Fourier coefficients of discrete difference of a square wave

Question

I have a discrete square wave $f(t)$ where $t \in \mathbb{N}$, of amplitude $A$, period $T$ and duty cycle $1/T$

$$ f(t) = \left\{\begin{matrix} A, & \mathrm{if}\;t=Tn\\ 0, & \mathrm{if}\;t\neq Tn \end{matrix}\right. \;,\; n \in \mathbb{N} $$

size = 23
T = 5
A = 10
t = np.arange(size)
f = np.zeros(size)
f[t%T==0] = A
fig, ax = plt.subplots(figsize=(12, 5))
ax.step(t, f)
ax.set(
    xticks=t,
    xlabel='time $t$',
    ylabel='amplitude $A$',
    title=f"Square wave of amplitude {A}, period {T} and duty cycle 1/{T}"
)
plt.show()

Given the Fourier series

$$ \mathcal{F}[g(t)] = a_0 + \sum_{h=1}^{\infty} a_h \cos(2 \pi h \nu \cdot t) - \sum_{h=1}^{\infty} b_h \sin(2 \pi h \nu \cdot t) $$

for this square wave

$$ a_0 = \frac{A}{T} $$

$$ a_h = \frac{2A}{\pi h} \sin \left( \frac{\pi h}{T} \right) $$

$$ b_h = 0 $$

so

$$ \mathcal{F}[f(t)] = \frac{A}{T} + \sum_{h=1}^{\infty} \frac{2A}{\pi h} \sin\left(\frac{\pi h}{T}\right) \cos\left(\frac{2 \pi h}{T} \cdot t\right) $$

where the Fourier coefficients of harmonic $h$ is $\frac{2A}{\pi h} \sin\left(\frac{\pi h}{T}\right)$.

The discrete difference of $f(t)$

$$ f'(t) = f(t) - f(t-1) $$

is the sum of two square waves $f(t)$ and $-f(t-1)$ with opposite amplitude and shifted by a sample unit

$$ f'(t) = \left\{\begin{matrix} +A, & \mathrm{if}\;t=Tn\\ -A, & \mathrm{if}\;t=Tn+1\\ 0, & \mathrm{otherwise} \end{matrix}\right. \;,\; n \in \mathbb{N} $$

t1 = t[1:]
f1 = np.diff(f)

fig, ax = plt.subplots(figsize=(12, 5))
ax.step(t1, f1)
ax.set(
    xticks=t,
    xlabel='time $t$',
    ylabel='amplitude $A$',
    title=f"Discrete difference of a square wave of amplitude {A}, period {T} and duty cycle 1/{T}"
)
plt.show()

We can thus say (is it correct?) that the Fourier series of $f'(t)$ is the sum of the Fourier series $\mathcal{F}[f(t)]$ and $\mathcal{F}[-f(t-1)]$

$$ \mathcal{F}[f'(t)] = \left[ \frac{A}{T} + \sum_{h=1}^{\infty} \frac{2A}{\pi h} \sin\left(\frac{\pi h}{T}\right) \cos\left(\frac{2 \pi h}{T} \cdot t\right) \right] - \left[ \frac{A}{T} + \sum_{h=1}^{\infty} \frac{2A}{\pi h} \sin\left(\frac{\pi h}{T}\right) \cos\left(\frac{2 \pi h}{T} \cdot (t-1)\right) \right] $$

$$ \mathcal{F}[f'(t)] = \sum_{h=1}^{\infty} \frac{2A}{\pi h} \sin\left( \frac{\pi h}{T}\right) \left[ \cos\left( \frac{2 \pi h}{T} \cdot t\right) - \cos\left( \frac{2 \pi h}{T} \cdot (t-1)\right) \right] $$

How can I get the Fourier coefficient of harmonic $h$ from $\mathcal{F}[f'(t)]$?

Is there a better way to express the Fourier series of $f'(t)$?

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EDIT #1

As @MattL. noticed (see comments) what I'm dealing with is actually a discrete-time signal, so a better representation of the signal would be

and for the discrete difference will look like

Answer 1

It's quite straightforward to express the Fourier coefficients of a shifted version of a function in terms of the Fourier coefficients of the original function.

If the Fourier series of a $T$-periodic function $f(t)$ is given by

$$f(t)=\sum_{k=-\infty}^{\infty}c_ke^{j2\pi kt/T}\tag{1}$$

then it follows that the Fourier series of a shifted version must be

$$\begin{align}f(t-t_0)&=\sum_{k=-\infty}^{\infty}c_ke^{j2\pi k(t-t_0)/T}\\&=\sum_{k=-\infty}^{\infty}c_ke^{-j2\pi kt_0/T}e^{j2\pi kt/T}\\&=\sum_{k=-\infty}^{\infty}\tilde{c}_ke^{j2\pi kt/T}\tag{2}\end{align}$$

Hence, the Fourier coefficients of $f(t-t_0)$ are given by

$$\tilde{c}_k=c_ke^{-j2\pi kt_0/T}\tag{3}$$

And, by linearity, the Fourier series coefficients of $g(t)=f(t)-f(t-t_0)$ are

$$d_k=c_k-\tilde{c}_k=c_k\big(1-e^{-j2\pi kt_0/T}\big)\tag{4}$$

---

For periodic discrete-time sequences we can use the Discrete Fourier Tansform (DFT), which is just a discrete Fourier series representation:

$$x[n]=\frac{1}{N}\sum_{k=0}^{N-1}X[k]e^{j2\pi nk/N}\tag{5}$$

with

$$X[k]=\sum_{k=0}^{N-1}x[n]e^{-j2\pi nk/N}\tag{6}$$

In a completely analogous manner to the continuous-time case, we can derive an expression for the DFT coefficients of the sequence $y[n]=x[n]-x[n-n_0]$, $n_0\in\mathbb{Z}$:

$$Y[k]=X[k]\big(1-e^{-j2\pi n_0k/N}\big)\tag{7}$$