Home » Control System – Steps For Solving Bode Plots – Electronicsguide4U? The 20 Detailed Answer

# Control System – Steps For Solving Bode Plots – Electronicsguide4U? The 20 Detailed Answer

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## Bode plot in control system

Bode plot in control system
Bode plot in control system

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Bode plot is a graph that represents a system frequency response.It is used to examine the stability of a closed loop control system.These are also known as logarithmic plot (because we draw these plots on semi-log papers) and are used for determining the relative stabilities of the given system.It is usually a mix of a a Bode phase plot, representing the frequency response phase shift and the Bode magnitude plot, depicting the magnitude of the frequency response gain.  We already know that , the system will go into uncontrolled oscillation if its an unstable system(when the mike in an auditorium is pointed or gets too close to a speaker with the resulting loud squeal from the system).Now let’s see the steps required for solving the Bode plots: Bring the given G(s)H(s) transfer function into standard time constant form. Replace all s by jω to get the frequency domain transfer function. Make a table of the standard factors present in the given transfer function. Now analyze the table of factors. Make the table for plot representing the magnitude variation. We can take help of this table to plot the magnitude curve . Select proper X and Y scales for the magnitude plot. Draw the lines of the corresponding factors from starting point to the corner frequency ( i.e the End point ). Note down the phase angle table and obtain the resultant phase angle by actual calculations. Prepare the table (containing phase angle variation) and then obtain resultant phase angle by actual calculation. Depending upon the highest frequency  term present , choose the starting point in the log scale . Plot the points as per phase angle table and then draw smooth curve for phase angle plot. Gain Margin calculations : From phase resultant curve find ‘ω’ where phase curve crosses -180 . This is called the phase crossover frequency i.e   . Draw the straight line from  upwards towards the magnitude plot , till it intersect magnitude curve. Call this point as G . The  difference between 0 dB and magnitude corresponding to point G is the gain margin. Phase Margin Calculations : From the magnitude resultant curve find where resultant crosses 0 dB . This is gain crossover frequency   . Draw the straight line from    downwards till it intersects the phase curve. Call this point as P. The distance between point P and -180  line is the phase margin. Only if both , gain margin and phase margin are positive , is the system stable i.e   <   . Now we shall see summary of Bode Magnitude  and Phase Plots of various terms : Sr no. Factor Magnitude (M) Phase(φ) 1 Constant gain k 20log|k| dB  Φ = 0  for k > 0 = -180  for k < 0 2 Poles at origin (integral) Straight line with -20n dB/dec slope and passing through [ω = 1 , 0 dB ] point Φ =  -90n 3 Zero at origin (derivative) Straight line with +20n dB/dec slope and passing through [ω = 1 , 0 dB ] point Φ =  90n 4 First order poles Line slopes are 1.       0 dB/dec (for  ω    ) 2.       -20dB/dec (for ω > ) Φ =  () 5 First order zeros   Line slopes are 1.       0 dB/dec (for  ω    ) 2.       20dB/dec (for ω >  ) Φ = ( ) 6 Second order poles   Line slopes are 1.       0 dB/dec (for  ω    ) 2.  -40dB/dec (for ω >  ) 3. Error detection of ω = as per Φ = 7 Second order zeros    Line slopes are 1. 0 dB/dec (for  ω    ) 2. +40 dB/dec (for ω >   ) 3.  Error detection of ω = as per Φ = Let’s make this topic clear by taking one simple example: A feedback system has : G(s)H(s) =    ,  we need to  draw bode plot and comment on stability. So, here we should follow the required steps as stated earlier in this post, Step 1 :  Bring the given G(s)H(s) transfer function into standard time constant form: G(s)H(s) =     =      =      , Step 2 :   Replace all s by jω to get the frequency domain transfer function: G(jω)H(jω) =      Step 3 : The following factors are present: Constant k = 60                                      . Pole at origin = 1/ jω                         . First order zero = (1 + jω/3). First order pole = 1/(1 + jω). First order pole = 1/(1 + jω/5). Step 4 :  Prepare table of factors : Sr no. Factor Magnitude curve Phase curve 1 K = 60 Straight line of 20log60 = 35.5 dB Φ =   0 2 1/ jω Slope of -20 dB/dec passing through 0 dB at ω = 1 rad/sec Φ =   -90 3 1/ ( 1 + jω) Line slopes are 1.       0 dB  (ω 1 ) 2.       -20 dB/dec (ω 1 ) 4 1/(1 + jω/5) Line slopes are 1.       0 dB  (ω  5 ) 2.       -20 dB/dec (ω  5 ) 5 1 + jω/3 Line slopes are 1.    0 dB  (ω   3 ) 2.    -20 dB/dec (ω 3 ) Step 5 : Magnitude plot table The product of different magnitude appears in any transfer function will be evaluated as the addition, while any division will be further evaluated as the substraction as we are dealing with a logarithmic scale. Sr no. Factors Resultant slope Start point (ω) End point (ω) 1 k Straight line of 35.5 dB 0.1 ∞ 2 1/ jω -20 dB/dec 0.1 1 3 1/ ( 1 + jω) -20 + (-20) =  -40 dB/dec 1 3 4 1/(1 +jω/5) -40 + 20 = -20 dB/dec 3 5 5 1 + jω/3 -20 + (-20) = -40 dB/dec 5 ∞ Step 6 : Phase plot table: For phase plot, we compute the angle contribution by indivual factors and then add each angle (at given ω) to get resultant phase. Mark these points in semilog paper and draw smooth phase graph. =  -90 –  –    + ω 1/ jω – – 0.1 -90 -5.7 -1.1 1.9 -94.9 0.5 – 90 -26.5 -5.7 9.46 -112.7 1 -90 -45 -11.3 18.4 -127.9 5 – 90 -78.69 -45 59 -154.79 10 – 90 -84.3 -63.4 73.3 -164.4 100 – 90 -89.4 -87.1 88.3 -178.2 500 – 90 -89.8 -89.4 89.6 -179.6 1000 – 90 -89.9 -89.71 89.8 -179.8 Step 7 : The bode plot is shown in below figure :  It is much event from the above plot that, by using this plot , we can easily comment on the stability of any controlling process without doing any  complex calculations.Bode plots thus represents the relative stability in the form of the phase margin and gain margin.Further it also addresses frequencies ranging  from low to high values.Hope you really enjoyed this post.In my next post we will see what a polar plot is and its significance in frequency response analysis.So stay tuned. report this ad

## What are the steps to be followed for construction of Bode plot?

Key Concept – To draw Bode diagram there are four steps:
1. Rewrite the transfer function in proper form.
2. Separate the transfer function into its constituent parts.
3. Draw the Bode diagram for each part.
4. Draw the overall Bode diagram by adding up the results from part 3.

## What is Bode plot in control system?

In electrical engineering and control theory, a Bode plot /ˈboʊdi/ is a graph of the frequency response of a system. It is usually a combination of a Bode magnitude plot, expressing the magnitude (usually in decibels) of the frequency response, and a Bode phase plot, expressing the phase shift.

## How do you find the phase angle of a Bode plot?

Bode analysis consists of plotting two graphs: the magnitude of Φ0(s) with s = jω, and the phase angle of Φ0(s) with s = jω, both plotted as a function of the frequency ω. Log scales are usually used for the frequency axis and for the magnitude of Φ0(jω). d B = 2 0 log 1 0 | Φ 0 ( j ω ) | .

## What is the control system?

A control system is a set of mechanical or electronic devices that regulates other devices or systems by way of control loops. Typically, control systems are computerized. Control systems are a central part of industry and of automation.

## What is phase control system?

Phase margin is defined as the amount of change in open-loop phase needed to make a closed-loop system unstable. The phase margin is the difference in phase between −180° and the phase at the gain cross-over frequency that gives a gain of 0 dB.

## What is feedback control system?

Advertisements. If either the output or some part of the output is returned to the input side and utilized as part of the system input, then it is known as feedback. Feedback plays an important role in order to improve the performance of the control systems.

## How do you analyze a Bode plot?

Bode plots show the frequency response, that is, the changes in magnitude and phase as a function of frequency. This is done on two semi-log scale plots. The top plot is typically magnitude or “gain” in dB. The bottom plot is phase, most commonly in degrees.

## Why Bode plot is used?

A Bode Plot is a useful tool that shows the gain and phase response of a given LTI system for different frequencies. Bode Plots are generally used with the Fourier Transform of a given system. An example of a Bode magnitude and phase plot set.

## What is GM and PM in Bode plot?

Gm and Pm of a system indicate the relative stability of the closed-loop system formed by applying unit negative feedback to sys , as shown in the following figure. Gm is the amount of gain variance required to make the loop gain unity at the frequency Wcg where the phase angle is –180° (modulo 360°).

## How do you find the phase angle of a control system?

Consider the open loop transfer function G(s)H(s)=1+sτ. For ω<1τ , the magnitude is 0 dB and phase angle is 0 degrees. For ω>1τ , the magnitude is 20logωτ dB and phase angle is 900. The following figure shows the corresponding Bode plot.

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### Rules for Constructing Bode Diagrams – Swarthmore College

To draw Bode diagram there are four steps: Rewrite the transfer function in proper form. Separate the transfer function into its constituent parts.

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