Integral of discrete function

Question

I started learning Discrete Fourier Transform. I'm not sure how they do translation from continuing transform to discrete transform. I attach appropriate link.
page 1. Translation from continuous transform to the discrete using property of the discrete function.

We could regard each sample $f[k]$ as an impulse having area $f[k]$. Then, since the integrand exists only at the sample points, $$ \begin{split} F(j\omega) &= \int_0^{(N-1)T} f(t)e^{-j\omega t} dt \\ &= f[0]e^{-j0} + f[1]e^{-j\omega T} + \ldots + f(N-1) e^{-j\omega (N-1)T} \\ &= \sum_{k=0}^{N-1} f[k] e^{-j\omega k T} \end{split} $$

I wonder if they are correct.

Answer 1

I don't know if this is the reply you are looking for, but I've found it useful (in improving my intuition) when learning probabilities and Fourier transforms to bring up analogies between "discrete calculus" and "continuous calculus".

Our base spaces for the sake of example are $N \to R$ (real sequences) and $R \to R$ (real functions).

a) For each, you have a derivative:

• $u'(n) = u_{n+1} - u_{n} = \frac{u(n+\Delta n) - u(n)}{\Delta n}$ where $\Delta n = 1$ is the input "increment". You get similar properties, ie, your sequence is increasing iff $u'(n) > 0$.

• $f'(x) = \lim_{\Delta x \to 0} \frac{f(x+\Delta x) - f(x)}{\Delta x}$, where $\Delta x$ is also some "increment" (although for the continuous case the word is ambiguous), arbitrarily close to 0. You could think of it as the "hyperreal" number $\epsilon$, a number such that for every non-zero positive real number $x$, $0 < \epsilon <x$. It's generally written $dx$ rather than $\Delta x$ in the continuous case.

b) Both also have an integral:

• $U(n) = \sum_{i = a}^{i = n} u(i) + C_a$, such that $U'(n) = u(n)$. Do note that you can very well define sequences (functions) from $Z \to R$ and have negative values for $n$. Also note that the constant $C_a$ is also present here and you have an analogy of the fundamental theorem of calculus you can use (look up "telescoping sums": https://mathworld.wolfram.com/TelescopingSum.html ).

• $F(x) = \int_{t=a}^{t=x} f(t) dt + C_a$, such that $F'(x) = f(x)$. You'll notice that the Riemann integral is technically defined as a limit starting from a discrete integral. $\int_{t=a}^{t=b} f(t) dt = \lim_{n \to \infty} \sum_{k=1}^{k=n} \Delta t * f(a + k\Delta t)$, where $\Delta t = \frac{b - a}{n} \to 0$ when $n \to \infty$. This is similar to understanding the continuous derivative as a limit starting from the discrete one.

c) And finally both lead to differential equations; one fundamental example:

• $u'(n) = \alpha u(n) \Leftrightarrow u(n+1) - u(n) = \alpha u(n) \Leftrightarrow u(n+1) = (\alpha + 1) u(n)$, a geometric series with solution $C(\alpha + 1)^n$, where $C = u_0$ is defined from the initial conditions. Do note that for $\alpha = 1$, $u(0) = 1$ you get that "the function which is its own discrete derivative is $2^n$ ", and $2$ is thus the discrete equivalent of $e$.
• $f'(x) = a f(x)$ has the famous solution $Ce^{ax}$, with $C$ being defined like $u_0$ from initial conditions. More specifically, if we know the solution to our differential equation verifies $f(x_0) = Ce^{ax_0} = y_0$ for some specific input $x_0$, then $C = y_0e^{-ax_0}$ and the solution can be written $y_0e^{a(x - x_0)}$. For $a = 1$, $f(0) = 1$, you get the classical exponential, "the function which is its own continuous derivative".