Control System Lectures - Bode Plots, Introduction

welcome back to control system lectures as always if you have any questions or comments on what I'm about to say please leave it in the comment section below and I'll address it in this lecture I want to present an introduction to frequency response methods and in particular I'm going to describe the bode plot how its generated and used to visualize the frequency response of a system across the entire spectrum I want to start by explaining another benefit of an LTI system if you design a linear time-invariant system then there's only a limited number of mathematical operations that you can use when building an LTI system you can only use one is multiplication by a constant the second is differentiation of the input signal the third is integration of the input signal and the fourth is adding two inputs together now you can always do division by multiplying by the reciprocal and you can also subtract by adding a negative number I might be wondering why it's important for me to start this lecture this way and why bring up the fact that the linear system can only consist of these six operations the answer is that when you subject a sinusoidal Zhen Shan and part of not changing shape means that it also won't change frequency a sine wave input will always generate a sine wave output of the same frequency in fact the only two differences possible between an input sine wave and the output sine wave that's generated is the amplitude which is the height of the sine wave and the phase that is shifting the sine wave and time in fact the sine wave is the only repeating waveform that doesn't change shape in a linear system you can see here graphically what it means to change an amplitude and also to shift in phase let me give you a quick example so you can see what I mean if we take an input sine wave and subjected to these two separate paths the upper path is just multiplying it by a constant two and the lower path integrating it and then summing them together produces the output if we make the input a pure sine wave of frequency 1/2 radians per second then we can just calculate mathematically with the output would be I'd like to make a quick note about frequency I'm going to refer to frequency in units of radians per second but this is proportional to Hertz so you can use either in this example alright back to solving this mathematically the output would then be the upper path which is a gain of two times the input plus the integral of the input with respect to time we can simplify this further by just taking the integral in the second half of the equation now in order to interpret this from here we're going to need some help from some trig identities now I didn't remember this I had to actually go look this up but this trig identity that I'm going to be using is a sine of X plus B cosine of X equals the square root of a squared plus B squared times the sine of X plus Rho where Rho is equal to the arc tangent of B over a now of course this is only true if a is greater than or equal to zero so let's apply this trig identity to our problem in our case a is equal to 2 and B is equal to negative 2 so the change in amplitude of the sine wave is just the square root of 2 squared plus 2 squared is the square root of 8 times the sine of 1/2 T plus the arctangent of 2 over -2 which is the arctangent of minus 1 if we write this in decimal format this is just equal to 2 point 8 3 times the sine of 1/2 t minus 0.78 5 radians now it should be obvious from this example that the sine wave that we input into the system is still a sine wave on the output however the amplitude has increased from 1 and the original input 22.8 3 that's the change in amplitude and the phase has been shifted by 0.785 radians notice also that the output has the same frequency as the input which is 1/2 radians per sec now you can draw the input and the output in the time domain and it would look something similar to the graph that we've done above and that would work if you were only concerned with the response to a single frequency however more often than not you're going to be interested in the whole frequency spectrum so how can you visualize the gain and phase shift across the entire spectrum well one way to do that is with the bode plot the bode plot allows frequency response information to be displayed graphically on two separate plots one for gain and one for phase both of which are plotted against frequency on a logarithmic horizontal axis but there's a caveat to plotting amplitude gain on a bode diagram we don't just plot the magnitude of the gain directly first we have to convert it into power and then we represent the power gain in decibels which is a logarithmic unit to convert the amplitude the decibels you first take the log base 10 of the ratio of amplitudes and then multiply it by 20 if you're confused about where this equation came from let me just give you a quick history on decibels when Bell Telephone laboratories was trying to quantify loss in audio levels through telephone circuits in the early 20th century they needed a unit of measurement and one that could cover a large range of audio power differences because the ear responds to sound pressure logarithmic lis using a logarithmic scale corresponds to the way humans perceive sound they created a measurement called the transmission unit tu where one tu equaled ten times the log base ten of the power loss and this was chosen because it approximately was the smallest power attenuation detectable to the average listener now to honor Bell Labs founder Alexander Graham Bell they later changed transmission unit to decibel or 1/10 of a bell where Bell is just the log base ten of the power ratio so to get from bells to decibels you have to multiply by ten also note that power and a sine wave is proportional to the square of the amplitude of the sine wave so now we can write the decibel equation with respect to amplitude squared which is 10 log base 10 of the amplitude squared and just because of the properties of logarithms the exponential to just becomes multiplication and you get 2 times 10 which is 20 so let's get back to plotting on the bode plot if we convert the game from amplitude and decibels we can plot the gain for that frequency Omega and for Omega of 1/2 we can calculate the gain to be approximately 9 decibels so on our gain plot we could find where 9 decibels would be on the vertical axis and where point 5 radians per second would be on the horizontal axis and put a little X there you can do the same for frequency but you first have to convert 0.785 radians into degrees because it's traditionally shown in degrees and that's negative 45 this is the frequency response for a single sine wave but remember we're interested in plotting the gain and phase across the entire spectrum so instead of running through the example above an infinite number of times each of the different frequency we could have just set it to a variable like Omega and solve the system for any frequency so if we had done that and we had a lot of time we could have plotted a gain and phase plot that looks something like this for our particular system and with the exception of being time consuming for this simple example solving the response to the sine wave input was relatively straightforward well except for having to remember that crazy trigonometry there is a way of solving this without having to remember those identities and still produce the same relationship with frequency and that's by referring to the system's transfer function if we take the Laplace transform of the system above the gain of 2 stays at 2 and the integral becomes 1 over s now writing the transfer function for a system like this can be done the output Y of T is equal to 2 times the input U of T plus U of T divided by s a few algebraic steps here and you can solve for y of T over U of T which is the transfer function which is equal to 2's plus 1 divided by s recall that the Laplace transform s is a complex variable Sigma plus J megha however when we're talking about frequency response of the system we are talking about the steady state response that is once all the transients have died out or in other words when the exponential term Sigma is zero so for steady state behavior s just becomes J Omega steady-state phase and gain can be calculated directly from a transfer function by setting s to J Omega that makes the transfer function become 2 plus 1 over Omega J where 2 is the real component and 1 over Omega is the imaginary component if we then plotted these real and imaginary components on a real and imaginary axis then we can calculate gain and phase geometrically from this plot so here our real component is 2 and for Omega of 1/2 from our example above 1 over Omega would also be well it's negative 2 I'm sorry I made a mistake here this 2 plus 1 over Omega should actually be a minus because J times J is negative 1 so it's negative 2 on the imaginary axis for Omega 0.5 and if we draw a line from the origin of the access to that point then the length of the line is the gain of the system and the length can be calculated by just taking the square root of the real component squared plus the imaginary component squared this is just the Pythagorean theorem the phase is the angle between that line and the positive real line you can solve for this using the arctangent I'm using arctangent 2 in this case only because it keeps track of the sign of the inputs and so you don't have to worry about whether a is positive or negative like we did up above so again the length of this line is the gained and the angle off the positive real line is the phase so you can see in this example that the real component doesn't change with frequency only the imaginary component and if we sweep through frequencies from near zero all the way up to near infinity you can get a visual understanding of what the gain in the phase are doing for this Euler system and Omega of near zero the imaginary component is almost infinite and at an Omega of near infinity the imaginary component is near zero and if you do that red line from the origin starting with Omega of zero down at near infinity which would be a near vertical line infinitely long you would see that they would get shorter and shorter and shorter and shorter and finally ending with length two if we go back up to our plot you can see that's exactly what the gain is doing the limit as frequency approaches infinity goes to 6 DB which is a gain of two and off to the left as it frequency approaches zero the gain goes off to infinity also if you look at phase here you can see that phase starts out at negative 90 degrees and then goes to zero degrees and again that's exactly what our plot is doing and that's how you can represent a large amount of frequency response data through a system in fact that's exactly how graphing program such as MATLAB would calculate the bode plot from a transfer function doing this by hand though still seems like it would take a long time but luckily there is yet another way to graph the frequency response across the spectrum and not only is this a much faster way but it also gives you an intuitive understanding into the frequency response of the system and you do this only by looking at the transfer function by following a few rules of thumb you can graph the bode plot by hand using straight line representations and I'll show you that in the next video for now as always thanks again for watching and don't forget to subscribe

Channel: Brian Douglas

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Keywords: Control System, Feedback, Closed Loop Control, Control Systems Lectures, Theory, Flight Controls, Education, Lecture, Lesson, Brian Douglas, Automatic Control, Control Theory, Control System Tutorial, Laplace Transform, LTI System, LTI, Linear Control, Linear, Time Invariant, Linear System, Control Example, example, simple, intuitive, Bode plot, bode diagram, bode, frequency response, Nyquist

Id: _eh1conN6YM

Length: 12min 44sec (764 seconds)

Published: Thu Oct 11 2012