Table 13.8 Completed Routh table for Example 13.6الجدول 13.8 جدول Routh المكتمل للمثال 13.6
s
5
s
5
s^(5) s^{5} (4) 2
(10) 5
(2) 1
s
4
s
4
s^(4) s^{4} 6
5
1
s
3
s
3
s^(3) s^{3}
−
|
2
5
6
5
|
6
=
3.33
−
2
5
6
5
6
=
3.33
-(|[2,5],[6,5]|)/(6)=3.33 -\frac{\left|\begin{array}{ll}2 & 5 \\ 6 & 5\end{array}\right|}{6}=3.33
−
|
2
1
6
1
|
6
=
0.66
−
2
1
6
1
6
=
0.66
-(|[2,1],[6,1]|)/(6)=0.66 -\frac{\left|\begin{array}{ll}2 & 1 \\ 6 & 1\end{array}\right|}{6}=0.66
−
|
2
0
6
0
|
6
=
0
−
2
0
6
0
6
=
0
-(|[2,0],[6,0]|)/(6)=0 -\frac{\left|\begin{array}{ll}2 & 0 \\ 6 & 0\end{array}\right|}{6}=0
s
2
s
2
s^(2) s^{2}
−
|
6
5
3.33
0.66
|
3.33
=
3.81
−
6
5
3.33
0.66
3.33
=
3.81
-(|[6,5],[3.33,0.66]|)/(3.33)=3.81 -\frac{\left|\begin{array}{cc}6 & 5 \\ 3.33 & 0.66\end{array}\right|}{3.33}=3.81
−
|
6
1
3.33
0
|
3.33
=
1
−
6
1
3.33
0
3.33
=
1
-(|[6,1],[3.33,0]|)/(3.33)=1 -\frac{\left|\begin{array}{cc}6 & 1 \\ 3.33 & 0\end{array}\right|}{3.33}=1
s
1
s
1
s^(1) s^{1}
−
|
3.33
0.66
3.81
1
|
3.81
=
0.214
−
3.33
0.66
3.81
1
3.81
=
0.214
-(|[3.33,0.66],[3.81,1]|)/(3.81)=0.214 -\frac{\left|\begin{array}{cc}3.33 & 0.66 \\ 3.81 & 1\end{array}\right|}{3.81}=0.214
−
|
3.33
0
3.81
0
|
3.81
=
0
−
3.33
0
3.81
0
3.81
=
0
-(|[3.33,0],[3.81,0]|)/(3.81)=0 -\frac{\left|\begin{array}{ll}3.33 & 0 \\ 3.81 & 0\end{array}\right|}{3.81}=0
s
0
s
0
s^(0) s^{0}
−
|
3.81
1
0.214
0
|
0.214
=
1
−
3.81
1
0.214
0
0.214
=
1
-(|[3.81,1],[0.214,0]|)/(0.214)=1 -\frac{\left|\begin{array}{cc}3.81 & 1 \\ 0.214 & 0\end{array}\right|}{0.214}=1
s^(5) (4) 2 (10) 5 (2) 1
s^(4) 6 5 1
s^(3) -(|[2,5],[6,5]|)/(6)=3.33 -(|[2,1],[6,1]|)/(6)=0.66 -(|[2,0],[6,0]|)/(6)=0
s^(2) -(|[6,5],[3.33,0.66]|)/(3.33)=3.81 -(|[6,1],[3.33,0]|)/(3.33)=1
s^(1) -(|[3.33,0.66],[3.81,1]|)/(3.81)=0.214 -(|[3.33,0],[3.81,0]|)/(3.81)=0
s^(0) -(|[3.81,1],[0.214,0]|)/(0.214)=1 | $s^{5}$ | (4) 2 | (10) 5 | (2) 1 |
| :--- | :--- | :--- | :--- |
| $s^{4}$ | 6 | 5 | 1 |
| $s^{3}$ | $-\frac{\left\|\begin{array}{ll}2 & 5 \\ 6 & 5\end{array}\right\|}{6}=3.33$ | $-\frac{\left\|\begin{array}{ll}2 & 1 \\ 6 & 1\end{array}\right\|}{6}=0.66$ | $-\frac{\left\|\begin{array}{ll}2 & 0 \\ 6 & 0\end{array}\right\|}{6}=0$ |
| $s^{2}$ | $-\frac{\left\|\begin{array}{cc}6 & 5 \\ 3.33 & 0.66\end{array}\right\|}{3.33}=3.81$ | $-\frac{\left\|\begin{array}{cc}6 & 1 \\ 3.33 & 0\end{array}\right\|}{3.33}=1$ | |
| $s^{1}$ | $-\frac{\left\|\begin{array}{cc}3.33 & 0.66 \\ 3.81 & 1\end{array}\right\|}{3.81}=0.214$ | $-\frac{\left\|\begin{array}{ll}3.33 & 0 \\ 3.81 & 0\end{array}\right\|}{3.81}=0$ | |
| $s^{0}$ | $-\frac{\left\|\begin{array}{cc}3.81 & 1 \\ 0.214 & 0\end{array}\right\|}{0.214}=1$ | | |
Table 13.9 Partial Routh table for Example 13.7الجدول 13.9 جدول المسار الجزئي للمثال 13.7
s
5
s
5
s^(5) s^{5} 1
5
6
s
4
s
4
s^(4) s^{4} (5) 1
(25) 5
(30) 6
s
3
s
3
s^(3) s^{3}
−
|
1
5
1
5
|
1
=
0
−
1
5
1
5
1
=
0
-(|[1,5],[1,5]|)/(1)=0 -\frac{\left|\begin{array}{ll}1 & 5 \\ 1 & 5\end{array}\right|}{1}=0
−
|
1
6
1
6
|
1
=
0
−
1
6
1
6
1
=
0
-(|[1,6],[1,6]|)/(1)=0 -\frac{\left|\begin{array}{ll}1 & 6 \\ 1 & 6\end{array}\right|}{1}=0 0
s^(5) 1 5 6
s^(4) (5) 1 (25) 5 (30) 6
s^(3) -(|[1,5],[1,5]|)/(1)=0 -(|[1,6],[1,6]|)/(1)=0 0 | $s^{5}$ | 1 | 5 | 6 |
| :--- | :--- | :--- | :--- |
| $s^{4}$ | (5) 1 | (25) 5 | (30) 6 |
| $s^{3}$ | $-\frac{\left\|\begin{array}{ll}1 & 5 \\ 1 & 5\end{array}\right\|}{1}=0$ | $-\frac{\left\|\begin{array}{ll}1 & 6 \\ 1 & 6\end{array}\right\|}{1}=0$ | 0 |
Solution حل
Start by forming the Routh table for the denominator of Equation 13.90 (see Table 13.9). In the second-row we multiply through by
1
/
5
1
/
5
1//5 1 / 5 for convenience. We stop at the third row, since the entire row consists of zeros, and use the following procedure.
ابدأ بإنشاء جدول روث لمقام المعادلة ١٣.٩٠ (انظر الجدول ١٣.٩). في الصف الثاني، نضرب العدد بـ {٠} لتسهيل العملية. نتوقف عند الصف الثالث، لأن الصف بأكمله يتكون من أصفار، ونتبع الإجراء التالي. First we return to the row immediately above the row of zeros and form an auxiliary polynomial, using the entries in that row as coefficients. The polynomial will start with the power of
s
s
s s in the label column and continue by skipping every other power of
s
s
s s . Thus, the polynomial formed for this example is
أولاً، نعود إلى الصف الذي يعلو صف الأصفار مباشرةً ونُشكِّل كثيرة حدود مساعدة، باستخدام مُدخلات ذلك الصف كمُعاملات. تبدأ كثيرة الحدود بقوة
s
s
s s
s
s
s s
P
(
s
)
=
s
4
+
5
s
2
+
6
P
(
s
)
=
s
4
+
5
s
2
+
6
P(s)=s^(4)+5s^(2)+6 P(s)=s^{4}+5 s^{2}+6