about the vertical axis, meaning $f(t) = f(-t)$, while odd functions are point-symmetric around the origin, or $f(-t) = -f(t)$. Examples are the cosine and sine functions, respectively. Moreover, cosine and sine are orthogonal (their inner product is zero). This means that even functions would cancel out the sine component, while odd functions cancel out the cosine component. For the 1D Fourier transform, this yields the following symmetries, always assuming that the input function $f(t)$ is real-valued:

One additional and important property is the linearity of the Fourier transform, meaning, it obeys the scaling and superposition principle. If a real-valued constant $a$ is given, and two functions $f(t)$ and $g(t)$ have the Fourier transforms of $F(\omega)$ and $G(\omega)$, then

It would be a nice exercise to prove this.