# Discrete Fourier Transform The discrete Fourier Transform is simply a change of coordinates. For example we have two rotated coordinate systems, the unprime system, and the prime system. We label the directions by numbers because they generalize to $N$ dimensions. ![\nTwo Coordinate Systems to Express a Vector $\bold v$ In](./Two_Coordinate_Systems.png "Two Coordinate Systems to Express a Vector $\bold v$ In")
We assume that the unprime coordinate system is spanned by orthogonal Cartesian vectors of N dimensions, whose elements are real numbers. The kth one will be denoted $\hat {\mathbf {k}}$ and so $\hat {\mathbf {k}} \cdot \hat {\mathbf {l}} = |\hat {\mathbf {k}}|^2\delta_{kl}$. The prime coordinates are spanned by an orthogonal basis set, not necessarilly normal (length one) denoted by $\hat {\mathbf {n'}}$, so $\hat {\mathbf {n'}} \cdot \hat {\mathbf {m'}} = |\hat {\mathbf {m'}}|^2\delta_{mn}$. The coordinates of the prime coordinate system are complex numbers. Most of us haven't done dot products with complex numbers. The property that $\mathbf{v} = |\mathbf{v}|^2$ is important and extending the dot product in a way that keeps that property and matches the convention others use is ${\mathbf {a}} \cdot {\mathbf {b}} = \sum_n a_n b_n^*$ where the star denotes complex conjugate. The vector $\mathbf b$, can be written in either prime or unprime coordinate systems. Assume we know the coordinates of $\mathbf v$ in the unprime coordinate system as well as the basis vectors, $\hat {\mathbf {k}}$ and $\hat {\mathbf {n'}}$. How do we find the prime coordinates of $\mathbf v$? $$\mathbf v = \sum_{k=0}^{N-1} v_k \hat {\mathbf {k}} = \sum_{n'=0}^{N-1} v_{n'} \hat {\mathbf {n'}}$$ Taking the inner product on both sides with the m'th unit vector, $$\mathbf v \cdot \hat {\mathbf {m'}} = \sum_{k=0}^{N-1} v_k \hat {\mathbf {k}} \cdot \hat {\mathbf {m'}} = \sum_{n'=0}^{N-1} v_{n'} \hat {\mathbf {n'}} \cdot \hat {\mathbf {m'}}$$ Using the orthogonality relationship for prime basis vectors, $$\mathbf v \cdot \hat {\mathbf {m'}} = \sum_{k=0}^{N-1} v_k \hat {\mathbf {k}} \cdot \hat {\mathbf {m'}} = \sum_{n'=0}^{N-1} v_{n'} |\hat {\mathbf {m'}}|^2 \delta_{n'm'}$$ and the sifting property of the Kroniger delta yields $$\mathbf v \cdot \hat {\mathbf {m'}} = \sum_{k=0}^{N-1} v_k \hat {\mathbf {k}} \cdot \hat {\mathbf {m'}} = v_{m'} |\hat {\mathbf {m'}}|^2$$ Dividing by $|\hat {\mathbf {m'}}|^2$ gives us what we were after. $$v_{m'} = \frac{\mathbf v \cdot \hat {\mathbf {m'}}} {|\hat {\mathbf {m'}}|^2} = \sum_{k=0}^{N-1}\frac { v_k \hat {\mathbf {k}} \cdot \hat {\mathbf {m'}}}{|\hat {\mathbf {m'}}|^2}$$ Now let's put this to use by applying this to the situation where $$\frac{\hat {\mathbf {m'}} \cdot \hat {\mathbf {k}}}{|\hat {\mathbf {m'}}|^2} = e^{j2 \pi n'k/N}$$ This is saying that the components of the prime vector, $\hat {\mathbf {m'}}$ along the unprime kth direction are proportional to $e^{j2\pi n'k/N}$. I leave it as an execrise for the reader to determine the constant of proportionality. If you want to tackle this, a useful hint is: $\hat {\mathbf {n'}} \cdot \hat {\mathbf {m'}} = \delta_{mn} |\hat {\mathbf {m'}}|^2$. (Hint: use $x = e^{j2 \pi k/N}$ and the geometric sum $\sum_{n'=0}^{N-1} x^{n'} = \frac{1-x^N}{1-x}$ when $n' \neq m'$.) From all this, you get $$v'_{m'} = \frac{\mathbf v \cdot \hat {\mathbf {m'}}} {|\hat {\mathbf {m'}}|^2} = \sum_{k=0}^{N-1}\frac { v_k \hat {\mathbf {k}} \cdot \hat {\mathbf {m'}}}{|\hat {\mathbf {m'}}|^2} = \sum_{k=0}^{N-1}\ v_k e^{-j2 \pi m'k/N}$$ The negative sign on the complex exponential comes from the complex conjugate operation for the second vector. ### Aside (This needs to be fixed or deleted, as the substitutions for the exponentials are not fixed after the last changes.) You can put this back into the original formula for $\mathbf v$ and get: $$\mathbf v = \sum_{n'=0}^{N-1} v_{n'} \hat {\mathbf {n'}} = \sum_{n'=0}^{N-1} \sum_{k=0}^{N-1} v_k e^{-j2 \pi n'k/N} \hat {\mathbf {n'}} = \sum_{n'=0}^{N-1} \sum_{k=0}^{N-1}\sum_{l=0}^{N-1} v_k e^{-j2 \pi n'k/N} e^{j2 \pi ln'/N} \hat {\mathbf {l}}$$ It is easy to do this for just one component of the prime coordinates. This is $$v_l = \mathbf {v} \cdot \mathbf {\hat l} = \sum_{n'=0}^{N-1}\sum_{k=0}^{N-1} v_k e^{-j2 \pi n'k/N} \hat {\mathbf {n'}} \cdot \hat {\mathbf {l}} = \sum_{n'=0}^{N-1}\sum_{l=0}^{N-1} v_k e^{-j2 \pi n'k/N} e^{j2 \pi ln'/N}$$ ## The Discrete Fourier Transform The kth element of a sequence of data taken by an A/D converter reading a signal, $h(t)$ at intervals $T$ seconds apart at times, $t \in \{0, T, 2T, 3T, ..., (N-1)T\}$ is $h(kT)$, known more simply as $h(k)$. The Discrete Fourier Transform (DFT) is a change of basis set similar to the one shown above where $m' = n$, $v'_{m'} = H(n)$ and $v_k = h(k)$. The DFT of $h(k)$ is known as $H(n)$ and has the same number of dimensions as the sampled time signal, $N$. $$n \in \{0, 1, 2, 3, ..., N-1\}$$ $$H(n) = \sum_{k=0}^{N-1} h(k) e^{-j2\pi nk/N}$$ The reverse transformation from $H(n)$ to $h(k)$ is given by the Inverse Discrete Fourier Transform: $$h(k) = 1/N\sum_{n=0}^{N-1} H(n) e^{j2\pi nk/N}$$ The DFT is not usually derived this way because it is actually an approximation to the continuous Fourier Transform as we shall see later this quarter. However, this should show you that it is a also a simple change of basis set to go from $h(k)$ to $H(n)$. The same vector $h(k)$ is just described in a different way. The notation used above, especially that for the components of the vectors will be very useful, because it allows us to easily work with $N$ dimensional vectors.

~~It is as if the $v_k$ above are like the $H(n)$ the basis vectors for the time coordinate system, $$\hat {\mathbf {n'}} = \sum_{l=0}^{N-1}e^{-j2\pi n'l/N} \hat {\mathbf {l}}$$ (the complex conjugate of what we did in the example above). Note this complex conjugate just changes $j$ for $-j$ in all the equations above, like changing the direction of rotation of the phasor. The $v'_{n'}$ are like are like the $h(k)$. The DFT was not derived this way because it is actually an approximation to the continuous Fourier Transform as we shall see this quarter. However, this should show you that it is a also a simple change of basis set to go from $h(k)$ to $H(n)$ and the vector stays the same. The only difference is how the vector is described. The notation used above, especially that for the components of the vectors will be very useful, because it allows us to easily work with $N$ dimensional vectors, and we need to work with those.~~

Notes for the explanation: Delete these after you are done with them. We could introduce $h(k) = IDFT(DFT(h(k)))$ and put the 1/N with the $h(k)$ instead of $H(n)$. I need to think about this some more. It sure looks like they just used $e^{-j2\pi nk/N}$ as the $\hat {\mathbf {n'}}$ basis and the prime coordinates were for frequency, and unprime were time. Explaining it this way seems a bit unnatural based on starting in time and going to frequency. What if we used a normalized basis for frequency, but not time?