Gene Surdotovich and I are hard at work preparing the 6th edition of *Intermediate Physics for Medicine and Biology*. One change compared to the 5th edition is that we are redrawing most of the figures using Mathematica. It’s a lot of work, but the revised figures look great and many are in color.

One advantage of redrawing the figures is that it forces us to rethink what the figure is all about and if it makes sense. This brings me to Figure 10.13 in the chapter about feedback. Specifically, it is from Section 10.6 about a negative feedback loop with two time constants. Without going into detail, let me outline what this figure is describing.

Chapter 10 centers around one particular feedback loop, relating the amount of carbon dioxide in your lungs (which we call *x*) to your breathing rate ( *y*). The faster you breath, the more CO2 you blow out of your lungs, so an increase in *y* causes a decrease in *x*. But your body detects when CO2 is building up and reacts by increasing your breathing rate, so an increase in *x* causes an increase in *y*. There is one additional parameter, your metabolic rate, *p*. If your metabolic rate increases, so does the amount of CO2 in your lungs.

Our book emphasizes mathematical modeling, so we develop a toy model of how *x* and *y* behave. We assume that initially *x* and *y* are in steady state for some *p*, and call these values *x*0, *y*0, and *p*0. At time *t* = 0, *p* increases from *p*0 to *p*0 + Δ *p*, which could represent you starting to exercise. How do *x* and *y* change with time? We define two new variables, *ξ* and *η*, that represent the deviation of *x* and *y* from their steady state values, so *x* = *x*0 + *ξ* and *y* = *y*0 + *η*. We then develop two differential equations for *ξ* and *η*,

The variables *ξ* and *η* have different time constants, *τ*1 and *τ*2. The parameters *G*1 and *G*2 are the “gains” of the system, determining how much *ξ* changes in response to *η*, and how much *η* changes in response to *ξ*. In our model, *G*1 is negative (an increase in breathing rate causes the amount of CO2 in the lungs to decrease) and *G*2 is positive (an increase in CO2 causes the rate of breathing to increase). The “open loop gain” of the feedback loop is the product *G*1*G*2. Finally, the constant *a* is simply a factor to get the units right.

All is good so far. But now let’s look at the 5th edition’s version of Fig. 10.13.

What’s wrong with it? First, the calculation uses a positive value of *G*1 and a negative value of *G*2, so it doesn’t correspond correctly to our model, which has negative *G*1 and positive *G*2. Second, the calculation uses Δ*p* = 0, so the steady state values of *x* and *y* don’t change and *ξ* and *η* both approach zero. That’s odd. I thought the whole point of the model was to look at how the system responds to changes in *p*. Finally, the initial values of *ξ* and *η* are not zero. What’s up with that? We know their values are zero for *t* < 0, when *x* = *x*0 and *y* = *y*0. How could they suddenly change at *t* = 0?

In the 6th edition, the new version of Figure 10.13 is going to look something like this:

The figure has color and switches from landscape to portrait orientation. Those changes are trivial. Here are the important differences:

- I made
*G*1 negative and*G*2 positive, like in our breathing model. Now an increase in CO2 causes the body to increase the breathing rate, rather than decrease it as in the 5thedition figure. - The parameter Δ
*p*is no longer zero. To be simple, I set*a*Δ*p*= 1. The person starts exercising at*t*= 0. - Because there is a change in metabolic rate, the new steady state values of
*ξ*and*η*are not zero. In fact, they are equal to*ξ*=*a*Δ*p*/(1-*G*1*G*2) and*η*=*G*2*a*Δ*p*/(1-*G*1*G*2). Notice how the factor of 1-*G*1*G*2 plays a big role. Since the product*G*1*G*2 is negative, this means that 1-*G*1*G*2 is a positive number greater than one. It’s in the denominator, so it makes*ξ*smaller. That’s the whole point. The feedback loop is designed to keep*ξ*from changing much. It’s a control system to suppress changes in*ξ*. To make life simple, I set*G*1 = −5 and*G*2 = 5 (the same values from the 5thedition except for the signs), so the open loop gain is 25 and the steady state value of*ξ*is only 1/26 of what it would be if no feedback were present (in which case,*ξ*would rise monotonically to one while*η*would remain zero). - The initial values of
*ξ*and*η*are now zero, so there is no instantaneous jump of these variables at*t*= 0.

When revising the edition of *IPMB*, I began wondering why Russ Hobbie and I never worried about the units for the time constants, the gains, *a*, or Δ*p*. This motivated me to write a new homework problem for the 6th edition, in which the student is asked to rewrite the model equations in nondimensional variables *Ξ*, *Η*, and *T* instead of *ξ*, *η*, and *t*. Interestingly, such a switch results in a pair of differential equations for *Ξ* and *Η* that depend on only two nondimensional parameters: the ratio of time constants and the open loop gain. So, our plot in the 5th edition has the qualitative behavior correct (except for the signs of *G*1 and *G*2). The system oscillates because the open loop gain is so high. The correct units for the various parameters would only rescale the horizontal and vertical axes.

Is the new version of Figure 10.13 in this blog post what you’ll see in the 6th edition of *IPMB*? I don’t know. I haven’t passed the figure by Gene yet, and he’s my Mathematica guru. He might make it even better.

What’s the moral of this story? THINK BEFORE YOU CALCULATE! That’s the motto I often would tell my students, but it applies just as well to textbook authors. The plot should not only be correct but also make physical sense. You should be able to explain what’s happening in words as well as pictures. If you can’t tell the story of what’s taking place by looking at the figure, something’s wrong.

Finally, is there really no physical problem that the original version of Fig. 10.13 describes? Actually, there is. Imagine you are resting throughout this “event”; you sit in your chair and don’t change your metabolic rate, so Δ*p* = 0 meaning *p* is the same before and after *t* = 0. However, at time *t* = 0, your “friend” sneaks up on you, shoves a fire extinguisher in front of your face, and gives you a quick, powerful blast of CO2. Except for the sign issue on *G*1 and *G*2, the original figure shows how your body would respond.

*Originally published at **http://hobbieroth.blogspot.com**.*