Project Alexandria

2.4b Feedback

3 min read

Feedback (Black's Formula)

Feedback refers to the process in which a portion (or all) of the output of a system is returned (fed back) to its input, such a system in positive feedback has the following block diagram:

\usepackage{amsmath}
\usetikzlibrary{positioning}
 
\begin{document}
\begin{tikzpicture}[scale=1.3, transform shape]
 % Main Node
 \node (s) {\(P(\omega)\)};
 \draw[thick] (1, -0.5) -- (-1, -0.5) -- (-1, 0.5) -- (1, 0.5) -- cycle;
 
 % Feedback Node
 \node[below=of s] (in) {\(K(\omega)\)};
 \draw[thick] (0.75, -2.15) -- (-0.75, -2.15) -- (-0.75, -1.15) -- (0.75, -1.15) -- cycle;
 
 % Arrows
 \draw[->, thick] (-4, 0) -- (-2.2, 0) node[at start, left]{$x(t)$};
 \draw[->, thick] (1, 0) -- (4, 0) node[at end, right]{$y(t)$};
 \draw[->, thick] (2, 0) -- (2, -1.65) |- (0.75, -1.65);
 \draw[->, thick] (-0.75, -1.65) -- (-2, -1.65) |- (-2, -0.2);
 
 \draw[thick, dashed] (-2.5, 1) -- (2.5, 1) -- (2.5, -2.5) -- (-2.5, -2.5) -- cycle;
 \node at (-2.25, -2.25) {H} ;
 
 \draw[thick] (-2,0) circle[radius=0.175] node{+};
 \draw[->, thick] (-1.825, 0) -- (-1.05, 0);
\end{tikzpicture}
\end{document}

where is the system being controlled (known as the plant), and is the device that adjusts the plant’s input based on the difference between desired and actual output, also known as the controller.

Harold Black’s solution to a feedback system was to instead apply negative feedback (i.e. when the adder on the left is negative) to a system. The transfer function of such a system can be expressed as the forward gain divided by 1 plus the loop gain, such that

Black argued that since positive feedback is unstable, negative feedback is a more stable alternative to traditional (positive) feedback.

Proof

We proceed by analyzing the block diagram for continuous-time LTI systems. If the input to the feedback system is , then the input to the system (since convolution is multiplication in the frequency domain) is

z(t) &= e^{j\omega t} + H(\omega)K(\omega)e^{j\omega t} \\ &= [1+H(\omega)K(\omega)]e^{j\omega t} \end{align}$$ Its output then becomes $$y(t)=[1+H(\omega)K(\omega)]P(\omega)e^{j\omega t}.$$ However, since $y(t) = H(\omega)e^{j \omega t}$, $$H(\omega)=\frac{P(\omega)}{1-K(\omega)P(\omega)}$$ Since in Black's solution, the adder is negative (the system is in negative feedback), this means that the loop gain is also negative. As such, $$H(\omega) = \frac{P(\omega)}{1 + K(\omega)P(\omega)}.$$ # Examples >[!example] Stabilizing an Unstable System > >Consider a current source in series with a capacitor. Such a system has the equation >$$i_{s}(t) = C\frac{d}{dt}v_{1}(t) \iff \frac{d}{dt}v_{1}(t) = \frac{1}{C}i_{s}(t).$$ >Let's consider the impulse response of this system. For $i_{s}(t)=\delta(t)$, >$$v_{1}(t) = h(t) = \frac{1}{C}u(t),$$ >since by definition $\int\delta(\tau)d\tau = u(t)$. However, this system is not BIBO stable; $\int^{\infty}_{-\infty}|h(\tau)|d\tau = \infty$. We can fix this by adding a resistor in parallel with the capacitor. Here, the voltage $v_{1}(t)$ is the same, but the current $i_{s}(t)$ is divided across the resistor and the capacitor. The differential equation for this new circuit is >$$i_{s}(t) = C\frac{d}{dt}v_{1}(t) + \frac{1}{R}v_{1}(t) \iff \frac{d}{dt}v_{1}(t) = \frac{1}{C}i_{s}(t) - \frac{1}{RC}v_{1}(t).$$ >Here, we essentially have a resistor in negative feedback, as seen in the differential equation above. The impulse response to $i_{s}(t) = \delta(t)$ of this circuit is >$$h(t) = \frac{1}{C}e^{-\frac{1}{RC}t}u(t),$$ >which is a standard solution to a first-order differential equation. This RC circuit is now BIBO stable, with a bound $\int^{\infty}_{-\infty}|h(\tau)|d\tau = R$ (sanity-check this on your own).

Graph View