Feedback

Let's consider the LTI system in Image 2. It's easy to see that the system is unstable. Those systems often represent real-world scenarios in which instabilities could be very dangerous like the Tacoma Narrows Bridge that collapsed in 1940 after the wind made it oscillate in a frequency where it resonated with the bridge's natural frequency. It could also represent systems where engineers are trying to control like on robotics and lasers where currents and voltages need to be very precise and can't vary a lot.

Image 1, Tacoma Narrows Bridge (1940)

Image 2, Unstable LTI system

So we often want systems to be stable, and one way we can do this is by adding the output of the system to its input and using this as the system input. This is known as a feedback loop and can be seen in Image 3.

Image 3, Feedback loop

To understand the feedback loop, it's useful to get its transfer function. For that, let's analyze the systems in Image 4.

Image 4, Closed loop systems

Those loops are very common in control theory where \(G(S)\) represents the system to be controlled, \(K\) is the controller, and \(H(S)\) is the sensor used to measure the output.

The transfer function of the system presented in the image is

\[\frac{Y(S)}{X(S)} = \frac{KG(S)}{1 + KG(S)H(S)}.\]

Show proof

Let's consider the signal \(E(S):\)

From it, we get the following equations:

\[\begin{align} E(S) &= X(S) - H(S)Y(S)\\ Y(S) &= KG(S)E(S). \end{align}\]

By replacing \(E(S)\) in the second equation with that in the first equation, we get

\[\begin{align} Y(S) &= KG(S)\left[ X(S) - H(S)Y(S) \right] \\ &=Y(S) + KG(S)H(S)Y(S) \\ &= KG(S)X(S) \\\\ Y(S)\left[1 + KG(S)H(S)\right] &= X(S)KG(S)\\\\ \frac{Y(S)}{X(S)} &= \frac{KG(S)}{1 + KG(S)H(S)}. \end{align}\]

By doing analogous process, we reach a transfer function of \(\frac{Y(S)}{X(S)} = \frac{KG(S)}{1 - KG(S)H(S)}\) for the system in (b).

What is the transfer function of the following system?

Show solution

By applying the formula previously mentioned, we get

\[\frac{\frac{3}{s-1}}{1 + \frac{3}{(s-1)(s-3)}} = \frac{\frac{3}{s-1}}{\hspace{2mm} \frac{s^2 - 7s + 3 + 3}{(s-1)(s-3)}\hspace{2mm} } = \frac{3(s-3)}{s^2 - 7s + 6}.\]

So,

\[\frac{Y(S)}{X(S)} = \frac{3s-9}{s^2 - 7s + 6}.\]

In electronics, feedback is often used as a means to deliver a voltage or current that will be more reliable and resistant to noise