Stabilność – kryterium Nyquista

Przykład:

Obiekt inercyjny piątego rzędu i regulator PD

$$G\left( s \right) = \frac{k}{\left( T_{1}s + 1 \right)\left( T_{2}s + 1 \right)\left( T_{3}s + 1 \right)\left( T_{4}s + 1 \right)\left( T_{5}s + 1 \right)}$$

R(s) = k_p(1+T_ds)

$$G_{\text{uo}}\left( s \right) = \frac{kk_{p}\left( 1 + T_{d}s \right)}{\left( T_{1}s + 1 \right)\left( T_{2}s + 1 \right)\left( T_{3}s + 1 \right)\left( T_{4}s + 1 \right)\left( T_{5}s + 1 \right)}$$

Dane:

k = 1,   T₁ = 1,   T₂ = 2,   T₃ = 3,   T₄ = 4,   T₅ = 5

$$G_{\text{uo}}\left( s \right) = \frac{k_{p}\left( 1 + T_{d}s \right)}{\begin{bmatrix} 120s^{5} + 274s^{4} + 225s^{3} + 85s^{2} + 15s + 1 \\ \end{bmatrix}}$$

$$G_{\text{uo}}\left( \text{jω} \right) = \frac{k_{p}\left( 1 + T_{d}\text{jω} \right)}{\begin{bmatrix} 120j\omega^{5} + 274\omega^{4} - 225\text{jω}^{3} - 85\omega^{2} + 15j\omega + 1 \\ \end{bmatrix}}$$

a = 274ω⁴ − 85ω² + 1

b = 120ω⁵ − 225ω³ + 15ω

$$G_{\text{uo}}\left( \text{jω} \right) = \frac{k_{p}\left( a - jb \right)}{\left( a^{2} + b^{2} \right)} \bullet \left( 1 + jT_{d}\omega \right)$$

$$G_{\text{uo}}\left( \text{jω} \right) = k_{p}\frac{a + bT_{d}\omega + jaT_{d}\omega - jb}{\left( a^{2} + b^{2} \right)}$$

$$G_{\text{uo}}\left( \text{jω} \right) = k_{p}\frac{a + bT_{d}\omega + j\left( aT_{d}\omega - b \right)}{\left( a^{2} + b^{2} \right)}$$

$$P_{\text{uo}}\left( \omega \right) = k_{p}\frac{a + bT_{d}\omega}{\left( a^{2} + b^{2} \right)}$$

$$Q_{\text{uo}}\left( \omega \right) = k_{p}\frac{aT_{d}\omega - b}{\left( a^{2} + b^{2} \right)}$$

Warunek granicy stabilności

$$\left\{ \begin{matrix} P_{\text{uo}}\left( \omega \right) = - 1 \\ Q_{\text{uo}}\left( \omega \right) = 0 \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p}\frac{a + bT_{d}\omega}{\left( a^{2} + b^{2} \right)} = - 1 \\ k_{p}\frac{aT_{d}\omega - b}{\left( a^{2} + b^{2} \right)} = 0 \\ \end{matrix} \right.\ $$

$$T_{d}\omega = \frac{b}{a}$$

$$k_{p}\frac{a + \frac{b^{2}}{a}}{\left( a^{2} + b^{2} \right)} = - 1$$

$$k_{p}\frac{1}{a} = - 1$$

k_p = −a

$$\left\{ \begin{matrix} k_{p} = - a \\ T_{d}\omega = \frac{b}{a} \\ \end{matrix} \right.\ $$

Szukamy ekstremum funkcji wzmocnienia

k_p = −(274ω⁴−85ω²+1)

(k_p)^′ = −(274•4ω³−170ω)

(k_p)^′ = −2ω(548ω²−85)

$$\omega = \sqrt{\frac{85}{548}}$$

$$\left\{ \begin{matrix} k_{p} = - a \\ T_{d}\omega = \frac{b}{a} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p} = - \left( 274\omega^{4} - 85\omega^{2} + 1 \right) \\ T_{d}\omega = \frac{120\omega^{5} - 225\omega^{3} + 15\omega}{274\omega^{4} - 85\omega^{2} + 1} \\ \end{matrix} \right.\ $$

Dla $\omega = \sqrt{\frac{85}{548}} = 0,3938$

$$\left\{ \begin{matrix} \left( k_{p} \right)_{\text{kr}} = 5,59 \\ \left( T_{d} \right)_{\text{kr}} = 3,04 \\ \end{matrix} \right.\ $$

Warunek: zapas modułu ΔL=ΔdB

ΔL = 20lgΔM = −Δ

$$\Delta M = 10^{- \frac{\Delta}{20}}$$

$$\left\{ \begin{matrix} P_{\text{uo}}\left( \omega \right) = - 1 + \Delta M \\ Q_{\text{uo}}\left( \omega \right) = 0 \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p}\frac{a + bT_{d}\omega}{\left( a^{2} + b^{2} \right)} = \Delta M - 1 \\ k_{p}\frac{aT_{d}\omega - b}{\left( a^{2} + b^{2} \right)} = 0 \\ \end{matrix} \right.\ $$

$$T_{d}\omega = \frac{b}{a}$$

$$k_{p}\frac{a + \frac{b^{2}}{a}}{\left( a^{2} + b^{2} \right)} = \Delta M - 1$$

$$k_{p}\frac{1}{a} = \Delta M - 1$$

k_p = −a(1−ΔM)

k_p = −(1−ΔM)(274ω⁴−85ω²+1)

$$\left\{ \begin{matrix} k_{p} = - a\left( 1 - \Delta M \right) \\ T_{d}\omega = \frac{b}{a} \\ \end{matrix} \right.\ $$

Szukamy ekstremum funkcji wzmocnienia

k_p = −(1−ΔM)(274ω⁴−85ω²+1)

(k_p)^′ = −(1−ΔM)(274•4ω³−170ω)

(k_p)^′ = −2ω(1−ΔM)(548ω²−85)

$$\omega = \sqrt{\frac{85}{548}}$$

$$\left\{ \begin{matrix} k_{p} = - a\left( 1 - \Delta M \right) \\ T_{d}\omega = \frac{b}{a} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p} = - \left( 1 - \Delta M \right)\left( 274\omega^{4} - 85\omega^{2} + 1 \right) \\ T_{d}\omega = \frac{120\omega^{5} - 225\omega^{3} + 15\omega}{274\omega^{4} - 85\omega^{2} + 1} \\ \end{matrix} \right.\ $$

Dla ΔL=6dB i $\omega = \sqrt{\frac{85}{548}} = 0,3938$

$$\left\{ \begin{matrix} k_{p} = 2,8 \\ T_{d} = 3,04 \\ \end{matrix} \right.\ $$

Warunek: zapas fazy Δϕ

$$\left\{ \begin{matrix} M\left( \omega \right) = 1 \\ \tan\text{Δφ} = \frac{Q\left( \omega \right)}{P\left( \omega \right)} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} \sqrt{{P\left( \omega \right)}^{2} + {Q\left( \omega \right)}^{2}} = 1 \\ \tan\text{Δφ} = \frac{Q\left( \omega \right)}{P\left( \omega \right)} \\ \end{matrix} \right.\ $$

a = 274ω⁴ − 85ω² + 1

b = 120ω⁵ − 225ω³ + 15ω

$$\left\{ \begin{matrix} k_{p}\sqrt{\frac{1 + \left( T_{d}\omega \right)^{2}}{a^{2} + b^{2}}} = 1 \\ \tan\text{Δφ} = \frac{aT_{d}\omega - b}{a + bT_{d}\omega} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p}^{2}\left( 1 + \left( T_{d}\omega \right)^{2} \right) = a^{2} + b^{2} \\ \operatorname{a\bullet tan}\text{Δφ} + bT_{d}\omega\tan\text{Δφ} = aT_{d}\omega - b \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p}^{2}\left( 1 + \left( T_{d}\omega \right)^{2} \right) = a^{2} + b^{2} \\ \operatorname{a\bullet tan}\text{Δφ} + b = T_{d}\omega\left( a - b\tan\text{Δφ} \right) \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p}^{2} = \frac{a^{2} + b^{2}}{1 + \left( T_{d}\omega \right)^{2}} \\ T_{d}\omega = \frac{a\tan\text{Δφ} + b}{a - \operatorname{b\bullet tan}\text{Δφ}} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p}^{2} = \frac{a^{2} + b^{2}}{1 + \left( \frac{a\tan\text{Δφ} + b}{a - \operatorname{b\bullet tan}\text{Δφ}} \right)^{2}} \\ T_{d}\omega = \frac{a\tan\text{Δφ} + b}{a - \operatorname{b\bullet tan}\text{Δφ}} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p}^{2} = \frac{a^{2} + b^{2}}{\frac{\left( a - b\tan\text{Δφ} \right)^{2} + \left( \operatorname{a\bullet tan}\text{Δφ} + b \right)^{2}}{\left( a - b\tan\text{Δφ} \right)^{2}}} \\ T_{d}\omega = \frac{a\tan\text{Δφ} + b}{a - \operatorname{b\bullet tan}\text{Δφ}} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p}^{2} = \frac{\left( a^{2} + b^{2} \right)\left( a - b\tan\text{Δφ} \right)^{2}}{\left( b\tan\text{Δφ} - a \right)^{2} + \left( \operatorname{a\bullet tan}\text{Δφ} + b \right)^{2}} \\ T_{d}\omega = \frac{a\tan\text{Δφ} + b}{a - \operatorname{b\bullet tan}\text{Δφ}} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p}^{2} = \frac{\left( a^{2} + b^{2} \right)\left( a - b\tan\text{Δφ} \right)^{2}}{\left( a^{2} + b^{2} \right)\left\lbrack 1 + \left( \tan\text{Δφ} \right)^{2} \right\rbrack} \\ T_{d}\omega = \frac{a\tan\text{Δφ} + b}{a - \operatorname{b\bullet tan}\text{Δφ}} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p} = \frac{a - b\tan\text{Δφ}}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}} \\ T_{d}\omega = \frac{a\tan\text{Δφ} + b}{a - \operatorname{b\bullet tan}\text{Δφ}} \\ \end{matrix} \right.\ $$

$$k_{p} = \frac{1}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}\left( 274\omega^{4} - 85\omega^{2} + 1 \right) - \frac{\tan\text{Δφ}}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}\left( 120\omega^{5} - 225\omega^{3} + 15\omega \right)$$

Liczymy ekstremum funkcji wzmocnienia

$$\left( k_{p} \right)^{'} = \frac{1}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}\left( 1096\omega^{3} - 170\omega \right) - \frac{\tan\text{Δφ}}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}\left( 600\omega^{4} - 675\omega^{2} + 15 \right)$$

Dla Δϕ=60 ̊

$$\left( k_{p} \right)^{'} = \frac{1}{2}\left( 1096\omega^{3} - 170\omega \right) - \frac{\sqrt{3}}{2}\left( 600\omega^{4} - 675\omega^{2} + 15 \right)$$

$$\left( k_{p} \right)^{'} = - 300\sqrt{3}\omega^{4} + 548\omega^{3} + 337,5\sqrt{3}\omega^{2} - 85\omega - 7,5\sqrt{3}$$

>> roots([-300*sqrt(3) 548 337.5*sqrt(3) -85 -sqrt(3)*7.5])

ans =

1.6657

-0.7298

0.2146

-0.0958

$$\left\{ \begin{matrix} k_{p} = \frac{a - b\tan\text{Δφ}}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}} \\ T_{d}\omega = \frac{a\tan\text{Δφ} + b}{a - \operatorname{b\bullet tan}\text{Δφ}} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p} = 2,076 \\ T_{d} = 3,36 \\ \end{matrix} \right.\ $$

Przykład:

Obiekt inercyjny piątego rzędu i regulator PD rzeczywisty

$$G\left( s \right) = \frac{k}{\left( T_{1}s + 1 \right)\left( T_{2}s + 1 \right)\left( T_{3}s + 1 \right)\left( T_{4}s + 1 \right)\left( T_{5}s + 1 \right)}$$

$$R\left( s \right) = k_{p}\left( 1 + \frac{T_{d}s}{Ts + 1} \right)$$

$$G_{\text{uo}}\left( s \right) = \frac{kk_{p}\left( 1 + \frac{T_{d}s}{Ts + 1} \right)}{\left( T_{1}s + 1 \right)\left( T_{2}s + 1 \right)\left( T_{3}s + 1 \right)\left( T_{4}s + 1 \right)\left( T_{5}s + 1 \right)}$$

Dane:

k = 1,   T₁ = 1,   T₂ = 2,   T₃ = 3,   T₄ = 4,   T₅ = 5

$$G_{\text{uo}}\left( s \right) = \frac{k_{p}\left( 1 + \frac{T_{d}s}{Ts + 1} \right)}{\begin{bmatrix} 120s^{5} + 274s^{4} + 225s^{3} + 85s^{2} + 15s + 1 \\ \end{bmatrix}}$$

$$G_{\text{uo}}\left( \text{jω} \right) = \frac{k_{p}\left( 1 + \frac{jT_{d}\omega}{1 + jT\omega} \right)}{\begin{bmatrix} 120j\omega^{5} + 274\omega^{4} - 225\text{jω}^{3} - 85\omega^{2} + 15j\omega + 1 \\ \end{bmatrix}}$$

a = 274ω⁴ − 85ω² + 1

b = 120ω⁵ − 225ω³ + 15ω

$$G_{\text{uo}}\left( \text{jω} \right) = \frac{k_{p}\left( a - jb \right)}{\left( a^{2} + b^{2} \right)} \bullet \left( 1 + \frac{jT_{d}\omega}{1 + j\text{Tω}} \bullet \frac{1 - jT\omega}{1 - jT\omega} \right)$$

$$G_{\text{uo}}\left( \text{jω} \right) = \frac{k_{p}\left( a - jb \right)}{\left( a^{2} + b^{2} \right)} \bullet \left( 1 + \frac{TT_{d}\omega^{2} + jT_{d}\omega}{1 + T^{2}\omega^{2}} \right)$$

$$G_{\text{uo}}\left( \text{jω} \right) = \frac{k_{p}\left( a - jb \right)}{\left( a^{2} + b^{2} \right)} \bullet \left( 1 + \frac{TT_{d}\omega^{2}}{1 + T^{2}\omega^{2}} + j\frac{T_{d}\omega}{1 + T^{2}\omega^{2}} \right)$$

$$G_{\text{uo}}\left( j\omega \right) = k_{p}\frac{a\left( 1 + \frac{TT_{d}\omega^{2}}{1 + T^{2}\omega^{2}} \right) + b\frac{T_{d}\omega}{1 + T^{2}\omega^{2}} + ja\frac{T_{d}\omega}{1 + T^{2}\omega^{2}} - jb\left( 1 + \frac{TT_{d}\omega^{2}}{1 + T^{2}\omega^{2}} \right)}{\left( a^{2} + b^{2} \right)}$$

$$G_{\text{uo}}\left( \text{jω} \right) = k_{p}\frac{a\left( 1 + \frac{TT_{d}\omega^{2}}{1 + T^{2}\omega^{2}} \right) + b\frac{T_{d}\omega}{1 + T^{2}\omega^{2}} + j\left( a\frac{T_{d}\omega}{1 + T^{2}\omega^{2}} - b\left( 1 + \frac{TT_{d}\omega^{2}}{1 + T^{2}\omega^{2}} \right) \right)}{\left( a^{2} + b^{2} \right)}$$

$$P_{\text{uo}}\left( \omega \right) = k_{p}\frac{a\left( 1 + \frac{TT_{d}\omega^{2}}{1 + T^{2}\omega^{2}} \right) + b\frac{T_{d}\omega}{1 + T^{2}\omega^{2}}}{\left( a^{2} + b^{2} \right)}$$

$$Q_{\text{uo}}\left( \omega \right) = k_{p}\frac{a\frac{T_{d}\omega}{1 + T^{2}\omega^{2}} - b\left( 1 + \frac{TT_{d}\omega^{2}}{1 + T^{2}\omega^{2}} \right)}{\left( a^{2} + b^{2} \right)}$$

Warunek granicy stabilności

$$\left\{ \begin{matrix} P_{\text{uo}}\left( \omega \right) = - 1 \\ Q_{\text{uo}}\left( \omega \right) = 0 \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p}\frac{a\left( 1 + \frac{TT_{d}\omega^{2}}{1 + T^{2}\omega^{2}} \right) + b\frac{T_{d}\omega}{1 + T^{2}\omega^{2}}}{\left( a^{2} + b^{2} \right)} = - 1 \\ k_{p}\frac{a\frac{T_{d}\omega}{1 + T^{2}\omega^{2}} - b\left( 1 + \frac{TT_{d}\omega^{2}}{1 + T^{2}\omega^{2}} \right)}{\left( a^{2} + b^{2} \right)} = 0 \\ \end{matrix} \right.\ $$

$$a\frac{T_{d}\omega}{1 + T^{2}\omega^{2}} = b\left( 1 + \frac{TT_{d}\omega^{2}}{1 + T^{2}\omega^{2}} \right)$$

$$a\frac{T_{d}\omega}{1 + T^{2}\omega^{2}} = b\frac{1 + T^{2}\omega^{2} + TT_{d}\omega^{2}}{1 + T^{2}\omega^{2}}$$

aT_dω = b(1+T²ω²+TT_dω²)

aT_dω = b + bT²ω² + bTT_dω²

bTT_dω² − aT_dω = −b − bT²ω²

T_dω(bTω−a) = −b(1+T²ω²)

$$T_{d}\omega = - \frac{b\left( 1 + T^{2}\omega^{2} \right)}{\left( bT\omega - a \right)}$$

$$k_{p}\frac{a\left( 1 + \frac{TT_{d}\omega^{2}}{1 + T^{2}\omega^{2}} \right) + b\frac{T_{d}\omega}{1 + T^{2}\omega^{2}}}{\left( a^{2} + b^{2} \right)} = - 1$$

$$k_{p}\frac{a\left( 1 + \frac{- T\omega\frac{b\left( 1 + T^{2}\omega^{2} \right)}{\left( bT\omega - a \right)}}{1 + T^{2}\omega^{2}} \right) - b\frac{\frac{b\left( 1 + T^{2}\omega^{2} \right)}{\left( bT\omega - a \right)}}{1 + T^{2}\omega^{2}}}{\left( a^{2} + b^{2} \right)} = - 1$$

$$k_{p}\frac{a\left( 1 - \frac{\text{bTω}}{\left( bT\omega - a \right)} \right) - b\frac{b}{\left( bT\omega - a \right)}}{\left( a^{2} + b^{2} \right)} = - 1$$

$$k_{p}\frac{a\frac{- a}{\left( bT\omega - a \right)} - b\frac{b}{\left( bT\omega - a \right)}}{\left( a^{2} + b^{2} \right)} = - 1$$

$$k_{p}\frac{{- a}^{2} - b^{2}}{\left( a^{2} + b^{2} \right)\left( bT\omega - a \right)} = - 1$$

$$k_{p}\frac{1}{\left( bT\omega - a \right)} = 1$$

k_p = bTω − a

$$\left\{ \begin{matrix} k_{p} = - a + bT\omega \\ T_{d}\omega = - \frac{b\left( 1 + T^{2}\omega^{2} \right)}{\left( bT\omega - a \right)} \\ \end{matrix} \right.\ $$

a = 274ω⁴ − 85ω² + 1

b = 120ω⁵ − 225ω³ + 15ω

Szukamy ekstremum funkcji wzmocnienia

k_p = −(274ω⁴−85ω²+1) + (120ω⁵−225ω³+15ω)Tω

k_p = 120Tω⁶ − (225T+274)ω⁴ + (15T+85)ω² − 1

(k_p)^′ = 720Tω⁵ − 4 • (225T+274)ω³ + 2 • (15T+85)ω

(k_p)^′ = 2ω[360Tω⁴−2•(225T+274)ω²+(15T+85)]

Δ = 4 • (225T+274)² − 4 • 360T(15T+85)

Δ = 4 • (50625T²+123300T+75076−5400T²−30600T)

Δ = 4 • (45225T²+92700T+75076) > 0 dla kazdego T

T	w1	(kp)kr	(Td)kr
0	0,3938	5,59	3,04
0,05	0,38862	5,46	2,97
0,1	0,38365	5,35	2,91
0,15	0,37890	5,24	2,84
0,2	0,37436	5,13	2,78
0,25	0,37003	5,04	2,71
0,3	0,36589	4,95	2,65
0,35	0,36192	4,86	2,59
0,4	0,35813	4,78	2,53
0,45	0,35449	4,71	2,48
0,5	0,35100	4,64	2,42
0,55	0,34765	4,58	2,36
0,6	0,34443	4,52	2,31
0,65	0,34134	4,46	2,26
0,7	0,33836	4,40	2,20
0,75	0,33550	4,35	2,15
0,8	0,33274	4,30	2,10
0,85	0,33008	4,26	2,05
0,9	0,32752	4,22	2,00
0,95	0,32505	4,18	1,95
1	0,32266	4,14	1,90

Wybrane parametry regulatora PDrz : k_p=2,5 T_d=2,7 T=0,25

Warunek: zapas fazy Δϕ

$$G_{\text{uo}}\left( \text{jω} \right) = \frac{k_{p}\left( a - jb \right)}{\left( a^{2} + b^{2} \right)} \bullet \left( 1 + \frac{TT_{d}\omega^{2}}{1 + T^{2}\omega^{2}} + j\frac{T_{d}\omega}{1 + T^{2}\omega^{2}} \right)$$

$$P_{\text{uo}}\left( \omega \right) = k_{p}\frac{a\left( 1 + \frac{TT_{d}\omega^{2}}{1 + T^{2}\omega^{2}} \right) + b\frac{T_{d}\omega}{1 + T^{2}\omega^{2}}}{\left( a^{2} + b^{2} \right)}$$

$$Q_{\text{uo}}\left( \omega \right) = k_{p}\frac{a\frac{T_{d}\omega}{1 + T^{2}\omega^{2}} - b\left( 1 + \frac{TT_{d}\omega^{2}}{1 + T^{2}\omega^{2}} \right)}{\left( a^{2} + b^{2} \right)}$$

$$M_{o}\left( \text{jω} \right) = \frac{1}{\sqrt{a^{2} + b^{2}}}$$

$$M_{R}\left( \text{jω} \right) = k_{p}\sqrt{\left( 1 + \frac{TT_{d}\omega^{2}}{1 + T^{2}\omega^{2}} \right)^{2} + \left( \frac{T_{d}\omega}{1 + T^{2}\omega^{2}} \right)^{2}}$$

$$\left\{ \begin{matrix} M\left( \omega \right) = 1 \\ \tan\text{Δφ} = \frac{Q\left( \omega \right)}{P\left( \omega \right)} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} M_{O}\left( \omega \right)M_{R}\left( \omega \right) = 1 \\ \tan\text{Δφ} = \frac{Q\left( \omega \right)}{P\left( \omega \right)} \\ \end{matrix} \right.\ $$

a = 274ω⁴ − 85ω² + 1

b = 120ω⁵ − 225ω³ + 15ω

$$\left\{ \begin{matrix} k_{p}\frac{1}{\sqrt{a^{2} + b^{2}}}\sqrt{\left( 1 + \frac{TT_{d}\omega^{2}}{1 + T^{2}\omega^{2}} \right)^{2} + \left( \frac{T_{d}\omega}{1 + T^{2}\omega^{2}} \right)^{2}} = 1 \\ \tan\text{Δφ} = \frac{a\frac{T_{d}\omega}{1 + T^{2}\omega^{2}} - b\left( 1 + \frac{TT_{d}\omega^{2}}{1 + T^{2}\omega^{2}} \right)}{a\left( 1 + \frac{TT_{d}\omega^{2}}{1 + T^{2}\omega^{2}} \right) + b\frac{T_{d}\omega}{1 + T^{2}\omega^{2}}} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p}^{2}\left( \left( 1 + \frac{TT_{d}\omega^{2}}{1 + T^{2}\omega^{2}} \right)^{2} + \left( \frac{T_{d}\omega}{1 + T^{2}\omega^{2}} \right)^{2} \right) = a^{2} + b^{2} \\ \operatorname{atan}\text{Δφ} + b\frac{T_{d}\omega}{1 + T^{2}\omega^{2}}\tan\text{Δφ} = a\frac{T_{d}\omega}{1 + T^{2}\omega^{2}} - b\left( 1 + \frac{TT_{d}\omega^{2}}{1 + T^{2}\omega^{2}} \right) \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p}^{2}\left( \left( 1 + \frac{TT_{d}\omega^{2}}{1 + T^{2}\omega^{2}} \right)^{2} + \left( \frac{T_{d}\omega}{1 + T^{2}\omega^{2}} \right)^{2} \right) = a^{2} + b^{2} \\ \operatorname{atan}\text{Δφ} - a\frac{T_{d}\omega}{1 + T^{2}\omega^{2}} + b\frac{T_{d}\omega}{1 + T^{2}\omega^{2}}\tan\text{Δφ} + b\frac{TT_{d}\omega^{2}}{1 + T^{2}\omega^{2}} = - b - a\tan\text{Δφ} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p}^{2}\left( \left( 1 + \frac{TT_{d}\omega^{2}}{1 + T^{2}\omega^{2}} \right)^{2} + \left( \frac{T_{d}\omega}{1 + T^{2}\omega^{2}} \right)^{2} \right) = a^{2} + b^{2} \\ T_{d}\omega\left\lbrack \operatorname{aT\omega tan}\text{Δφ} - a + b\tan\text{Δφ} + bT\omega \right\rbrack = - \left( b + a\tan\text{Δφ} \right)\left( 1 + T^{2}\omega^{2} \right) \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p}^{2}\left( \left( 1 + \frac{TT_{d}\omega^{2}}{1 + T^{2}\omega^{2}} \right)^{2} + \left( \frac{T_{d}\omega}{1 + T^{2}\omega^{2}} \right)^{2} \right) = a^{2} + b^{2} \\ T_{d}\omega = \frac{- \left( b + a\tan\text{Δφ} \right)\left( 1 + T^{2}\omega^{2} \right)}{b\tan\text{Δφ} - a + T\omega\left( b + \operatorname{atan}\text{Δφ} \right)} \\ \end{matrix} \right.\ $$

$$k_{p}^{2}\left( \left( 1 - \frac{\text{Tω}\left( b + a\tan\text{Δφ} \right)}{b\tan\text{Δφ} - a + T\omega\left( b + \operatorname{atan}\text{Δφ} \right)} \right)^{2} + \left( \frac{b + a\tan\text{Δφ}}{b\tan\text{Δφ} - a + T\omega\left( b + \operatorname{atan}\text{Δφ} \right)} \right)^{2} \right) = a^{2} + b^{2}$$

$$k_{p}^{2}\left( \left( \frac{b\tan\text{Δφ} - a}{b\tan\text{Δφ} - a + T\omega\left( b + \operatorname{atan}\text{Δφ} \right)} \right)^{2} + \left( \frac{b + a\tan\text{Δφ}}{b\tan\text{Δφ} - a + T\omega\left( b + \operatorname{atan}\text{Δφ} \right)} \right)^{2} \right) = a^{2} + b^{2}$$

$$k_{p}^{2}\frac{\left( b\tan\text{Δφ} - a \right)^{2} + \left( b + a\tan\text{Δφ} \right)^{2}}{\left\lbrack b\tan\text{Δφ} - a + T\omega\left( b + \operatorname{atan}\text{Δφ} \right) \right\rbrack^{2}} = a^{2} + b^{2}$$

$$k_{p}^{2}\frac{\left( a^{2} + b^{2} \right)\left( 1 + \left( \tan\text{Δφ} \right)^{2} \right)}{\left\lbrack b\tan\text{Δφ} - a + T\omega\left( b + \operatorname{atan}\text{Δφ} \right) \right\rbrack^{2}} = a^{2} + b^{2}$$

$$k_{p}^{2}\frac{\left( 1 + \left( \tan\text{Δφ} \right)^{2} \right)}{\left\lbrack b\tan\text{Δφ} - a + T\omega\left( b + \operatorname{atan}\text{Δφ} \right) \right\rbrack^{2}} = 1$$

$$k_{p} = \frac{b\tan\text{Δφ} - a + T\omega\left( b + \operatorname{atan}\text{Δφ} \right)}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}$$

$$k_{p} = \frac{b\tan\text{Δφ} + bT\omega + a\operatorname{T\omega tan}\text{Δφ} - a}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}$$

$$k_{p} = \frac{\tan\text{Δφ}}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}b + \frac{T}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}b\omega + \frac{\operatorname{Ttan}\text{Δφ}}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}a\omega - \frac{1}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}a$$

$$k_{p} = \frac{\tan\text{Δφ}}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}\left( 120\omega^{5} - 225\omega^{3} + 15\omega \right) + \frac{T}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}\left( 120\omega^{6} - 225\omega^{4} + 15\omega^{2} \right) + \frac{\operatorname{Ttan}\text{Δφ}}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}\left( 274\omega^{5} - 85\omega^{3} + \omega \right) - \frac{1}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}\left( 274\omega^{4} - 85\omega^{2} + 1 \right)$$

$$k_{p} = \frac{120T}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}\omega^{6} + \frac{\tan\text{Δφ}}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}\left( 120 + 274T \right)\omega^{5} - \frac{1}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}\left( 274 + 225T \right)\omega^{4} - \frac{\tan\text{Δφ}}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}\left( 225 + 85T \right)\omega^{3} + \frac{1}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}\left( 85 + 15T \right)\omega^{2} + \frac{\tan\text{Δφ}}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}\left( 15 + T \right)\omega - \frac{1}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}$$

Szukamy ekstremum funkcji wzmocnienia:

$$\left( k_{p} \right)^{'} = \frac{720T}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}\omega^{5} + \frac{\tan\text{Δφ}}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}5\left( 120 + 274T \right)\omega^{4} - \frac{1}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}4\left( 274 + 225T \right)\omega^{3} - \frac{\tan\text{Δφ}}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}3\left( 225 + 85T \right)\omega^{2} + \frac{1}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}2\left( 85 + 15T \right)\omega + \frac{\tan\text{Δφ}}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}\left( 15 + T \right)$$

Dla Δϕ=60 ̊

$$\left( k_{p} \right)^{'} = 360T\omega^{5} + \frac{5\sqrt{3}}{2}\left( 120 + 274T \right)\omega^{4} - \left( 548 + 450T \right)\omega^{3} - \frac{3\sqrt{3}}{2}\left( 225 + 85T \right)\omega^{2} + \left( 85 + 15T \right)\omega + \frac{\sqrt{3}}{2}\left( 15 + T \right)$$

$$\left\{ \begin{matrix} k_{p} = \frac{b\tan\text{Δφ} - a + T\omega\left( b + \operatorname{atan}\text{Δφ} \right)}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}} \\ T_{d}\omega = \frac{- \left( b + a\tan\text{Δφ} \right)\left( 1 + T^{2}\omega^{2} \right)}{b\tan\text{Δφ} - a + T\omega\left( b + \operatorname{atan}\text{Δφ} \right)} \\ \end{matrix} \right.\ $$

a = 274ω⁴ − 85ω² + 1

b = 120ω⁵ − 225ω³ + 15ω

T	w1	kp	Td
0,05	0,21350	2,06	3,33
0,1	0,21180	2,05	3,26
0,15	0,21000	2,03	3,19
0,2	0,20840	2,02	3,13
0,25	0,20670	2,00	3,07
0,3	0,20520	1,99	3,01
0,35	0,20370	1,98	2,95
0,4	0,20220	1,97	2,89
0,45	0,20070	1,96	2,82
0,5	0,19930	1,95	2,76

Przykład:

Obiekt inercyjny piątego rzędu i regulator PID

$$G\left( s \right) = \frac{k}{\left( T_{1}s + 1 \right)\left( T_{2}s + 1 \right)\left( T_{3}s + 1 \right)\left( T_{4}s + 1 \right)\left( T_{5}s + 1 \right)}$$

$$R\left( s \right) = k_{p}\left( 1 + \frac{1}{T_{i}s} + T_{d}s \right)$$

$$G_{\text{uo}}\left( s \right) = \frac{kk_{p}\left( 1 + \frac{1}{T_{i}s} + T_{d}s \right)}{\left( T_{1}s + 1 \right)\left( T_{2}s + 1 \right)\left( T_{3}s + 1 \right)\left( T_{4}s + 1 \right)\left( T_{5}s + 1 \right)}$$

Dane:

k = 1,   T₁ = 1,   T₂ = 2,   T₃ = 3,   T₄ = 4,   T₅ = 5

$$G_{\text{uo}}\left( \text{jω} \right) = \frac{k_{p}\left( 1 + \frac{1}{T_{i}\text{jω}} + T_{d}\text{jω} \right)}{\begin{bmatrix} 120j\omega^{5} + 274\omega^{4} - 225\text{jω}^{3} - 85\omega^{2} + 15j\omega + 1 \\ \end{bmatrix}}$$

a = 274ω⁴ − 85ω² + 1

b = 120ω⁵ − 225ω³ + 15ω

$$G_{\text{uo}}\left( \text{jω} \right) = \frac{k_{p}\left( a - jb \right)}{\left( a^{2} + b^{2} \right)} \bullet \left( 1 + j\left( T_{d}\omega - \frac{1}{T_{i}\omega} \right) \right)$$

$$G_{\text{uo}}\left( \text{jω} \right) = k_{p}\frac{a + b\left( T_{d}\omega - \frac{1}{T_{i}\omega} \right) + ja\left( T_{d}\omega - \frac{1}{T_{i}\omega} \right) - jb}{\left( a^{2} + b^{2} \right)}$$

$$G_{\text{uo}}\left( \text{jω} \right) = k_{p}\frac{a + b\left( T_{d}\omega - \frac{1}{T_{i}\omega} \right) + j\left( \text{aT}_{d}\omega - \frac{a}{T_{i}\omega} - b \right)}{\left( a^{2} + b^{2} \right)}$$

$$P_{\text{uo}}\left( \omega \right) = k_{p}\frac{a + bT_{d}\omega - \frac{b}{T_{i}\omega}}{\left( a^{2} + b^{2} \right)}$$

$$Q_{\text{uo}}\left( \omega \right) = k_{p}\frac{{- b + aT}_{d}\omega - \frac{a}{T_{i}\omega}}{\left( a^{2} + b^{2} \right)}$$

Warunek granicy stabilności

$$\left\{ \begin{matrix} P_{\text{uo}}\left( \omega \right) = - 1 \\ Q_{\text{uo}}\left( \omega \right) = 0 \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p}\frac{a + bT_{d}\omega - \frac{b}{T_{i}\omega}}{\left( a^{2} + b^{2} \right)} = - 1 \\ k_{p}\frac{{- b + aT}_{d}\omega - \frac{a}{T_{i}\omega}}{\left( a^{2} + b^{2} \right)} = 0 \\ \end{matrix} \right.\ $$

$$T_{d}\omega - \frac{1}{T_{i}\omega} = \frac{b}{a}$$

$$k_{p}\frac{a + b\frac{b}{a}}{\left( a^{2} + b^{2} \right)} = - 1$$

$$k_{p}\frac{1}{a} = - 1$$

$$\left\{ \begin{matrix} k_{p} = - a \\ T_{d}\omega - \frac{1}{T_{i}\omega} = \frac{b}{a} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p} = - a \\ T_{d}\omega - \frac{b}{a} = \frac{1}{T_{i}\omega} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p} = - a \\ \frac{1}{T_{i}\omega} = T_{d}\omega - \frac{b}{a} \\ \end{matrix} \right.\ $$

$$\frac{k_{p}}{T_{i}\omega} = b - aT_{d}\omega$$

$$\frac{k_{p}}{T_{i}} = b\omega - aT_{d}\omega^{2}$$

Szukamy ekstremum wyznaczonej funkcji z parametrem T_d

a = 274ω⁴ − 85ω² + 1

b = 120ω⁵ − 225ω³ + 15ω

$$\left( \frac{k_{p}}{T_{i}} \right)^{'} = \left\lbrack \left( 120\omega^{5} - 225\omega^{3} + 15\omega \right)\omega - T_{d}{\left( 274\omega^{4} - 85\omega^{2} + 1 \right)\omega}^{2} \right\rbrack^{'}$$

$$\left( \frac{k_{p}}{T_{i}} \right)^{'} = \left( \left( 120\omega^{6} - 225\omega^{4} + 15\omega^{2} \right) - T_{d}\left( 274\omega^{6} - 85\omega^{4} + \omega^{2} \right) \right)^{'}$$

$$\left( \frac{k_{p}}{T_{i}} \right)^{'} = \left( 120\omega^{6} - 225\omega^{4} + 15\omega^{2} - 274T_{d}\omega^{6} + 85T_{d}\omega^{4} - T_{d}\omega^{2} \right)^{'}$$

$$\left( \frac{k_{p}}{T_{i}} \right)^{'} = \left( \left( 120 - 274T_{d} \right)\omega^{6} + \left( 85T_{d} - 225 \right)\omega^{4} + \left( 15 - T_{d} \right)\omega^{2} \right)^{'}$$

$$\left( \frac{k_{p}}{T_{i}} \right)^{'} = 2\omega\left\lbrack 3\left( 120 - 274T_{d} \right)\omega^{4} + 2\left( 85T_{d} - 225 \right)\omega^{2} + \left( 15 - T_{d} \right) \right\rbrack$$

=4[(85T_d−225)²−3(120−274T_d)(15−T_d)]

=4[7225T_d²−38250T_d+50625−5400+360T_d+12330T_d−822T_d²]

=4(6403T_d²−25560T_d+45225)

$$\omega_{1}^{2} = \frac{- 85T_{d} + 225 - \sqrt{6403T_{d}^{2} - 25560T_{d} + 45225}}{3\left( 120 - 274T_{d} \right)}$$

$$\omega_{2}^{2} = \frac{- 85T_{d} + 225 - \sqrt{6403T_{d}^{2} - 25560T_{d} + 45225}}{3\left( 120 - 274T_{d} \right)}$$

Td	w1	kp	Ti
0	0,18513	1,59	6,25
0,05	0,18637	1,62	6,300496
0,1	0,18763	2,26	10,64036
0,15	0,18893	2,22	9,954836
0,2	0,19024	2,17	9,374234
0,25	0,19159	2,13	8,82347
0,3	0,19297	2,09	8,352568
0,35	0,19437	2,06	7,916507
0,4	0,19581	2,02	7,511543
0,45	0,19727	1,98	7,134455
0,5	0,19877	1,94	6,795203
0,55	0,20029	-445,58	18,31182
0,6	0,20185	-6000,31	-9,62861

Warunek: zapas modułu ΔL=ΔdB

ΔL = 20lgΔM = −Δ

$$\Delta M = 10^{- \frac{\Delta}{20}}$$

$$\left\{ \begin{matrix} P_{\text{uo}}\left( \omega \right) = - 1 + \Delta M \\ Q_{\text{uo}}\left( \omega \right) = 0 \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p}\frac{a + bT_{d}\omega - \frac{b}{T_{i}\omega}}{\left( a^{2} + b^{2} \right)} = \Delta M - 1 \\ k_{p}\frac{{- b + aT}_{d}\omega - \frac{a}{T_{i}\omega}}{\left( a^{2} + b^{2} \right)} = 0 \\ \end{matrix} \right.\ $$

$$T_{d}\omega - \frac{1}{T_{i}\omega} = \frac{b}{a}$$

$$k_{p}\frac{a + b\frac{b}{a}}{\left( a^{2} + b^{2} \right)} = \Delta M - 1$$

$$k_{p}\frac{1}{a} = \Delta M - 1$$

$$\left\{ \begin{matrix} k_{p} = a\left( \Delta M - 1 \right) \\ T_{d}\omega - \frac{1}{T_{i}\omega} = \frac{b}{a} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p} = a\left( \Delta M - 1 \right) \\ T_{d}\omega - \frac{b}{a} = \frac{1}{T_{i}\omega} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p} = a\left( \Delta M - 1 \right) \\ \frac{1}{T_{i}\omega} = T_{d}\omega - \frac{b}{a} \\ \end{matrix} \right.\ $$

$$\frac{k_{p}}{T_{i}\omega} = \left( b - aT_{d}\omega \right)\left( 1 - \Delta M \right)$$

$$\frac{k_{p}}{T_{i}} = \left( b\omega - aT_{d}\omega^{2} \right)\left( 1 - \Delta M \right)$$

Szukamy ekstremum wyznaczonej funkcji z parametrem T_d

a = 274ω⁴ − 85ω² + 1

b = 120ω⁵ − 225ω³ + 15ω

$$\left( \frac{k_{p}}{T_{i}} \right)^{'} = \left( 1 - \Delta M \right)\left\lbrack \left( 120\omega^{5} - 225\omega^{3} + 15\omega \right)\omega - T_{d}{\left( 274\omega^{4} - 85\omega^{2} + 1 \right)\omega}^{2} \right\rbrack^{'}$$

$$\left( \frac{k_{p}}{T_{i}} \right)^{'} = {\left( 1 - \Delta M \right)\left( \left( 120\omega^{6} - 225\omega^{4} + 15\omega^{2} \right) - T_{d}\left( 274\omega^{6} - 85\omega^{4} + \omega^{2} \right) \right)}^{'}$$

$$\left( \frac{k_{p}}{T_{i}} \right)^{'} = \left( 1 - \Delta M \right)\left( 120\omega^{6} - 225\omega^{4} + 15\omega^{2} - 274T_{d}\omega^{6} + 85T_{d}\omega^{4} - T_{d}\omega^{2} \right)^{'}$$

$$\left( \frac{k_{p}}{T_{i}} \right)^{'} = {\left( 1 - \Delta M \right)\left( \left( 120 - 274T_{d} \right)\omega^{6} + \left( 85T_{d} - 225 \right)\omega^{4} + \left( 15 - T_{d} \right)\omega^{2} \right)}^{'}$$

$$\left( \frac{k_{p}}{T_{i}} \right)^{'} = 2\omega\left( 1 - \Delta M \right)\left\lbrack 3\left( 120 - 274T_{d} \right)\omega^{4} + 2\left( 85T_{d} - 225 \right)\omega^{2} + \left( 15 - T_{d} \right) \right\rbrack$$

=4[(85T_d−225)²−3(120−274T_d)(15−T_d)]

=4[7225T_d²−38250T_d+50625−5400+360T_d+12330T_d−822T_d²]

=4(6403T_d²−25560T_d+45225)

$$\omega_{1}^{2} = \frac{- 85T_{d} + 225 - \sqrt{6403T_{d}^{2} - 25560T_{d} + 45225}}{3\left( 120 - 274T_{d} \right)}$$

$$\omega_{2}^{2} = \frac{- 85T_{d} + 225 - \sqrt{6403T_{d}^{2} - 25560T_{d} + 45225}}{3\left( 120 - 274T_{d} \right)}$$

Td	w1	kp	Ti
0	0,18513	0,79	6,25
0,05	0,18637	0,81	6,300496
0,1	0,18763	1,13	10,64036
0,15	0,18893	1,11	9,954836
0,2	0,19024	1,08	9,374234
0,25	0,19159	1,06	8,82347
0,3	0,19297	1,05	8,352568
0,35	0,19437	1,03	7,916507
0,4	0,19581	1,01	7,511543
0,45	0,19727	0,99	7,134455
0,5	0,19877	0,97	6,795203

Warunek: zapas fazy Δϕ

$$G_{\text{uo}}\left( \text{jω} \right) = k_{p}\frac{a + b\left( T_{d}\omega - \frac{1}{T_{i}\omega} \right) + j\left( \text{aT}_{d}\omega - \frac{a}{T_{i}\omega} - b \right)}{\left( a^{2} + b^{2} \right)}$$

$$P_{\text{uo}}\left( \omega \right) = k_{p}\frac{a + bT_{d}\omega - \frac{b}{T_{i}\omega}}{\left( a^{2} + b^{2} \right)}$$

$$Q_{\text{uo}}\left( \omega \right) = k_{p}\frac{{- b + aT}_{d}\omega - \frac{a}{T_{i}\omega}}{\left( a^{2} + b^{2} \right)}$$

$$M_{o}\left( \text{jω} \right) = \frac{1}{\sqrt{a^{2} + b^{2}}}$$

$$M_{R}\left( \text{jω} \right) = k_{p}\sqrt{1 + \left( T_{d}\omega - \frac{1}{T_{i}\omega} \right)^{2}}$$

$$\left\{ \begin{matrix} M\left( \omega \right) = 1 \\ \tan\text{Δφ} = \frac{Q\left( \omega \right)}{P\left( \omega \right)} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} M_{O}\left( \omega \right)M_{R}\left( \omega \right) = 1 \\ \tan\text{Δφ} = \frac{Q\left( \omega \right)}{P\left( \omega \right)} \\ \end{matrix} \right.\ $$

a = 274ω⁴ − 85ω² + 1

b = 120ω⁵ − 225ω³ + 15ω

$$\left\{ \begin{matrix} \frac{k_{p}}{\sqrt{a^{2} + b^{2}}}\sqrt{1 + \left( T_{d}\omega - \frac{1}{T_{i}\omega} \right)^{2}} = 1 \\ \tan\text{Δφ} = \frac{- b + a\left( T_{d}\omega - \frac{1}{T_{i}\omega} \right)}{a + b\left( T_{d}\omega - \frac{1}{T_{i}\omega} \right)} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} \frac{k_{p}}{\sqrt{a^{2} + b^{2}}}\sqrt{1 + \left( T_{d}\omega - \frac{1}{T_{i}\omega} \right)^{2}} = 1 \\ a\tan\text{Δφ} + b\tan\text{Δφ}\left( T_{d}\omega - \frac{1}{T_{i}\omega} \right) = - b + a\left( T_{d}\omega - \frac{1}{T_{i}\omega} \right) \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} \frac{k_{p}}{\sqrt{a^{2} + b^{2}}}\sqrt{1 + \left( T_{d}\omega - \frac{1}{T_{i}\omega} \right)^{2}} = 1 \\ \left( b\tan\text{Δφ} - a \right)\left( T_{d}\omega - \frac{1}{T_{i}\omega} \right) = - b - a\tan\text{Δφ} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} \frac{k_{p}}{\sqrt{a^{2} + b^{2}}}\sqrt{1 + \left( T_{d}\omega - \frac{1}{T_{i}\omega} \right)^{2}} = 1 \\ T_{d}\omega - \frac{1}{T_{i}\omega} = - \frac{b + a\tan\text{Δφ}}{b\tan\text{Δφ} - a} \\ \end{matrix} \right.\ $$

$$\frac{k_{p}}{\sqrt{a^{2} + b^{2}}}\sqrt{1 + \left( \frac{b + a\tan\text{Δφ}}{b\tan\text{Δφ} - a} \right)^{2}} = 1$$

$$\frac{k_{p}}{\sqrt{a^{2} + b^{2}}}\sqrt{\frac{{\left( a - b\tan\text{Δφ} \right)^{2} + \left( b + a\tan\text{Δφ} \right)}^{2}}{\left( b\tan\text{Δφ} - a \right)^{2}}} = 1$$

$$\frac{k_{p}}{\sqrt{a^{2} + b^{2}}}\sqrt{\frac{\left( a^{2} + b^{2} \right)\left( 1 + \left( \tan\text{Δφ} \right)^{2} \right)}{\left( b\tan\text{Δφ} - a \right)^{2}}} = 1$$

$$k_{p} = \frac{b\tan{\text{Δφ} - a}}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}$$

$$\left\{ \begin{matrix} k_{p} = \frac{b\tan\text{Δφ} - a}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}} \\ T_{d}\omega - \frac{1}{T_{i}\omega} = - \frac{b + a\tan\text{Δφ}}{b\tan\text{Δφ} - a} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} k_{p} = \frac{b\tan\text{Δφ} - a}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}} \\ \frac{1}{T_{i}\omega} = T_{d}\omega + \frac{b + a\tan\text{Δφ}}{b\tan\text{Δφ} - a} \\ \end{matrix} \right.\ $$

$$\frac{k_{p}}{T_{i}\omega} = \left( T_{d}\omega + \frac{b + a\tan\text{Δφ}}{b\tan\text{Δφ} - a} \right)\frac{b\tan{\text{Δφ} - a}}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}$$

$$\frac{k_{p}}{T_{i}\omega} = \frac{1}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}\left\lbrack T_{d}\omega\left( b\tan\text{Δφ} - a \right) + b + a\tan\text{Δφ} \right\rbrack$$

$$\frac{k_{p}}{T_{i}} = \frac{1}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}\left\lbrack T_{d}b\omega^{2}\tan\text{Δφ} - T_{d}a\omega^{2} + \text{bω} + \text{aω}\tan\text{Δφ} \right\rbrack$$

Szukamy ekstremum wyznaczonej funkcji z parametrem T_d

a = 274ω⁴ − 85ω² + 1

b = 120ω⁵ − 225ω³ + 15ω

$$\left( \frac{k_{p}}{T_{i}} \right)^{'} = \frac{1}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}\left\lbrack T_{d}\left( 120\omega^{5} - 225\omega^{3} + 15\omega \right)\omega^{2}\tan\text{Δφ} - T_{d}\left( 274\omega^{4} - 85\omega^{2} + 1 \right)\omega^{2} + \left( 120\omega^{5} - 225\omega^{3} + 15\omega \right)\omega + \left( 274\omega^{4} - 85\omega^{2} + 1 \right)\omega\tan\text{Δφ} \right\rbrack^{'}$$

$$\left( \frac{k_{p}}{T_{i}} \right)^{'} = \frac{1}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}\left\lbrack 120{T_{d}\omega}^{7}\tan\text{Δφ} - 225{T_{d}\omega}^{5}\tan\text{Δφ} + 15T_{d}\omega^{3}\tan\text{Δφ} - 274T_{d}\omega^{6} + 85T_{d}\omega^{4} - {T_{d}\omega}^{2} + 120\omega^{6} - 225\omega^{4} + 15\omega^{2} + 274\omega^{5}\tan\text{Δφ} - 85\omega^{3}\tan\text{Δφ} + \omega\tan\text{Δφ} \right\rbrack^{'}$$

$$\left( \frac{k_{p}}{T_{i}} \right)^{'} = \frac{1}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}\left\lbrack 120{T_{d}\omega}^{7}\tan\text{Δφ} + \left( 120 - 274T_{d} \right)\omega^{6} + \left( 274 - 225T_{d} \right)\omega^{5}\tan\text{Δφ} + \left( 85T_{d} - 225 \right)\omega^{4} + \left( 15T_{d} - 85 \right)\omega^{3}\tan\text{Δφ} + \left( {15 - T}_{d} \right)\omega^{2} + \omega\tan\text{Δφ} \right\rbrack^{'}$$

$$\left( \frac{k_{p}}{T_{i}} \right)^{'} = \frac{1}{\sqrt{1 + \left( \tan\text{Δφ} \right)^{2}}}\left\lbrack 840{T_{d}\omega}^{6}\tan\text{Δφ} + 6\left( 120 - 274T_{d} \right)\omega^{5} + 5\left( 274 - 225T_{d} \right)\omega^{4}\tan\text{Δφ} + 4\left( 85T_{d} - 225 \right)\omega^{3} + 3\left( 15T_{d} - 85 \right)\omega^{2}\tan\text{Δφ} + 2\left( {15 - T}_{d} \right)\omega + \tan\text{Δφ} \right\rbrack$$

Dla Δϕ=60 ̊

$$\left( \frac{k_{p}}{T_{i}} \right)^{'} = \left\lbrack 420{{\sqrt{3}T}_{d}\omega}^{6} + \left( 360 - 822T_{d} \right)\omega^{5} + \sqrt{3}\left( 685 - 562,5T_{d} \right)\omega^{4} + \left( 170T_{d} - 450 \right)\omega^{3} + \sqrt{3}\left( 22,5T_{d} - 127,5 \right)\omega^{2} + \left( 15 - T_{d} \right)\omega + 0,5\sqrt{3} \right\rbrack$$

Td	w1	kp	Ti
0	0,096	1,172	14,734
0,05	0,09580	1,15	11,90651
0,5	0,10020	1,23	12,10053
1	0,10580	1,34	12,27725
1,5	0,11230	1,45	12,39409
2	0,11970	1,58	12,40172
2,5	0,12820	1,72	12,28026
3	0,13740	1,86	11,95301
3,5	0,14740	2,00	11,43656
4	0,15700	2,13	10,68797
4,5	0,16630	2,24	9,815442
5	0,17490	2,32	8,891611
5,5	0,18260	2,39	7,985072
6	0,18940	2,43	7,143023
6,5	0,19540	2,46	6,388137
7	0,20050	2,49	5,723418
7,5	0,20510	2,50	5,147402
8	0,20910	2,51	4,6493

Stabilność – kryterium Hurwitza

Badanie stabilności układu regulacji na podstawie transmitancji operatorowej.

$$G\left( s \right) = \frac{k}{a_{n}s^{n} + a_{n - 1}s^{n - 1} + ...\ldots\ldots + a_{3}s^{3} + a_{2}s^{2} + a_{1}s + a_{0}}$$

Macierz Hurwitza

$$\Delta_{n} = \left| \begin{matrix} \begin{matrix} \begin{matrix} a_{n - 1} & a_{n} & 0 \\ a_{n - 3} & a_{n - 2} & a_{n - 1} \\ a_{n - 5} & a_{n - 4} & a_{n - 3} \\ \end{matrix} \\ \begin{matrix} a_{n - 7} & a_{n - 6} & a_{n - 5} \\ a_{n - 9} & a_{n - 8} & a_{n - 7} \\ \ldots & \ldots & \ldots \\ \end{matrix} \\ \end{matrix} & \begin{matrix} 0 \\ a_{n} \\ \begin{matrix} a_{n - 2} \\ a_{n - 4} \\ \begin{matrix} a_{n - 6} \\ \ldots \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} & \begin{matrix} \begin{matrix} 0 & \ldots \\ \end{matrix} \\ \begin{matrix} 0 & \ldots \\ \end{matrix} \\ \begin{matrix} \begin{matrix} a_{n - 1} & \ldots \\ \end{matrix} \\ \begin{matrix} a_{n - 3} & \ldots \\ \end{matrix} \\ \begin{matrix} \begin{matrix} a_{n - 5} & \ldots \\ \end{matrix} \\ \begin{matrix} \ldots & \ldots \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \right|$$

1. Warunek konieczny - wszystkie wyrazy równania charakterystycznego istnieją i są dodatnie

a_n > 0,    a_n − 1 > 0,   ...……   a₃ > 0,    a₂ > 0      a₁ > 0,      a₀ > 0

1. Warunek dostateczny – wszystkie podwyznaczniki wyznacznika głównego macierzy Hurwitza są większe od zera

Δ_n > 0,    Δ_n − 1 > 0,   ...……   Δ₃ > 0,    Δ₂ > 0      Δ₁ > 0

Jeżeli któryś ze współczynników równania charakterystycznego jest ujemny lub równy zero albo któryś z podwyznaczników jest ujemny lub równy zero to układ jest niestabilny.

Ponieważ Δ_n = a₀Δ_n − 1 i Δ₁ = a_n − 1, to niecelowe jest sprawdzanie dodatności tych wyznaczników. Konieczne jest sprawdzanie dodatności wyznaczników od Δ₂ do Δ_n-1.

Jeżeli któryś z podwyznaczników jest równy zero to układ jest na granicy stabilności – oscylacje o stałej amplitudzie.

Kryterium Hurwitza nie podaje żadnych progów stabilności.

Przykład:

$$G\left( s \right) = \frac{k}{\left( T_{1}s + 1 \right)\left( T_{2}s + 1 \right)\left( T_{3}s + 1 \right)\left( T_{4}s + 1 \right)\left( T_{5}s + 1 \right)}$$

Dane:

k = 1,   T₁ = 1,   T₂ = 2,   T₃ = 3,   T₄ = 4,   T₅ = 5

$$G\left( s \right) = \frac{1}{120s^{5} + 274s^{4} + 225s^{3} + 85s^{2} + 15s + 1}$$

a₅ = 120,    a₄ = 274,     a₃ = 225,    a₂ = 85      a₁ = 15,      a₀ = 1

$$\Delta_{5} = \left| \begin{matrix} \begin{matrix} \begin{matrix} a_{4} & a_{5} & 0 \\ a_{2} & a_{3} & a_{4} \\ a_{0} & a_{1} & a_{2} \\ \end{matrix} \\ \begin{matrix} 0 & 0 & a_{0} \\ 0 & 0 & 0 \\ \end{matrix} \\ \end{matrix} & \begin{matrix} 0 \\ a_{5} \\ \begin{matrix} a_{3} \\ a_{1} \\ \begin{matrix} 0 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} & \begin{matrix} \begin{matrix} 0 \\ \end{matrix} \\ \begin{matrix} 0 \\ \end{matrix} \\ \begin{matrix} \begin{matrix} a_{4} \\ \end{matrix} \\ \begin{matrix} a_{2} \\ \end{matrix} \\ \begin{matrix} \begin{matrix} a_{0} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \right|$$

$$\Delta_{5} = \left| \begin{matrix} \begin{matrix} \begin{matrix} 274 & 120 & 0 \\ 85 & 225 & 274 \\ \begin{matrix} 1 \\ 0 \\ 0 \\ \end{matrix} & \begin{matrix} 15 \\ 0 \\ 0 \\ \end{matrix} & \begin{matrix} 85 \\ 1 \\ 0 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} & \begin{matrix} 0 \\ 120 \\ \begin{matrix} 225 \\ 15 \\ \begin{matrix} 0 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} & \begin{matrix} \begin{matrix} 0 \\ \end{matrix} \\ \begin{matrix} 0 \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 274 \\ \end{matrix} \\ \begin{matrix} 85 \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 1 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \right|$$

$$\Delta_{2} = \left| \begin{matrix} 274 & 120 \\ 85 & 225 \\ \end{matrix} \right| = 274 \bullet 225 - 120 \bullet 85 = 51450$$

$$\Delta_{3} = \left| \begin{matrix} \begin{matrix} \begin{matrix} 274 & 120 & 0 \\ 85 & 225 & 274 \\ \begin{matrix} 1 \\ \end{matrix} & \begin{matrix} 15 \\ \end{matrix} & \begin{matrix} 85 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \right| = 274 \bullet 225 \bullet 85 + 1 \bullet 120 \bullet 274 - 274 \bullet 15 \bullet 274 - 85 \bullet 120 \bullet 85 = 3279990$$

$$\Delta_{4} = \left| \begin{matrix} \begin{matrix} \begin{matrix} 274 & 120 & 0 \\ 85 & 225 & 274 \\ \begin{matrix} 1 \\ 0 \\ \end{matrix} & \begin{matrix} 15 \\ 0 \\ \end{matrix} & \begin{matrix} 85 \\ 1 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} & \begin{matrix} 0 \\ 120 \\ \begin{matrix} 225 \\ 15 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \right| = \left( - 1 \right)^{1 + 1} \bullet 274 \bullet \left| \begin{matrix} \begin{matrix} \begin{matrix} 225 & 274 & 120 \\ 15 & 85 & 225 \\ \begin{matrix} 0 \\ \end{matrix} & \begin{matrix} 1 \\ \end{matrix} & \begin{matrix} 15 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \right| + \left( - 1 \right)^{1 + 2} \bullet 120 \bullet \left| \begin{matrix} \begin{matrix} \begin{matrix} 85 & 274 & 120 \\ 1 & 85 & 225 \\ \begin{matrix} 0 \\ \end{matrix} & \begin{matrix} 1 \\ \end{matrix} & \begin{matrix} 15 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \right| =$$

$$\Delta_{4} = 274 \bullet \left| \begin{matrix} \begin{matrix} \begin{matrix} 225 & 274 & 120 \\ 15 & 85 & 225 \\ \begin{matrix} 0 \\ \end{matrix} & \begin{matrix} 1 \\ \end{matrix} & \begin{matrix} 15 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \right| - 120 \bullet \left| \begin{matrix} \begin{matrix} \begin{matrix} 85 & 274 & 120 \\ 1 & 85 & 225 \\ \begin{matrix} 0 \\ \end{matrix} & \begin{matrix} 1 \\ \end{matrix} & \begin{matrix} 15 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \right| =$$

Δ₄ = 274(225•85•15+15•120−225•225−15•274•15) − 120(85•85•15+120−85•225−274•15) = 38102400

Układ otwarty jest stabilny, wszystkie warunki są spełnione

Badanie granicy stabilności układu zamkniętego:

$$G\left( s \right) = \frac{1}{120s^{5} + 274s^{4} + 225s^{3} + 85s^{2} + 15s + 1}$$

R(s) = k_p

$$G_{z}\left( s \right) = \frac{k_{p}\frac{1}{120s^{5} + 274s^{4} + 225s^{3} + 85s^{2} + 15s + 1}}{1 + k_{p}\frac{1}{120s^{5} + 274s^{4} + 225s^{3} + 85s^{2} + 15s + 1}}$$

$$G_{z}\left( s \right) = \frac{k_{p}}{120s^{5} + 274s^{4} + 225s^{3} + 85s^{2} + 15s + 1 + k_{p}}$$

a₅ = 120,    a₄ = 274,     a₃ = 225,    a₂ = 85      a₁ = 15,      a₀ = 1 + k_p

$$\Delta_{5} = \left| \begin{matrix} \begin{matrix} \begin{matrix} a_{4} & a_{5} & 0 \\ a_{2} & a_{3} & a_{4} \\ a_{0} & a_{1} & a_{2} \\ \end{matrix} \\ \begin{matrix} 0 & 0 & a_{0} \\ 0 & 0 & 0 \\ \end{matrix} \\ \end{matrix} & \begin{matrix} 0 \\ a_{5} \\ \begin{matrix} a_{3} \\ a_{1} \\ \begin{matrix} 0 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} & \begin{matrix} \begin{matrix} 0 \\ \end{matrix} \\ \begin{matrix} 0 \\ \end{matrix} \\ \begin{matrix} \begin{matrix} a_{4} \\ \end{matrix} \\ \begin{matrix} a_{2} \\ \end{matrix} \\ \begin{matrix} \begin{matrix} a_{0} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \right|$$

$$\Delta_{5} = \left| \begin{matrix} \begin{matrix} \begin{matrix} 274 & 120 & 0 \\ 85 & 225 & 274 \\ \begin{matrix} 1 + k_{p} \\ 0 \\ 0 \\ \end{matrix} & \begin{matrix} 15 \\ 0 \\ 0 \\ \end{matrix} & \begin{matrix} 85 \\ 1 + k_{p} \\ 0 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} & \begin{matrix} 0 \\ 120 \\ \begin{matrix} 225 \\ 15 \\ \begin{matrix} 0 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} & \begin{matrix} \begin{matrix} 0 \\ \end{matrix} \\ \begin{matrix} 0 \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 274 \\ \end{matrix} \\ \begin{matrix} 85 \\ \end{matrix} \\ 1 + k_{p} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \right|$$

$$\Delta_{2} = \left| \begin{matrix} 274 & 120 \\ 85 & 225 \\ \end{matrix} \right| = 274 \bullet 225 - 120 \bullet 85 = 51450$$

$$\Delta_{3} = \left| \begin{matrix} \begin{matrix} \begin{matrix} 274 & 120 & 0 \\ 85 & 225 & 274 \\ \begin{matrix} 1 + k_{p} \\ \end{matrix} & \begin{matrix} 15 \\ \end{matrix} & \begin{matrix} 85 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \right| = 274 \bullet 225 \bullet 85 + \left( 1 + k_{p} \right) \bullet 120 \bullet 274 - 274 \bullet 15 \bullet 274 - 85 \bullet 120 \bullet 85 = 5240250 + \left( 1 + k_{p} \right)32880 - 1126140 - 867000 = 3247110 + \left( 1 + k_{p} \right)32880$$

k_p ≥ −97, 756

$$\Delta_{4} = \left| \begin{matrix} \begin{matrix} \begin{matrix} 274 & 120 & 0 \\ 85 & 225 & 274 \\ \begin{matrix} 1 + k_{p} \\ 0 \\ \end{matrix} & \begin{matrix} 15 \\ 0 \\ \end{matrix} & \begin{matrix} 85 \\ 1 + k_{p} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} & \begin{matrix} 0 \\ 120 \\ \begin{matrix} 225 \\ 15 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \right| = \left( - 1 \right)^{1 + 1} \bullet 274 \bullet \left| \begin{matrix} \begin{matrix} \begin{matrix} 225 & 274 & 120 \\ 15 & 85 & 225 \\ \begin{matrix} 0 \\ \end{matrix} & \begin{matrix} 1 + k_{p} \\ \end{matrix} & \begin{matrix} 15 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \right| + \left( - 1 \right)^{1 + 2} \bullet 120 \bullet \left| \begin{matrix} \begin{matrix} \begin{matrix} 85 & 274 & 120 \\ 1 + k_{p} & 85 & 225 \\ \begin{matrix} 0 \\ \end{matrix} & \begin{matrix} 1 + k_{p} \\ \end{matrix} & \begin{matrix} 15 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \right| =$$

$$\Delta_{4} = 274 \bullet \left| \begin{matrix} \begin{matrix} \begin{matrix} 225 & 274 & 120 \\ 15 & 85 & 225 \\ \begin{matrix} 0 \\ \end{matrix} & \begin{matrix} 1 + k_{p} \\ \end{matrix} & \begin{matrix} 15 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \right| - 120 \bullet \left| \begin{matrix} \begin{matrix} \begin{matrix} 85 & 274 & 120 \\ 1 + k_{p} & 85 & 225 \\ \begin{matrix} 0 \\ \end{matrix} & \begin{matrix} 1 + k_{p} \\ \end{matrix} & \begin{matrix} 15 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \right| =$$

Δ₄ = 274(225•85•15+15•120(1+k_p)−225•225(1+k_p)−15•274•15) − 120(85•85•15+120(1+k_p)(1+k_p)−85•225(1+k_p)−274•15(1+k_p)) = 274(286875+1800(1+k_p)−50625(1+k_p)−61650) − 120(108375+120(1+k_p)(1+k_p)−19125(1+k_p)−4110(1+k_p)) = 274(225225−48825(1+k_p)) − 120(108375+120(1+k_p)(1+k_p)−23235(1+k_p)) = 48706650 − 10589850(1+k_p) − 14400(1+k_p)(1+k_p)

1 + k_p = 4, 571     i       1 + k_p = −734

Wzmocnienie krytyczne wynosi (k_p)_kr = 3, 571