1
\$\begingroup\$

I didn't understand the yellow-marked sentence clearly, and even I did, what does it have to do with the conclusion that follows it? (marked in green). Thanks in advance..

Source: Modern control engineering 5th ed, chapter 2, state-space, page 30

**Modern control engineering 5th ed, chapter 2, state-space, page 30**

\$\endgroup\$
4
  • 2
    \$\begingroup\$ What is the source for this document and on what page is the quote. Context is needed. \$\endgroup\$
    – Andy aka
    Dec 4 '18 at 12:20
  • \$\begingroup\$ It is a value that you use to compare against when making PID calculations for instance. The other terms are current as at this instant and have no history component. \$\endgroup\$
    – KalleMP
    Dec 4 '18 at 12:50
  • \$\begingroup\$ @Andy aka It's contained in the post now.. \$\endgroup\$
    – Yuri
    Dec 4 '18 at 13:27
  • \$\begingroup\$ @KalleMP, thank you my friend, but how to use them as memory devices? \$\endgroup\$
    – Yuri
    Dec 4 '18 at 13:28
3
\$\begingroup\$

An integrator has its current state (value) because of the past values.

That is kind of a memory: It memorizes the sum of the "past".

The green sentence follows from the yellow sentence because the yellow sentence defines the integrators as state, and the green sentence just repeats that.

\$\endgroup\$
2
  • \$\begingroup\$ Oh, I see, thanks a lot. But why does the number of state variables equal the number of integrator outputs here ? \$\endgroup\$
    – Yuri
    Dec 4 '18 at 13:29
  • 1
    \$\begingroup\$ because ... yellow sentence. \$\endgroup\$ Dec 4 '18 at 13:31
0
\$\begingroup\$

That's because when there are no past values it becomes equal to the present value which is represented by the present state variables. So it is how many ever times you integrate that many times you will get a present state variable

\$\endgroup\$

This site is temporarily in read only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .