### Automatics

**Chapter 13. Differential Equations**

**Chapter 13.1 Introduction**

This is the first approach to

**differential equations**. We will analyze the process of filling the

**tank without a hole**and with a

**hole in the bottom**. Then simple

**differential equations**will appear. There will be a lot of descriptions and the topic is not the easiest one. But try to face it. If there are still problems, raise the white flag and read on from

**chapter 14**. Only then treat the transfer function

**G(s)**as “something” that converts input

**x(t)**to output

**y(t)**.

**Chap. 13.2 Filling the tank without a hole**

**Flow**

Fig. 13-1

Fig. 13-1

**Q=1 liter/sec**means that

**1 liter**of water

**flows**out of the tap

**in 1 second**.

Generally –

**Flow**is the volume of liquid that flowed out of the tap

**in 1 second**. At the same time, it is the volume of liquid

**V(t)**by which the tank was enriched in

**1 second**. So we can easily calculate the volume

**V(t**) after time

**t**or the height

**h(t)**. But only if the flow is

**constant**over time! So we do not turn the valve, as we have done many times with the slider.

Fig. 13-2

Fig. 13-2

**Fig. 13-2a**is the flow formula, provided that we do not turn the valve. So it’s constant. And when it’s not constant? Then the approximate formula for

**instantaneous flow**appears (

**Fig. 13-2b**), where

**ΔV**is the volume of liquid in the tank that flowed out in time

**Δt**. This formula is the more accurate the shorter the time

**Δt**. Then the flow

**Q(t)**becomes more and more similar to the

**derivatives**shown in

**Fig. 13-2c**or

**Fig. 13-2d**. These are the simplest forms of

**differential equations**.

Fig. 13-3

Fig. 13-3

**Fig. 13-3a**is a differential equation whose “known” is the flow function

**Q(t)**and the unknown is the volume function

**V(t)**. We do not yet know what the

**V(t)**time charte is, but we know that its derivative from

**V(t)**is the

**Q(t)**flow. How to find the

**V(t)**function that satisfies this equation?

In the case when

**Q(t)**is constant, as in

**Fig. 13-1 Q=1liter/sec**, the matter is trivial.

**V(t)=Q*t**. A typical linear function.

**V(t=0)=0**,

**V(t=1 sec)=1**liter,

**V(t=2 sec)=2 liters**, etc.

And when

**Q(t)**is time-varying? Then we solve the differential equation in

**Fig. 13-3a**by integrating both sides.

**Fig. 13-3b**is the “integrated on both sides” equation

**Fig. 13-3a**

**Fig. 13-3c**is the result of “bilateral integration”. Red arrows and

**Fig. 12-7c**

**from chapter. 12**tell why.

By the way. Already in elementary school, in tasks like “A train goes from

**city A**to

**B**at a constant speed…” you were solving differential equations! The velocity

**V**is the

**derivative**of the distance

**S**. When the

**velocity**is constant, the formula

**S=V*t**is a solution to the differential equation

**S'(t)=V**. But let’s go back to the tank without a hole as an example of the simplest differential equation

**The formula**

Fig. 13-4

Fig. 13-4

**V'(t)=Q(t**) is in other words “instantaneous flow

**Q(t)**derivative of flow

**V(t)**“, more figuratively – volume growth rate

**V'(t**). When you fill the bathtub, the speed of increasing the volume

**V'(t)**is more associated with the speed of increasing the level in the bathtub

**h'(t)**. Let’s try to simplify the general formula

**V'(t)=Q(t**) as much as possible.

Assuming that:

– the bathtub is a rectangular prism with an area of

**1 m²**

– at the time

**t=0**, we quickly turned on the

**tap**and from then on there is a constant flow

**Q=1 liter/sec**

– the bathtub is governed by the differential equation

**h'(t)=1.25(t)**

where

**1.25(t)**is the flow unit step from

**0 to 1.25 liters/sec**, and more generally

**y'(t)=x(t)**where

**x(t)**is the input and

**y(t)**is the output.

Our goal is to find a

**h(t)**function that satisfies this equation. Let’s integrate both sides of the equation as in

**Fig. 13-3c**

Fig. 13-5

Fig. 13-5

**Fig. 13-5a**– differential equation describing the filled tank, where the output is the water level

**h(t)**

**Fig. 13-5b**– integration of both sides of the version from

**Fig. 13-5a**

**Fig. 13-5c**– solution as a result of the above integration

**Fig. 13-5d**– definition of the flow step

**1.25(t**) as the

**input signal**

**Fig. 13-5d**– definition of the solution as an increasing sawtooth of the level

**h(t)**

Let’s solve the differential

**Fig.13-5a**equation using the free

**Scilab**program.

**Fig. 13-6**From

**chapter 12**, e.g.

**Fig. 12-14**, we know that the output of the

**integral unit**is the

**integral of the input**and vice versa, the

**inpu**t is the

**derivative of the output**. Therefore, we conclude that

**x(t)=y'(t)**, i.e.

**1.25(t)=h'(t)**. Thus, the scheme in

**Fig. 13-6**solves the differential equation

**h'(t)=1.25(t)**. Let’s check it:

At the moment

**t=0**, the valve opened immediately. From now on, the flow

**Q(t)=1.25 liter/sec**is constant. The

**time chart**shows that the water level rises at a constant speed in accordance with the formula

**h(t)=1.25*t(t)**, where the level is given

**in millimeters**. e.g. In

**10**seconds, the water level in the tank is

**h=12.5 mm**.

*****

Let me remind you that

**for t<0**(negative)

**x(t)=y(t)=0**.

*****Don’t worry about the strange size of the tank in which

**S=1m2**and the level h(t) will rise by approx.

**12.5mm**.

**Chapter 13.3 Filling a tank with a hole k=1****Fig. 13-7**Unlike

**Fig. 13-4**, the water level will rise more and more slowly due to the outflow of

**Q2(t)**. Previously, the rate of level rise

**h'(t)**was proportional to the input flow

**Q(t)**, so now it will be proportional to the flow

**difference Q1(t)-Q2(t)**. We know that the higher the water level, the greater the outflow. So

**Q2(t)=k*h(t)**. Question. How

**k**depends on the area of the hole? Sure, the

**bigger**the hole, the bigger

**k**! When

**k=0**, there is no hole and there is the previous case, i.e.

**Fig. 13-4**.

Let’s assume that the size of the hole corresponds to

**k=1**.

The figure shows that the above tank is described by the differential equation:

**1.25(t) = h'(t) + h(t)**

and more generally:

**x(t)=y'(t)+y(t)**

Let’s try to predict which

**h(t)**function satisfies this equation? I can’t seem to get on my fingers anymore. Integrating the two sides of the equation, as for the equation

**x'(t)=y(t)**, will not help either. There are methods, of course, but we don’t know them yet. Maybe we can at least partially predict how the

**h(t)**level behaves over time?

Certainly for

**t=0 h(t)=0**. Then

**1.25=h'(t)**.

**Slope**(derivative!) of the

**h(t)=1.25**. In a moment

**h(t)**is no longer zero and

**h'(t)=1.25(t)-h(t)**. The slope

**h(t)**will

**decrease**a bit. The level is rising a little slower, and the outflow of

**Q2(t)**will increase a bit. And when will the level settle? Then when what goes in, goes out. That is, when

**Q2(t) = Q1(t)**. When the level is fixed,

**h'(t)=0**and

**h(t)=1.25 mm**. Doesn’t this remind us of the

**inertial unit**?

Let’s solve the

**differential equation**using a

**diagram**with an

**integrating unit**. So we will use the Xcos application of the free SCILAB program.

**Fig. 13-8**For each integral term

**1/s**, its

**input**is the

**derivative**of the output. And it doesn’t matter if the input is a single signal as in

**Fig. 13-6**or a difference of

**2 signals**as in

**Fig. 13-8**. It can even be any function of several signals, e.g. a

**product**. Therefore, the

**integrating unit**is ideal for solving such

**differential equations**where

**y(t)**and its derivative

**y'(t)**occur.

So

**h'(t)=1.25(t)-h(t)**otherwise

**1.25(t)=h'(t)+h(t)**and generally

**x(t)=y'(t)+y(t)**

The scheme will give

**y(t)**where the equation

**x(t)=y(t)+y'(t)**is satisfied at any time.

It seems that the

**tank**with the hole is an

**inertial unit**. The graph shows that the time constant

**T=1 sec**. After

**6…7**seconds, the tank will reach a constant level

**h=1.25 mm**.

Notice that the equation

**1.25(t)=h'(t)+h(t)**holds true all the time. At the beginning, the derivative

**h'(t)**is the largest and equals

**1.25 mm/sec**and the level

**h(0)=0**. Then the slope decreases and the level increases. In steady state, the slope, i.e.

**h'(t)=0**and

**h(t)=1.25**.

A little comment.

After approx.

**7 sec**. the water level was set at

**h=1.25 mm**. Very small this level, there is only

**1.25mm**at the bottom! Apparently the hole is so big that the bottom will only become wet. Let’s double the area of the hole. So let’s say

**k=0.5**. Will the level be

**twice**as large?

**Chapter 13.4 Filling a tank with a smaller hole k=0.5****Fig. 13-9**Now the hole is smaller (

**k=0.5**) and the bath is described by the differential equation:

**1.25(t)=h'(t)+0.5*h(t)**

and more generally:

**x(t)=y'(t)+0.5*y(t)**

Compare with the previous tank and find the difference.

We will solve the differential equation using a scheme with an

**integrating unit**

**F****ig. 13-10**As we expected, the fixed level will be

**2**times higher. Here

**y=2.5 mm**. As for the dynamics, the system is still inertial but has a time constant twice as large ->

**T=2 sec.**

**Note**

We assumed that the water outflow is proportional to the level

**Q2(t)=k*h(t)**. Indeed, when I pull out the plug in the bathtub, the water flows the fastest at the beginning, then (when the level is lower) it flows slower! However, this is only a first approximation! In fact, the

**Q2(t)**outflow is proportional to the

**root**of the

**level**! So let’s check out this more accurate mathematical model.

**Chapter 13.5 Filling the tank with a hole k=1-more accurate model****Fig. 13-11**The above equations show that the

**Q2**outflow is proportional to the root of

**h(t)**. The resulting equation is an example of a non-linear differential equation, usually more difficult to solve than the previously considered

**linear**ones. Fortunately for

**Xcos**, this is no problem.

**Fig. 13-12**At first glance, it looks like an

**inertial unit**. But he is not! Compared to

**chapter 13.3**, where

**k=1**was also present, the liquid reached a higher fixed level

**h=1.5625 mm**. This level can also be determined theoretically. Then the inflow of

**Q1(t)**is equal to the outflow of

**Q2(t)**.

One more thing. Remember that I assumed a step

**x(t)=1.25(t)**instead of

**x(t)=1(t)**as usual. Why? Because for

**x(t)=1(t)**the levels determined for the tank “with a hole” and “without” are exceptionally the same

**h=1mm!**. The reader might think that this is the case for any

**x(t).**But it was just a pure coincidence!

**Chapter 13.6 More on differential equations**

In a **tank without a hole**, we learned the simplest differential equation **1.25(t)=h'(t)**

A **tank with a hole** is described by a more difficult differential equation **1.25(t)= h'(t)+h(t)**

The last equation can be generalized to** x(t) = a1*y'(t) + a0*y(t)** where:

–** x(t)** – input signal – equivalent to step **1.25(t)****– y'(t)** – derivative of the output signal – equivalent of** h'(t)****– y(t)** – output signal – equivalent of **h(t)****– a0, a1** constant coefficients – in equation **1.25(t) = h'(t) + h(t)**–> **a1=a0=1**The equation can be further generalized to a higher degree of derivative – e.g.

**degree 3**.

**– x(t) = a3*y”‘(t) + a2*y”(t) +a1*y'(t) +a0*y(t)**

**a0, a1, a2, a3**are

**constant**coefficients.

Since we’re generalizing so well, the derivatives can also be on the left-hand side, e.g.

**– b3*x”'(t) + b2*x”(t) + b1*x'(t) + b0*x(t)= a3*y”'(t) + a2*y”(t) + a1*y'(t) + a0*y(t)**

An example of a **non-linear differential equation** was a more accurate model of a tank with a hole from Chapter **13.5** or, for example,**y(t)*x”(t) + t*x(t) = y(t)”+x(t)*y'(t) + y(t)**

We will not deal with such equations. Therefore, **each differential** equation will be a** linear differential equation** for us.

And how to solve them? So how to find the function** y(t)** knowing the function **x(t)**? Turns out you’ve done it many times in this course. The input signal **x(t)** was most often a **unit step 1(t)**, less often a l**inear** or **quadratic signa**l.

Then **Xcos** based on:

– block diagram

– input signal x(t)

created the appropriate linear differential equation, which he solved by calculating y(t) and showed on the **time char**t. It is known that differential equations can also be solved on paper. Mathematicians have been doing it for a long time, before there were **computers** and **Xcos**.

In the next chapter, we will deal with **operatotional calculus** as a tool for solving **linear differential equations**.