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Table of contents
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Table of contents

0:00
Начало
0:05
О методе оценки устойчивости критерием Рауса
1:44
Способ заполнения таблицы Рауса
7:35
Задача 1 на оценку устойчивости САУ критерием Рауса
13:19
Задача 2 на оценку устойчивости САУ критерием Рауса
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теория автоматического управления
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  • ruRussian
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00:00:06
Good afternoon Dear viewers of the channel, we
00:00:08
continue to study the theory of
00:00:10
automatic control and today’s
00:00:13
video is devoted to another
00:00:16
algebraic stability criterion,
00:00:18
this is the Rous stability criterion. It should be
00:00:21
noted that this criterion was
00:00:23
formulated
00:00:25
a little earlier than the
00:00:28
Hurwitz criteria are known to all by the
00:00:31
outstanding English mathematician and
00:00:34
scientist Edward Rous, this criterion is
00:00:38
also included in program for studying the theory of
00:00:41
automatic control therefore
00:00:42
Let's look at it in detail within the framework of
00:00:45
this video
00:00:48
in
00:00:49
1875 Edward John Rounds proposed
00:00:52
this criterion
00:00:54
for study, we are given the
00:00:57
transfer function of the system,
00:01:00
this criterion is determined through the
00:01:03
characteristic polynomial That is, through the
00:01:05
polynomial which we have in the
00:01:07
denominator of the transfer function Like this
00:01:10
it looks like our characteristic
00:01:13
polynomials, we need to compile a
00:01:15
raus table based on it in accordance with the
00:01:18
compiled table, we formulate the
00:01:22
following raus criterion for
00:01:26
symptomatic stability of a linear self-propelled control system,
00:01:28
it is necessary enough that all
00:01:31
elements of the first column of the raus table
00:01:33
are positive. There are several
00:01:36
variations of this criterion, but let's
00:01:39
look at it in the framework of this video exactly
00:01:42
in this formulation,
00:01:45
now let's look at How the
00:01:47
Rous table is
00:01:51
filled out in accordance with the
00:01:54
characteristic polynomial,
00:01:57
it looks like this and Let's
00:01:59
discuss how to remember the first two
00:02:03
lines of this table, that is, these two. They are
00:02:06
completely formed as in the
00:02:08
Hurwitz criterion, that is, the first table we
00:02:11
fill in at even powers
00:02:17
with coefficients at even powers S
00:02:20
That is, if I start from the first, then
00:02:22
we step over the subsequent ones and so on,
00:02:25
that is, here they are, then the second line
00:02:29
is filled in with coefficients at odd
00:02:31
powers, let’s say this, we start
00:02:34
with the second, step over the next element
00:02:36
and so on, and so on to the next line,
00:02:39
these are these, these are filled in according to a
00:02:43
certain rule in accordance with the
00:02:45
following formulas Let's discuss them
00:02:47
in order to find element B1, we
00:02:50
always
00:02:52
need to form a construction such that
00:02:56
one is divided by minus the first element of the
00:02:59
previous line, here it is, it will always be
00:03:03
found to calculate the coefficients in
00:03:05
this line further We need a
00:03:08
second-order determinant that is formed from the
00:03:10
elements of the previous two lines, namely
00:03:13
this is the construction that
00:03:16
is located on B1 and in these
00:03:18
determinants for the coefficients B1
00:03:21
b2b3 and so on the
00:03:22
elements of the first column of these
00:03:26
second-order determinants they
00:03:28
will never change, that is, they there
00:03:30
will be these elements, they
00:03:33
will always be here, this construction here
00:03:34
will not change at all for elements
00:03:37
B1 b2b3, only these
00:03:41
two elements will change, forming the second column
00:03:44
of this determinant, which determines
00:03:47
as follows: if we consider
00:03:48
B1, then for it these will change
00:03:51
if B 2
00:03:52
then these two diagonally adjacent
00:03:56
we take if B3 then these look B2 the
00:04:01
obligatory element as I already said
00:04:03
will be construction 1 divide minus the first
00:04:08
element of the previous line And this is the
00:04:10
previous line here it is and the first
00:04:12
element and N -1 that is one divided by
00:04:14
minus a N -1 Here it is Next comes a
00:04:18
second-order determinant in which the
00:04:21
first column of this determinant does not
00:04:24
change, so it is formed from the
00:04:26
previous two first elements of the
00:04:31
previous lines, as here
00:04:35
further if B2 then we obliquely
00:04:37
consider the neighboring one, that is, we do not take
00:04:40
those elements that are located above
00:04:43
B2 itself because they have already participated
00:04:47
in the formation of B1 That is, if B2 then
00:04:49
we take these neighboring ones
00:04:52
further B3 this design
00:04:55
remains unchanged for us and B3
00:04:59
diagonally adjacent these elements are minus
00:05:03
6
00:05:04
and so on B4 B5 if necessary is calculated
00:05:09
as follows, as described
00:05:12
C1 C2 C3 is formed in a completely
00:05:15
similar way, but for their
00:05:18
formation the
00:05:20
previous two lines are needed for C
00:05:23
1C 23, that is, the construction for calculating
00:05:27
these coefficients is also invariably the
00:05:31
required element is 1 divide
00:05:34
minus the first element of the previous line
00:05:36
for C1 the previous line is already
00:05:38
line B1 b2b3 here is the first element B1
00:05:42
Here it is, then the
00:05:45
previous
00:05:48
elements in the third and second line they
00:05:54
will always form the first column
00:05:58
of this determinant changes these are the ones
00:06:01
that diagonally if C is
00:06:03
one then an -3 B2 C2 let's say necessarily
00:06:09
this is not we definitely forget, we don’t
00:06:12
forget the
00:06:13
first column which will not change for
00:06:16
all coefficients C1 c23 and so on in
00:06:20
this line
00:06:22
2
00:06:24
elements change in the second in the second column for
00:06:27
these elements C2 then we are already considering
00:06:30
these two sc3 then these two, that is,
00:06:34
C3 we have this construction does not change
00:06:38
changes C3 obliquely these two
00:06:42
elements, that is, an-7 -7 B4 and so on,
00:06:47
if necessary, calculate 4 5 and so on
00:06:51
It should be noted that the number of rows of the
00:06:54
Rous table is one greater than the order of the system
00:06:57
If the system is of third order,
00:07:00
then the number of rows in the table rows will be
00:07:04
4
00:07:06
and the number of columns As the
00:07:09
algorithm shows us is
00:07:13
one less. That is, if a
00:07:16
third order system then there will be 4 rows and
00:07:21
two columns. If a 4
00:07:24
order system there will be 5 rows there will be 3 columns and
00:07:30
so on. Let's look at all this using the
00:07:33
following example.
00:07:36
Let's solve the following problem example
00:07:38
first we are given such a system, that is,
00:07:42
such a transfer function, we need to
00:07:45
determine the stability of the system according to the
00:07:47
criteria,
00:07:48
we write out the
00:07:51
characteristic equation separately,
00:07:54
this
00:07:55
equation is in the denominator of the
00:07:58
transfer functions and we need to
00:08:00
compile a table of raus,
00:08:02
as we said, the first two
00:08:04
elements are compiled according to the coefficients for even and
00:08:08
odd powers S the first line
00:08:11
is filled in. Elementary, that is, we take a
00:08:13
two, step over the next element
00:08:16
four, a one, we write down
00:08:19
further with odd powers five,
00:08:23
we step over a two and 0
00:08:26
5 2 0 now unknown coefficients B1
00:08:29
b2b3 C1 C2 C3 D1 D2 D3 we need to
00:08:34
calculate
00:08:36
the system, we have 4 orders, so here we have
00:08:40
we have 5 rows and three columns,
00:08:45
element B1 is considered as we
00:08:48
stipulated earlier, that is, the obligatory
00:08:51
element here is this
00:08:53
construction 1 divide by minus the first
00:08:56
element of the previous line with the daughter of
00:08:58
the previous line we have
00:08:59
five Next comes the second-
00:09:03
order determinant where invariably
00:09:06
in the first column of this the determinant
00:09:08
will be the elements of the previous lines, that is,
00:09:11
2 and 5. Here they are written and diagonally, if
00:09:16
we are looking for B1, then these ones
00:09:20
standing opposite
00:09:24
are written down, we
00:09:25
calculate the determinant, that is, 4 minus 20
00:09:29
minus 16, divide this by -5, we
00:09:32
get 32, then we count B2, that is,
00:09:37
this here is the construction of us invariably,
00:09:39
respectively, these elements
00:09:41
Define the second order are also
00:09:43
invariable if we count B2 then
00:09:47
obliquely we take these two elements
00:09:50
as the second column of this
00:09:52
determinant
00:09:55
we consider our determinant we multiply by -1
00:09:58
divide by 5 we get one
00:10:01
further B3 That is, as we said this
00:10:05
construction does not change here and for B3
00:10:08
here there is nothing else diagonally
00:10:12
except
00:10:13
imaginary zeros, so here
00:10:15
the element B3 is equal to zero,
00:10:19
then we formed the
00:10:23
third row of the Rous table like this, now
00:10:25
it remains to calculate C1 c23 which
00:10:29
we calculate as follows, also the
00:10:31
previous row here is
00:10:34
this one line and here we have a
00:10:37
construction that will be the next one divided
00:10:40
by minus the previous element the first
00:10:43
element of the previous line that is, 32
00:10:45
is here
00:10:48
so here such a construction
00:10:50
will be
00:10:52
to form the first column in this
00:10:56
determinant we need the previous two
00:10:58
lines here they are
00:11:01
532 will invariably form
00:11:04
the first here column and diagonally C1 two
00:11:08
one will form the
00:11:12
next column of this determinant,
00:11:15
open the determinant, multiply by
00:11:18
minus 1 divide 3.2 we get
00:11:21
this number, then C2 this is as we
00:11:25
said, the design does not change, but for
00:11:27
C2 obliquely the neighboring elements for the
00:11:31
second column are 0 we bring them here we
00:11:34
transfer, we
00:11:35
open
00:11:38
the determinant, we multiply everything we need, it
00:11:39
turns out 0 C3 is also 0 because there are
00:11:44
also zeros here
00:11:46
and we already fill out the fourth row of the
00:11:49
Rous table. Now all that remains is to calculate
00:11:52
D1 D2 D3 D1 we also count only for D1 the
00:11:57
previous two lines are
00:12:00
these
00:12:02
minus 1 divided by the first element of the
00:12:05
previous line And this is this
00:12:08
number
00:12:09
0.4375 this will be this coefficient
00:12:13
which will not change
00:12:15
these two elements they will always
00:12:18
form the first column of this
00:12:20
determinant
00:12:22
and D1 diagonally adjacent this is 10 we
00:12:26
move here we
00:12:28
open the determinant we calculate everything we
00:12:31
need we got it one further D2
00:12:35
0 and d30 why because
00:12:40
in the determinants there will always be rows
00:12:44
and columns with zeros in this way we
00:12:48
filled out the
00:12:49
entire table of rows, calculated and we see
00:12:54
that the first column consists entirely of
00:12:58
positive elements This suggests
00:13:00
that
00:13:02
our system is symptomatically stable and
00:13:05
the conclusion is we do according to the
00:13:07
Routh criterion is such that with stability in
00:13:11
this way we solve problems
00:13:15
related to the Routh criterion,
00:13:20
let’s solve one more problem to consolidate
00:13:23
example 2 we are given a system with a
00:13:27
third-order transfer function and we
00:13:30
need to determine the stability of the system according to the
00:13:32
Routh criterion, we write out the
00:13:34
characteristic equation separately
00:13:36
system and you need to create a table
00:13:42
third-order system Therefore, the
00:13:44
table will have 4 lines and two columns, the
00:13:48
first two lines are filled in
00:13:51
similar to the Hurwitz criterion, that is,
00:13:54
first, with odd powers, let’s say 5,
00:13:57
we step over one element 1 5 1 then
00:14:01
with even 2 and 2 we fill in B1,
00:14:06
we calculate Using the formulas which were
00:14:09
described for the algorithm, that is, the
00:14:11
obligatory element will be one divided
00:14:14
by minus the first element of the previous
00:14:15
line And here is two,
00:14:18
so we wrote it here, the first column
00:14:21
of the determinant will be determined from the
00:14:23
previous lines necessarily
00:14:25
52 they will not change and then the
00:14:28
diagonally located neighboring elements
00:14:32
are 1-2 will be the
00:14:34
next column of this determinant,
00:14:36
open the determinant, it is equal to -4
00:14:40
B2 Similarly, this construction
00:14:43
will not change
00:14:46
for B2 diagonally, we have here
00:14:48
Imaginary zeros, so we wrote them here
00:14:51
and this element is equal to zero,
00:14:54
fill in the third line of the row table,
00:14:57
then calculate C1
00:15:00
for it
00:15:02
one divide on the minus, the first element of the
00:15:06
previous line is -4, so he
00:15:08
wrote down the
00:15:10
previous two elements in this column
00:15:13
to form the first column of determinants
00:15:18
and diagonally for C1 2.0 which we
00:15:21
move here and calculate that C2 C1
00:15:24
is equal to 2 Similarly, we calculate C2 this
00:15:29
construction does not change for C2
00:15:32
Imaginary here we
00:15:35
move the zeros here and C2 is equal to 0.
00:15:39
Thus, we have filled in all the elements of the
00:15:41
Rous table and we see that the first
00:15:45
column of the Rous table contains a
00:15:49
negative number,
00:15:51
that is, among the elements of the first column of the
00:15:53
Rous table there are negative
00:15:55
elements. This indicates that according to the
00:15:57
Rous criterion, this system is not
00:16:00
stable the system is not stable conclusion for
00:16:04
this problem
00:16:06
So, within the framework of this video, we
00:16:09
studied the algebraic criterion of Routh I
00:16:13
hope that this video was useful
00:16:14
Thank you all for your attention, subscribe
00:16:17
to the channel, leave comments and
00:16:20
likes, success to all

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