# Understanding Block Diagrams in Control Setups

Working on some automation for a Boeing 747 autopilot, I am trying to wrap my head around what sections of block diagrams like this one mean:

I understand that something is being done to the yaw rate signal, but what specifically does the second block in this diagram mean? What would be the mathematical operation occurring here? Is it some kind of lag filter?

Thanks!

• Are you familiar with the concepts of Laplace tranfform, and transfer function used to describe LTI systems?
– AJN
Apr 4 at 14:20
• Unfortunately not really, those concepts are above my head. The closest thing I understand currently is simple mathematical integration. Apr 4 at 19:27
• If you really want to understand this stuff, then you need to study the following math: differential equations, the Laplace transform, and dynamic systems analysis (i.e., control systems). If you take an engineering course in control systems it'll cover just enough of the Laplace Transform to get you through the class. Apr 5 at 2:57
• Understood, thank you! I'll do some reading and learning and come back to this. Apr 5 at 11:40

but what specifically does the second block in this diagram mean?

A block diagram like the one in the question represents the relation between output of a system and its input. you can think of the variable $$s$$ as the derivative operator $$s$$ can also be thought of as the frequency variable $$j\omega$$. Then $$\frac{output}{input} = \frac{o}{i} = \frac{5}{0.4s + 1}\\ o \times (0.4s + 1) = 5 \times i\\ 0.4 \times \frac{d\ o(t)}{dt} + 1 \times o(t) = 5 \times i(t)$$

Once the input $$i(t)$$ is specified, the above differential equation can be solved to find the output $$o(t)$$.

What would be the mathematical operation occurring here? Is it some kind of lag filter?

In this case it is a low pass filter which passes the low frequencies through (with gain $$\approx 5$$), attenuates high frequency signals, and in the process creates phase lag on the output signal compared to the input. Without further context, we don't know if the engineer's original intention was signal attenuation with phase lag as side effect or vice versa.

• Interesting, so from a purely mathematical perspective the final output of this part of the block diagram is just 5 times the input at any given instant? I feel like I must be misunderstanding. Apr 4 at 19:30
• @ChristopherHelton: no, you can't just multiply by 5. As AJN points out, "the above differential equation can be solved" -- that's more than just multiplying by 5. Apr 5 at 14:58
• @AJN I deleted my comment and replaced it -- while it looked like I was trying to refute what you said, I meant to cite it to help Mr. Helton understand that you can't just multiply by a constant. Apr 5 at 14:59
• @TimWescott I had mistaken the no as applying to the bold text.
– AJN
Apr 6 at 3:39

Once again, agree with Tim! Here is some more detail.

Solving dynamic systems control problems like this is most easily done by transforming the problem from the time domain (which is where newton's equations live with time = t as the relevant variable) to the frequency domain (where frequency = s is the relevant variable), also known as the s domain.

That transformation is called the Laplace transform. Applying it to the state equation for a dynamical system (either one long n-th order equation or a matrix of n oneth-order equations) is a skill which can be taught to upper-division engineering students, believe it or not. The matrix form is particularly handy because computers are really good at inverting matrices.

In the s domain, differentiation is algebraically handled as multiplication and integration is handled as division, making these operations quick and easy to solve by hand. Having done so once for any particular class of dynamic system topology, you have therefore solved it for all members of that class and are thereby entitled to take the rest of the afternoon off and go home and read magazines.

• Thanks for the clarifications. I'll have to do more learning on differentiation and transfer functions before I think I'll truly understand, but I will come back to this after doing some learning! Thank you! Apr 5 at 11:43

The block on the left is a source of a measurement of the yaw rate, in degrees per second.

The block on the right has the transfer function $$\frac{5}{0.4 s + 1} \tag 1$$

This is Laplace-domain notation for a first-order low-pass filter with a DC gain of 5 and a time constant of 0.4 seconds.