Current source and switched capacitors in parallel

A capacitor is an electronic device able to store electrical energy in an electrical field. Usually, the capacitor is defined in its most simple version as a device with two plates with area \(A\), separated by air (or any other dielectric material) a distance \(d\).

Parallel plate capacitor.svg
By inductiveload – own drawing, done in Inkscape 0.44, Public Domain, Link

If a current source is forced through the capacitor, the electrons (charge) will be deposited in one of the plates, creating in turn a electrical field across them. There won’t be any effective charge transference from one plate to the other because the space between them is filled with a dielectric material (non conductive). However, the electrical field across it can force the repulsion or attraction of charge at the other side of the plate.

Capacitor schematic with dielectric.svg
By Papa November – self-made SVG version of Image:Dielectric.png, incorporating Image:Capacitor schematic.svg as its base., CC BY-SA 3.0, Link

A capacitor is characterized by its capacitance. The capacitance is measured in Farads (F) and defines the ratio between the amount of charge needed to increase one volt at the terminals of the capacitor.

\[ C= \frac{Q}{V} \]

Therefore, a capacitor with 1 F will need 1 Coulomb (1 C) of charge to set 1 V across its terminals. Remember, that 1 C represents the amount of energy transported by a constant current of 1 A in 1 second.

Behavior of a capacitor connected to a DC current source

Let’s see what happens when we connect a DC current source to a capacitor. Transforming a little bit the previous expression, we can obtain:

\[ C = \frac{Q}{V}  \Rightarrow V = \frac{Q}{C} \]

As \(Q = \int{i\left(t\right) dt}\), we can get the voltage across the capacitor as a function of the time and the current:

\[ V\left(t\right) = \frac{1}{C} \int{i\left(t\right) dt} \]

Therefore, if we inject a constant current of \(1~mA\) for \(1~ns\), the voltage \(V_1\left(t\right)\) will be as follows:

In this example, what would be the charge store in the capacitor at time \(t_1 = 1~ns\)?

\[ C = Q/V \Rightarrow Q = C \cdot V = 1~pF ·  1V = 1\cdot10^{-12}~C \]

Now, what would be the value of the voltage \(V_1\) if at time \(t_1 = 1~ns\) a second capacitor \(C_2\) (discharged) of \(2~pF\) of capacitance?

At that very moment in which the second capacitor is connected in parallel, the charge in \(C_1\) will be distributed between \(C_1\) and \(C_2\).

The effective capacitance now is \(3~pF\) (\(1~pF  + 2~pF\)). Remember that using the capacitance definition, we got:

\[ V = \frac{Q}{C_p} = \frac{Q}{C_1 + C_2} = \frac{1\cdot 10^{-12} C}{1\cdot10^{-12}~F + 2\cdot10^{-12}~F} = 0.333~V \]

Therefore, as the capacitance now has increased and the charge remains the same (although shared between \(C_1\) and \(C_2\)), according to the previous expression the voltage has to drop.

If we would like to know the charge stored in every individual capacitor at time \(t_1 = 1~ns\), we could compute it as:
\[ Q = C \cdot V\]
\[ Q_1\left(t_1 = 1~ns\right) = C_1 \cdot V_1\left(t_1 = 1~ns\right) = 1~pF \cdot 0.333~V = 3.33\cdot10^{-13}~C\]
\[ Q_2\left(t_1 = 1~ns\right) = C_2 \cdot V_1\left(t_1 = 1~ns\right) = 2~pF \cdot 0.333~V = 6.66\cdot10^{-13}~C\]

The result is consistent with the previous calculation stating that the amount of charge at \(t_1=1~ns\) was \(Q = 10^{-12}~C\).

Finally, let’s see what happens if at time \(t_2 = 2~ns\) the capacitor \(C_2\) is again disconnected and only capacitor \(C_1\) is connected to the current source.

At time \(t_2 = 2~ns\) the voltage \(V_1\) will have increased to:

\[V\left(t_2 = 2~ns\right) = V_1\left(t_{1}^{+}\right) + \frac{1}{C_1 + C_2} \int_{t_1 = 1~ns}^{t_2 = 2~ns}{i\left(t\right) dt} \]

\[V\left(t_2 = 2~ns\right) = 0.333 V + \frac{1}{1~pF + 2~pF} \int_{t_1 = 1~ns}^{t_2 = 2~ns}{1~mA~dt} = 0.333~V +0.333~V = 0.666~V\]

Now the charge will be distributed as follows:

\[ Q_1\left(t_2 = 2~ns\right) = C_1 \cdot V_1\left(t_1 = 2~ns\right) = 1~pF \cdot 0.666~V  = 6.66\cdot10^{-13}~C\]
\[ Q_2\left(t_2 = 2~ns\right) = C_2 \cdot V_1\left(t_1 = 2~ns\right) = 2~pF \cdot 0.666~V = 1.22\cdot10^{-12}~C\]

If \(C_2\) is disconnected, the charge stored in it will be lost and won’t be redistributed towards \(C_1\). Therefore, the total amount of charge in the system at \(t_2 = 2~ns\) will only be that on \(C_1\).

Now, the voltage increment at the same rate as in the period \(0 < t < t_1 = 1~ns\).

Therefore, if we plot the voltage from \(t=0~ns\) to \(t_3 = 3~ns\), we would get the following voltage profile for \(V_1\):

Permanent record

The freedom of a country can only be measured by its respect for the rights of its citizens, and it’s my conviction that these rights are in fact limitations of state power that define exactly where and when a government may not infringe into that domain of personal or individual freedoms that during the American Revolution was called «liberty» and during the Internet Revolution is called «privacy».

Edward Snowden, Permanent Record (2019)

Romancero bot: converting article text into voice with Amazon Polly and sending it to Telegram

The following Python script extracts the content of an article using Readibility (ported to Python), converts it to voice using the Amazon Polly service and finally sends the audio as a voice note to a given user in Telegram using Telethon (Telegram client for Python).

For running the script you will need to install the following Python packages:

Also, you will need to create a AWS account. If you already have an AWS account, make sure that you have a user created in the IAM Management Console with the following permissions:

User permissions for creating Polly jobs and accessing/writing files in a S3 bucket.

When creating this user, make sure you write down its ID access key and its secret access key. You will need them to configure the aws-cli client.

With this Amazon credentials, you can configure the AWS client by executing the following command:

In this step, you will need to fulfill the details with the user credentials you wrote down when creating it.

Now, you will need to create a Telegram API ID. For this, you can go to the Telegram section «Create an Application«. After following the steps described in the official documentation, you will obtain an API ID (it’s a number) and a API hash (it’s a string).

With these steps already completed, you can place all the needed details in the script and run it.

The user you specified will receive a message like this:

Parser in PHP using regular expressions

You can use regular expressions in PHP with the function preg_match ( string $pattern , string $subject [, array &$matches [, int $flags = 0 [, int $offset = 0 ]]] ) . Only the first two paremeters are mandatory and they are the regex and the string where you want to search respectively.

In case of finding a result, preg_match() returns an array where the item at index 0 is the whole match. From 1 onwards they are placed the different groups of your regular expressions (in case there is any). If no match is found, preg_match() returns null.

One of the details that must be taken into account when using regular expressions on PHP is that they must be enclosed by forward slashes (/), e.g. $multiline_meaning_re = '/^([A-za-z ,"().\';:]+)/'; . This regular expression matches any string with any set of letters, spaces, commas, double and single quotes, parenthesis points, colon and/or semicolon.

As a complete example, the following snippet opens a file, parses it to look for English idioms and uploads all of them a MySQL database.

You can find further information about the preg_match() in the PHP official documentation.

Introduction to D3 (Data-Driven Document)

One of the first questions you may ask yourself when getting introduced in D3 is: why are we using selectAll(‘html-tag-name’) method if there is no item to select of that type?

First, let’s see an example of the situation we are talking about:

In the previous example, the only existing HTML tag is <body> . We select body ( d3.select('body') ) and then we perform the .selectAll('h2') . On its own, it makes no sense as the method won’t return any value since no <h2>  tag exists. Nevertheless, it will make sense if we keep looking at the following code.

After the select .selectAll('h2')  we attach the existing dataset to the selected items ( .data(dataset) ). Then, we use the enter()  method, which gives meaning to the previous .selectAll('h2') . When using enter() , D3 looks for the number of selected items to bind them with the data. In case of having not enough items in the selection, the enter()  method will create them.

Therefore, as .selectAll('h2')  was empty and the dataset  variable contains 9 elements, it will iterate the code 9 times. In case of having already created some  <h2>  elements, it will simply fulfill the HTML code the necessary iteration to cover all the dataset  elements. Remember that who does this iteration is  the data() method.  It parses the data set, and any method that’s chained after data() is run once for each item in the data set.

You can find more information in the official documentation at the D3js.org website.

Scales

In D3 there exists the Scale function to change the value of the data set so that it can fit in the screen. Two important methods are range() and  domain(). The domain method covers the set of input values whereas the range function convers the set of output values. Let’s see an example:

From freeCodeCamp:

The domain()  method passes information to the scale about the raw data values for the plot. The range()  method gives it information about the actual space on the web page for the visualization.

Send audio from mobile phone (Android/iOS) to Raspberry

Platform used: Raspberry Pi 3 B+
Bluetooth module: built-in
First, update firmware to make sure you have latest version:

Now, install all the needed pulse-audio packages:

Configure the bluetooth in the Raspberry:

In your mobile phone, search for the raspberry bluetooth signal and pair to it:

Now, you can try to play some audio in your mobile and it should be reproduced in the the Raspberry.
In my case, audio was driven out to the HDMI. In case you want to switch it through the Jack 3.5 mm output you can run:

Then, select Advanced Options > Audio > Force 3.5 mm (‘headphone’) jack
With this all set, you can stream any audio such as the built-in music player, Spotify or YouTube to the Raspberry and from it to the connected speakers or HDMI display.

Finally, as the device has been stored as a trusted device, every time the Raspberry is booted, you won’t need to repeat this process. It will be so easy as connecting your mobile phone to the available raspberry bluetooth signal.

If sound sounds distorted, try restarting pulseaudio with:

Handling with several devices

If you pair several devices, only first connected device will be able to stream audio. If I connect my computer to the raspberry and then I try to connect my mobile phone (both previously paired and trusted), phone connection will fail. First, I’ll need to disconenct my computer and only then I’ll be able to successfully connect to the raspberry with my mobile phone.

Web scraper with Scrapy

Basis in a vector space

A vector space basis is the skeleton from which a vector space is built. It allows to decompose any signal into a linear combination of simple building blocks, namely, the basis vectors. The Fourier Transform is just a change of basis.

A vector basis is the linear combination of a set of vector that can write any vector of the space.

\[ w^{(k)} \leftarrow \text{basis} \]

The canonical basis in \(\mathbb{R}^2\) are:

\(e^{(0)} = [1, 0]^T \ \ e^{(1) } = [0,1]^T \)

Nevertheless, there are more basis for \(\mathbb{R}^2\):

\(e^{(0)} = [1, 0]^T \ \ e^{(1) } = [1,1]^T \)

This former basis is not linearly independent as information of \(e^{(0)}\) is inside \(e^{(1)}\).

Formal definition

H is a vector space.

W is a set of vectors from H such that \(W = \left\{ w^{(k)}  \right\} \)

W is a basis of H if:

  1. We can write \( \forall x \in H\): \( x = \sum_{k=0}^{K-1} \alpha_k w^{(k)}, \ \ \alpha_k \in \mathbb{C} \)
  2. \( \alpha_k  \) are unique, namely, there is linear independence in the basis, as a given point can only be expressed in a unique combination of the basis.

Orthogonal basis are those which inner product returns 0:

\( \left \langle w^{(k)}, w^{(n)} \right \rangle = 0, \ \ \text{for } k \neq n \)

In addition, if the self inner product of every basis element return 1, the basis are orthonormal.

How to change the basis?

An element in the vector space can be represented with a new basis computing the projection of the current basis in the new basis. If \(x\) is a vector element and is represented with the vector basis \(w^{(K)}\) with the coefficients \(a_k\), it can also be represented as a linear combination of the basis \(v^{(k)}\) with the coefficients \( \beta_k\). In a mathematical notation:

\[ x = \sum_{k=0}^{K-1} \alpha_k w^{(k)} = \sum_{k=0}^{K-1} \beta_k v^{(k)} \]

If \(\left\{ v^{(k)} \right\}\) is orthonormal, the new coefficients \(\beta_k\) can be computed as a linear combination of the previous coefficients and the projection of the new basis over the original one:

\[\beta_h = \left \langle v^{(h)}, x \right \rangle = \left \langle v^{(h)}, \sum_{k=0}^{K-1} \alpha_k w^{(k)} \right \rangle = \sum_{k=0}^{K-1} \alpha_k \left\langle v^{(h)}, w^{(k)} \right \rangle \]

This operation can also be represented in a matrix form as follows:

\[ \beta_h = \begin{bmatrix}
c_{00} & c_{01} & \cdots & c_{0\left(K-1 \right )}\\
& & \vdots & \\
c_{\left(K-1 \right )0} & c_{\left(K-1 \right )01} & \cdots & c_{\left(K-1 \right )\left(K-1 \right )}
\end{bmatrix}\begin{bmatrix}
\alpha_0 \\
\vdots \\
\alpha_{K-1}
\end{bmatrix} \]

This operation is widely used in algebra. A well-known example of a change of basis could be the Discrete Fourier Transform (DFT).

Inner product in vector space

The inner product is an operation that measures the similarity between vectors.  In a general way, the inner product could be defined as an operation of 2 operands, which are elements of a vector space. The result is a scalar in the set of the complex numbers:

\[ \left \langle \cdot, \cdot \right \rangle : V \times V \rightarrow \mathbb{C}  \]

Formal properties

For \(x, y, z \in V\) and \(\alpha \in \mathbb{C}\), the inner product must fulfill the following rules:

To be distributive to vector addition:

\( \left \langle x+y, z \right \rangle = \left \langle x, z \right \rangle + \left \langle y, z \right \rangle \)

Conmutative with conjugate (applies when vectors are complex):

\( \left \langle x,y \right \rangle  = \left \langle y, x \right \rangle^* \)

Distributive respect scalar multiplication:

\(  \left \langle \alpha x, y \right \rangle =  \alpha^* \left \langle x, u \right \rangle \)

\(  \left \langle  x, \alpha y \right \rangle =  \alpha \left \langle x, u \right \rangle \)

The self inner product must be necessarily a real number:

\(  \left \langle  x, x \right \rangle \geq 0 \)

The self inner product can be zero only when the element is the null element:

\( \left \langle x,x \right \rangle = 0 \Leftrightarrow x = 0 \)

Inner product in \(\mathbb{R}^2 \)

The inner product in \( \mathbb{R}^2\) is defined as follows:

\( \left \langle x, y \right \rangle = x_0 y_0 + x_1 y_1 \)

In self inner product represents the squared norm of the vector:

\( \left \langle x, x \right \rangle = x^2_0 + y^2_0 = \left \| x \right \|^2 \)

Inner product in finite length signals

In this case, the inner product is defined as:

\[ \left \langle x ,y \right \rangle = \sum_{n= 0}^{N-1} x^*[n] y[n] \]

Properties of vector spaces

Vector spaces must meet the following rules:
Addition to be commutative:
\( x + y = y + x \)

Addition to be distributive:
\( (x+y)+z = x + (y + z) \)

Scalar multiplication to be distributive with respect to vector addition:
\( \alpha\left(x + y \right) = \alpha x + \alpha y\)

Scalar multiplication to be distributive with respect to vector the addition of field scalars:
\( \left( \alpha + \beta \right) x = \alpha x + \beta y \)

Scalar multiplication to be associative:
\( \alpha\left(\beta x \right) = \left(\alpha \beta \right) x \)

It must exist a null element:
\( \exists 0 \in V \ \ | \ \ x + 0 = 0 + x = x \)

It must exist an inverse element for every element in the vector space:
\( \forall x \in V \exists (-x)\ \ | \ \ x + (-x) = 0\)