The post Inverse Laplace Transform of a Transfer Function Using Matlab appeared first on Electrical Academia.
]]>Let’s find out Inverse Laplace of the following function
\[X(s)\frac{10{{s}^{2}}+20s+40}{{{s}^{3}}+12{{s}^{2}}+47s+60}=\frac{Numerator}{Denumerator}\]
Let’s write a little code in Matlab now:
%% % Calculate Inverse Laplace of a function using Matlab %Let's write numerator and denumerator from the given transfer function num = [10 20 40]; % Numerator Coefficients den = [1 12 47 60]; %Denumerator Coefficients % "residue" command is used to do Partial Fraction Operation &; % returns "residue", and "Poles" and direct term of the partial fraction % expansion % Write "help residue" in Maltab GUI to get better insight [Residue,Poles,Direct_Term] = residue(num,den)
Results:
Here, we get the following results:
Residue =
95.0000
-120.0000
35.0000
Poles =
-5.0000
-4.0000
-3.0000
Direct_Term =
[]
Using above mentioned results, let’s write the partial fraction expansion of the function X(s)
\[X(s)=\frac{95}{s+5}-\frac{120}{s+4}+\frac{35}{s+3}\]
From partial fraction expression, we can easily write the inverse Laplace transform as:
$x(t)=95{{e}^{-5t}}-120{{e}^{-4t}}+35{{e}^{-3t}}$
In order to convert partial fraction expansion back to the original function, we can use the following Matlab command:
\[[b,a]\text{ }=\text{ }residue(r,p,k)\]
Here, we have b (coefficients for numerator) and a (coefficients for denumerator) to write the function like X(s).
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]]>The post Fourier series of a Square Wave using Matlab appeared first on Electrical Academia.
]]>Let’s assume we have a square wave with following characteristics:
$\begin{align} & Period=2ms \\ & Peak-to-Peak\text{ }Value=2\text{ }V \\ & Average\text{ }Value=0\text{ }V \\\end{align}$
So, we can express it as:
\[\begin{align} & x(t)=\frac{4}{\pi }\sum\limits_{n=1}^{\infty }{\frac{1}{(2n-1)}\sin \left[ (2n-1)2\pi {{f}_{o}}t \right]}\text{ }\cdots \text{ }(1)\text{ } \\ & and\text{ }assume \\ & {{f}_{0}}=500Hz \\\end{align}\]
If g(t) is given by
$g(t)=\frac{4}{\pi }\sum\limits_{n=1}^{12}{\frac{1}{(2n-1)}\sin \left[ (2n-1)2\pi {{f}_{o}}t \right]}\text{ }\cdots \text{ }(2)\text{ }$
Now, we will write a Matlab code for g(t) between 0 and 4ms with an interval of 0.05 ms to demonstrate that g(t) is a decent approximation of original function x(t).
% Fourier Series Expansion for Square Wave %% Parameters as mentioned in text f = 500; % Frequecny C = 4/pi; % Constant Value dt = 5.0e-05; % Interval between teo time steps tpts = (4.0e-3/5.0e-5) + 1; % Total points "(final point-initial point)/Interval+1% for n = 1: 12 % Values we are considering to approximate Fourier Seires instead of infinity as given in original function x(t) for m = 1: tpts % Here, we'll consider all "t" points to cover "from 0 to 4ms interval" s1(n,m) = (4/pi)*(1/(2*n - 1))*sin((2*n - 1)*2*pi*f*dt*(m-1)); % Approximate Fourier Series g(t) end end for m = 1:tpts a1 = s1(:,m); % VERY IMPORTANT ! Here, we are assigning a1 for each single column (total 81) a2(m) = sum(a1); % Here, we are summing up the whole column to one single value (adding all 12 values in one column) end % Now, we have a row vector "a2" with total values "81" f1 = a2'; % Here, we have final values "f1" (total 81 points) as transpose of a2 computed above t = 0.0:5.0e-5:4.0e-3; % it's already given in text (0 to 4ms with interval of 0.05 ms) %% Plotting results plot(t,f1) xlabel('Time, s') ylabel('Amplitude, V') title('Fourier Series Expansion')
Results:
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]]>The post Ladder Diagram | Schematic Diagram | Wiring Diagram appeared first on Electrical Academia.
]]>Wiring diagrams tend to show a close representation of the interior position of electrical components in a control cabinet and/or circuit. Sometimes wire diagrams can closely represent a picture. The only difference is components are represented by electrical symbols whether they are NEMA standard or IEC standard symbols, whereas a pictorial will have a more realistic representation of an electrical component.
Wiring diagrams can be very accurate for depicting equipment layout. All connection wires are shown and connected from one component to another. Wiring diagrams are used widely by electricians when connecting electrical or electronic equipment and by technicians when maintaining equipment.
Figure 1. Fan Wiring Diagram
Schematic diagrams do not often show the physical relationship or layout of the components in the circuit; however, schematic diagrams are useful for understanding the sequence of operations or the operation of the circuit.
Schematic diagrams are not intended to illustrate the physical size or appearance of the device nor the location. In troubleshooting, electrical schematics are essential because it enables a technician to trace the circuit and its function without regard to the actual location or physical size of the component.
Figure 2. Schematic view of a Simple Circuit
A ladder diagram is used to point out relationships between circuit components, not the actual location of the components. Ladder diagrams provide a fast and easy understanding of the connection of electrical components in a circuit or operation.
Figure 3. Ladder Diagram
The arrangement of symbols in ladder diagram should promote clarity and understanding. Graphic symbols, abbreviations, and device designations are drawn per industry standards. The circuit should indicate the most direct path of logical sequence. Lines between the symbols can be horizontal or vertical but should be drawn to minimize lines from crossing each other.
Ladder diagrams should not be confused with a one-line diagram. A one line diagram has only one line between individual components. Ladder diagrams; however, often show multiple lines leading to and from components whether they are series or parallel connections.
The ladder diagram, shown in figure 4, is easy to read since there are only two basic parts- the rails and the rungs. The rails are the two dark vertical lines that represent the power source to the control circuit. The control circuit voltage is usually rated at 12V-120V depending on the rated values of the loads connected in the circuit. The rungs are the horizontal lines that illustrate how the control devices and loads are interconnected to make up the control circuit.
1.A ladder diagram translates similarly to a book. Read the ladder diagram from left to right then up to down in order to understand the sequence of the operation.
2. The loads in a ladder diagram are always and shall be connected in parallel on the rungs.
3. The load is the last component connected to the right side of the rail unless there is a protective contact that opens the circuit in the case of an overload event.
4. Control devices are connected between the left side of the rail and the load or other control devices and the load. Control devices are never to be connected from the left rail to the right rail. The result of this type of connection will result in a short circuit.
Figure 4. Ladder Diagram
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]]>The post Bode Plot Example | Bode Diagram Example MATLAB appeared first on Electrical Academia.
]]>In this article, Bode Plot of Simple Phase-Lag Network (First Order System) is obtained using Matlab. In order to draw Bode Plot, we need transfer function from which we deduce the equations for Magnitude and Phase.
\[G(s)=\frac{1}{2s+1}\]
Function in the frequency domain can be written as:
\[G(s)=\frac{1}{2j\omega +1}\]
From above expression, we can deduce the corner frequency or break point as;
\[\omega =\frac{1}{2}\]
For Magnitude Plot:
When ω<<1 (very very small value), then
$G(s)\approx 1$
So, for very small value of ω, log magnitude of the transfer function would be;
\[|G(j\omega ){{|}_{dB}}=20\text{ }log|G(j\omega )|=20\text{ }log(1)=0\]
Hence, magnitude response would be constant below break point.
When ω>>1 (very very large value), then
\[G(s)\approx \frac{1}{2j\omega }\]
So, for very large value of ω, log magnitude of the transfer function would be;
\[|G(j\omega ){{|}_{dB}}=20\text{ }log|G(j\omega )|=20\text{ }log\left| \frac{1}{|2j\omega |} \right|=20\text{ }log\left( \frac{1}{2\omega } \right)=20log\left| 1 \right|-20log\left| 2\omega \right|=-20log(2\omega )\]
So, above the break point, the magnitude plot would be a straight line with -20 dB/decade slope
Now, phase of the transfer function G(s) can be calculated as;
\[\angle G(j\omega )=0-{{\tan }^{-1}}(\omega T)={{\tan }^{-1}}(\omega T)\]
For Phase Plot:
When ω is very very small (ω≈0), then
\[\angle G(j\omega )H(j\omega )={{0}^{\centerdot }}\]
When ω is very very large (ω→∞), then
\[\angle G(j\omega )H(j\omega )=-{{90}^{\centerdot }}\]
Bode Plot Example Matlab Code
Here, we implemented the bode-plot for the comprehensive understanding of the readers.
% Bode Plot for Phase-Lag Network Example clc % Transfer function K = [1]; T = 2; num = [K]; den = [T 1]; H = tf(num, den) % Bode Plot grid on bode(H) grid %For Asymptotic Plot % num=[1]; % den=[2 1]; % bode_asymptotic(num,den);
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]]>The post Bode Plot MATLAB | Bode Plot Transfer Function appeared first on Electrical Academia.
]]>H.W. Bode introduced a method to present the information of a polar plot of a transfer function GH(s), actually the frequency response GH (jω), as two plots with the angular frequency were at the common axis. The first plot shows the magnitude of the transfer function as a function of ω, and the second plot shows the phase as a function of ω. This pair of plots is referred to as Bode Plot or Bode Diagram.
The magnitude of the transfer function is expressed in decibels (dB), the phase in degrees and the common parameter of frequency is plotted on a logarithmic scale in radians. At times, the magnitude of a transfer function is referred to as gain and the corresponding plot as a gain plot.
In bode-plot, low-frequency asymptote (that is ω<<1/T) and high-frequency asymptote (that is ω>>1/T) cut off at 0 decibels (dB) line where ω=1/T, that is the frequency called corner frequency or break point.
The Bode plot or diagram of a transfer function can be constructed by combining the transfer functions of following elementary factors.
We will discuss above elementary factors one by one:
A constant K may be considered as complex number expressed in polar form with magnitude K and angle 0^{.} if K is positive or -180^{.} if K is negative. For gain factor K, the bode-plot is obtained as:
If the open loop gain is
$G\left( j\omega \right)H\left( j\omega \right)=K$
Then it’s Magnitude (dB) is
\[{{\left| G\left( j\omega \right)H\left( j\omega \right) \right|}_{dB}}=20log\left| K \right|=\text{Constant}\]
And its Phase is
\[\angle G(j\omega )H(j\omega )=\left\{ \begin{matrix} {{0}^{\centerdot }} & K>0 \\-{{180}^{\centerdot }} & K<0 \\\end{matrix} \right.\]
The value in decibels is positive when the magnitude of the constant is greater than 1 and negative when less than 1.A number equal to 1 has a value of 0 in decibels.
The log-magnitude plot for a gain factor K is a straight horizontal line with the magnitude of 20logK decibels. By changing the value of gain K in the transfer function brings up or brings down the log-magnitude curve by the proportionate amount. The gain factor K has no effect on the phase curve of the bode plot.
Here, we implemented the bode plot of gain factor K for the comprehensive understanding of the readers.
% Bode Plot for Constant Gain Factors K=4,10,12 clear all;close all;clc num1 = 4; den = [1]; sys1 = tf(num1,den); grid; bodeplot(sys1) hold on num2 = 10; den = [1]; sys2 = tf(num2,den); grid; bodeplot(sys2) hold on num3 = 20; den = [1]; sys3 = tf(num3,den); grid; bodeplot(sys3) grid on hold off legend('K=4','K=10','K=12','Orientation','horizontal');
Fig.1: Plot for Gain Factor K
Pure integrator and differentiator are represented by transfer function 1/s and s respectively. Bode plots are obtained from the corresponding frequency response function 1/jω and jω.
For differentiator, the bode-plot is obtained as:
If the open loop gain is
$G\left( j\omega \right)H\left( j\omega \right)=j\omega $
Then it’s Magnitude (dB) is
\[{{\left| G\left( j\omega \right)H\left( j\omega \right) \right|}_{dB}}=20log\left| j\omega \right|=20log(\omega )\]
And it’s Phase is
\[\angle G(j\omega )H(j\omega )=\angle j\omega ={{90}^{\centerdot }}\]
When we draw bode-plot for differentiator, we can observe that magnitude plot is a straight line with a slope of +20 dB/decade. Whereas, the phase plot is a straight line with angle 90^{o}.
Here, we implemented the Bode plot of a differentiator for the comprehensive understanding of the readers.
% Bode Plot for Differentiator clear all;close all;clc % Transfer function K = [1 0]; T = 1; num = [K]; den = [T]; H = tf(num, den) % Bode Plot grid on bode(H);
Fig.2: Plot for Differentiator
Similarly, for Integrator, the bode-plot is obtained as:
If the open loop gain is
$G\left( j\omega \right)H\left( j\omega \right)={}^{1}/{}_{j\omega }$
Then it’s Magnitude (dB) is
${{\left| G\left( j\omega \right)H\left( j\omega \right) \right|}_{dB}}=20log\left| \frac{1}{j\omega } \right|=20log(1)-20log(\omega )=-20log(\omega )$
And its Phase is
\[\angle G(j\omega )H(j\omega )=\angle \frac{1}{j\omega }=-{{90}^{\centerdot }}\]
When we draw bode-plot for the integrator, we can observe that magnitude plot is a straight line with a slope of -20 dB/decade. Whereas, the phase plot is a straight line with angle -90^{o}.
Here, we implemented the bode plot of Integrator for the comprehensive understanding of the readers.
% Bode Plot for Integrator clear all;close all;clc % Transfer function K = [1]; T = 1; num = [K]; den = [T 0]; H = tf(num, den) % Bode Plot grid on bode(H)
Fig.3: Plot for Integrator
These bode diagrams arise from phase-lag or phase-lead networks whose respective transfer functions are 1/Ts+1 and Ts+1.
For a simple phase-lag network, the bode-plot is obtained as:
If the open loop gain is
\[G\left( j\omega \right)H\left( j\omega \right)=\frac{1}{1+j\omega T}\]
Where T is a real constant.
Then it’s Magnitude (dB) is
\[{{\left| G\left( j\omega \right)H\left( j\omega \right) \right|}_{dB}}=20log\left| \frac{1}{1+j\omega T} \right|=20log\left| 1 \right|-20log\left| 1+j\omega T \right|=-20log(\sqrt{{{1}^{2}}+{{\omega }^{2}}{{T}^{2}}})\]
For Magnitude Plot
For low frequency asymptote (when s→0)
When ω<<1/T (very very small, like approaches to zero), then magnitude would be:
\[|G(j\omega )H(j\omega ){{|}_{dB}}=20\text{ }log(1)=0dB\]
Which means that magnitude plot would be a straight line at 0 dB at low frequency (ω<<1/T).
Similarly, for high-frequency asymptote (when s→∞)
When ω>>1/T (very very big, like approaches to infinity), then magnitude would be:
\[|G(j\omega )H(j\omega ){{|}_{dB}}=-20\text{ }log(\omega T)\]
Which means that magnitude plot would be a straight line with -20 dB/decade slope at high frequency (ω>>1/T).
And its Phase is
\[\angle G(j\omega )H(j\omega )=\angle \frac{1}{1+j\omega T}=-{{\tan }^{-1}}(\omega T)\]
For Phase plot
When ω<<0.1/T, then
\[\angle G(j\omega )H(j\omega )={{0}^{\centerdot }}\]
Which means that phase plot would be a straight line at 0^{. }until ^{ }ω=0.1/T.
When ω=0.1/T, then
\[\angle G(j\omega )H(j\omega )=-{{45}^{\centerdot }}\]
Which means that phase plot would be a straight line with -45^{.} /decade slope until ^{ }ω=10/T.
When ω>>10/T, then
\[\angle G(j\omega )H(j\omega )=-{{90}^{\centerdot }}\]
Which means that phase plot would be a straight line at -90^{o}.
Here, we implemented the Bode plot of a Simple Phase Lag network for the comprehensive understanding of the readers.
% Bode Plot for Phase-Lag Network clear all;close all;clc % Transfer function K = [1]; T = 1; num = [K]; den = [T 1]; H = tf(num, den) % Bode Plot grid on bode(H) grid
Fig.4: Plot for Simple Phase-Lag Network
Similarly, for a simple phase-lead network, the bode-plot is obtained as:
If the open loop gain is
$G\left( j\omega \right)H\left( j\omega \right)=1+j\omega T$
Where T is a real constant.
Then it’s Magnitude (dB) is
\[{{\left| G\left( j\omega \right)H\left( j\omega \right) \right|}_{dB}}=20log\left| 1+j\omega T \right|=20log(\sqrt{{{1}^{2}}+{{\omega }^{2}}{{T}^{2}}})\]
For magnitude plot
For low frequency asymptote (when s→0)
When ω<<1/T (very very small, like approaches to zero), then magnitude would be:
\[|G(j\omega )H(j\omega ){{|}_{dB}}=20\text{ }log(1)=0dB\]
Which means that magnitude plot would be a straight line at 0 dB at low frequency (ω<<1/T).
Similarly, for high-frequency asymptote (when s→∞)
When ω>>1/T (very very big, like approaches to infinity), then magnitude would be:
\[|G(j\omega )H(j\omega ){{|}_{dB}}=20\text{ }log(\omega T)\]
Which means that magnitude plot would be a straight line at with +20 dB/decade slope at high frequency (ω>>1/T).
And its Phase is
\[\angle G(j\omega )H(j\omega )=\angle 1+j\omega T={{\tan }^{-1}}(\omega T)\]
For Phase plot
When ω<<0.1/T, then
\[\angle G(j\omega )H(j\omega )={{0}^{\centerdot }}\]
Which means that phase plot would be a straight line at 0^{. }until ^{ }ω=0.1/T.
When ω=0.1/T, then
\[\angle G(j\omega )H(j\omega )=-{{45}^{\centerdot }}\]
Which means that phase plot would be a straight line with +45^{.} /decade slope until ^{ }ω=10/T.
When ω>>10/T, then
\[\angle G(j\omega )H(j\omega )={{90}^{\centerdot }}\]
Which means that phase plot would be a straight line at 90^{o}.
Here, we implemented the bode plot of a Simple Phase Lead network for the comprehensive understanding of the readers.
% Bode Plot for Phase-Lead Network clc % Transfer function K = [1]; T = 1; num = [K 1]; den = [T]; H = tf(num, den) % Bode Plot grid on bode(H) grid
Fig.5: Plot for Simple Phase-Lead Network
The transfer function for typical quadratic function can be written as
\[G\left( j\omega \right)H\left( j\omega \right)=\frac{\omega _{n}^{2}}{{{s}^{2}}+2\zeta {{\omega }_{n}}s+\omega _{n}^{2}}\]
The Magnitude Plot is obtained from
\[|G(j\omega )H(j\omega ){{|}_{dB}}=20\log \left| \frac{\omega _{n}^{2}}{{{\left( j\omega \right)}^{2}}+2\zeta {{\omega }_{n}}\left( j\omega \right)+\omega _{n}^{2}} \right|\]
\[|G(j\omega )H(j\omega ){{|}_{dB}}=20\log \left| \frac{1}{{{\left( {}^{j\omega }/{}_{{{\omega }_{n}}} \right)}^{2}}+2\zeta {{\omega }_{n}}\left( {}^{j\omega }/{}_{{{\omega }_{n}}} \right)+1} \right|\]
The above equation can be simplified as
\[|G(j\omega )H(j\omega ){{|}_{dB}}=-20~log{{\left[ {{\left( 1-{{\left( {}^{\omega }/{}_{{{\omega }_{n}}} \right)}^{2}} \right)}^{2}}+{{\left( {}^{2\zeta \omega }/{}_{{{\omega }_{n}}} \right)}^{2}} \right]}^{{}^{1}/{}_{2}}}\]
And the Phase Plot is obtained from
\[\angle G(j\omega )H(j\omega )=-ta{{n}^{-1}}\left( \frac{{}^{2\zeta \omega }/{}_{{{\omega }_{n}}}}{1-{{\left( {}^{\omega }/{}_{{{\omega }_{n}}} \right)}^{2}}} \right)\]
In above expression, the damping ratio ζ is a parameter and the frequency is normalized as ω/ ω_{n }.
The bode-plot shown in the following figure was obtained from Matlab Software. Asymptotic approximations were drawn by two lines: one for a value of ω/ ω_{n }<<1 and one for ω/ ω_{n }>>1._{ }
At low frequencies, that is, for values for ω such that ω<< ω_{n}, the magnitude in decibels can be approximated by
\[|G(j\omega )H(j\omega ){{|}_{dB}}=-20\log {{1}^{{}^{1}/{}_{2}}}=0\]
Similarly, for higher frequencies that value of magnitude is approximated by
\[|G(j\omega )H(j\omega ){{|}_{dB}}=-20\log {{\left[ {{\left( -{{\left( {}^{\omega }/{}_{{{\omega }_{n}}} \right)}^{2}} \right)}^{2}} \right]}^{{}^{1}/{}_{2}}}=-40\log {}^{\omega }/{}_{{{\omega }_{n}}}\]
Which is a line with slope of -40 dB/decade on semi-log paper. These asymptotes meet when ω= ω_{n }=1; this point is called a break frequency or break point.
Similarly, for phase plot
When ω<< ω_{n }, then
$\angle G(j\omega )H(j\omega )={{0}^{\centerdot }}$
Which means that phase plot would be a straight line at 0^{o}.
When ω>> ω_{n }, then
$\angle G(j\omega )H(j\omega )=-{{180}^{\centerdot }}$
Which means that phase plot would be a straight line at 180^{o}.
When ω= ω_{n }, then
$\angle G(j\omega )H(j\omega )=-{{90}^{\centerdot }}$
Which means that phase plot would be a straight line with -90^{.} /decade slope.
Here, we implemented the bode plot of a second-order network for the comprehensive understanding of the readers.
%Bode-Plot for Second-Order System clear all;close all;clc fn = 1; % Define natural frequency (Hz) wn = 2*pi*fn; % Natural frequency conversion in rad/s Zeta = 0.05; % Damping Factor Num = [0 0 wn^2]; % Numerator Den = [1 2*Zeta*wn wn^2]; % Denominator Gs= tf(Num,Den) % Transfer Function bode(Gs); %Bode-Plot grid on
Fig.6: Plot for Second Order System
There are certain terms, which we need to familiar with to fully understand the bode plot.
It is the frequency, where phase shift becomes -180^{o}.
It is the frequency at which amplitude ratio becomes 1 or log modulus of transfer function becomes 0.
The characteristic equation of a typical system can be written as,
$1+GH\left( s \right)=0$
Where
$GH\left( s \right)=-1$
And in frequency domain,
$GH\left( j\omega \right)=-1$
How far -1 is from open loop transfer function GH (jω) measures the stability of a system. This distance can be measured in terms called phase margin and gain margin.
Phase and gain margin are usually measured from open loop response and cannot be obtained from the frequency response of a closed loop system directly.
It can be described as an increase in the open-loop system gain |GH (jω)| when system phase is at 180^{.} Which will cause marginal stability of a system.
In order to measure the gain margin,
In order to measure the phase margin,
For more comprehensive understanding, see the solved Example for Bode Plot using Matlab.
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]]>The post State Space Representation and Example appeared first on Electrical Academia.
]]>For state models, the equations are set up into a set of first-order differential equations in terms of chosen state variables, and the outputs are stated in these same state variables. Since the variables elimination is not an integral part of this method, state models can be obtained easily. State models should be inferred from system equations directly.
Consider a system described by nth-order differential equation.
\[\frac{{{d}^{n}}w}{d{{t}^{n}}}+{{a}_{n}}\frac{{{d}^{n-1}}w}{d{{t}^{n-1}}}+.\ldots +{{a}_{2}}\frac{dw}{dt}+{{a}_{1}}w=r\text{ (1)}\]
A state mode for this system is not unique but depends on the choice of a set of state variables x_{1}(t), x_{2}(t)… x_{n} (t). One possible choice is the following;
${{x}_{1}}=w$
${{x}_{2}}=\dot{w}$
${{x}_{n}}=\overset{(n-1)}{\mathop{w}}\,$ (n represents system order)
Directly from these definitions the following equations are obtained;
$\overset{.}{\mathop{{{x}_{1}}}}\,={{x}_{2}}$
$\overset{.}{\mathop{{{x}_{2}}}}\,={{x}_{3}}$
${{\overset{.}{\mathop{x}}\,}_{n-1}}={{x}_{n}}$
And substitution of the definitions into (1) yields
${{\overset{.}{\mathop{x}}\,}_{n}}=-{{a}_{1}}{{x}_{1}}-{{a}_{2}}{{x}_{2}}-\ldots -{{a}_{n}}{{x}_{n}}+r$
Together these form a set of first-order differential equations. The output w can also be expressed in terms of the state variables;
${{x}_{1}}=w$
In vector matrix form these equations can be arranged as follows;
\[\left[ \begin{matrix} \begin{matrix} \overset{\centerdot }{\mathop{{{x}_{1}}}}\, \\ \overset{\centerdot }{\mathop{{{x}_{2}}}}\, \\\end{matrix} \\ \begin{align} & \vdots \\ & \vdots \\\end{align} \\ \overset{\centerdot }{\mathop{{{x}_{n}}}}\, \\\end{matrix} \right]=\left[ \begin{matrix} 0 & \begin{matrix} 1 & 0 & \cdots \\\end{matrix} & 0 \\ \begin{matrix} 0 \\ \vdots \\ 0 \\\end{matrix} & Remaining~Entries & \begin{matrix} 0 \\ \vdots \\ 1 \\\end{matrix} \\ -{{a}_{1}} & \begin{matrix} -{{a}_{2}} & \cdots \\\end{matrix} & -{{a}_{n}} \\\end{matrix} \right]\left[ \begin{matrix} {{x}_{1}} \\ {{x}_{2}} \\ \vdots \\ \begin{matrix} \vdots \\ {{x}_{n}} \\\end{matrix} \\\end{matrix} \right]+\left[ \begin{matrix} 0 \\ \vdots \\ \vdots \\ \begin{matrix} 0 \\ 1 \\\end{matrix} \\\end{matrix} \right]r\]
$w=\left[ \begin{matrix} 1 & 0 & \cdots & 0 \\\end{matrix} \right]\left[ \begin{matrix} {{x}_{1}} \\ {{x}_{2}} \\ \vdots \\ {{x}_{n}} \\\end{matrix} \right]$
The standard form of a state-space model is as follows:
$\dot{x}=Ax+Bu~\text{ }\left( state~equation \right)$
$y=Cx+Du~\text{ }\left( output~equation \right)$
Here x is the state vector, the vectors of the state variables,
y, the output vector,
u, the control vector,
A, the system matrix,
In the preceding example the control vector is the scalar function r and the output vector the scalar function w. It may be seen that
$x=\left[ \begin{matrix} \begin{matrix} {{x}_{1}} \\ {{x}_{2}} \\\end{matrix} \\ \vdots \\ {{x}_{n}} \\\end{matrix} \right]=\left[ \begin{matrix} \begin{matrix} w \\ {\dot{w}} \\\end{matrix} \\ \vdots \\ {{w}^{\left( n-1 \right)}} \\\end{matrix} \right]$
$B=\left[ \begin{matrix} \begin{matrix} 0 \\ \vdots \\\end{matrix} \\ 0 \\ 1 \\\end{matrix} \right]$
$A=\left[ \begin{matrix} 0 & \begin{matrix} 1 & 0 & \cdots \\\end{matrix} & 0 \\ \begin{matrix} 0 \\ \vdots \\ 0 \\\end{matrix} & Remaining~Entries & \begin{matrix} 0 \\ \vdots \\ 1 \\\end{matrix} \\ -{{a}_{1}} & \begin{matrix} -{{a}_{2}} & \cdots \\\end{matrix} & -{{a}_{n}} \\\end{matrix} \right]$
$C=\left[ \begin{matrix} 1 & 0 \\\end{matrix}~~~\begin{matrix} \cdots & 0 \\\end{matrix} \right]$
$D=0$
This form of A is a companion matrix.
Some of the advantages of the state variables or state space approach are as follows;
Let’s use the following electrical network to illustrate the formulation of a state-space equation.
Fig.1: RLC Electrical Network viewed as a Process
To illustrate that a given set is not unique two different sets of state variables will be used. As a rule of thumb, the order of a system equals the number of energy storage elements, especially in electrical networks.
The plant or process is an electrical circuit, and the control variables u is the applied voltage e (t). The immediate step is to determine the order of the system which in this case is 2, corresponding to the 2 independent energy storage elements, the capacitor, and the inductor.
Select the voltage v and the current i as the state variables; the state equations become
\[\frac{dv}{dt}=\frac{i}{C}~~\text{ (1a)}\]
And
\[\frac{di}{dt}=-\frac{R}{L}i-\frac{1}{L}v+\frac{1}{L}e\text{ (1b)}\]
By the definition of stat variables, we write
\[x=\left[ \begin{matrix} {{x}_{1}} \\ {{x}_{2}} \\\end{matrix} \right]=\left[ \frac{v}{i} \right]\]
And express Eqs. (1a,1b) as
$\dot{x}=Ax+Bu$
Where
\[A=\left[ \begin{matrix} 0 & \frac{1}{C} \\ -\frac{1}{L} & -\frac{R}{L} \\\end{matrix} \right]\]
\[b=\left[ \begin{matrix} 0 \\ \frac{1}{L} \\\end{matrix} \right]\]
\[u=e\]
The use of lowercase b designates a vector. The symbol u is not boldface since we have a scalar for the control force. The output, then, is v and Ri or x_{1} and Rx_{2}, which in matrix from becomes.
$C=Cx$
Where
$c=\left[ \begin{align} & {{c}_{1}} \\ & {{c}_{2}} \\\end{align} \right]$
$C=\left[ \begin{matrix} 1 & 0 \\ 0 & R \\\end{matrix} \right]$
For this choice, the two state variables can easily be determined from the output vectors; that is,
${{x}_{1}}={{c}_{1}}$
\[{{x}_{2}}=\frac{{{c}_{2}}}{R}\]
The equation for the system in figure 1 can be written as
\[e=Ri+L\frac{di}{dt}+v\]
And using
\[i=C\frac{dv}{dt}\]
We have
\[e=RC\frac{dv}{dt}+LC\frac{{{d}^{2}}v}{d{{t}^{2}}}+v\]
By the use of the Laplace Transform, we have
\[\frac{V\left( s \right)}{E\left( s \right)}=\frac{{}^{1}/{}_{LC}}{{{s}^{2}}+{}^{Rs}/{}_{L}+{}^{1}/{}_{LC}}\]
In order to have a numerical result, select R=6Ω, L=1H, and C=0.2F.then
\[\frac{V\left( s \right)}{E\left( s \right)}=\frac{5}{{{s}^{2}}+6s+5}=\frac{5}{\left( s+1 \right)\left( s+5 \right)}\]
By expanding this transfer function into partial fraction, we have
\[\frac{V\left( s \right)}{E\left( s \right)}=\frac{1.25}{s+1}-\frac{1.25}{s+5}\]
From which
\[V\left( s \right)=\frac{1.25~E\left( s \right)}{s+1}-\frac{1.25~E\left( s \right)}{s+5}\]
Now, we define
\[{{x}_{1}}\left( s \right)=\frac{1.25~E\left( s \right)}{s+1}\]
\[{{x}_{2}}\left( s \right)=-\frac{1.25~E\left( s \right)}{s+5}\]
These equations become
$s{{X}_{1}}\left( s \right)=-{{x}_{1}}\left( s \right)+1.25~E\left( s \right)\text{ (2a)}$
\[\] ${{X}_{2}}\left( s \right)=-5{{X}_{2}}\left( s \right)-1.25~E\left( s \right)\text{ 2(b)}$
The inverse Laplace Transforms of Equations 2(a) and 2(b) yield
${{\overset{.}{\mathop{x}}\,}_{1}}=-{{x}_{1}}+1.25~e$
$\overset{.}{\mathop{{{x}_{2}}}}\,=-5{{x}_{2}}-1.25~e$
Which in state-space form become
$\dot{x}=Ax+bu$
Where
$A=\left[ \begin{matrix} -1 & 0 \\ 0 & -5 \\\end{matrix} \right]$
$~b=\left[ \begin{matrix} 1.25 \\ -1.25 \\\end{matrix} \right]$
$u=e$
Since
$V\left( s \right)={{X}_{1}}\left( s \right)+{{X}_{2}}\left( s \right)$
We have
$v={{x}_{1}}+{{x}_{2}}$
Or
$c=\left[ 1~~~1 \right]x=Cx$
Where
$c=v$
This example illustrates that the choice of state variables to represent a system is not unique and that the choice of states does not always have a simple physical interpretation. Also shown is that the choice is not completely arbitrary.
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]]>The post Root Locus Method | Root Locus Matlab appeared first on Electrical Academia.
]]>Consider as a standard form for root locus construction, the open-loop transfer function given by;
\[GH(s)=K\frac{(s+{{z}_{1}})(s+{{z}_{2}})\cdots (s+{{z}_{z}})}{(s+{{p}_{1}})(s+{{p}_{2}})\cdots (s+{{p}_{p}})}\]
Where there are z finite zeros and p finite poles of GH(s). We write the characteristic equation for the system given below;
Fig.1: Closed-Loop System
\[\frac{C(s)}{R(s)}=\frac{G(s)}{1+G(s)H(s)}\]
The characteristic equation of this closed loop system would be
$1+G(s)H(s)=0$
Or
$G(s)H(s)=-1$
The closed loop poles are the values of s that satisfy both of the following conditions:
Angle Condition:
$\angle G(s)H(s)=\pm {{180}^{\centerdot }}(2k+1)\text{ (k=0,1,3,}\cdots \text{)}$
Magnitude Condition:
$\left| G(s)H(s) \right|=1$
Now, we shall proceed to set up a set of guidelines for achieving two goals:
1. The number of branches of root locus is equal to the number of closed loop poles, generally the number of poles of GH(s).
Each branch contains one closed-loop pole for any particular value of K
2. Each branch starts at an open-loop pole of GH(s) (when K=0) and ends at a zero of GH(s) (when K=∞)
3. A locus will exist at a point on the real axis whenever the sum of the number of poles and zeros on the axis to the right of the point in question is odd.
Each pole or zero to the right of a test point s_{1}, on the real axis has an angle of 180 degrees associated with it. When an odd number of poles and zeros exist to the right of s_{1}, the angle of GH(s) will be an odd multiple of 180 degrees.
4. The locus is symmetrical with respect to a real axis.
The fact arises from the properties of the roots of an algebraic equation with real coefficients; complex roots must occur in a complex conjugate pair.
5. The sum of the roots of a characteristic equation is equal to the negative of the coefficient of s^{n-1 }where n is the order of the equation.
This guideline assumes the characteristic equation is expressed as
${{s}^{n}}+{{a}_{n-1}}{{s}^{n-1}}+{{a}_{n-2}}{{s}^{n-2}}+\cdots +{{a}_{1}}s+{{a}_{0}}=0$
6. When a branch of locus approaches an infinity, it approaches an asymptote whose direction is
\[{{\theta }_{A}}=\pm \frac{(2k+1)\pi }{{{n}_{p}}-{{n}_{z}}}\text{ (k=0,}\pm \text{1,}\pm \text{2,}\cdots \text{)}\]
Where n_{p} and n_{z} are open-loop poles and zeros of GH(s).
7. The asymptotes intercept the real axis at a point given by
\[{{\sigma }_{A}}=\frac{(sum\text{ of poles})-(sum\text{ of zeros})}{(\text{number of poles})-(\text{number of zeros})}\]
8. The point at which the root locus leaves (or enters ) the real axis is found by cut and try.
9. The angle at which the locus leaves a complex pole is determined by applying the 180-degree criterion.
${{x}_{1}}+{{x}_{2}}={{180}^{\centerdot }}$
10. The angle by which the locus enters a complex zero is determined by the 180-degree criteria.
11.There is a limited stability where the root locus crosses the imaginary axis.
Find the loci of the closed loop system poles for the following system
First, the open-loop pole-zero pattern is plotted, consisting of poles at the origin and at -a, as shown in the figure:
There are two poles, so two-locus branches, starting at 0 and -1 for K=0.
Since there are no open-loop zeros, there must be two asymptotes, from guideline 6, they will be at +90^{o} and -90^{o}.
From guideline 7, these asymptotes will intersect the real axis at
\[{{\sigma }_{A}}=\frac{0-a-0}{2-0}=-\frac{a}{2}\]
By guideline 3, the real axis between 0 and -a is part of the locus, because it lies to the left of one (an odd number) pole on this axis.
Now, we will talk about root locus using Matlab. Here, we will draw the root locus of the following transfer function:
\[GH(s)=\frac{s+2}{s(s+1)}\]
This is an open loop transfer function, however, in order to plot Root Locus (as we mentioned in the beginning), we need Closed Loop Transfer Function Poles, which we can obtain from the following characteristics equation:
\[1+KGH(s)=1+K\frac{s+2}{s(s+1)}=0\]
The characteristics equation can be re-written as:
${{s}^{2}}+(1+K)s+2K=0$
Now, we will use the coefficients of the above characteristic equation in order to plot the Root Locus using Matlab.
% Matlab Code to Compute Root Locus % for 1 + K(s+2)/(s(s+1)) clear all; close all;clc % Numerator and Denumerator for Open Loop Transfer Function num = [1 2]; den = [1 1 0]; %Display Open-Loop Transfer Function disp(['L(s) = ', poly2str(num,'s'), '/', poly2str(den, 's')]); disp(' '); figure %Open-Loop Zeros, Plot with o's olzeros = roots(num) %Open-Loop Poles, Plot with x's olpoles = roots(den) % Plotting Open-Loop Zeros and Poles plot(real(olzeros),imag(olzeros),'ro'); hold on; plot(real(olpoles),imag(olpoles),'rx'); % Plot Closed-Loop Poles over Various Values of K for K = 0.1:0.1:10, % Root of Characteristics Equation "1+KGH(s)" closed_poles = roots([1, 1+K, 2*K]); plot(real(closed_poles),imag(closed_poles),'bx'); end hold off; xlabel('Real Axis'); ylabel('Imaginary Axis'); title('Root Locus for Closed Loop Transfer Function');
Result
Fig: Root Locus for Closed Loop Transfer Function at Various Values of Gain K
The above root locus graph shows the Open Loop transfer function Poles ( at -1 and 0 with blue *) and Zeros (at -2 with red o) as well as Closed Loop transfer function Poles at varying values of gain parameter K from 0.01 to 10.
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]]>The post Transient Response | First and Second Order System Transient Response appeared first on Electrical Academia.
]]>Damping Oscillation: A typical Transient Response Example
For a system with transfer function G(s), whether open loop or closed loop and input R(s), the output is
\[C\left( s \right)=\text{ }G\text{ }\left( s \right)\text{ }R\left( s \right)\]
For distinct poles, whether real or complex, the partial fraction expansion of C(s) and the corresponding solution c(t) are, from
\[C\left( s \right)=\frac{{{K}_{1}}}{\left( s+{{p}_{1}} \right)}+\frac{{{K}_{2}}}{\left( s+{{p}_{2}} \right)}+\ldots +\frac{{{K}_{n}}}{\left( s+{{p}_{n}} \right)}\text{ (1)}\]
$c\left( t \right)={{K}_{1}}\exp \left( -{{p}_{1}}t \right)+{{K}_{2}}\exp \left( -{{p}_{2}}t \right)+\ldots +{{K}_{n}}\exp \left( -{{p}_{n}}t \right)$
The denominator of C(s)= G(s)R(s) and its partial faction expansion contain terms due to the poles of input R(s) and those of the system G(s). The terms due to R(s) yield the forced solution whereas the system poles give the transient solution, and this is the part of the response into which more insight is needed.
This is the most important characteristic of the transient response. For a system to be useful, the transient solution must decay to zero. This leads to the following definition,
The fundamental stability theorem can be formulated by examination of (1). If any system pole –p_{i} is positive or has a positive real part, then the corresponding exponential grows, so the system is unstable. A positive real part means that the pole lies in the right half of the s-plane. Hence;
A system is stable if and only if all the system poles lie in the left half of the s plane.
The first order system shown in the following figure is very common for analysis purposes in control system.
Fig. 1: First Order System
For a step input R(s) =1/s,
\[C(s)=\frac{{}^{1}/{}_{T}}{s(1+{}^{1}/{}_{T})}=\frac{{{K}_{1}}}{s}+\frac{{{K}_{2}}}{s+{}^{1}/{}_{T}}\]
${{K}_{1}}={{\left. \frac{{}^{1}/{}_{T}}{s+{}^{1}/{}_{T}} \right|}_{s=0}}=1$
${{K}_{2}}={{\left. \frac{{}^{1}/{}_{T}}{s} \right|}_{s=-{}^{1}/{}_{T}}}=-1$
Hence the transient response would be,
In the above transient response, first term indicates the forced solution because of the input while the second term indicates the transient solution, because of the system pole. Figure 2 demonstrates this transient (second term) and c(t). It can be clearly seen in figure 2 (a) that the transient is a decaying exponential; if the response takes long to decay, then the system’s overall response is slow, hence we can say that the response decay speed is of significant importance. The speed of decay is usualy measured in terms of time constant.
Fig. 2(a): Step Response of Simple Lag Network
Fig. 2(b): Step Response of Simple Lag Network
Because e^{-t/T}= e^{-1} when t=T, it can be observed that:
c (T)=1-0.368=0.632
The values at t=T provide one point for sketching the curves in figures (2a,2b). Also the curves are initially tangent to the dashed lines, since
\[\frac{d}{dt}\left( {{e}^{-{}^{t}/{}_{\tau }}} \right){{|}_{t=0}}=-\frac{1}{T}{{e}^{-{}^{t}/{}_{T}}}{{|}_{t=0}}=-\frac{1}{T}\]
These two facts provide a good sketch for the response.
Now consider the correlation between this response and the pole position at s=-1/T in Fig .1. The purpose of developing such insight is that it will permit the nature of the transient response of a system to be judged by inspection of the pole-zero pattern.
For the simple-lag network, two characteristics are crucial:
As discussed, for stability, the system pole -1/T must lie the left half of the s-plane, since otherwise the transient e^{-t/T} grows instead of decays as t increases.
In order to speed up the system response (that is by reducing its time constant T), the pole -1/T must be moved on the left side of the s-plane.
This very common transfer function to represent the second order system can be reduced to the standard form
\[G\left( s \right)=\frac{\omega _{n}^{2}}{{{s}^{2}}+2\zeta {{\omega }_{n}}s+\omega _{n}^{2}}\]
Where
ω_{n }= undraped natural frequency
ζ = damping ratio
For a unit step input which is R(s) =1/s, the transform of the output would be
\[C\left( s \right)=\frac{\omega _{n}^{2}}{s({{s}^{2}}+2\zeta {{\omega }_{n}}s+\omega _{n}^{2})}\]
The characteristic equation would be
${{s}^{2}}+2\zeta {{\omega }_{n}}s+\omega _{n}^{2}=0$
These system poles depend on ζ:
$\zeta >1:overdamped:{{s}_{1,2}}=-\zeta {{\omega }_{n}}\pm {{\omega }_{n}}\sqrt{{{\zeta }^{2}}-1}$
$\zeta =1:critically~damped:{{s}_{1,2}}=-{{\omega }_{n}}$
$\zeta <1:underdamped:{{s}_{1,2}}=-\zeta {{\omega }_{n}}\pm j{{\omega }_{n}}\sqrt{1-{{\zeta }^{2}}}$
Fig.3: System Poles Quadratic Lag
Figure 3 shows the s-plane for plotting the pole positions.
$\left| {{s}_{1,2}} \right|={{\left[ {{\left( \zeta {{\omega }_{n}} \right)}^{2}}+{{\left( {{\omega }_{n}}\sqrt{1-{{\zeta }^{2}}} \right)}^{2}} \right]}^{{}^{1}/{}_{2}}}={{\omega }_{n}}$
From the geometry in figure 3, it is seen also that
$Cos\left( \theta \right)=\frac{\zeta {{\omega }_{n}}}{{{\omega }_{n}}}=\zeta $ Hence,
The damping ratio $\zeta =\cos \phi $ , where ∅ is the position angle of the poles with the negative real axis.
This is the time constant in seconds for the amplitude of oscillation to decay to e^{-1} of its initial value:${{e}^{-\zeta {{\omega }_{n}}t}}={{e}^{-1}}$, Hence
\[T=\frac{1}{\zeta {{\omega }_{n}}}\]
Analogous to the simple lag, the amplitude decays to 2% of its initial value in 4T seconds. It is again important to determine the correlation between dynamic behavior and the pole positions in the s-plane in Figure 3:
The real part $-\zeta {{\omega }_{n}}$ of the poles must be negative for the transient to decay; that is, the poles must lie in the left half of the s-plane.
To avoid excessive overshoot and unduly oscillatory behavior, damping ratio ζ must be adequate. Since ζ = Cosϕ, the angle ϕ may not be close is 90^{o}.
The time constant is reduced (that is, the speed of decay of the transient is increased) by increasing by increasing the negative real part of the pole position.
The speed of response of the system is increased by increasing the distance ω_{n} of the poles to the origin.
This equals the distance of the poles to the origin. Moving the poles out radially (with ζ constant) increases the speed of response while the percentage overshoot remains constant.
This frequency also called the resonance frequency or damped natural frequency equals the imaginary part of the pole positions.
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]]>The post Physical Quantities and Units | Physical Quantity Definition appeared first on Electrical Academia.
]]>A physical quantity is characterized by defining how it is measured or by expressing how it is computed from other measurements. For instance, distance and time are expressed by defining methods for evaluating them, while we express average speed by stating that it is computed as distance traveled divided up by time covered.
Four fundamental physical quantities (length, time, mass, and electric current) are utilized in order to express almost all physical quantities.
Physical quantities are expressed in terms of units, which are standard values. For instance, the length of a race (a physical quantity) can be defined in units of meters. Without standard units, it would be highly challenging to define and equate measured values in a purposeful way.
There are two leading units systems practiced globally: SI units (i.e. metric system) and English units (i.e. imperial system). English units were historically used by the British Empire and are still extensively utilized in the United States. Almost every other country on the planet practices SI units. The acronym “SI” is inferred from the French word Système International.
Physical Quantity SI Unit Symbol
Length Meter m
Time Second s
Current Ampere A
Admittance Siemen S
Angle Radian Rad
Angular Velocity Radian per second Rad/s
Apparent power Volt-Ampere VA
Area Square meter ${{m}^{2}}$
Capacitance Farad F
Charge Coulomb C
Conductance Siemen S
Electric Field Intensity Volt per Meter V/m
Electric Flux Coulomb C
Electric Flux Density Coulomb per Meter Square $\frac{C}{{{m}^{2}}}$
Energy Joule J
Force Newton N
Frequency hertz Hz
Impedance Ohm $\Omega $
Inductance Henry H
Magnetic Flux intensity Ampere per Meter A/m
Magnetic Flux Weber Wb
Magnetic Flux Density Tesla T
Magneto motive Force Ampere-Turn At
Moment of Inertia Kilogram Meter Square $kg.{{m}^{2}}$
Power Watt W
Pressure Pascal Pa
Reactance Ohm $\Omega $
Reactive Power Volt-Ampere Reactive VAR
Resistance Ohm $\Omega $
Resistivity Ohm-Meter $\Omega m$
Susceptance Siemen S
Torque Newton-Meter Nm
Voltage Volt V
Volume Meter Cube ${{m}^{3}}$
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]]>The post Nyquist Theorem | Nyquist Stability Criterion appeared first on Electrical Academia.
]]>The Nyquist criterion is a frequency domain tool which is used in the study of stability. To use this criterion, the frequency response data of a system must be presented as a polar plot in which the magnitude and the phase angle are expressed as a function of frequency.
In 1932, H. Nyquist used a theorem by Cauchy regarding the function of complex variables to develop a criterion for the stability of the system. Cauchy’s theorem is concerned with mapping contours from one complex plane to another. Since we are interested in the presence of roots of the characteristic equation in the right half of the s-plane, we will be mapping contours that enclose the right half-plane. The characteristics equation
$1+GH(s)=0$
Is a function F(s) of the complex variable s set equal to zero, that is
$F(s)=1+GH(s)=0$
Above equation can also be expressed as
\[F(s)=\frac{K(s+{{z}_{1}})(s+{{z}_{2}})\cdots (s+{{z}_{n}})}{(s+{{p}_{1}})(s+{{p}_{2}})\cdots (s+{{p}_{m}})}=0\]
Where the -z_{i}’s are the roots of the characteristics equation and the –p_{i}’s are the poles of the corresponding open loop transfer function GH(s). We first map contours of the s-plane into the (1+GH)-plane, then for convenience, we omit the 1 of (1+GH) and map the contours into the GH-plane. The resulting contours provide information on the existence of roots that have positive real part, that is, that are located in the right- half plane.
A theorem by Cauchy can provide information about the number of zeros of a function F(s) having positive real parts. In our study of stability, we will be applying the theorem to the characteristic equation of the system expressed as
$F(s)=1+GH(s)=0\text{ (1)}$
Where F(s) is often given as
\[F(s)=\frac{K(s+{{z}_{1}})(s+{{z}_{2}})\cdots (s+{{z}_{n}})}{(s+{{p}_{1}})(s+{{p}_{2}})\cdots (s+{{p}_{m}})}=0\text{ (2)}\]
As applied to equation (1) and (2), Cauchy’s theorem is as follows:
If :
Then,
The corresponding contour (or image) in the F(s)-plane encircles the origin N=Z-P times in the clockwise direction.
The theorem is summarized by
N= Z-P (3)
Where
P= the number of poles of F(s)
Z=the number of zero (roots) of F(s)
N=the number of clockwise encirclements of the origin in the F(s)-plane.
It is convenient to map with GH(s) rather than 1+ GH(s) =F(s). For the GH(s) mapping, equation (3) applies, if we let N=the number of clockwise encirclements of the -1 point in the GH-plane. For the majority of the applications, we know GH(s) in the factored form:
$GH(s)=\frac{K(s+{{s}_{1}})(s+{{s}_{2}})\cdots }{(s+{{p}_{1}})(s+{{p}_{2}})\cdots }$
Where –s_{i}’s are the zeros of the open-loop transfer function. By finding the image in the GH-plane rather than F(s) plane, we avoid adding 1 to each of the computation.
The stability of a system is determined by searching the right-half of the s-plane for zero (roots) of the characteristics equation:
$F(s)=1+GH(s)=0$
If there are no roots in the right half plane, then the system is stable.
Now we shall establish the procedure for searching the right-half plane and relating the stability of the system to the polar plot. The positive jω-axis in the s-plane, used for the polar plot, will be expanded to a contour enclosing the entire right half-plane.
The contour, shown in the following figure, consists of three parts.
Fig.1: Nyquist Contour
The complete contour, enclosing the entire right half-plane, is called the Nyquist contour.
A smaller radius could be used if we knew that there were no roots of the characteristics equation beyond the semi-circle.
We use GH(s)-plane, which is
$F(s)-1=GH(s)$
To map the Nyquist contour of the s-plane into the GH-plane. This change in the mapping function is convenient because we usually know GH(s) in the factored form as,
$GH(s)=\frac{K(s+{{s}_{1}})(s+{{s}_{2}})\cdots }{(s+{{p}_{1}})(s+{{p}_{2}})\cdots }$
The Nyquist criterion for stability can now be stated:
When the Nyquist contour is mapped into the GH-plane by the open-loop transfer function GH(s) of a system, then one of the following two cases applies:
When no poles of GH(s) are in the right half of the s-plane, the corresponding feedback control system is stable if and only if the control of the image of the Nyquist contour does not circle the (-1+j0) point in the GH-plane.
When the number of poles of GH(s) in the right half of the s-plane is not zero, the corresponding feedback control system is stable if and only if the contour of the image of the Nyquist contour encircles the (-1+j0) point in a counter-clockwise direction by an amount equal to the number of poles of GH(s) with positive real parts.
The basis of these two cases is:
N=Z-P
For case 1, where P=0 we have N=Z, that is, the number of zeros is equal to the number of encirclements. For a stable system, N must be zero.
For case 2, N=Z-P results in N=-P, since Z=0 for a stable system.
A Nyquist plot, known as a polar plot, is a frequency response of a linear system, which means we replace s with jω in a transfer function G(s).
In order to draw a polar plot for transfer function G (jω) for the frequency range 0 to infinity, we need to be recognizant of the following four important points;
Imaginary (G(jω))=0
Real (G(jω))=0
Consider the first-order system
\[G(s)=\frac{1}{1+sT}\]
Where T is the time constant.
Let’s represent function G(s) in terms of frequency response G (jω) by substituting s = jω:
$G(j\omega )=\frac{1}{1+j\omega T}$
The magnitude |G (jω)| is obtained as:
$\left| G(j\omega ) \right|=\frac{1}{\sqrt{1+{{\omega }^{2}}{{T}^{2}}}}$
The phase of G (jω), expressed by φ, is calculated as:
$\angle G(j\omega )=-arctan(\omega T)$
Start of plot where ω=0
$\left| G(j\omega ) \right|=\frac{1}{\sqrt{1+0}}=1$
$\phi ={{\tan }^{-1}}\left( \frac{0}{1} \right)=0$
End of plot where ω=∞
$\left| G(j\omega ) \right|=\frac{1}{\sqrt{1+\infty }}=0$
$\phi ={{\tan }^{-1}}\left( \frac{-\infty }{1} \right)=-{{90}^{\centerdot }}$
The mid part of plot where ω = 1/T
$\left| G(j\omega ) \right|=\frac{1}{\sqrt{1+1}}=\frac{1}{\sqrt{2}}$
$\phi ={{\tan }^{-1}}\left( \frac{-1}{1} \right)=-{{45}^{\centerdot }}$
Let’s combine the above points in a table,
Value of ω | $\left| G(j\omega ) \right|$ | $\angle G(j\omega )$ |
ω=0 | 1 | 0 |
ω = 1/T | $\frac{1}{\sqrt{2}}$ | $-{{45}^{\centerdot }}$ |
ω=∞ | 0 | $-{{90}^{\centerdot }}$ |
So, for the first order system, the polar plot will look like:
Fig.2: Polar Plot for First-Order System
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