Characterizing Step Responses of RC and RLC Circuits: Damped & Series-Connected Capacitors, Lab Reports of Mechanical Engineering

Download Characterizing Step Responses of RC and RLC Circuits: Damped & Series-Connected Capacitors and more Lab Reports Mechanical Engineering in PDF only on Docsity! © Dr. Seeler Department of Mechanical Engineering Fall 2004 Lafayette College ME 479: Dynamic Systems, Controls and Mechatronics Laboratory Lab 1: First and Second Order Step Responses, Electromagnetism, and Induction. OBJECTIVES 1. Review oscilloscope operation. 2. Engineering Modeling: Measure the transient response of first order electrical systems (R-C circuits) to step changes in applied voltage. Apply techniques for estimating the time constant to determine an elemental parameter of a first order dynamic system. 3. Electromagnetism and Induction: Measure the transient response of second order electrical systems (R-L-C circuits) to step changes in applied voltage. Apply techniques for estimating the damping ratio and natural frequency to determine an elemental parameter of a second order dynamic system. THE VALUE OF PLANNING AND PREPARATION Since this is the beginning of the last year of your undergraduate education, it is now time to begin your transition from an engineering student to practicing engineer. There is good news and bad news about real world engineering practice. The good news is there is very little, if any, calculus and no exams in the real world. The bad news is that the real world doesn’t tolerate shoddy work. 95% complete is unacceptable. A C+ is also unacceptable. Planning is key in engineering practice. Good planning prevents problems. Engineering students, on the other hand, do not have to plan many things. Students who foul up their time management through poor planning can generally scramble and “complete” an assignment by submitting something of poor quality. School rewards good exam takers, more or less independent of their work habits. The real world rewards careful, organized, and hard working, competent people. If you are asked to developed a plan, predict a value, estimate a parameter, etc. as a pre- lab, it must be completed and presented in an acceptable problem set format BEFORE LAB. Unsatisfactory performance in lab, specifically coming to lab ill-prepared is unacceptable and you will be required to repeat the three hour lab. You must complete all of the labs to complete this course. When I interact with you at the lab bench, I will encourage you to apply your engineering skills to understanding and debugging relatively simple systems. The better prepared you are, the more fun lab will be. READING In ME 479, you will be expected to have read and digested all relevant material BEFORE LAB. In your engineering practice, you will be expected to educate yourself to maintain and extend your skills and understanding. If you are among those who have not 2 developed an effective and efficient method for reading and understanding an engineering text, now is the time. WRITING A significant difference between school and the real world is that students enjoy a great deal of anonymity. Privacy laws prevent disclosure of a student’s performance (grades). Professional practice is public. When you begin your professional practice, you will make an immediate transition from thinking, writing, estimating, calculating, designing, and analyzing in private to performing the same tasks in public. Everyone working with you, and many people you have never met, will have seen your work and formed an opinion about its timeliness, cost, and quality. Accordingly, in ME 479, you will be asked on occasion to share your work with others for their critique. Initially, just presenting work publicly, not to mention having it critiqued, can be surprisingly difficult for some people. It is essential that an engineer be able to present his or her ideas publicly and be thick-skinned enough to accept constructive criticism of their work. Who would hire an engineer, manager, etc. who insisted on working in private and could not accept constructive suggestions? All important engineering work is presented on paper, much of it in writing. Practicing engineers spend about half their time writing and it is a very important part of their jobs. Written work submitted in ME 479 should be in memorandum form and in a “technical writing” style that is direct and succinct. Memorandum style is not a license of poor structure or grammatical errors. Now is the time to start developing a professional writing style. Memo Content and Style You will be asked to “Report and interpret your results in a brief memo.” What do that mean? First, re-read the Lab handout to remind yourself of what you were trying to accomplish. Just what were the results you were asked to achieve? Assume that you are writing to a peer in your company who has performed similar measurements. Also, assume your reader knows how to use the equipment and how to reduce the data. Save yourself the effort of regurgitating the Lab handout. Simply reference the lab handout, i.e. “The procedure of Lab X for Y yielded the following results:” “Brief memo” means to the point. It will not be graded by the pound. Do not limit yourself to a paragraph format. A memo format is an opportunity to use to use tables and figures to present data or results when appropriate. Use lists to present ranked or sequential items. Avoid a narrative of the form “We did this, then we did that.” If you need to report the results of a number of unsuccessful attempts, consider using a table in which you summarize the pertinent information. The memos should be as terse as possible without being “choppy”. Words are the worse way of expressing quantitative results. Interpret your results. Do not apologize for “bad” data. Be matter-of-fact. Unless you can identify a specific error, do not attribute unexpected or inconsistent results to “measurement error”. Do not dismiss apparently inconsistent results. Data are the data. Physical reality is much more complicated than presented in sophomoric textbooks. That is not to say that calculations are error free. When asked to interpret your results, you need to judge how reasonable the results are compared to the theoretical prediction. If your find a substantial 5 WaveStar has a graphical user’s interface. The left-hand panel, resembles a directory tree. When you first bring up the program, the left panel contains the items “local” and your scope’s name. Click on the plus sign next to your scope’s name to expand the directory tree. You will see the tools that can be used to transfer data from the scope to the computer or transfer settings and commands from the computer to the scope. The right-hand panel remains blank until you open a file or create a new file. Click on File, select New Data Sheet, select YT Sheet (the data we will display are “y-t” data, where the y axis of the plot is voltage and t is time) and click OK. The right panel will have a blank replica of an oscilloscope screen. A trace is added to the screen by dragging and dropping a Wave Form from the left panel. A dialog box appears when you drag and drop a wave form onto the screen. Select “Link”. Data are transferred from the scope to the computer over a serial cable, which is too slow for the computer to mirror the real time image of the oscilloscope. WaveStar must work with data stored in the scope’s buffer when the scope is stopped either manually or with the trigger. Our interest will be analyzing transient responses, which must be captured anyway, so the “off- line” data transfer will not be a problem. The data are transferred from the scope to the computer when you click on the right-pointing arrow at the top of the computer’s image of the oscilloscope screen. Once the data have been transfer to the computer, they can be analyzed by the WaveStar’s measurement functions. You will find that the these algorithms cannot replace a trained eye, so you will often need to plot the data and identify values yourself. The buffer data can be transferred as arrays or vectors to other Windows based software. Our computers have Excel and Mathcad installed. We cannot transfer all 2500 data points from WaveStar to Mathcad directly. We need to use the intermediate step of creating an Excel file. In WaveStar, open the Edit menu and click on Copy. Open an Excel workbook and paste into the first cell. You will see the data with a text “header” in the first two rows. Delete the text header and save the Excel workbook. Start Mathcad. Select the Insert menu. Click on “Data” and then “File Input”. A dialog box will appear. Select Microsoft Excel for the file format and then browse for your file. Once you have plotted the file you can “right-click” within the plot to bring up Mathcad’s “Trace” function which produces a crosshairs cursor, similar to the oscilloscope you used in Instrumentation. REVIEW OF THE DYNAMIC RESPONSE OF PHYSICAL SYSTEMS The following remarks are limited to physical systems that can be described by linear ordinary differential equations with constant coefficients. The mathematical of Classical Control Theory depends on superposition. Unfortunately, superposition does not apply to non-linear systems (where doubling the input does not double the output) or systems with non-constant coefficients (where the system's response to identical inputs depends on when they are applied). Consequently, control of non-constant and non-linear systems is graduate material. If the system is described by linear ordinary differential equations with constant coefficients then superposition applies and the response can be broken down into two components: 6 1. The "natural" response due to disturbing the system from its former state. The natural response is the solution of the homogeneous system equation, as we will review. The natural response is commonly but erroneously called the transient response because it generally, but needn't always, die away. 2. The "forced" response due to the system following the input (the forcing function). The forced response may be either steady, when the system is subjected to a step change (a constant), or periodic, when the system is forced or driven by a periodic input. If a system is driven by a periodic forcing function, such as a sine wave, the system will ultimately reach "steady-state" when each cycle of its response is identical to the proceeding cycle. The shape and magnitude of the steady-state response may be very different than the periodic input, but the output will always have the same frequency as the input as the system tries to follow the forcing function. First Order Step Responses The step response of a first order system is non-oscillatory. First order systems have only one independent energy storage element. A first order system’s step response results from monotonic energy transfer across the system boundary into or out of the one independent energy storage. It is useful to “normalize” step responses. The time axis is “normalized” by dividing time by the time constant τ. The vertical axis can be normalized by dividing the output variable V(t) by the range ∆V. The response of all first order systems subjected to a step input can be described by a two normalized curves with the percent change in the output variable plotted against time measured in units of time constants. There are two fundamental first order step responses are: 7 1. Decay to Zero: t V(t) V e − = ∆ τ where 0V V∆ = 2. Stable Growth from Zero: t V(t) V(1 e ) − = ∆ − τ where SSV V∆ = Note that the growth and decay curves are mirror images of one another. Other possible first order step responses are described by the growth and decay curves offset vertically by adding a constant to the growth and decay functions. 10 Poles of the Underdamped and Overdamped Second Order Systems whose Step Responses are Plotted Above. Although the oscillatory underdamped response requires more mathematics to describe, it is easier to characterize because its features are distinctive. The non-oscillatory overdamped step response ranges from almost oscillatory “critically damped” case to a pair of widely spaced real poles which may be adequately described as a first order system consisting of just the right most “dominant” pole. We will make extensive use of block diagrams in ME 478 and ME 479. Recall that the lines in a block diagram represent the Laplace transforms of time variables and the blocks are operators. Block diagrams of the two functional forms needed to describe second order systems are: Overdamped Second Order System Underdamped Second Order System We will excite the systems with step inputs of magnitude M, su(t) M u (t)= The Laplace transform is: 11 { } { } { }s s 1 Mu(t) M u (t) M u (t) M s s = = =L L L = Applying this input U(s) to the overdamped transfer function yields: ( )( ) ( )( ) Y(s) M A M AU(s) Y(s) U(s) s s a s a s s a s a ⎛ ⎞⎛ ⎞ = =⎜ ⎟⎜ ⎟ ⎜ ⎟+ + + +⎝ ⎠ ⎝ ⎠ = and to the underdamped transfer function yields: ( )2 2 2 2n n n n Y(s) M A M AU(s) Y(s) U(s) s s 2 s s s 2 s ⎛ ⎞⎛ ⎞ = =⎜ ⎟⎜ ⎟ + ζω + ω + ζω + ω⎝ ⎠ ⎝ ⎠ = An abbreviated Laplace transform table is attached. Transform pair 17, below, is an overdamped, non-oscillatory second order unit step response. Transform pair 24 is an underdamped, oscillatory second order unit step response. No. f(t) F(s) 17 ( )at bt1 11 be aeab a b − −⎡ ⎤+ −⎢ ⎥−⎣ ⎦ ( )( ) 1 s s a s b+ + 24 ( )n t 2n211 e sin 1 t1 −ζω− ω − ζ + φ − ζ 2 d1 1 1tan tan− − ⎛ ⎞⎛ ⎞ω − ζ ⎜ ⎟φ = =⎜ ⎟⎜ ⎟ ⎜ ⎟σ ζ⎝ ⎠ ⎝ ⎠ ( ) 2 n 2 2 n ns s 2 s ω + ζω + ω Since these are unit step responses and our step inputs will have some magnitude M other than unity, we will need to scale the responses multiplying by the magnitude M of our step input. We will also need to scale the unit step responses to correct the numerator term so that we have an exact match with the transform pair in the table. In the overdamped case, we simply factor out the amplitude constant A: ( )( ) ( )( )OverDamped M A 1Y(s) M A s s a s b s s a s b ⎛ ⎞ = = ⎜ ⎟⎜ ⎟+ + + +⎝ ⎠ In the underdamped case, we multiply and divide by the 2nω to match the Laplace transform table: 12 ( ) ( ) 2 2 n n 2 22 2 2 2 n nn n n n M A M AY(s) s s 2 s s s 2 s ⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ω ω⎜ ⎟ ⎜ ⎟= =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ω ω+ ζω + ω + ζω + ω⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ Characterizing Second Order Systems from their Step Responses There are two ways to characterize an underdamped second order step response. One characterization is used for “system identification”, which means to determine the system’s transfer function. The damping ratio ζ and undamped natural frequency ωn are determined so that we can establish the characteristic function and, hopefully, the entire transfer function of the system which produced the step response. The other characterization of an underdamped step response is purely descriptive. In practice, you may find occasionally that you need run a step response test twice to get both the dynamic response and the steady-state response data needed. It can be impossible to amplify a signal on the oscilloscope sufficiently to see the detail of the dynamic response without pushing part of the trace off the screen. Likewise, decreasing the time per division to stretch the time axis may make it impossible to catch the entire transient period. Running the same test more than once is always a good idea since we need to know how repeatable the data are. Describing an Underdamped Step Response for a Performance Specification There is a standard way to describe a underdamped step response which is used to specify the required performance of a system. If our purpose is to characterize an oscillatory step response without determining the system’s transfer function, then we measure and report time and amplitude values of the response at the points shown in the figure below. Note that settling time is defined in terms of an envelope comprised of lines parallel to the steady-state value and spaced at either +/- 2 % or +/- 5% of the steady-state value. The settling time is the time at when the trace enters the settling time envelope and stays in it. 15 Time (sec) Amplitude Amplitude Minus Steady-State Value Peak 1 0.2 23.0 8.0 Peak 2 0.6 17.2 2.2 dT 0.60 0.20 0.40sec= − = d d 2 rad 2 rad rad15.7 T 0.4sec sec π π ω = = = 2 2 1 8.0ln 2 1 2.2 0.2 1 8.04 ln 2 1 2.2 ⎛ ⎞ ⎜ ⎟− ⎝ ⎠ζ = = ⎛ ⎞⎛ ⎞π + ⎜ ⎟⎜ ⎟− ⎝ ⎠⎝ ⎠ ( ) d n 2 2 15.7 rad rad16 sec sec1 1 0.2 ω ω = = = − ζ − ( )( ) ( )22 2 2 2n ns 2 s s 2 0.2 16 s 16 s 6.4s 256+ ζω + ω = + + = + + Using the form: ( ) 2 n 2 2 n n MY s K s s 2 s ⎛ ⎞ω⎛ ⎞= ⎜ ⎟⎜ ⎟ + ζω + ω⎝ ⎠ ⎝ ⎠ where M is the magnitude of the step input and K is a yet to be determined gain, so far we know: ( ) 2 20 256Y s K s s 6.4s 256 ⎛ ⎞ ⎛= ⎜ ⎟ ⎜ + +⎝ ⎠ ⎝ ⎞ ⎟ ⎠ We determine the gain of the transfer function from the final value of the response. From the Laplace transform pair, we see that the steady-state value of the time domain unit step response is 1 because the real exponential n te−ζω approaches zero as time increases. 16 24 ( )n t 2n211 e sin 1 t1 −ζω− ω − − ζ ζ + φ 2 d1 1 1tan tan− − ⎛ ⎞⎛ ⎞ω − ζ ⎜ ⎟φ = =⎜ ⎟⎜ ⎟ ⎜ ⎟σ ζ⎝ ⎠ ⎝ ⎠ ( ) 2 n 2 2 n ns s 2 s ω + ζω + ω Our data has a steady-state value of 15 for a step input with a magnitude of 20. Hence: 15 20K= → K 0.75= and the transfer function is: ( ) ( ) ( ) ( )( )2n 2 2 2 2 n n Output s Y s 0.75 256 192K MInput s s 2 s s 6.4s 256 s 6.4s 256 s ⎛ ⎞ω = = = =⎜ ⎟+ ζω + ω + + + +⎛ ⎞ ⎝ ⎠⎜ ⎟ ⎝ ⎠ Overdamped System Identification An overdamped step response consists of an exponential decay from a positive value and an exponential decay from an negative value superimposed on a step. An overdamped step response is more difficult to characterize than an underdamped step response because its features are less distinct. There are two iterative methods. The older method uses a family of normalized response curves. This method is now obsolete as a technique for deriving a transfer function, but it retains some usefulness in providing initial values for the computer-based iteration. Step Response Plots Method The output variable axis of second order step response plots is the observed or actual response y(t) normalized (or divided by) the steady-state value yss(t). Consequently, the normalized steady-state value of the all curves is 1. The time axis is normalized by multiply time by the “undamped natural frequency ωn” of the system: Normalized nt t= ω which yields a dimensionless value since dn 21 ω ω = − ζ and d d 2 rad T π ω = . Td is the damped period and carries units of time. Radians are the dimensionless ratio of arc length divided by radius. The damping ratio ζ is also dimensionless. Although this scaling makes sense for an oscillatory system, it is a misnomer for the step response of an overdamped system. The step response has no undamped natural frequency ωn 17 since it doesn’t oscillate. The value used as the undamped natural frequency of an overdamped system is calculated from the standard form for an underdamped system: 2 2 n ns bs c s 2 s 2+ + = + ζω + ω The method is a follows: Pick a point on the actual response and determine its amplitude y(t) and time t. Normalize the amplitude y(t) by dividing it by the steady-state value. Refer to the response plots. Estimate (or guess) a value of the damping ratio ζ which characterizes the actual system and find the response curve closest to that value. Using the step response curve which corresponds to the estimated damping ratio ζ, find the normalized time ωnt which corresponds to the normalized amplitude. You can now calculate a value of undamped natural frequency ωn. Now test your values. Pick a different point on the normalized step response curve and determine its normalized amplitude and normalized time. Return to the actual overdamped response. Multiply the normalized amplitude by the steady-state value. Divide the normalized time by ωn. It the resulting point is “reasonably” close to the actual overdamped response curve, then you have a valid estimate of ζ. If it isn’t, refine your estimate and try again. If this method seems very approximate, it is. It is typical of how engineering analyses were performed in the bad-old-days before calculators and personal computers. Computer-Based, Iterative Method A computer-based, iterative method is more practical than using normalized response curves. The first step is to identify, if possible, whether the response is an underdamped step response, and described by the Laplace transform pair No. 24, or an overdamped step response described by Laplace transform pair No. 17. If the damping ratio ζ  is not close to 1, then it is easy to distinguish between the two types of responses, since very underdamped systems are very oscillatory and very overdamped systems “creep” towards steady-state. If ζ is near 1, then it is difficult to distinguish between underdamped and overdamped systems visually. If you can’t tell which model to use, then guess. If you find it impossible to iterate to a “reasonable” fit to the data, then try the other model. Having picked a step response model, the next step is to choose initial values of the parameters to begin the iteration. It can be very time consuming to blindly iterate to fit three unknowns. It usually saves a significant amount of time to use an approximate technique, such as interpolating from Response Curves, to provide the initial values for the iteration. 20 Now estimate the second pole. A useful rule-of-thumb is that if the secondary pole is an order of magnitude or larger than the dominant pole, then its effects (other than the amplitude scaling) can be neglected. (This rule-of-thumb only holds for systems without zeros.) ( )b 10 a− ≤ − This gives us an upper bound for -b: ( )( )b 10 3.2 32− ≤ − = − We now have enough values to make a first estimate of the amplitude term A using the steady- state value of the step response, y(ss) = 5.0: { } at t M A 1y(ss) lim y(t) 1 b e a b a b − →∞ = = + − bt 0 a e −−( )0 M Aa b ⎡ ⎤ =⎢ ⎥⎣ ⎦ y(ss)a bA M = ( )( )( )5.0 3.2 32 A 2 20 6= = Now we are ready to begin iterating our model using Mathcad. It is very helpful to plot points from the observed response on the same graph as the Mathcad model. Mathcad will plot functions and vectors on the same graph. The initial and final values and three to five additional points from the observed response will give us sufficient accuracy: 21 First Iteration Model Not bad. In many cases, this model would be adequate and there would be no need for the iteration. However, it takes little time to iterate to a model which is more precise. The technique is to shift the poles alone the real axis of the s-plane to adjust their “speed” or time constants. Values of σ closer to the imaginary axis have larger time constants. Consequently, moving a pole to the right “slows” the pole. Values of σ further from the imaginary axis toward 22 negative infinity are have smaller time constants. Consequently, moving a pole to the left makes it “faster”. Inspecting the fit of the model to the data, we see that the faster pole (s = -b) is a bit too fast since the model plots to the left (at earlier time) of the data. Recall that we used an upper bound estimate for s = -b. Referring to the s-plane, we will slow s = -b down by moving it to the right. Let’s try splitting the interval between s = -32 and s = -3 and use s = -18 = -b. We need to recalculate the amplitude term A using b = 18: ( )( ) ( )( )( )y(ss) a b 5.0 3.2 18 A 1 M 20 = = 4= Second Iteration Model 25 We improved the fit and the beginning and pick up three points at the end of the response. We need to slow down the faster pole to pick up points at the beginning. Our two poles are s = -a = -5 and s = -b = -18. Let’s move –b half way toward –a and use s = -b = -12. Recalculating A: ( )( ) ( )( )( )y(ss) a b 5.0 5 12 A 1 M 20 = = 5= Fifth Iteration Model 26 The faster pole still looks too fast since we are to the left of the data points (at earlier time) except for the origin, which, of course, doesn’t count. Let’s again reduce the spacing between the poles by half and move the faster pole from s = -12 to s = -8.5. The amplitude factor is now: ( )( ) ( )( )( )y(ss) a b 5.0 5 8.5 A 10.6 M 20 = = = Final Model 27 We have a good curve fit for the step response. To derive the transfer function of the system, we must remember to divide the Laplace transform of the output Y(s): ( )( )OverDamped M AY(s) s s a s b = + + by the Laplace transform of the input: MU(s) s = which yields: ( )( ) ( )( ) ( )( ) M A s s a s bOutput(s) Y(s) A 10.6 G(s) MInput(s) U(s) s a s b s 5 s 8.5 s + + = = = = ≡ + + + +⎛ ⎞ ⎜ ⎟ ⎝ ⎠ RC CIRCUITS We will characterize step responses RC circuits in order to determine the values of unknown parameters. Our output variable will be the voltage across the capacitor. In ME 352, we worked with this model of an RC circuit: An RC Circuit Model The ME 352 model assumed an ideal voltage source. An ideal source would allow us to control power variable as we wish while supplying as much of the conjugate power variable as the system draws. If we had an ideal voltage source, we would have the same result if we gave the system a step input in voltage up from 0 to V or down from V to 0 independent of magnitude of the current demanded and the time scale of the system. In ME 479, we will work with a real DC power supply rather than ideal voltage source. Our DC power supply has dynamic characteristics since it is a dynamic energetic system. In order to remove the dynamic response 30 Nodes: 1R i i= 1 1R C R i i i= + 2 2 2R C i i= Elements: 112 1 R v R i= 223 2 R v R i= 1 2g C 1 dv i C dt = 2 3g C 2 dv i C dt = Energy: 1 2System C C = +E E E 1 2 C 1 1 C v 2 =E 2g 2 2 C 2 1 C v 2 =E 3g Reduction: Input v1g, Output v3g 12 2gv v v= + 11 R 2gv R i v= + 11 R 23 3gv R i v v= + + 1 21 R 2 R 3g v R i R i v= + + 1 21 R 2 C 3g v R i R i v= + + 1 3g 1 R 2 2 3g dv v R i R C v dt = + + ( )1 2 3g1 C R 2 2 3g dv v R i i R C v dt = + + + 2 2g 3g 1 1 1 C 2 2 3g dv dv v R C R i R C v dt dt = + + + ( )23 3g 3g 3g 1 1 1 2 2 2 3g d v v dv dv v R C R C R C v dt dt dt + = + + + ( ) ( )23 3g 3g1 1 1 2 2 2 3g d v v dv v R C R C R C v dt dt + = + + + ( )3g 3g231 1 1 1 1 2 2 2 3g dv dvdvv R C R C R C R C v dt dt dt = + + + + ( ) ( )22 R 3g1 1 1 1 1 2 2 2 3g d R i dv v R C R C R C R C v dt dt = + + + + ( )2C 3g1 2 1 1 1 1 2 2 2 3g di dv v R R C R C R C R C v dt dt = + + + + ( ) 2 3g 3g 1 2 1 2 1 1 1 2 2 2 3g2 d v dv v R R C C R C R C R C v dt dt = + + + + System equation Unit check: [ ] vR i ⎡ ⎤= ⎢ ⎥⎣ ⎦ [ ] i tC v ⎡ ⎤= ⎢ ⎥⎣ ⎦ 31 [ ] ( ) 2 3g 3g 1 2 1 2 1 1 1 2 2 2 3g2 d v dv v R R C C R C R C R C v dt dt ⎡ ⎤ ⎡ ⎤ ⎡ ⎤= + + +⎢ ⎥ ⎢ ⎥ + ⎣ ⎦⎢ ⎥ ⎣ ⎦⎣ ⎦ [ ] [ ]2 2 2 v vv R C RC v t t ⎡ ⎤ ⎡ ⎤= +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ + [ ] [ ] 2 2 2 v i t v v i t vv v i v t i v t ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎡ ⎤= +⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎣ ⎦⎢ ⎥⎣ ⎦ + [ ] [ ] [ ] [ ]v v v v= + + Units check. forced, atural, homogenous solution has complex roots. Solving the characteristic equation: Although this system is second order, it is impossible for the system have an oscillatory step response. Recall that the characteristic equation of a system with an oscillatory un n 2as bs c 0+ + = using the quadratic equation yields: 2 1 2 b b 4acs ,s 2a − ± − = For the characteristic equation of this RC circuit to have complex roots: 2b 4ac< substituting: the same type. scillatory internal energy transfer requires two different energy storage modes. f the electrical roperties, temperature in place of voltage, and heat flow rate in place of current. ( )21 1 1 2 2 2 1 2 1 2R C R C R C 4R R C C+ + < Although evaluating this inequality is not difficult, it is involved since the parameters have no values, other than being limited to positive real numbers. We will not evaluate this inequality here. However, if you are interested in the problem, ask your instructor for a copy. The technique is to expand the left side to a summation and then establish a set of inequalities by using one left-hand term at a time. The set of inequalities provides limiting values which are impossible to satisfy simultaneously, proving that the inequality is false and the characteristic equation cannot have complex roots. The physical reason that this system does not have an oscillatory step response is that the two independent energy storage elements are O The RC circuit with two capacitors in parallel is the basis of a number of important engineering models. Transient heat conduction through a homogenous material is represented by the analogous model with thermal resistance and heat capacitance in place o p 32 It is worthwhile evaluating the step response of this system using Laplace transforms to lustrate a common error. First create the transfer function: il { } ( ) 2 3g 3g 1 2 1 2 1 1 1 2 2 2 3g2 d v dv v R R C C R C R C R C v dt dt ⎧ ⎫⎪ ⎪= + + +⎨ ⎬ ⎪ ⎪⎩ ⎭ L L + ( ) ( ) ( ) ( ) (21 2 1 2 3g 1 )1 2 2 2 3g 3gV s R R C C s V s R C R C R C sV s V s= + + + + 1 ( ) ( ) ( )21 2 1 2 1 1 1 2 2 2 3gV s R R C C s R C R C R C s 1 V s⎡ ⎤= + + + +⎣ ⎦ ( ) ( ) ( ) 3g 2 1 2 1 2 1 1 1 2 2 2 V s 1 V s R R C C s R C R C R C s 1 = + + + + sing the following parameter values R1 = R2 = 1KΩ, C1 = C2 = 1 µF: U ( ) ( ) ( )( )( )( ) ( )( ) ( )( ) ( )( )( ) 3g 2 V s 1 V s 1K 1K 1 F 1 F s 1K 1 F 1K 1 F 1K 1 F s 1 = Ω Ω µ µ + Ω µ + Ω µ + Ω µ + ( ) ( ) ( ) ( ) 3g 6 2 3 V s 1 V s 1 10 s 3 10 s 1− − = × + × + We know this system is overdamped, since it cannot oscillate. Consequently, we will use ( )( ) 3gV (s)Output(s) Y(s) A G(s) Input(s) U(s) V(s) s a s b = = = ≡ + + We will use the quadratic equation to find the roots of the characteristic equation. Although the quadratic equation will correctly determine the roots –a and –b regardless of the form of the characteristic equation, we may have a gain error unless we clear the s2 term of its ultiplying constant. m ( ) ( ) ( ) ( ) 6 3g 2 36 2 3 V s 1 1 V s s 3 10 s 1 101 10 s 3 10 s 1− − × = = + × + ×× + × + 6 10 = )( ) ( )6 2 31 10 s 3 10 s 1 0− −× + × + → ( ) ( ) (( ) 26 6 1 2 6 3 10 3 10 4 1 10 s ,s 2 1 10 − − − − × ± × − × = × 6− 35 Nodes: 1R i i= 1 1R C i i= 1 2C R i i= 2 2R C i i= Elements: 112 1 R v R i= 234 2 R v R i= 1 23 C 1 dvi C dt = 2 4g C 2 dv i C dt = Energy: 1 2System C C = +E E E 1 2 C 1 1 C v 2 =E 23 2 2 C 2 1 C v 2 =E 4g Reduction: Input v1g, Output v4g 1g 12 23 34 4gv v v v v= + + + 1 21g 1 R 23 2 R 4gv R i v R i v= + + + 1 2R R1g 4g23 1 2 di didv dvdvR R dt dt dt dt dt = + + + 1 1 2R C R1g 4g1 2 1 di i didv dv R R dt dt C dt dt = + + + 2 2 2C C C1g 4g 1 2 1 di i didv dv R R dt dt C dt dt = + + + 2 2 1g 4g 4g 4g 4g2 1 2 2 22 2 1 dv d v dv d v dvCR C R C dt dt C dt dt dt = + + + ( ) 2 1g 4g 4g2 1 2 2 2 1 dv d v dvCR R C 1 dt dt C dt ⎛ ⎞ = + + +⎜ ⎟ ⎝ ⎠ Although the highest order derivative of the output variable is second order, the system equation is actually first order since there is no zero order term output variable term v4g. Integrating both sides with respect to time yields the first order system equation. ( ) ( ) 4g 21g 1 2 2 4g 1 dv Cd dv R R C 1 dt dt dt C ⎛ ⎞⎛ ⎞ = + + +⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ v ( ) ( ) 4g 21g 1 2 2 4g 1 dv Cd dv dt R R C 1 v dt dt dt dt C ⎛ ⎞⎛ ⎞ = + + +⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ ∫ ∫ ( ) 4g 21g 1 2 2 4g 1 dv Cv R R C 1 v dt C ⎛ ⎞ = + + +⎜ ⎟ ⎝ ⎠ Physically, the two capacitors are dependent energy storages because they share the same current. It is impossible to place a charge in one without also charging the other. From the equation list: 36 1 2C C i i= , 1 23 C 1 dvi C dt = , and 2 4g C 2 dv i C dt = . Therefore: 4g23 2 1 dvdv C dt C dt ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ RLC CIRCUITS The dynamic response of an RLC circuit is not intuitive because most electromagnetic phenomena are outside of the realm of our senses. Electrical inductance in particular makes the physics of an RLC circuit obscure. It helps to use the analogy between the mechanical behavior of fluid systems and the electromagnetic behavior of circuits. A capacitor stores energy in its electric field as measured by the voltage across the capacitor. Although fluid capacitors may store energy by lifting fluid against gravity or in the strain of the fluid or the container the fluid is in, all three modes of energy storage are functions of the fluid pressure. Consequently, fluid capacitors and electrical capacitors are directly analogous. They both store energy as a function of their across variable. An electrical inductance stores energy a the magnetic field created by moving charge or current. Changes in current, and, consequently, in the magnetic field created by the current, induce voltages on conductors. There are mathematically analogous aspects of fluid systems, since there is kinetic energy stored in the mass of moving fluid. Therefore, pressure is analogous to voltage and volume flow rate is analogous to current. As you recall, from ME 352, these are pairs of “across” and “through” variables in the linear graph system. The analogous energy storage modes lead to analogous dynamic responses. In a fluid system, there must be a pressure drop in the direction of flow to accelerate fluid. and a pressure rise in the flow direction to decelerate it. Analogously, there must be a voltage drop across an inductor to start current flow and a voltage rise to stop it. The oscillation of voltage in an electric circuit results from energy flow back and forth between the capacitor and the inductor. This oscillation is analogous to the fluctuation of pressure due to water sloshing back and forth in the equivalent fluid system. The key concept is that if a system has an oscillation, then it has energy flowing back and forth between two independent energy storage elements. Inductors and capacitors are independent energy storage elements because they have different energy storage variables. Here is a brief explanation of the response of the series RLC circuit shown below to a step change of input voltage. Assume that there is no energy stored in the circuit at time t=0. When the circuit is first closed the voltage difference from node 2 to node 3 across the inductor starts the flow of current through the inductor, creating a magnetic field. The capacitor develops a voltage across node 3 to ground as current flows through the inductor into the capacitor. The rise in voltage across the capacitor slows the current through the inductor. However, a voltage rise must be created across the inductor to stop the current. The voltage rise is created by the energy flowing out of the magnetic field surround the inductor. The voltage rise created by the collapsing magnetic field of the inductor causes an overshoot above the voltage of the source. The capacitor stores charge at that voltage. Current then flows back towards the source, decreasing the voltage in the capacitor. The voltage drop across the inductor from node 2 to node 3 eventually becomes large, reversing the flow of current again. Hence, there is an oscillation. 37 Series RLC Circuit We will investigate the step response of a circuit comprised of a 1 or 2 microfarad capacitor and an inductor we will make by winding wire around a steel washer. The output will be the voltage across the capacitor C which we will observe on the oscilloscope. A model of the circuit we will use is shown below. RLC Circuit Model Including the Oscilloscope Resistive and Capacitive Input Impedance Although the circuit will not have a discrete resistor in it, there is finite resistance in the wire and in the contacts between the components. We will lump these resistances into Rcoil. Note that node 2 is a fictitious node needed to give the coil both resistance and induction. There is no physical location in the circuit that corresponds to node 2. Node 2 exists in the sense that the voltage drop across the coil results from two phenomenon, resistance and inductance: 13 12 23 coil coil resis tance induc tan ce v v v v v= + ≡ + We will calculate the resistance of the wire using published values for its resistivity or conductivity per length and compare that value to the value we calculate from the step response of the RLC circuit. Capacitance C will be either 1 µF or 2 µF. Using 1 µF and recalling that a pico is 10 -12: 40 The effect of strengthening the field by shaping the wire is taken to its limit with toroidal coil, since the magnetic field lines never leave the coil. Useful analogies for the shape of magnetic fields are “potential flows” in which a quantity such as a fluid or heat flows from a high potential (pressure or temperature) to a low potential. An example of a potential flow is the seepage flow of a fluid through a porous medium, such as water seeping through sand. Seepage flows are slow, so the kinetic energy of the flow is negligibly small. The equivalent of the magnetic North and South poles would be a fluid source at high pressure and a fluid “sink” or drain at low pressure. Although the flow the analogy provides the shape of magnetic field lines, it cannot provide their direction of the flow. All seepage flow is down the pressure gradient. Fluid flow lines do not form closed loops since there cannot be seepage flow from low pressure to high pressure. Accepting that limitation, the flow analogy useful enough that it is incorporated into magnetic terminology. The second way we can increase the strength of a magnetic field is by placing a “ferromagnetic” material in the field. Ferromagnetic materials, iron, cobalt, nickel and their alloys, significantly enhance the strength of an applied magnetic field by the magnetic dipoles of the magnetic domains within the ferromagnetic material orienting in response to an applied field. This property is called permeability, for the ease with which the magnetic field can permeate a volume, as given the symbol µ (Greek for m). Ferromagnetic materials can be thought of as not only intensifying the magnetic field but also channeling it, since the magnetic field outside of the ferromagnetic material is generally negligibly small compared to the field inside the ferromagnetic material. A bar magnetic is a good example. The magnetic field lines within the bar are straight and tightly packed. In the surrounding space, they are further apart and curved. The engineering approximation that all of the magnetic filed is confined to the ferromagnetic materials in a device greatly simplifies calculations while maintaining engineering accuracy. Although permeability is usually referred to as a “material” property, that is not strictly correct since a magnetic field can permeate a vacuum in which there is no material. The permeability of free space (a vacuum) is given the symbol µ0 and has the value of 4π x 10-7 henrys per meter. The term 4π is introduced for convenience when performing calculations using solid angles and spherical surfaces. A solid angle is analogous to a radian. Where a radian is the ratio of an arc length to the circumference of a circle, a solid angle is the ratio of the area of a portion of a sphere to the entire area of a sphere. The circumference of a circle equals πD or 2πR. The area of a sphere equals πD2 or 4πR2. The “relative permeability” of a material, µr, is defined as: r 0 µ µ ≡ µ Values of the relative permeability of electrical steels range from 2,000 to 6, 000. The relative permeability of other ferromagnetic materials typically range up to 30,000 but can exceed 100,000. The permeability of air is approximately equal to the permeability of free space. 41 The relationship between the magnetic flux density, B and the vector field H, called the “applied magnetic field intensity”, are related by the magnetic permeability, µ: = µB H (excluding the effect of permanent magnets on B), where µ scales H to yield B. Why work with B and H rather than just B? Why add the complication of another vector field if B and H are just scalar multiples of each other? Although we will make the engineering approximation that there is a value of permeability for a given material, in fact the relationship between the applied magnetic field intensity H and the resulting magnetic flux density B is non- linear and multi-valued. The contribution of ferromagnetic materials to the magnetic flux density B eventually is also limited. The limit, called “saturation”, is reached when all of the domains are oriented to the applied magnetic field. Any further increase in the applied magnetic field intensity H does increase the magnetic flux density B, but the incremental increase is only at the permeability of free space, µ0, as if the ferromagnetic material were not present. Electrical Induction As you recall from ME 352, the magnetic energy storage property of electrical circuit elements is described by inductance L in henrys. The inductance L is a measure of how large a magnetic flux a given current i can create. The turns of wire which form the coil are said to be “coupled” to the magnetic flux they create. A Torroidal Inductor This “coupling” is called the “flux linkage” and is given symbol λ (Greek for l as in linkage). The “flux linkage” λ between a magnetic flux φ and a coil is proportional to the 42 number of turns of wire in the coil, N, the flux flows through or “threads” (as in threading the eye of a needle) or “links” (as in links of a chain): Nλ = φ If the relationship the magnetic flux φ threading a coil and current i through a coil is linear, then there is a linear relationship between the flux linkage λ and current i which defines the parameter inductance L: Liλ = Differentiating this expression with respect to time yields: d dL dt dt iλ = Recall from ME 352 that the a voltage v12 develops across an electrical inductor in response to a time varying current i. The voltage acts to maintain the current through the coil. Consequently, voltage v12 has the opposite sign of the rate of change of current di dt : 12 div L dt = − The two differential equations above reduce to a relationship between the time rate of change of the flux linkage λ and the resulting voltage v12. 12 d v dt λ = − We will make our own toroidal inductors by wrapping copper wire around steel washers. The inductors will be used in an RLC circuit in which the only unknown is the value of the inductor. We will capture a step response of the RLC circuit and, from the frequency and damping ratio of the underdamped response, calculate the inductance L. Using the calculated inductance, we will estimate the permeability µ of the steel washers we used for the core. Transformers Electrical transformers consist of two or more coils of wire wrapped around the same ferromagnetic core. The same magnetic flux φ flows through both coils. By convention, the input coil is called the primary and the output coil is called the secondary. Although schematics show the primary and secondary coils at different locations on the ferromagnetic core, in practice both coils are wound together. The number of turns in the primary and secondary windings are N1 and N2, respectively. 45 LAB 1 PRELAB: 1. Derive the system equations for the following circuits where the output variable of interest is the voltage across the capacitor: Circuit A Circuit B LAB 1 PROCEDURE: 1 Familiarization with the Oscilloscope. Use the function (signal) generator to produce a periodic signal. (a) Take turns with your lab partner exploring the functions of the oscilloscope, capturing and measuring the voltages and frequencies and periods of the signals, with and without the Math functions. (b) When you are comfortable with the scope, bring up the Tektronix Instrument Manager software on the computer and establish communication with the scope. Capture a “y-t” signal on the scope and transfer the data to the computer. Use the measurement functions of WaveStar to characterize the signal. (c) Copy the data from WaveStar to Excel and then import the data into Mathcad. Plot the data in Mathcad. (d) In Mathcad, use a range variable calculation to create a running average of 5 data points to smooth the data. Be careful not to run over the beginning or end of the data vector and remember that the ORIGIN (or lowest value of a vector subscript) is 0, not 1. Plot the original and smoothed data. 46 2. RC Circuit: (a) Construct the RC circuit shown above with C = 1 µF and R1= 1 MΩ  Measure its time constant. How does it compare with the calculated time constant? Replace R1= 1 MΩ with R1 = 1 KΩ Does that change the relationship between the measured and calculated time constants? If so, why? (b) Replace the capacitor with one provided by your instructor (keep R1= 1 KΩ). Determine the value of the “mystery” capacitor. 3. RLC Circuit with the I-Made-It-Myself Toroidal Inductor: (a) Measure the resistance of 2 meters of 24 gauge copper wire. (b) Create an I-Made-It-Myself toroidal inductor by wrapping 10 turns of the wire around the steel washer provided. (c) Create the RLC circuit shown above using the toroidal inductor. Use two 1 µF capacitors in parallel for the capacitance C. (d) Capture the step response of the RLC circuit using the procedure of Part 2 above. Determine the damping ratio, damped period, damped frequency, and ideal undamped natural frequency of the circuit. Estimate the current during an oscillation using a finite difference approximation for the rate of change of voltage across the capacitance: C dvi C dt = → C vi C t ∆ ≈ ∆ (e) Determine the inductance Lcoil of the toroidal inductor. (f) Repeat Parts (b) though (e) for 20 turns of wire and 40 turns of wire. How does the inductance vary with the number of turns? 47 (g) Repeat Parts (b) through (f) for a toroidal inductor with a core of two steel washers. How does the inductance vary with the number of washers? (h) Time permitting, estimate the permeability of the steel of the washers.