How To Find The Closed Loop Transfer Function

Closed-Loop Stability

Tony Roskilly , Rikard Mikalsen , in Marine Systems Identification, Modeling and Control, 2015

Case: 2nd-society arrangement

Consider a institute with open-loop transfer function $\frac{0.75}{s(s+\mathrm{one})}$ which nosotros desire to control with a proportional controller, G, equally shown in Effigy 5.iv.

Effigy five.4. Closed-loop system with proportional control.

The closed-loop transfer function is

$\frac{Y(s)}{U(\mathrm{south})}=\frac{K\frac{0.75}{s(s+1)}}{\mathrm{i}+\mathrm{Thou}\frac{0.75}{s(s+1)}}=\frac{0.75K}{{\mathrm{south}}^2+s+0.75K}.$

The closed-loop poles are found by solving the characteristic equation:

$s^{\mathrm{two}}+\mathrm{southward}+0.75K=0\Rightarrow \mathrm{due\; south}=\frac{-1\pm \sqrt{\mathrm{ane}-3\mathrm{Thou}}}{\mathrm{two}}.$

We run across that if (1 − 3Yard) < 0, the roots will be complex. So nosotros have

$s=\left\{\begin{array}{ll}\frac{-1\pm \sqrt{\mathrm{ane}-\mathrm{iii}K}}{\mathrm{two}}& \text{if}K\le \frac{1}{\mathrm{iii}}\\ \frac{-1\pm \mathrm{j}\sqrt{3\mathrm{Yard}-\mathrm{one}}}{2}& \text{if}\mathrm{1000}>\frac{1}{\mathrm{three}}(\text{remember that}\mathrm{j}=\sqrt{-1}).\end{array}\right.$

If One thousand = 0, the poles are at 0 and − one. As G increases, the pole at null becomes more negative and the pole at − 1 becomes more than positive (while $K<\frac{\mathrm{i}}{\mathrm{iii}}$ ). In other words, the two poles movement toward each other.

At $\mathrm{Grand}=\frac{\mathrm{ane}}{3}$ , both poles are at − 0.5 (since the square root term becomes zero). For $K>\frac{\mathrm{i}}{3}$ , the poles become complex, with constant real value − 0.v and opposite imaginary values increasing in magnitude with increasing Grand. These values are known as a complex conjugate pair. Figure v.v shows the root locus plot for the organisation.

Effigy five.five. Root locus plot for $K(s)=\frac{0.75}{\mathrm{south}(s+1)}$ .

From the above assay we can meet that for $\mathrm{Yard}>\frac{\mathrm{ane}}{\mathrm{iii}}$ , the system will become oscillatory and the frequency will increase with increasing Chiliad. The oscillations will be damped since the real part of s (i.e., σ) is negative. If the root locus were to get beyond the y-axis into the right-paw plane (i.e., σ > 0), we would know that the system would be unstable for that value of K.

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On Robust Fault-Tolerant Command of Missile Control System

Xiangchong Liu , ... Hongcai Zhang , in Error Detection, Supervision and Safety of Technical Processes 2006, 2007

(2) stability command over plant time-varying parameters

The 2nd trouble is how to maintain the stability of the system as the parameters of the plant change.

Define the shut loop transfer function is One thousand_i . i = 1, two. G_i is G ₁ and K _two respectively at time t _ane and t _two.

(9) $G_i=\frac{(K_{3}s+\mathrm{i}+{\mathrm{One\; thousand}}_a{\mathrm{Grand}}_1)\cdot G_a}{{\mathrm{Thou}}_{3}s+\mathrm{ane}+{\mathrm{Yard}}_a{K}_{\mathrm{i}}+(K_{\mathrm{three}}s+1)G_a{\mathrm{Grand}}_{bi}{\mathrm{Thousand}}_{\mathrm{south}}{G}_{k2}(\mathrm{ane}+G_a{G}_1)}$

Co-ordinate to Nyquist stability police force, equation (10) tin can be obtained (Kathryn 50. et al., 1991, Caracciolo R., et al., 2005, Olivier V., et al., 2003).

(x) $|{\mathrm{Fifty}}_{\mathrm{two}}\left(j\omega \right)-L_{\mathrm{one}}\left(j\omega \right)|<|L_1\left(j\omega \right)+1|,\omega \in R$

(11) is obtained by putting (four) into (10).

(xi) $\begin{array}{c}\begin{array}{c}\begin{array}{c}|\frac{(K_{3}s+\mathrm{ane})\cdot G_a{G}_{b\mathrm{two}}{\mathrm{Chiliad}}_{\mathrm{due\; south}}{G}_{k\mathrm{ii}}}{K_{3}s+1+G_a{K}_{\mathrm{i}}}-\frac{(G_{3}s+1)\cdot G_a{G}_{b1}{\mathrm{Thousand}}_{\mathrm{due\; south}}{\mathrm{1000}}_{k2}}{K_3\mathrm{southward}+\mathrm{one}+{\mathrm{Thou}}_a{K}_1}|\\ <|\frac{(K_3\mathrm{southward}+\mathrm{i})\cdot G_a{\mathrm{Grand}}_{b1}{G}_s{\mathrm{1000}}_{\mathrm{chiliad}2}}{{\mathrm{One\; thousand}}_3\mathrm{due\; south}+\mathrm{ane}+G_a{G}_1}+1|,\omega \in R\end{array}\\ \text{i.e.}\left|\frac{(K_{3}s+\mathrm{i})\cdot {\mathrm{1000}}_a{G}_s{K}_2}{K_{3}s+\mathrm{ane}+G_a{K}_1}(G_{b2}-G_{b\mathrm{i}})\right|\end{array}\\ <|\frac{({\mathrm{Grand}}_{3}s+\mathrm{one})\cdot G_a{\mathrm{Thousand}}_{b1}{G}_{\mathrm{due\; south}}{K}_2}{K_3\mathrm{southward}+1+{\mathrm{Yard}}_a{K}_1}+1|,\omega \in R,\end{array}$

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Stability of Feedback Control Systems

Nicolae Lobontiu , in System Dynamics for Technology Students (2d Edition), 2018

12.1.2 MIMO Systems

The airtight-loop transfer office matrix of a MIMO control system is given in Eqs. (xi.119) and (11.115) for a nonunity feedback, namely:

(12.half-dozen) $\left[G_{C\mathrm{50}}\left(\mathrm{south}\right)\right]={\left({\left[G\left(\mathrm{south}\right)\right]}^{-\mathrm{i}}+\left[H\left(s\right)\right]\right)}^{-1}={\left({\left(\left[G_p\left(s\right)\right]·\left[G_a\left(s\right)\right]·\left[G_c\left(s\right)\right]\right)}^{-1}+\left[H\left(s\right)\right]\right)}^{-1}$

For a unity-feedback system [G _CL (s)] is obtained from Eq. (12.half-dozen) taking into account that the feedback transfer function matrix is the identity matrix: [H(s)]   =   [I]. The airtight-loop transfer role matrix of Eq. (12.half-dozen) is calculated based on the following adjoint matrix and determinant:

(12.7) $\left[G_{C\mathrm{50}}\left(s\right)\right]=\frac{\mathrm{adj}\left({\left[\mathrm{1000}\left(\mathrm{south}\right)\right]}^{-1}+\left[H\left(s\right)\right]\right)}{\det \left({\left[G\left(s\right)\right]}^{-1}+\left[H\left(s\right)\right]\right)}$

As a result, the closed-loop poles are constitute by solving the characteristic equation of Eq. (12.vii), which is:

(12.viii) $\det \left({\left[G\left(s\right)\right]}^{-\mathrm{ane}}+\left[H\left(\mathrm{southward}\right)\right]\right)=0$

Example 12.4

A unity-feedback MIMO control system is formed of a proportional controller and a plant whose transfer function matrices are $\left[{\mathrm{Yard}}_c\left(\mathrm{due\; south}\right)\right]=\left[\begin{array}{cc}\mathrm{i}& 0\\ 0& \mathrm{five}\end{array}\right];\left[G_p\left(s\right)\right]=\left[\begin{array}{cc}\frac{1}{{\mathrm{south}}^2+3s+2}& \frac{2\mathrm{south}+\mathrm{ane}}{{\mathrm{south}}^2+\mathrm{three}s+2}\\ \frac{2s+\mathrm{ane}}{s^2+3s+\mathrm{two}}& \frac{s-1}{{\mathrm{south}}^2+\mathrm{iii}s+2}\end{array}\right]$ . Analyze the stability of this system.

Solution

The following MATLAB code is used hither:

>> syms due south

>> gc=[i,0;0,5];

>> den=sˆ2+3∗s+2;

>> gp=[ane/den,(2∗due south+1)/den;(2∗south+1)/den,(s-1)/den];

>> g=gp∗gc;

>> det(inv(thou)+eye(two))% eye(2) is the 2x2 [H]=[I] identity matrix

The last command returns the determinant expression of Eq. (12.eight), which is as well the characteristic expression:

(12.9) $\det \left({\left[G\left(\mathrm{southward}\right)\right]}^{-1}+\left[I\right]\right)=\frac{-s^4-11s^3-4{\mathrm{southward}}^{\mathrm{two}}+\mathrm{five}s+\mathrm{fourteen}}{5\left(4{\mathrm{south}}^{\mathrm{ii}}+3\mathrm{south}+\mathrm{ii}\right)}$

Solving the characteristic equation corresponding to Eq. (12.9)—numerator equals zero—results in the post-obit airtight-loop poles (roots of the characteristic equation):

- 10.5895 + 0.0000i

1.0697 + 0.0000i

- 0.7401 + 0.8296i

- 0.7401 - 0.8296i

which are obtained using, for instance, the MATLAB control:

>> roots([-1,-11,-four,5,14])

Because of the real positive root, the system is unstable.

State-Space Stability

As discussed in Affiliate 8 , a transfer part matrix can be obtained from a land-space model. The closed-loop transfer office matrix is expressed in terms of the land-space matrices [ A], [B], [C], and [D] as in Eq. (8.14):

(12.10) $\left[{\mathrm{1000}}_{C\mathrm{50}}\left(s\right)\right]=\left[C\right]·{\left(s·\left[I\right]-\left[A\right]\right)}^{-\mathrm{one}}·\left[B\right]+\left[D\right]$

The closed-loop transfer-function matrix [One thousand _CL (south)] of Eq. (12.x) tin can be written as:

(12.11) $\begin{array}{c}\left[{\mathrm{Chiliad}}_{CL}\left(s\right)\right]=\left[C\right]·\frac{\mathrm{adj}\left(\mathrm{due\; south}·\left[I\right]-\left[A\right]\right)}{\det \left(s·\left[I\right]-\left[A\right]\right)}·\left[B\right]+\left[D\right]\\ =\frac{1}{\det \left(s·\left[I\right]-\left[A\right]\right)}·\left(\left[C\right]·\mathrm{adj}\left(\mathrm{south}·\left[I\right]-\left[A\right]\right)·\left[B\right]+\det \left(s·\left[I\right]-\left[A\right]\right)·\left[D\right]\right)\end{array}$

which indicates that the closed-loop poles are the roots of the feature equation:

(12.12) $\det \left(s·\left[I\right]-\left[A\right]\right)=0$

At the same time, the eigenvalues λ of a foursquare matrix [A] are connected to the respective eigenvectors {y} every bit

(12.13) $\left(\mathrm{\lambda }·\left[I\right]-\left[A\right]\right)·\left\{y\right\}=0$

Eq. (12.13) is the curtailed form of a system of homogeneous equations with the unknown being the components of the eigenvector {y}—its solution is nontrivial when the organization determinant is zippo:

(12.14) $\det \left(\mathrm{\lambda }·\left[I\right]-\left[A\right]\right)=0$

Comparing Eqs. (12.12) and (12.14) shows that the closed-loop poles of the feedback system are identical to the eigenvalues of the state matrix [A] that corresponds to the land-space model of the same arrangement. As a consequence, the stability of a feedback system, which is formulated in state space, tin can be studied by solving for or discussing the nature of matrix [A] eigenvalues as per Eq. (12.fourteen).

Instance 12.five

Analyze the stability of a feedback control organisation modeled in country-space course whose state matrix is $\left[A\right]=\left[\begin{array}{ccc}1& 0& \mathrm{two}\\ -3& -5& \mathrm{i}\\ 1& 4& 1\end{array}\right]$ .

Solution

According to Eq. (12.14), the eigenvalue equation is:

(12.15) $\det \left(\mathrm{\lambda }·\left[I\right]-\left[A\right]\right)={\mathrm{\lambda }}^{\mathrm{iii}}+\mathrm{three}{\mathrm{\lambda }}^2-15\mathrm{\lambda }+23$

whose roots are −six.086, ane.543   ±   1.183·j. These eigenvalues are too obtained using the MATLAB code:

>> a=[ane,0,2;-3,-five,1;one,4,i];

>> eig(a)

Because the eigenvalues are too the closed-loop poles of a land-space organisation and because two eigenvalues are positive, it follows that two closed-loop poles are located in the RHP and one in the LHP—equally a result, this arrangement is unstable.

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Automatic control systems

C.F. Beards BSc, PhD, C Eng, MRAeS, MIOA , in Engineering Vibration Analysis with Application to Control Systems, 1995

5.5 Arrangement TRANSFER FUNCTIONS

The block diagram of any linear closed loop system incorporating negative feedback and having i input and one output variable can be reduced to the form shown in Fig. 5.34.

Fig. v.34. Closed loop arrangement block diagram.

The input and output variables have Laplace transforms θ_i(s) and θ_o(s) respectively, and the forward path of the system has a transfer role (TF) Φ_o(s ). This is the open loop transfer function (OLTF) since it describes the behaviour of the organization with the feedback loop open.

When the loop is closed, the input to Φ_o(s) is the error signal θ₀(south) − θ_one(s) and thus

that is

${\theta }_o\left(\mathrm{southward}\right)/{\theta }_i\left(\mathrm{southward}\right)={\varphi }_o\left(s\right)/\left(1+{\varphi }_o\right(\mathrm{due\; south}\left)\right).$

This equation determines the overall behaviour of the system when the loop is closed and ${\varphi }_o\left(\mathrm{southward}\right)/\left(1+{\varphi }_o\right(s\left)\right)$ is accordingly known equally the airtight loop transfer function (CLTF) denoted by ${\varphi }_c\left(s\right)$

Case 58

Detect the CLTF for the system shown below in block form.

The signal leaving the kickoff junction is

${\theta }_i-{\varphi }_{\mathrm{half-dozen}}{\theta }_o.$

The bespeak leaving the 2d junction is

$[{\theta }_i-{\varphi }_6{\theta }_0]\left({\varphi }_1+\varphi \right)-{\varphi }_{\mathrm{five}}{\theta }_o/{\varphi }_4.$

Thus

$\left[\right({\theta }_i-{\varphi }_{\mathrm{half-dozen}}{\theta }_o\left)\right({\varphi }_{\mathrm{i}}+{\varphi }_2)-{\varphi }_5{\theta }_o/{\varphi }_{\mathrm{four}}]{\varphi }_{\mathrm{iii}}{\varphi }_4={\theta }_o.$

Hence the CLTF,

$\frac{{\theta }_o}{{\theta }_1}=\frac{\left({\varphi }_1+{\varphi }_{\mathrm{ii}}\right){\varphi }_3{\varphi }_4}{1+{\varphi }_3{\varphi }_5+{\varphi }_3{\varphi }_4{\varphi }_6\left({\varphi }_1+{\varphi }_2\right)}.$

For an electric position servo used for controlling the athwart position of a turntable the block diagram model is every bit shown in Fig. v.35.

Fig. 5.35. Electric position servo.

This block diagram can be simplified as shown in Fig. 5.36.

Fig. 5.36. Cake diagram.

From Fig. 5.36,

(five.10) ${\theta }_o=({\mathrm{Yard}}_1{\theta }_i-{\mathrm{Grand}}_3{\theta }_o)(G{M}_2/(J{\mathrm{south}}^2+cs\left)\right).$

If the OLTF and CLTF are to be determined the block diagram is required in the form of Fig. 5.34. The OLTF,

${\varphi }_o\left(\mathrm{southward}\right)={\theta }_o/({\theta }_i-{\theta }_o)$

and CLTF,

${\varphi }_c\left(\mathrm{southward}\right)={\theta }_o/{\theta }_i.$

From Equation (5.ten),

$\left(\right(J{s}^2+cs)/(M{K}_2\left)+K_3\right){\theta }_o=K_1{\theta }_i.$

Hence

${\varphi }_c\left(s\right)=\frac{\mathrm{One\; thousand}{\mathrm{1000}}_{\mathrm{ane}}{K}_2}{J{s}^2+cs+K{K}_{\mathrm{ii}}{\mathrm{Yard}}_3}.$

and

${\varphi }_o\left(\mathrm{southward}\right)=\frac{G{K}_1{K}_{\mathrm{ii}}}{J{s}^{\mathrm{two}}+cs+M{\mathrm{Grand}}_2{K}_3-\mathrm{Thousand}{K}_1{K}_{\mathrm{ii}}}.$

It can be seen from the expression for Φ_c(south) that the frequency equation is

$J{s}^2+cs+\mathrm{Grand}{\mathrm{Thou}}_{\mathrm{ii}}{\mathrm{Thou}}_3=0$

or

${\mathrm{southward}}^2+cs/J+\mathrm{1000}{\mathrm{Thou}}_2{K}_3/J=0$

since the values of s which satisfy this equation make Φ_c(s) = ∞. These values tin can be denoted by p ₁ and p ₂ where

$(s-p_1)(s-p_2)=0,$

$p_1=a+jb\text{and}{p}_{\mathrm{ii}}=a-jb.$

Now

$s=-c/\mathrm{two}J\pm j\surd (G{\mathrm{Thousand}}_2{M}_3/J-(c/2J\left){}^2\right)$

and so

$a=-c/2J\text{and}b=\surd (G{K}_{\mathrm{two}}{K}_3/J-(c/2J\left){}^2\right).$

These roots tin can be plotted on the s-plane as shown in Fig. five.37, as One thousand increases from goose egg.

Fig. 5.37. s-plane.

For an oscillatory response b > 0, that is,

$G{K}_{\mathrm{ii}}{K}_3>\frac{c^2}{4J}.$

The frequency equation of a system governs its response to a stimulus because its roots are the same as the roots of the complementary role. If the roots lie on the left-hand side of the s-plane the response dies away with time and the system is stable. If the roots lie on the correct-paw side of the s-plane the response grows with time and the organisation is unstable.

If general whatever transfer office Φ(s) has the course

$\varphi \left(\mathrm{southward}\right)=\frac{K(s^m+{\alpha }_1{s}^{\mathrm{chiliad}-1}+{\alpha }_2{s}^{\mathrm{thou}2}+.\cdot .\cdot .\cdot {\propto }_m)}{\left({\mathrm{southward}}^n+{\beta }_{\mathrm{ane}}{\mathrm{due\; south}}^{n-I}+{\beta }_2{s}^{n-\mathrm{ii}}+{\beta }_n\right)}\text{and}n\ge m$

with a prepare of roots $s_1,{\mathrm{south}}_2,s_{\mathrm{iii}},\dots ,s_n$ Thus

$\varphi \left(s\right)=\frac{\mathrm{Chiliad}(\mathrm{southward}-z_1)(s-z_2.).\cdot .\cdot .\cdot (\mathrm{southward}-z_{\mathrm{1000}})}{(\mathrm{due\; south}-p_{\mathrm{i}})(s-p_2)(s-p_n)},$

where the values of $s=p_{\mathrm{one}},p_2,p_n$ are those which brand Φ(s) = ∞ and are called poles, and the values of $s=z_{\mathrm{i}},z_2,z_m$ are those which make Φ(s) = 0 and are called zeros. Hence the poles of Φ_c(s) are the natural frequencies of the organisation.

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Frequency Domain Analysis

Tony Roskilly , Rikard Mikalsen , in Marine Systems Identification, Modeling and Command, 2015

6.5.i K circles

Consider the closed-loop transfer function

$G_{\mathrm{CL}}(\mathrm{southward})=\frac{KM(\mathrm{south})}{1+KG(\mathrm{south})}.$

The closed-loop magnitude is given past

$M(j\omega )=\frac{|KM(j\omega )|}{|\mathrm{ane}+K\mathrm{1000}(j\omega )|}.$

Consider now writing the open-loop transfer office in terms of rectangular coordinates: KG(jω) = x + jy. Substituting this gives

$M=\frac{|x+jy|}{|\mathrm{i}+\mathrm{10}+jy|}={\left(\frac{x^2+y^2}{{(x+1)}^{\mathrm{ii}}+y^2}\right)}^{\mathrm{ane}/2},$

or

$K^{\mathrm{ii}}=\frac{{\mathrm{10}}^2+y^2}{{(\mathrm{ten}+1)}^2+y^2}.$

This tin can be rearranged to give

${\left(x+\frac{{\mathrm{Thousand}}^{\mathrm{ii}}}{{\mathrm{Yard}}^2-\mathrm{one}}\right)}^{\mathrm{ii}}+y^2={\left(\frac{\mathrm{One\; thousand}}{{\mathrm{Thou}}^2-1}\right)}^{\mathrm{two}},$

which is the equation of a family of circles depending upon the value of M. The heart of the circle is given past

$x_0=\frac{-M^2}{M^2-1},y_0=0$

and the radius is

$r_0=\left|\right.\frac{M}{K^2-1}\left|\right..$

This is illustrated in Effigy vi.15. The M circles are to the left of x = −0.5 for M > 1, and to the right of x = −0.5 for M < i. When M = one, the circle becomes the straight line at x = −0.five.

Figure 6.15. M circles.

The utilize of 1000-circles can be seen from a simple example. Figure half dozen.sixteena shows the polar frequency response plot with superimposed M circles, and Figure half-dozen.sixteenb shows the respective Cartesian plot. Every bit tin can be seen, it is possible to read off the values of Chiliad _max and ω _K directly from the polar frequency response.

Figure 6.sixteen. The use of Yard circles for evaluating closed-loop system response. (a) Polar plot. (b) Cartesian plot.

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Modeling of digital control systems

M. Sami Fadali , Antonio Visioli , in Digital Control Technology (3rd Edition), 2020

three.7 The closed-loop transfer role

Using the results of Section 3.5, the digital control organisation of Fig. iii.1 yields the airtight-loop block diagram of Fig. 3.14. The block diagram includes a comparator, a digital controller with transfer function C(z), and the ADC-analog subsystem-DAC transfer function G _ZAS (z). The controller and comparator are actually computer programs and supersede the computer block in Fig. iii.1. The cake diagram is identical to those usually encountered in southward-domain analysis of analog systems, with the variable s replaced by z . Hence, the closed-loop transfer function for the organization is given past

Figure iii.14. Block diagram of a single-loop digital command organization.

(three.34) ${\mathrm{Chiliad}}_{cl}\left(z\right)=\frac{C\left(z\right){\mathrm{One\; thousand}}_{ZA\mathrm{South}}\left(z\right)}{1+C\left(z\right)G_{ZAS}\left(z\right)}$

and the closed-loop characteristic equation is

(3.35) $\mathrm{ane}+C\left(z\right){\mathrm{Grand}}_{ZAS}\left(z\right)=0$

The roots of the equation are the closed-loop organization poles, which can be selected for desired time response specifications as in due south-domain design. Earlier we discuss this in some detail, we first examine alternative system configurations and their transfer functions.

When deriving airtight-loop transfer functions of other configurations, the results of Department three.4 must be considered advisedly, as seen from Example three.8.

Example iii.8

Find the Laplace transform of the analog and sampled output for the cake diagram of Fig. 3.15.

Figure iii.xv. Block diagram of a system with sampling in the feedback path.

Solution

The analog variable 10(t) has the Laplace transform

$X\left(\mathrm{south}\right)=H\left(\mathrm{south}\right)G\left(s\right)D\left(s\right)E\left(s\right)$

which involves three multiplications in the s-domain. In the time domain, ten(t) is obtained after 3 convolutions.

From the block diagram

$E\left(s\right)=R\left(s\right)-X^{\ast }\left(\mathrm{due\; south}\right)$

Substituting in the Ten(s) expression, sampling then gives

$\mathrm{Ten}\left(s\right)=H\left(s\right)\mathrm{One\; thousand}\left(\mathrm{due\; south}\right)D\left(s\right)\left[R\left(s\right)-X^{\ast }\left(\mathrm{due\; south}\right)\right]$

Thus, the impulse-sampled variable 10∗(t) has the Laplace transform

$X^{\ast }\left(\mathrm{southward}\right)={\left(H\mathrm{Grand}DR\right)}^{\ast }\left(\mathrm{southward}\right)-{\left(H\mathrm{Thousand}D\right)}^{\ast }\left(\mathrm{due\; south}\right)X^{\ast }\left(\mathrm{southward}\right)$

where, as in the first part of Case 3.1, several components are no longer separable. These terms are obtained as shown in Example 3.ane by inverse Laplace transforming, impulse sampling, and then Laplace transforming the impulse-sampled waveform.

Side by side, we solve for X ^∗(south)

$X^{\ast }\left(s\right)=\frac{{\left(HGDR\right)}^{\ast }\left(s\right)}{1+{\left(HGD\right)}^{\ast }\left(s\right)}$

and and so Eastward(s)

$E\left(s\right)=R\left(\mathrm{southward}\right)-\frac{{\left(HGDR\right)}^{\ast }\left(s\right)}{1+{\left(HGD\right)}^{\ast }\left(\mathrm{southward}\right)}$

With some feel, the last ii expressions can be obtained from the block diagram directly. The combined terms are clearly the ones not separated by samplers in the block diagram.

From the block diagram, the Laplace transform of the output is Y(s)   = 1000(southward)D(due south)E(s). Substituting for E(south) gives

$Y\left(\mathrm{due\; south}\right)=G\left(s\right)D\left(\mathrm{southward}\right)\left[R\left(\mathrm{south}\right)-\frac{{\left(HMDR\right)}^{\ast }\left(\mathrm{due\; south}\right)}{1+{\left(HGD\right)}^{\ast }\left(s\right)}\right]$

Thus, the sampled output is

$Y^{\ast }\left(\mathrm{southward}\right)={\left(\mathrm{Thousand}DR\right)}^{\ast }\left(s\right)-{\left(GD\right)}^{\ast }\left(\mathrm{due\; south}\right)\frac{{\left(HGDR\right)}^{\ast }\left(s\right)}{1+{\left(H\mathrm{Thou}D\right)}^{\ast }\left(\mathrm{due\; south}\right)}$

With the transformation z   =   east ^st , nosotros can rewrite the sampled output as

$Y\left(z\right)=\left(GDR\right)\left(z\right)-\left(GD\right)\left(z\right)\frac{\left(H\mathrm{Chiliad}DR\right)\left(z\right)}{\mathrm{one}+\left(HGD\right)\left(z\right)}$

The terminal equation demonstrates how for some digital systems, no expression is bachelor for the transfer function excluding the input. However, the preceding system has a airtight-loop characteristic equation similar to (3.35) given by 1+(HGD) (z)   = 0. This equation can be used in design, as in cases where a closed-loop transfer function is defined.

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Time- and Frequency-Domain Controls of Feedback Systems

Nicolae Lobontiu , in System Dynamics for Applied science Students (2d Edition), 2018

thirteen.2.2 MIMO Feedback Systems by the Transfer Function Matrix and Country-Space Methods

The steady-state error vector of a nonunity-feedback MIMO system is the difference between the input (reference) vector and the transformed output (command) or back vector:

(13.68) $\left\{e\left(t\right)\right\}=\left\{r\left(t\right)\right\}-\left\{b\left(t\right)\right\}$

The supposition in Eq. (13.68) is that {r(t)} and {b(t)} have the aforementioned number of components. The steady-country fault vector is calculated based on the last-value theorem equally the limit of {e(t)} when time grows to infinity as:

(xiii.69) $\begin{array}{l}\left\{\mathrm{due\; east}\left(\infty \right)\right\}=\underset{t\to \infty }{\lim }\left\{\mathrm{eastward}\left(t\right)\right\}=\underset{s\to 0}{\lim }\left[s·\left\{\mathrm{Eastward}\left(s\right)\right\}\right]=\underset{\mathrm{southward}\to 0}{\lim }\left[s·(\{R\left(\mathrm{due\; south}\right)\}-\left\{B\left(\mathrm{due\; south}\right)\right\})\right]\\ =\underset{s\to 0}{\lim }\left[\mathrm{southward}·(\left[I\right]-[H\left(s\right)]·\left[G_{C\mathrm{Fifty}}\left(\mathrm{southward}\right)\right])·\left\{R\left(\mathrm{due\; south}\right)\right\}\right]\\ \mathrm{where}\left[G_{CL}\left(s\right)\right]={(\left[I\right]+[G\left(s\right)]·\left[H\left(s\right)\right])}^{-\mathrm{ane}}·\left[G\left(\mathrm{due\; south}\right)\right]\mathrm{with}\left[G\left(s\right)\right]=\left[{\mathrm{Thou}}_p\left(s\right)\right]·\left[{\mathrm{1000}}_a\left(s\right)\right]·\left[G_c\left(\mathrm{due\; south}\right)\right]\end{array}$

The airtight-loop and feed-forward transfer function matrices of Eq. (13.69) are derived in Eqs. (11.110) and (11.113). Eq. (13.69) shows that the steady-state errors depend on both the feedback organisation and the input. As seen in Eq. (13.69) , the steady mistake vector of a MIMO feedback system requires knowledge of the closed-loop transfer function matrix [ G _CL (s)] and reference vector {R(s)}.

For unity-feedback systems, the feedback transfer function matrix is the identity matrix, and the time-domain error is:

(13.lxx) $\left\{e\left(t\right)\right\}=\left\{r\left(t\right)\right\}-\left\{c\left(t\right)\right\}$

The corresponding steady-country transfer function is calculated from Eq. (13.69) by just using [H(south)]   =   [I].

Note that the closed-loop transfer function matrix can also be obtained from an existing country-space model in terms of the matrices [A], [B], [C], and [D], as discussed in Chapter 12, namely:

(13.71) $\left\{C\left(s\right)\right\}=\left[{\mathrm{Yard}}_{CL}\left(\mathrm{south}\right)\right]·\left\{R\left(s\right)\right\}\mathrm{with}\left[M_{CL}\left(\mathrm{south}\right)\right]=\left[C\right]·{\left(s·\left[I\right]-\left[A\right]\right)}^{-1}·\left[B\right]+\left[D\right]$

Instance xiii.nine

A two-variable feedback system is controlled equally shown in the block diagram of Figure xiii.24. The plant transfer functions are Thousand _p11(s)   =   0.1/(s  +   0.two), Yard _p12(southward)   =   0.5/(s  +   0.2), G _p21(s)   =   0.2/(south  +   0.2), K _p22(southward)   =   0.3/(s  +   0.two). Find the gains of ii proportional controllers G _ci(southward)   = K ₁ and G _c2(s)   = K ₂ that volition produce a second-degree closed-loop characteristic polynomial, with a damping ratio ξ   =   0.seven and a natural frequency ω _n   =   10   rad/s. Plot the 2 controlled time responses c _one(t) and c ₂(t) corresponding to r ₁(t)   =   1 and r _ii(t)   =   1 and calculate the resulting steady-land errors e _ane(∞) and e ₂(∞).

Figure thirteen.24. Block Diagram of a Negative-Feedback MIMO Control System.

Solution

The feedforward transfer function matrix is:

(thirteen.72) $\left[\mathrm{Grand}\left(s\right)\right]=\left[G_p\left(\mathrm{southward}\right)\right]·\left[{\mathrm{1000}}_c\left(s\right)\right]=\left[\begin{array}{cc}0.1/\left(\mathrm{due\; south}+\mathrm{0.two}\right)& \mathrm{0.five}/\left(s+0.2\right)\\ 0.2/\left(s+0.2\right)& \mathrm{0.iii}/\left(s+0.2\right)\end{array}\right]·\left[\begin{array}{cc}{K}_1& 0\\ 0& M_2\end{array}\right]$

every bit in Eq. (11.115) with [Grand_a (due south)] = [I ]. The closed-loop transfer function matrix of this unity-feedback system is calculated every bit in Eq. (eleven.118) resulting in:

(thirteen.73) $\left[G_{CL}\left(\mathrm{south}\right)\right]={(\left[I\right]+[\mathrm{Yard}\left(\mathrm{southward}\right)])}^{-1}·\left[G\left(\mathrm{southward}\right)\right]=\left[\begin{array}{cc}{\mathrm{Yard}}_1·\left(10\mathrm{due\; south}-\mathrm{seven}{K}_2+\mathrm{ii}\right)/D_{CL}\left(\mathrm{due\; south}\right)& 10K_2·\left(5s+1\right)/D_{CL}\left(s\right)\\ 4{\mathrm{Yard}}_{\mathrm{one}}·\left(\mathrm{v}s+1\right)/D_{C\mathrm{Fifty}}\left(s\right)& K_2·\left(\mathrm{xxx}\mathrm{southward}-\mathrm{vii}{\mathrm{1000}}_1+6\right)/D_{CL}\left(\mathrm{southward}\right)\end{array}\right]$

where the closed-loop characteristic polynomial is:

(xiii.74) $D_{CL}\left(s\right)=100s^2+\mathrm{x}\left(K_1+3K_{\mathrm{ii}}+4\right)·s+2K_1+\mathrm{six}{K}_{\mathrm{ii}}-7K_1·K_{\mathrm{two}}+\mathrm{iv}$

Comparison of this polynomial to the standard class south ²

+

 

2ξ·ω _n ·s

+

 

(ω _{due north} )^two results in:

(13.75) $2\mathrm{\xi }·{\mathrm{\omega }}_n=\frac{\mathrm{ten}\left(K_1+\mathrm{three}{\mathrm{Thou}}_2+4\right)}{100}=\frac{K_1+\mathrm{iii}{K}_2+4}{10};{\mathrm{\omega }}_{\mathrm{northward}}^2=\frac{2{\mathrm{One\; thousand}}_1+\mathrm{half\; dozen}{K}_2-7{\mathrm{Thou}}_{\mathrm{one}}·G_2+\mathrm{four}}{100}$

Solving Eq. (13.75) with ξ

 

=

 

0.7, ω _n

=

 

10

 

rad/s, yields two sets of values for the proportional gains: Chiliad ₁

=

 

161.763, K _two

=

 

−8.587 and K ₁

=

 

− 25.763, One thousand _ii

=

 

53.921. The plots of the controlled signals c ₁ and c ₂ are plotted in Figure 13.25. They resulted by applying the MATLAB stride command to the corresponding transfer functions of the closed-loop transfer function matrix of Eq. (13.73). The steady-country errors are calculated based on Eq. (13.69), which simplifies to:

(13.76) $\left\{e\left(\infty \right)\right\}=\underset{\mathrm{south}\to 0}{\lim }(\left[I\right]-[M_{CL}\left(\mathrm{due\; south}\right)])·\left\{\begin{array}{l}1\\ \mathrm{i}\end{array}\right\}$

For the showtime set of values of the proportional proceeds, the two steady-country errors are e _i(∞)

 

=

 

0.0038 and e _two(∞)

 

=

 

−0.032, whereas for the second prepare, the errors are east _i(∞)

 

=

 

−0.012 and e ₂(∞)

 

=

 

0.0056.

Effigy 13.25. Unit-Pace Time Response of MIMO Feedback System: (a) Thousand _ane  =   161.763, M ₂  =   –viii.587; (b) K ₁  =   –25.763, K ₂  =   53.921.

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Stability

Derong Liu , in The Electrical Engineering Handbook, 2005

Example ane

A system with (airtight-loop) transfer part is given by: $H\left(s\right)=\frac{\mathrm{ane}}{\left(s+1\right)\left(s+2\right)\left(\mathrm{south}+3\right)},$ which will be stable. A system with transfer office is given past: $H\left(\mathrm{south}\right)=\frac{s+1}{\left(s+1\right)\left(s^2+\mathrm{ii}s+3\right)},$ which will be unstable. A system with transfer function is given by: $H\left(\mathrm{south}\right)=\frac{3\left(s+3\right)}{\left(\mathrm{due\; south}+2\right)\left({\mathrm{southward}}^2+\mathrm{v}\right)},$ which volition exist unstable or marginally stable.

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Command strategies of air current energy conversion organisation-based doubly fed induction generator

Boaz Wadawa , ... Smail Sahnoun , in Renewable Energy Systems, 2021

9.2.half dozen Modeling and synthesis of the DC omnibus PI and the network filter

The control of the DC bus voltage around the capacitance (C) in Fig. 9.1 is based on the expression of the following energy balance (Wadawa et al., 2019):

(9.39) $P_R=P_c+P_g$

where P _g is the power transmitted or received to the grid by the inverter; P _R is the agile ability of the bus on the rotor side of the DFIG; and P _c is the power stored by the capacitor C.

(9.40) $P_c=V_{dc}{i}_c$

Five _DC and i _c are the voltage across the capacitor and the current in the capacitor, respectively.

While, using the law of meshing, the matrix of the voltages at the terminals of the filter of RF resistors and LF inductances of the chief Fig. 9.1, tin can be written:

(9.41) $\left(\begin{array}{c}{V}_{F1}\\ 5_{F2}\\ 5_{F3}\end{array}\right)=R_F\left(\begin{array}{c}{i}_1\\ i_2\\ i_3\end{array}\right)+L_F\frac{d}{dt}\left(\begin{array}{c}{i}_1\\ i_2\\ i_{\mathrm{iii}}\end{array}\right)+\left(\begin{array}{c}{5}_{ps1}\\ {\mathrm{Five}}_{p\mathrm{southward}2}\\ V_{p\mathrm{due\; south}3}\end{array}\right)$

Five _F123 and Five _ps123 are the iii-phase input and output voltages at the filter terminals and i ₁₂₃ is the three-phase currents through the filter.

Applying Park'due south transform to Eq. (ix.13) and so rearranging the latter yield the equation for the control law of the active and reactive currents i _fd and i _fq of the filter.

(9.42) $\{\begin{array}{l}{i}_{fd}=\frac{1}{\left(R_F+{\mathrm{Fifty}}_F.s\right)}\left(V_{Fd}+L_F{\omega }_s{i}_{fq}-V_{\mathrm{south}d}\right)\hfill \\ i_{fq}=\frac{1}{\left(R_F+{\mathrm{50}}_F.s\right)}\left(V_{Fq}-L_F{\omega }_s{i}_{fd}-V_{sq}\right)\hfill \end{array}$

where Five _fdq and V _sdq are voltages at the filter terminals in the marking (d, q).

The active and reactive powers are written equally follows:

(9.43) $\{\begin{array}{c}{P}_f=V_s{i}_{fq}\\ Q_f=V_{\mathrm{south}}{i}_{fd}\end{array}$

When Eqs. (ix.39)–(ix.43) are rearranged using the IVC principle according to the steps presented in Section ix.two.iv.ii higher up, the DC bus filter control model configuration in Fig. 9.xi is obtained. Precisely, the latter represents the block diagram of the PI control model of the DC bus voltage, combined with the IVC of the agile and reactive currents through the filter to the grid.

Figure 9.xi. DC bus voltage control model with the indirect vector control of active and reactive filter currents.

9.two.6.ane Synthesis of the PI controller of your DC bus voltage (Nazari et al., 2017)

The DC double-decker voltage regulation law is shown in Fig. 9.12.

Effigy 9.12. DC motorcoach voltage regulation loop (V _DC ).

The open- and closed-loop transfer functions in the diagram in Fig. 9.12 are divers past the expressions:

(9.44) ${\mathrm{TFOL}}_{V_{\mathrm{dc}}}=\frac{{\mathrm{Grand}}_p+\frac{{\mathrm{Thousand}}_i}{\mathrm{due\; south}}}{Cs}$

(9.45) ${\mathrm{TFCL}}_{V_{\mathrm{dc}}}=\frac{\frac{K_p\mathrm{due\; south}+{\mathrm{Thousand}}_i}{C}}{{\mathrm{south}}^{\mathrm{two}}+\frac{{\mathrm{Thou}}_p\mathrm{southward}}{C}+\frac{{\mathrm{Grand}}_i}{C}}$

Past identifying ${\mathrm{TFCL}}_{V_{\mathrm{dc}}}$ to a second-gild filter, we deduce:

The parameters K _p , One thousand _i , τ _r , and ɛ of the PI controller are calculated and presented in Table A2.

9.2.vi.2 Overview of the PI filter electric current controllers ifd and ifq

The synthesis of regulation by IVC of the filter currents follows the same steps as those gear up out in Section 9.2.4.ii of the DFIG above. Except that here, the coupling term between the i _fd and i _fq filter currents is of the form (Lω).

(nine.46) $\{\begin{array}{c}{K}_i=C{\omega }_0^{\mathrm{ii}}\\ K_p=2\xi {\omega }_{0}C\end{array}$

The parameters K _p , K _i , and τ _r of the PI controller are calculated and presented in Table A2.

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Noise in the Luenberger Observer

George Ellis , in Observers in Control Systems, 2002

vii.2.ane.2 Comparison to Traditional (Nonobserver) Systems

The noise sensitivity of the traditional and observer-based control systems can be compared by analyzing the differences of their respective transfer functions. The traditional control system with a noisy sensor is shown in Figure vii-4.

Figure 7-4. Cake diagram of traditional (nonobserver) system.

The dissonance sensitivity of the actual state,C(S), is written out from Mason's signal flow graphs:

(7.10) $\frac{C\left(s\right)}{N\left(\mathrm{due\; south}\right)}=\frac{G_C\left(\mathrm{southward}\right)\times G_{PC}\left(s\right)\times G_P\left(s\right)}{1+M_C\left(\mathrm{southward}\right)\times G_{PC}\left(s\right)\times G_P\left(\mathrm{south}\right)\times G_s\left(s\right)}.$

Rearranging Equation 7.ten to isolate the airtight-loop transfer function ¹ yields:

(7.eleven) $\begin{array}{l}\frac{C\left(s\right)}{\mathrm{Due\; north}\left(s\right)}=\frac{{\mathrm{1000}}_C\left(s\right)\times K_{PC}\left(s\right)\times G_P\left(s\right)}{1+K_C\left(s\right)\times {\mathrm{Grand}}_{PC}\left(s\right)\times {\mathrm{One\; thousand}}_P\left(\mathrm{due\; south}\right)\times {\mathrm{Grand}}_{\mathrm{south}}\left(\mathrm{due\; south}\right)}\\ \frac{C\left(s\right)}{\mathrm{Due\; north}\left(s\right)}=G_s^{-1}\left(s\right)\times G_{CL}\left(s\right).\end{array}$

At frequencies well below the control-constabulary bandwidth, the open-loop proceeds will boss the "i" in the denominator and the dissonance susceptibility will be:

(7.12) $\frac{C\left(s\right)}{\mathrm{North}\left(s\right)}=K_s^{-1}\left(s\right).$

A similar result occurs when the observer-based transfer role of Equation seven.8 is evaluated beneath the observer bandwidth (where Chiliad_OLPF (S) ≈ ane) and the command-police force bandwidth (where the airtight-loop control-law response = 1). Equation 7.8 reduces to:

(vii.13) $\frac{C\left(\mathrm{due\; south}\right)}{N\left(s\right)}=G_{SE\mathrm{southward}t}^{-1}\left(s\right).$

So, at low frequencies, the dissonance susceptibility of the two systems is well-nigh the same. All the same, in that location is a significant difference when the transfer functions are evaluated at higher frequencies. In the traditional system, at frequencies higher than the control-law bandwidth, the "1" in the denominator of Equation seven.11 dominates and the racket sensitivity reduces to:

(7.14) $\begin{array}{l}\\ \frac{C\left(s\right)}{\mathrm{North}\left(s\right)}={\mathrm{Chiliad}}_C\left(\mathrm{due\; south}\right)\times {\mathrm{1000}}_{PC}\left(\mathrm{southward}\right)\times {\mathrm{Yard}}_P\left(\mathrm{due\; south}\right).\end{array}$

Nonetheless, the observer-based control system produces a different upshot. Above the command-law bandwidth, where "i" dominates the denominator of Equation 7.eight, that equation reduces to:

(7.15) $\begin{array}{l}\\ \frac{C\left(\mathrm{south}\right)}{N\left(s\right)}={\mathrm{Yard}}_{SE\mathrm{due\; south}t}^{-1}\left(s\right)\times {\mathrm{Grand}}_{OLPF}\left(s\right)\times {\mathrm{Chiliad}}_C\left(s\right)\times G_{PC}\left(s\right)\times {\mathrm{Grand}}_P\left(\mathrm{southward}\right).\end{array}$

Comparing the dissonance susceptibility indicated past Equations 7.14 and 7.15, the primary difference is the appearance of the term G ^−one _SEst (S) in the observer-based arrangement. Since G_SEst (South) is ordinarily a term that attenuates at high frequency,1000 ⁻¹ _SEst(S) will normally amplify at loftier frequency. Above the sensor bandwidth and below the observer bandwidth, the observer-based organization will be noisier by an amount approximately equal to the attenuation provided past the sensor.

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