This example can be developed on the blackboard for 10 to 15 minutes.

## Example: Fourier $d_n$ coefficients of an asymmetrical square pulse train

### Problem Statement

Find the coefficients for the Fourier series of the signal $f(t)$

### Approach

#### Pulse Train

The signal is a pulse train, as in previous example, but in this case, the pulse occupies one-fifth of the train; or in other words, one-tenth of the period.

#### Generating Function

\begin{equation} f(t) = \pi \left( \frac{t-nT}{T/10}\right) \end{equation}

#### Spectrum

For this example, finding the coefficients for the Fourier serier of the signal $d_n$ will be equivalent to represent the spectrum of the signal.

### Solution

By only varying the integration limits, we can follow the calculations from the first example, where the description of the steps can be read.

\[\begin{aligned} d_n &= \frac{1}{T} \int_{-\infty}^{\infty} f(t) e^{-j 2\pi n f_0 t} \, dt \\ &= \frac{1}{T} \int_{-T/20}^{T/20} \pi \left( \frac{t}{T/2} \right) e^{-j 2\pi n f_0 t} \, dt \\ &= \frac{1}{T} \cdot \frac{-1}{j 2\pi n f_0} \left[ e^{-j 2 \pi n f_0 t} \right]_{-T/20}^{T/20} \\ &= \frac{1}{T} \cdot \frac{-1}{j 2\pi n f_0} \left[ e^{\frac{-j 2 \pi n f_0 T}{10}} - e^{\frac{+j 2 \pi n f_0 T}{4}} \right] \\ &= \frac{e^{\frac{j \pi n}{10}} - e^{\frac{-j \pi n}{10}}}{j 2\pi n} \\ &= \frac{\cos\left(\frac{\pi n}{10}\right) + j \sin\left(\frac{\pi n}{10}\right) - \cos\left(\frac{-\pi n}{10}\right) - j \sin\left(\frac{-\pi n}{10}\right)}{j 2\pi n} \\ &= \frac{\cos\left(\frac{\pi n}{10}\right) + j \sin\left(\frac{\pi n}{10}\right) - \cos\left(\frac{\pi n}{10}\right) + j \sin\left(\frac{\pi n}{10}\right)}{j 2\pi n} \\ &= \frac{j 2 \sin\left(\frac{\pi n}{10}\right)}{j 2\pi n} = \frac{\sin\left(\frac{\pi n}{10}\right)}{\pi n} = \frac{1}{10} \frac{\sin\left(\frac{\pi n}{10}\right)}{\frac{\pi n}{10}} = \frac{1}{10} sinc_n\left(\frac{n}{10}\right) \end{aligned}\]### Conclussions

#### Spectrum Morphology

The amplitude spectrum in the last figure although with a lower level (the maximum amplitude value is 0.1), logically has the same shape as the corresponding pulse train with the same high and low duration; however, the density of spectral lines has increased. The width of the main lobe has also increased, which could mean that if a single harmonic was enough to reconstruct the wide pulse with good approximation, 10 harmonics would be needed to reconstruct the narrow pulse.

#### Bandiwdth

Narrow pulses require more \textbf{bandwidth} to be reconstructed. An amplifier for \textit{fast} signals requires greater bandwidth than an amplifier for \textit{slow} signals, even if they are signals of the same frequency.

#### Limit case $T \simeq \infty$

If $T$ increases to become infinite, which would be equivalent to a signal with a single pulse, then the spectrum density would increase until it becomes continuous.