Control System Analysis, Transfer function

Open loop and closed loop transfer function – examples

To understand the concept of Transfer Function in open loop, or on the contrary, closed loop, we use a block diagram of a closed loop system, Figure 1:

null

Figure 1

Where G (s) is the transfer function of the plant and H (s) is the transfer function of the sensor. The sensor generates a signal B (s) that is fed back to the summing point, where it is compared with the reference signal R (s), generating a signal called the error signal E(s). Applying block algebra to Figure 1 we can clearly see that the output signal C (s) can be obtained by multiplying E (s) by G (s):

nullThat is to say:null

The function G (s) of equation (1) is known as the direct path transfer function (quotient between the output and the error signal):

null

Again applying block algebra to Figure 1 we can see that the feedback signal B (s) can be obtained by multiplying C (s) by H (s), that is:

nullThat is:null

The product G(s)H (s) from equation (2) is known as the open-loop transfer function (quotient between the feedback signal and the error signal):

null

Important notes:

  • If the transfer function H (s) of the feedback path (FT of the sensor) is equal to one, H(s)=1, only in this case, the closed-loop transfer function is equal to the transfer function direct path;
  • The direct path transfer function G (s) is also known simply as the Direct Transfer Function.

That is, if the system is represented by the DB in Figure 2:

null

Figure 2

Then the direct transfer function G(s) is also the open-loop function.

Once again, applying block algebra to Figure 1 we can see that the output signal C (s) can be obtained by multiplying G (s) by E (s), that is:

nullSolving for C (s), we obtain that:

nullFrom where:

null

The function C (s) / R (s) of equation (3) is known as the closed-loop transfer function (quotient between the output signal and the input signal):

null

Important note: Equation (3) allows us to obtain the Laplace transform of the output for any input, once we know what the closed-loop transfer function is, by:

null

Example:

 

I suggest to visit: Effect of adding a zero to a control system

Source:

  1. Katsuhiko Ogata, Ingeniería de Control Moderno, páginas 65-66.

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Control System Analysis, PID Control

Effect of adding a zero – example – Control System Design

Effects of the addition of zeros: the addition of a zero to the open loop transfer function has the effect of pulling the rootlocus to the left, with which the system tends to be more stable, and also accelerates the settlement of the response. The effect of such control is to introduce a degree of foresight to the system and speed up the transient response.

To illustrate this effect, let’s look at the following example:

Example 1

Suppose we are in the presence of a system with an unstable plant. An example of such a situation is the following:

null

Where G (s) is the transfer function of the plant and H (s) is the transfer function of the sensor used to assemble the closed-loop system, as shown in Figure 1:

null

Figure 1

We know from Block Algebra and Transfer Function theory that the open loop transfer function of this system Gd (s) is:

null

We also know that the rootlocus is drawn with the open loop transfer function of this system Gd (s), for which we can use the following command in Matlab:

null

null

Graph 1

Analysis: In graph 1 we can see that the system is unstable for all positive values ​​of gain K. That is, if we move through the blue and green curves, varying the value of K, as in graph 2, where K1 = 0.143; K2 = 3.66 and K1 = 30.5, respectively, we see that the poles of the system are located on the right side of the s-plane, and therefore it is an unstable system:

null

Let’s apply the principle of addition of a zero to the open loop transfer function for this case. We are going to add a zero at s = -0.5 (Figure 2), therefore the Gd (s) of the system is:

null

null

Figure 2

Let’s see the effect of adding a zero to the system by:

null

null

Graph 3

Analysis: In graph 3 we see that the LGR of the system has shifted to the left and that the system is stable for any positive value of the gain k, that is, that all the poles of the closed-loop system are located on the left side from plane s (Graph 4), an essential condition for the system to be stable:

null

Graph 4

Source:

Katsuhiko Ogata, Modern Control Engineering, pages 442-443.

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Fourier Transform

Fourier transform for relevant signals – Matlab plot

  • We now consider the exponential signal:

null

Where b is a real constant, and U(t) is a unit step. Take into account that x(t)= U(t) when b=0. For any value of b, the Fourier transform X(ω) of x(t) is given by:

null

Because:

null

Equation (1) is expressed as:

null

null

null

null

The graph of is called the Continuous Amplitude Spectrum of x(t), and the graph of is called the Continuous Phase Spectrum of x(t):

null

null

Both graphs can be generated using the following command in Matlab:

>> w=0:0.2:50;
>> b=10;
>> X=1./(b+j*w);
>> subplot(211), plot (w,abs(X));%gráfica de magnitud de X
>> subplot(212), plot (w,angle(X));%gráfica del ángulo de X

null

Sources:

  1. Fundamentos_de_Señales_y_Sistemas_usando la Web y Matlab
  2. Análisis de sistemas lineales asistido con Scilab, Ebert Brea.
  3. Analisis_de_Sistemas_Lineales
  4. Oppenheim – Señales y Sistemas

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Block Diagram, Control System Analysis, Transfer function

Mass-spring-damper rotational system. Problems solved. Catalog 3

The transfer function of a rotational Mass-Spring-Damper System. 

In this PDF guide, the Transfer Function of the exercises that are most commonly used in the mass-spring-damper system classes that are in turn part of control systems, signals and systems, analysis of electrical networks with DC motor, is determined. electronic systems in mechatronics, etc. It is a good resource to also learn how to obtain the block diagram of the system, or the representation in state variables. Request via email – WhatsApp. Payment is provided by PayPal, Credit or debit card. Cost: € 15

Below, the statements of problems solved in this guide.

1. Given the System of Figure 22, find the transfer function Θ(s)/T(s).

null

2. Given the System of Figure 23, find the transfer function Θ(s)/T(s).

null

3. Given the System of Figure 24, find the transfer functions Θ1(s)/T(s)  and Θ2(s)/T(s).

null

4. Find the state-space representation of the system of the previous exercise, Figure 24, taking Θ1(t)  as the output and T(t) as the input. Build the block diagram of the system and determine the transfer function Θ1(s)/T(s).

5. Given the System of Figure 26, find the transfer function ΘL(s)/Tm(s).

null

6. Given the System of Figure 27, find the transfer functions Θ1(s)/Tm(s) and Θ2(s)/Tm(s).

null

7. Given the System of Figure 28, find the transfer functions Θ1(s)/T(s) and Θ2(s)/T(s).

null

8. Given the System of Figure 29, find the transfer function Θ2(s)/T(s).

null

9. Given the System of Figure 30, find the transfer functions  Θ1(s)/T(s) and Θ2(s)/T(s) . Consider: k1=9, k2=3 N-m/rad, b1=8, b2=1 N-m-s/rad, J1=5, J2=3 Kg-m2.

null

10. Find the state-space representation of the system of the previous exercise, Figure 30, taking Θ2(t) as the output and T(t) as the input. Direct Transform the state-space representation obtained into the transfer function Θ2(s)/T(s). Consider k1=9, k2=3 N-m/rad, b1=8, b2=1 N-m-s/rad, J1=5, J2=3 Kg-m2.

11. Find the state-space representation of the system in Figure 32, taking Θ2(t)  as the output and T(t) as the input. Directly, using Matlab, Transform the state-space representation obtained into the transfer function  Θ2(s)/T(s). Consider k1= k2=1 N-m/rad, b1= b2=1 N-m/rad, J=1 Kg-m2.

null

12. Given the System of Figure 33, find the transfer functions  Θ1(s)/Tm(s) and Θ2(s)/Tm(s).

null

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Mass-spring-damper rotational system. Problems solved. Catalog 3

In this PDF guide, the Transfer Function of the exercises that are most commonly used in the mass-spring-damper system classes that are in turn part of control systems, signals and systems, analysis of electrical networks with DC motor, is determined. electronic systems in mechatronics, etc. It is a good resource to also learn how to obtain the block diagram of the system, or the representation in state variables. Request via email – WhatsApp

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Control System Analysis, Transfer function

Mass-spring-damper Problems solved. Catalog 2

The transfer function of a Mass-Spring-Damper System. 

In this PDF guide, the Transfer Function of the exercises that are most commonly used in the mass-spring-damper system classes that are in turn part of control systems, signals and systems, analysis of electrical networks with DC motor, is determined. electronic systems in mechatronics, etc. It is a good resource to also learn how to obtain the block diagram of the system, or the representation in state variables. Request via email – WhatsApp. Payment is provided by PayPal, Credit or debit card. Cost: € 15

Below, the statements of problems solved in this guide.

1. Given the System of Figure 12, find the transfer functions Y1(s)/U(s) and Y2(s)/U(s).

null

2. Given the System of Figure 13, find the transfer functions Y1(s)/U(s) and Y2(s)/U(s).

null

3. Given the System of Figure 14, find the transfer functions X1(s)/U(s)  and  X2(s)/U(s).

null

4. Given the System of Figure 15, find the transfer function X2(s)/U(s). Consider k1=1, k2= 15 N/m, b1=4, b2= 16 N-s/m, m1= 8, m2=3  Kg.

null

5. Given the System of Figure 16, find the transfer function X3(s)/U(s). Consider k1=5, k2= 4, k3= 4  N/m, b1=2, b2= 2, b3= 3  N-s/m, m1= 4, m2=5, m3=5  Kg.

null

6.Given the System of Figure 17, find the transfer function X1(s)/U(s). Consider k1=k2= 1 N/m, b1= b2= b3= 1  N-s/m, m1= 2, m2=1, m3=1  Kg. The same exercise is solved with state variables in the next number..

null

7. Find the state-space representation of the system of the previous exercise, Figure 17, taking x1(t) as the output and u(t) as the input. Transform the state-space representation obtained in the transfer function X1(s)/U(s). Consider k1=k2= 1 N/m, b1= b2= b3= 1  N-s/m, m1= 2, m2=1, m3=1  Kg.

8. Given the System of Figure 19, find the transfer function Yh(s)/fup(s). Consider kh=7, ks=8, kave=5  N/m, bf=3, bh= 10  N-s/m, mh=1, mf=2 Kg.

null

9. Given the System of Figure 20, find the transfer functions X2(s)/U(s) and X3(s)/U(s). Consider k1=1, k2=2, k3=3, k4=4 N/m, b1=2,b2= 1,b3= 3 N-s/m, m1=2,m2=1,m3=3  Kg.

null

10. Find the state-space representation of the system in Figure 21, taking x3(t) as the output and u(t) as the input. Transform the state-space representation obtained in the transfer function X3(s)/U(s). Consider k=2 N/m, b1=b2=b3=b4=b5=1 N-s/m, m1=2,m2=1,m3=1  Kg.

null

11. Determine the transfer function X1(s)/ F(s) and the block diagram of the system in Figure 22:

null

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Mass-spring-damper system. Problems solved. Catalog 2

In this PDF guide, the Transfer Function of the exercises that are most commonly used in the mass-spring-damper system classes that are in turn part of control systems, signals and systems, analysis of electrical networks with DC motor, is determined. electronic systems in mechatronics, etc. It is a good resource to also learn how to obtain the block diagram of the system, or the representation in state variables. Request via email – WhatsApp. Payment is provided by PayPal, Credit or debit card. Cost: € 5

€15,00

You can be interested in:

Elaborado por Prof. Larry Francis Obando – Technical Specialist – Educational Content Writer – Twitter: @dademuch

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Bode Diagram, Control System Analysis, Transfer function

Obtaining Transfer Function from Bode Diagram

Bode plots are a convenient presentation of the frequency response data for
the purpose of estimating the transfer function. These plots allow parts of the
transfer function to be determined and extracted, leading the way to further
refinements to find the remaining parts of the transfer function.

Although experience and intuition are invaluable in the process, the following steps are still offered as a guideline:

1. Look at the Bode magnitude and phase plots and estimate the pole-zero configuration of the system. Look at the initial slope on the magnitude plot to determine system type. Look at phase excursions to get an idea of the difference between the number of poles and the number of zeros.
2. See if portions of the magnitude and phase curves represent obvious first- or second-order pole or zero frequency response plots.
3. See if there is any telltale peaking or depressions in the magnitude response plot that indicate an underdamped second-order pole or zero, respectively.
4. If any pole or zero responses can be identified, overlay appropriate ±20 or ±40 dB/decade lines on the magnitude curve or ±45°/decade lines on the phase curve and estimate the break frequencies.For second-order poles or zeros, estimate the damping ratio and natural frequency from the standard curves.
5. Form a transfer function of unity gain using the poles and zeros found. Obtain the frequency response of this transfer function and subtract this response from the previous frequency response (Franklin, 1991). You now have a frequency response of reduced complexity from which to begin the process again to extract more of the system’s poles and zeros. A computer program such as MATLAB is of invaluable help for this step.

Example

Find the transfer function of the subsystem whose Bode plots are shown in Figure 1:

null

Figure 1

Let us first extract the underdamped poles that we suspect, based on the peaking in the magnitude curve.We estimate the natural frequency to be near the peak frequency, or approximately 5 rad/s. From Figure 1, we see a peak of about 6.5 dB, which translates into a damping ratio of about ζ=0,24. The unity gain second-order function is thus:

null

The frequency response plot of this function is made and subtracted from the previous
Bode plots to yield the response in Figure 2:

null

Figure 2

Overlaying a -20 dB/decade line on the magnitude response and a -45°/decade line on the phase response, we detect a final pole. From the phase response, we estimate the break frequency at 90 rad/s. Subtracting the response of G2(s)=90/(s+90) from the previous response yields the response in Figure 3.

null

Figure 3

Figure 3 has a magnitude and phase curve similar to that generated by a lag function. We draw a -20 dB/decade line and fit it to the curves. The break frequencies are read from the figure as 9 and 30 rad/s. A unity gain transfer function containing a pole at -9 and a zero at -30 is G3(s)=0.3(s+30)/(s+9). Upon subtraction of G1(s)G2(s)G3(s), we find the magnitude frequency response flat ±1 dB and the phase response flat at -3± 5°. We thus conclude that we are finished extracting dynamic transfer functions as:

null

It is interesting to note that the original curve was obtained from the function:

null

Sources:

  1. Modern_Control_Engineering, Ogata 4t
  2. Control Systems Engineering, Nise
  3. Sistemas de Control Automatico, Kuo

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Bode Diagram

The Bode Diagrams – Plotting of the frequency response of a control system.

The Bode Diagrams is the plotting of the frequency response of a system with separate magnitude and phase plots. The log-magnitude and phase frequency response curves as functions of log ω  are called Bode plots or Bode diagrams. Sketching Bode plots can be simplified because they can be approximated as a sequence of straight lines. Straight-line approximations simplify the evaluation of the magnitude and phase frequency response.

When plotting separate magnitude and phase plots, the magnitude curve can be plotted in decibels (dB) vs. log ω, where dB = 20 log M. Meanwhile, the phase curve is plotted as phase angle vs. log ω.

Example

Plot The Bode Diagram for the frequency response of the characterized by the system Transfer Function G(s):

null

null

Basic Factors of G(jω)H(jω)

The main advantage of using a logarithmic trace is the relative ease of plotting the frequency response curves. The basic factors that are frequent in an arbitrary transfer function G(jω)H(jω) are:

  1. The gain K
  2. Integral and derivative factors null,
  3. First order factors null,
  4. The quadratic factors null.

Once we become familiar with the logarithmic traces of these basic factors, it is possible to use them in order to construct a composite logarithmic trace for any form of G(jω)H(jω), plotting the curves for each factor and adding individual curves in graphical form, since adding the logarithms of the gains corresponds to multiplying them among themselves.

The process of obtaining the logarithmic trace is further simplified by asymptotic approximations for the curves of each factor.

The gain K. A number greater than the unit has a positive value in decibels, while a number smaller than the unit has a negative value.

The logarithmic magnitude curve for a constant gain K is a horizontal line whose magnitude is 20 log K decibels. The phase angle of the gain K is zero. The effect of varying the gain K in the transfer function is that the logarithmic magnitude curve of the transfer function is raised or lowered by the corresponding constant amount, but does not affect the phase curve.

Integral and derivative factorsnull(Poles and Zeros in the origin). The logarithmic magnitude of l/ in decibels is:

null

The phase angle of l/ is constant and equal to -90 °.

In Bode’s Diagram, the frequency ratios are expressed in terms of octaves or decades. An octave is a frequency band from ω1 to 2ω1, where ω1 is any frequency. A decade is a frequency band from ω1 to 10ω1, where, again, ω1 is any frequency. (On the logarithmic scale of the semi-logarithmic paper, any given frequency ratio is represented by the same horizontal distance. For example, the horizontal distance of ω=1  to ω=10 is equal to that of ω=3  to ω=30.

If the logarithmic magnitude of -20 log ω dB is plotted against ω on a logarithmic scale, a line is obtained. To draw this line, we need to locate a point (0 dB, ω= 1) on it. Given that:

null

The slope m of the line for l/ is:

null

The phase angle of the factor l/ is constant and equal to -90°.

Smilarly:null

The slope m of the line for is:null

The phase angle of the factor  is constant and equal to -90°.

The following figure shows frequency response curves for l/ and , respectively.

null

Note that both logarithmic quantities become equal to 0 dB at ω=1.

Therefore, if the transfer function contains the factor (l/)n or ()n, the logarithmic magnitude becomes, respectively, in:

nullOr well:

null

Therefore, the slopes of the logarithmic magnitude curves for the factors (l/)n and ()n are -20n dB/decade und 20n dB/decade, respectively.

The phase angle of (l/)n is equal to -90°n over the entire frequency range, while the phase angle of ()n is equal to 90°n over the entire frequency range. The magnitude curves will pass through the point (0 dBω= 1).

First order Factorsnull. The logarithmic magnitude of l/(1+jωT) in decibels is:

null

For low frequencies, such that ω<<1/T, the logarithmic magnitude is approximated by:

null

Therefore, the logarithmic magnitude curve for low frequencies in this factor is the constant 0 dB line. For high frequencies, such that:

null

The latter is an approximate expression for the high frequency range. At ω=1/T, the logarithmic magnitude is equal to 0 dB; at ω=10/T, the logarithmic magnitude is -20 dB. Therefore, the value of -20 log ωT dB decreases by 20 dB for all decades of ω. Thus, for ω>>1/T, the logarithmic magnitude curve is a straight line with a slope of -20dB/decade (or -6 dB/octave).

Our analysis shows that the logarithmic representation of the frequency response curve of the factor l/(1+jωT) is approximated by two asymptotes (straight lines), one of which is a straight line of 0 dB for the frequency range 0<ω<1/T and the other is a straight line with a slope of -20 dB/decade (or -6 dB/octave) for the frequency range 1/T<ω<∞. The frequency in which the two asymptotes meet is called the corner frequency or cutoff frequency. For the factor l/(1+jωT), the frequency ω=1/T is the corner frequency, since at that point both asymptotes have the same value.

null

An advantage of Bode Diagrams is that, for reciprocal factors, for example, the factor 1+jωT, the logarithmic magnitude and phase angle curves only need to change sign. Therefore, the slope of the high frequency asymptote of 1+jωT is 20 dB/decade, and the phase angle varies from 0° to 90° as the frequency ω increases from zero to infinity, as can be seen in the following figure:

null

Quadratic Factorsnull. Control systems usually have quadratic factors of the form:null

If ζ>1, this quadratic factor is expressed as a product of two first-order factors with real poles. If 0<ζ<1, this quadratic factor is the product of two complex conjugate factors.

The asymptotic frequency response curve for null is obtained as follows. Given that:

null

For low frequencies such that ω<<ωn, the logarithmic magnitude becomes:

null

Therefore, the low frequency asymptote is a horizontal line at 0 dB. For high frequencies such that ω>>ωn, the logarithmic magnitude becomes:

null

The equation for the high frequency asymptote is a line with a slope of -40dB/decade, given that:

null

The high frequency asymptote intersects the low frequency at ω=ωn, since at this frequency:

null

This frequency ωn is the corner frequency for the quadratic factor considered.

null

Sources:

  1. Modern_Control_Engineering, Ogata 4t
  2. Control Systems Engineering, Nise
  3. Sistemas de Control Automatico, Kuo

Literature review by:

Prof. Larry Francis Obando – Technical Specialist – Educational Content Writer

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Control System Analysis, The Nyquist Criteria

Stability via the Nyquist Diagram – The Nyquist Criteria

The Nyquist criterion can tell us if the system is stable or unstable by determining how many closed-loop poles are in the right half-plane of the closed-loop system of Figure 1:

null

Figure 1

Consider the contour A defined in s-plane of Figure 2:

null

Figure 2

If a contour, A, that encircles the entire right half-plane of the root-locus of the system determined by the characteristic equation 1+ G(s)H(s), is mapped through G(s)H(s), then the number of closed-loop poles, Z, in the right half-plane equals the number of open-loop poles, P, that are in the right half-plane minus the number of counterclockwise revolutions, N, around -1 of the mapping; that is, Z:

null

Thus, to reach stability, Z must be equal to zero.

This mapping is called the Nyquist diagram, or Nyquist plot, of G(s)H(s).

To understand the Nyquist criteria for stability, we must first establish four important concepts that will be used during its application:

(1) the relationship between the poles of 1+ G(s)H(s) and the poles of G(s)H(s); (2) the relationship between the zeros of 1+ G(s)H(s) and the poles of the closed-loop transfer function (3) the concept of mapping points; and (4) the concept of mapping contours.

We could demonstrate that the poles of 1+ G(s)H(s) are the same as the
poles of G(s)H(s), the open-loop system, and (2) the zeros of  1+ G(s)H(s) are the
same as the poles of closed-loop transfer function of the system.

Example:

null

Step 1. Find the open-loop transfer function G(s)H(s) of the system.

Consider the closed-loop control system as follows:

null

The system characteristic equation is as follows:

null

The factor form of this characteristic equation is:

null

To determine the previous factor form:

null

Where the open-loop transfer function G(s)H(s) of the system is:

null

Step 2. Use Command Window of Matlab to draw the Nyquist Diagram, applying the following commands:

>> s=tf(‘s’);

>> G=10/(s^3+2*s^2+5*s);

>> nyquist(G);

null

null

We can see at the previous Diagram that for:

null

To reach stability, Z must be equal to zero:

null

Recalling that the poles of 1+ G(s)H(s), are the same as the poles of G(s)H(s), the open-loop system, we can determine P, the number of open-loop poles enclosed by the contour A from:

null

null

A detour around the poles on the contour is required:

null

In the Nyquist Diagram obtained for the system of Task 2, the point -1+j0 is highlighted in red:

null

We can see that N=0, so:

null

However, the Nyquist diagram intersects the real axis at -1+j0. Hence, according to the Nyquist Criteria, the system is marginally stable.

Sources:

  1. Modern_Control_Engineering, Ogata 4t
  2. Control Systems Engineering, Nise
  3. Sistemas de Control Automatico, Kuo

Literature review by:

Prof. Larry Francis Obando – Technical Specialist – Educational Content Writer

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Escuela de Ingeniería Electrónica de la Universidad Simón Bolívar, USB Valle de Sartenejas.

Escuela de Ingeniería Eléctrica de la Universidad Central de Venezuela, UCV CCs

Escuela de Turismo de la Universidad Simón Bolívar, Núcleo Litoral.

Contacto: España. +34633129287

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Control System Analysis, The Nyquist Criteria

Sketching the Nyquist Diagram

The Nyquist diagram is also known as “The Polar Trace” of a transfer function G(jω), is a graph of the magnitude of G(jω) against the phase angle of G(jω) in polar coordinates, according to ω variables of zero to infinity. Therefore, The Nyquist diagram is the geometric place of vectors:

null
conforming ω modified from zero to infinity. Note that, in polar graphs, the phase angles are positive (negative) if they are measured counterclockwise (clockwise) from the positive real axis.

The following Figure shows an example of a Nyquist Diagram:

null

All points of the polar trace of G(jω) represent the end point of a vector at a given value of ω. The projections of G(jω) on the real and imaginary axes are their real and imaginary components. The magnitude and phase angle of G(jω) must be calculated directly for each frequency ω in order to construct polar traces.

Conceptually, the Nyquist diagram is plotted by substituting the points
of the contour that encloses the right half-plane into the function G(s)H(s).  This process is called mapping. Next Figure shows the process of mapping:

null

Consider the closed-loop control system of Figure 1:

null

Figure 1

Thus, in the Nyquist diagram, the contour that encloses the right half-plane, shown in Figure 2, can be mapped through the function G(s)H(s), derived from Figure 1,  by substituting points along the contour into G(s)H(s):

null

Figure 2

If a contour, A, that encircles the entire right half-plane of the root-locus of the system determined by the characteristic equation 1+ G(s)H(s), is mapped through G(s)H(s), then the number of closed-loop poles, Z, in the right half-plane equals the number of open-loop poles, P, that are in the right half-plane minus the number of counterclockwise revolutions, N, around -1+j0 of the mapping; that is, Z:

null

Thus, Z must be equal to zero to reach stability.

This mapping is called the Nyquist diagram, or Nyquist plot, of G(s)H(s). For more information an examples, see: The Nyquist Criteria

 

Example

Consider the control system whose block diagram and diagram are shown in the following Figure 3:

null

Figure 3

Conceptually, the Nyquist diagram is plotted by substituting the points of the contour shown in Figure 4(a) into G(s)H(s):

null

Each Pole and Zero term of G(s) shown in Figure 3(b) is a vector in Figure 4(a) and 4(b). The resultant vector, , found at any point along the contour is in general the product of the Zero vectors divided by the product of the Pole vectors (see Figure 4(c)). Thus, the magnitude of the resultant is the product of the Zero lengths divided by the product of the Pole lengths, and the angle of the resultant is the sum of the Zero angles minus the sum of the Pole angles.

null

Figure 4

The mapping from point A to point C can also be explained analytically. From
A to C the collection of points along the contour is imaginary. Hence, from A to C,
G(s)H(s)=G(s)*1=G(s)=G(jω), or from Figure 3(b):

nullAt zero frequency:

null

Thus, the Nyquist diagram starts at 50/3 at an angle of 0°. As ω increases the real part remains positive, and the imaginary part remains negative.

At null the real part becomes negative. At null, the Nyquist diagram crosses the negative real axis since the imaginary term goes to zero. The real value at the axis crossing, point Q in Figure 4(c), is -0.874. Continuing toward , the real part is negative, and the imaginary part is positive. At infinite frequency:

nullor approximately zero at 90°.

Around the infinite semicircle from point C to point D shown in Figure 4(b), the vectors rotate clockwise, each by 180°. Hence, the resultant undergoes a counterclockwise rotation of 3×180, starting at point C’ and ending at point D’ of Figure 4(c).

Nyquist diagram with Matlab

Consider the following open loop transfer function:

null

To create the Nyquist Diagram of the system, use the following commands in the command window of Matlab:

>> s=tf(‘s’)

>> G=1/(s^2+0.8*s+1)

>> nyquist(G)

This line of commands yields:

null

We can obtain information of points of interest in the Nyquist Diagram by cliking once over the contour. This yields:

null

Sources:

  1. Modern_Control_Engineering, Ogata 4t
  2. Control Systems Engineering, Nise
  3. Sistemas de Control Automatico, Kuo

Literature review by:

Prof. Larry Francis Obando – Technical Specialist – Educational Content Writer

I solve problems!!

WhatsApp:  +34633129287  Immediate Attention!!

Twitter: @dademuch

Copywriting, Content Marketing, Tesis, Monografías, Paper Académicos, White Papers (Español – Inglés)

Escuela de Ingeniería Electrónica de la Universidad Simón Bolívar, USB Valle de Sartenejas.

Escuela de Ingeniería Eléctrica de la Universidad Central de Venezuela, UCV CCs

Escuela de Turismo de la Universidad Simón Bolívar, Núcleo Litoral.

Contacto: España. +34633129287

Caracas, Quito, Guayaquil, Cuenca. 

WhatsApp:  +34633129287

Twitter: @dademuch

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Block Diagram, Control System Analysis

Definition of Electromechanical System

“Electromechanical Systems are those hybrid systems of mechanical and electrical variables.” Applications for electromechanical components cover a broad spectrum, from control systems for robots and star-trackers, to household appliances and hard disk position controls on a computer, or the control of DC motors in air conditioning systems for residential installations.

gettyimages-155388818-1024x1024
Detail of copper winding, stack and shaft of a electric permeant magnet motor for home appliances.

Figure 2.1 shows an electromechanical drive system. It consists of a power and energy source, a gate circuit for the converter, electronic converters (rectifier, inverter, electronic power controller), current sensors (shunts, current transformer, Hall sensor), voltage sensor (divider voltage, potential transformer), speed sensors (tachometers) and displacement sensors (encoders), three-phase rotary machines, gearboxes and specific loads (pump, fan, car, etc.). In Figure 2-1 all components, with the exception of gears, are represented by a Transfer Function (output variables as a function of time), while the gearbox is represented by a Characteristic Function (Xout output variable depending on the input variable Xin)

The electric machine is perhaps the best example of an electromechanical device because of the frequency with which it is used in numerous applications of daily life. An electric machine is a device that can convert mechanical energy into electrical energy (a hydroelectric plant, for example), or convert electrical energy into mechanical energy (a motor).

For the study of electromechanical systems from the point of view of control engineering, we have decided to focus our attention on DC motors, especially armature-controlled DC servo motors, as they are components intensively used in emerging industries that combine electromechanical engineering with Telematics, as is the case with Robotics and Drones technology. And because, precisely, these areas, together with that of electric vehicles and industry 4.0, are initiating a paradigm shift in all areas of life.

null

We are dedicated to developing the mathematical model of an electromechanical system with DC motor, as well as the characteristics of this system when it is part of an open loop or closed loop control system (Servomotors). We also provide numerous examples of how to determine and use the Transfer Function of an electromechanical system to analyze its stability and its response over time (transient and steady state).

And gradually we will cover these industries more specifically, with great potential for innovation and future labor demand.

NEXT:

Sources:

  1. Control Systems Engineering, Nise
  2. Sistemas de Control Automatico Benjamin C Kuo
  3. Modern_Control_Engineering, Ogata 4t
  4. Libro Rashid – Power Electronic Handbook p 663-666
  5. Getty Images

Literature review by:

Prof. Larry Francis Obando – Technical Specialist – Educational Content Writer

WhatsApp:  +34633129287  

Twitter: @dademuch

Copywriting, Content Marketing, Tesis, Monografías, Paper Académicos, White Papers (Español – Inglés)

Escuela de Ingeniería Eléctrica de la Universidad Central de Venezuela, UCV CCs

Escuela de Ingeniería Electrónica de la Universidad Simón Bolívar, USB Valle de Sartenejas.

Escuela de Turismo de la Universidad Simón Bolívar, Núcleo Litoral.

Contacto: España. +34633129287

Caracas, Quito, Guayaquil, Cuenca. 

WhatsApp:  +34633129287   +593998524011  

Twitter: @dademuch

FACEBOOK: DademuchConnection

email: dademuchconnection@gmail.com