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Unformatted text preview: Electrical Engineering Applications with the TI-89 David R. Voltmer Rose-Hulman Institute of Technology Terre Haute, Indiana Mark A. Yoder Rose-Hulman Institute of Technology Terre Haute, Indiana 00afront.doc . Karen Pressnell Revised: 02/08/99 7:51 AM Printed: 03/25/99 1:44 PM Page 1 of 2 Important notice regarding book materials Texas Instruments makes no warranty, either expressed or implied, including but not limited to any implied warranties of merchantability and fitness for a particular purpose, regarding any programs or book materials and makes such materials available solely on an “as-is” basis. In no event shall Texas Instruments be liable to anyone for special, collateral, incidental, or consequential damages in connection with or arising out of the purchase or use of these materials, and the sole and exclusive liability of Texas Instruments, regardless of the form of action, shall not exceed the purchase price of this book. 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Send inquiries to this address: Texas Instruments Incorporated 7800 Banner Drive, M/S 3918 Dallas, TX 75251 Attention: Manager, Business Services Copyright © 1999 Texas Instruments Incorporated. Except for the specific rights granted herein, all rights are reserved. Printed in the United States of America. ISBN: 1-886309-25-6 FRISBEE is a registered trademark of WHAM-O. Maple is a registered trademark of Waterloo Maple, Inc. Mathcad is a registered trademark of MathSoft, Inc. Mathematica is a registered trademark of Wolfram Research, Inc. MATLAB is a registered trademark of The Math Works, Inc. TI-GRAPH LINK is a trademark of Texas Instruments Incorporated. We invite your comments and suggestions about this book. Call us at 1 - 8 0 0 - T I - C A R E S or send e-mail to t i - c a r e s @ t i . c o m . Also, you can call or send e-mail to request information about other current and future publications from Texas Instruments. Visit the TI World Wide Web home page. The web address is: h t t p : / / w w w . t i . c o m / c a l c / 00afront.doc . Karen Pressnell Revised: 01/28/99 10:39 AM Printed: 01/28/99 10:46 AM Page 2 of 2 Contents Do This First Chapter 1: Topic 1: Topic 2: Topic 3: Chapter 2: Topic 4: Topic 5: Topic 6: Topic 7: Topic 8: Chapter 3: Topic 9: Topic 10: Topic 11: Topic 12: Chapter 4: Topic 13: Topic 14: Topic 15: Topic 16: Topic 17: Topic 18: Chapter 5: Topic 19: Topic 20: Topic 21: Topic 22: Topic 23: Topic 24: Topic 25: Topic 26: Chapter 6: Topic 27: Topic 28: Topic 29: Topic 30: Topic 31: 1 DC Circuit Analysis 11 Nodal Equations Using solve( ) Nodal Equations with Voltage Sources Nodal Equations Using simult( ) 11 12 14 Transient Circuit Analysis: Symbolic 17 RC First-Order Circuit Graphing First-Order Solutions First-Order Circuit with an Initial Condition First-Order Circuit with a Time Varying Source RLC Second-Order Circuit 17 18 19 21 23 Transient Circuit Analysis: Numeric 27 RLC Circuit: Direction Field RLC Circuit: Time Domain RLC Circuit: Multiple Initial Conditions RLC Circuit: Adjusting the Circuit Parameters 27 30 30 31 Steady-State Circuit Analysis and Filer Design 33 Phasor Analysis Graphing Frequency Response Filter Design Overview Butterworth Filter Chebyshev Filter Logarithmic Frequency Plots 33 35 38 39 42 44 Power Engineering 47 Phasor Algebra Average Power Complex Power Power Factor Power Factor Correction Using Impedances Power Factor Correction Using Power Triangle Y-∆ and ∆-Y Transformations Unbalanced Three-Phase Systems 47 49 51 53 54 55 56 57 Laplace Analysis: The s-domain 61 RLC Circuit Critical Damping Poles and Zeros in the Complex Plane Frequency Response 3D Poles and Zeros 61 64 65 68 69 iii 00bfront.doc . Karen Pressnell Revised: 01/29/99 2:31 PM Printed: 01/29/99 2:32 PM Page iii of 4 Chapter 7: The Convolution Topic 32: Convolution Integral Topic 33: Piecewise Convolution Topic 34: Graphing Piecewise Convolution Results Chapter 8: Fourier Series Topic 35: Square Wave: Computing the Coefficients Topic 36: Square Wave: Constructing the Wave from the Coefficients Chapter 9: Topic 37: Topic 38: Topic 39: Topic 40: Topic 41: Topic 42: Topic 43: Topic 44: 83 85 86 87 88 90 91 92 Gradient Surface Normal Divergence Curl Laplacian Line Integrals Surface Integrals Separation of Variables 3D Potential Graphs Relaxation Method 3D Graphs of Tabular Data Characteristic Impedance Reflection Coefficient Phase Shift Input Impedance/Admittance VSWR Impedance Matching Chapter 13: Antennas Topic 62: Incremental Dipole Topic 63: Antenna Patterns Topic 64: Phased Arrays Chapter 14: Manipulating Lab Data: The Diode Topic 65: The Diode Equation Topic 66: Lab Data Chapter 15: Financial Calculations Topic 67: Topic 68: Topic 69: Topic 70: Simple Interest Compound Interest Loans Annuities Index iv 00bfront.doc . 77 80 83 Chapter 12: Transmission Lines Topic 56: Topic 57: Topic 58: Topic 59: Topic 60: Topic 61: 77 Coordinate Systems and Coordinate Transformations Vector Components Angle between Vectors Parallel and Perpendicular Vectors Rectangular to Cylindrical Vector Transformation Cylindrical to Rectangular Vector Transformation Rectangular to Spherical Vector Transformation Spherical to Rectangular Vector Transformation Chapter 11: Electromagnetics Topic 52: Topic 53: Topic 54: Topic 55: 71 71 75 Vectors Chapter 10: Vector Calculus Topic 45: Topic 46: Topic 47: Topic 48: Topic 49: Topic 50: Topic 51: 71 Karen Pressnell Revised: 01/29/99 2:31 PM Printed: 01/29/99 2:32 PM Page iv of 4 95 95 97 98 99 100 101 103 105 105 109 112 115 117 117 121 122 122 123 124 127 127 130 132 135 135 137 141 141 142 143 144 147 Preface To Students: This book is written for electrical engineering students. It is a collection of examples that show how to solve many common electrical engineering problems using the TI-89. It is not a textbook; if you do not know how to solve the problem, look it up in your textbook first. If you do know how to solve the problem, this book will show you how to use the TI-89 to get the answer with more insight and less tedium. We show you how to use the TI-89 in class, in lab, on homework, and so forth. Many of you may now use Mapleê, Mathematicaê, MATLABê, Mathcadê, or other symbolic or numeric software. You will be pleasantly surprised to find that the TI-89 can solve many of the same problems as the big boys, but it will boot up in only a second or two, it rarely crashes, it fits your pocket book (even if you have a small one), and can fit in your pocket (if you have a big one). You should find this book easy to use. Although we show how to use many of the features of the TI-89, we assume you already know your way around it. First read Do This First, then jump to the section discussing the problem you want to solve. To Instructors: Read the To Students section. When writing this book, we resisted the temptation to show how the TI-89 can be used to solve problems in ways that differ from standard electrical engineering texts. Although it has the power and ability to approach many problems in new ways, that was not our focus. Our focus is to help students learn the basic material better by showing them how to use the TI-89 to do the tedious things so they don’t get lost in the details. Our approach was best summed up by Gottfried Wilhelm Leibniz when, in the 17th century, he said, “It is unworthy of excellent men to lose hours like slaves in the labor of calculation.” — David Voltmer — Mark Yoder v 00bfront.doc . Karen Pressnell Revised: 01/29/99 2:31 PM Printed: 01/29/99 2:32 PM Page v of 4 About the Authors DAVID VOLTMER (AKA Smilin’ Dave) loves teaching electrical engineering at Rose-Hulman Institute of Technology. He claims to be good in the areas of electromagnetics, microwaves, antennas, communications and design. A few of his many projects designed to assist student learning include PC-based instruments, SPICE48, and Visual Electromagnetics (VEM). The writing of this book was accompanied by the sounds of clawhammer banjo music and with regular training breaks for long-distance cycling. MARK A. YODER, Associate Professor of Electrical and Computer Engineering, received his B.S. degree in 1980 and Ph.D. in 1984, both in Electrical Engineering and both at Purdue University. While there, he did research in speech and in image processing, in addition to studying the dynamics of a disk passing through a non-viscous medium (that is, playing a lot of Frisbeeê). In 1988 he discovered that teaching was where the fun is, so he headed for Rose-Hulman in Terre Haute, Indiana. Here he pioneered the use of symbolic algebra systems in electrical engineering education and helped develop a class on computer vision. He developed an expert system for diagnosing a fiber-optic communications system for the International Centers for Communication Technology, and he hopes to work on computer assisted Bible translation. He has co-authored a book on digital signal processing for sophomores. Dr. Yoder’s biography is not complete without mention of his family. His wife Sarah has her Ph.D. in Electrical Engineering from Purdue. They have nine children aged 15, 15, 12, 10, 8, 6, 4, 2, and -0.5 – two boys, six girls, and one on the way. vi 00bfront.doc . Karen Pressnell Revised: 01/29/99 2:31 PM Printed: 01/29/99 2:32 PM Page vi of 4 Features Used cSolve( ), ’, “, Í, abs( ), angle( ), Numeric Solver, when( ), log( ), DelVar, DrawFunc, DrawInv, NewProb, a, ¥#, ¥%, ¥$ Chapter 4 Setup ¥1, NewFold steady, . setMode(“Angle”,“Degree”) Steady-State Circuit Analysis And Filter Design This chapter shows how the TI-89 implements phasors to perform sinusoidal steady-state analysis. The focus is on how to enter and display complex numbers. This chapter also shows a typical steady-state application—how to use the Numeric Solver to find the required order of lowpass Butterworth and Chebyshev filters in making a standard “handbook” filter design. Topic 13: Phasor Analysis Given the circuit shown in Figure 1, find v, the voltage across the current source. Figure 1. A circuit in steady-state The first step is to convert the actual circuit to its phasor equivalent. The circuit shown in Figure 2 includes these conversions. Figure 2. The phasor equivalent of the Figure 1 circuit Only one nodal equation is needed to solve for v − 8∠0°+ v v v + + =0 10 6 + j8 − j5 © 1999 TEXAS INSTRUMENTS INCORPORATED 04steady.doc . Karen Pressnell Revised: 01/25/99 9:25 AM Printed: 01/28/99 8:40 AM Page 33 of 14 34 ELECTRICAL ENGINEERING APPLICATIONS WITH THE TI-89 1. Clear the TI-89 by pressing 2 ˆ 2:NewProb ¸. 2. Be sure the Complex Format mode is set to Rectangular. Be sure the Graph mode is set to Function. 3. Enter the equation as shown in screen 1. · 8 « v e 10 « v e c 6 « 2 ) 8 d « v e c · 2 ) 5 d Á 0 § n1 4. (1) Note: The usual imaginary number j used in electrical engineering is entered as i which is 2 ). Using solve(n1,v) will return “false” since it is valid for real solutions only. To get a complex solution, enter cSolve( ) as shown in screen 2. (2) Note: To enter cSolve(, press „ A:Complex 1:cSolve( or ½ cSolve(. 5. Phasors are expressed as a magnitude at an angle, M∠θ. There are a couple of ways to obtain this form. The first way is to use the functions abs( ) (top of screen 3) and angle( ) (middle of screen 3). In this example, ¥ ‘ is used to get the approximate values for the second angle( ) command (bottom of screen 3). This shows that the phasor form of the voltage is 40∠ -36.87 in Degree mode. (3) Note: To see the values on the right end of a solution line in the history section, press C to get to the line and then press B to move to the right. Note: Press 3 to switch to degree mode if it isn’t already set. If in radian mode the angle would have been given in radians. The second approach is to put the TI-89 in Polar mode. Press 3 and select Complex Format 3:Polar (screen 4). (4) 6. Using cSolve( ) (and with ¥ ‘ for a second approximate solution) gives the same results in the polar mode as abs( ) and angle( ) in the rectangular mode, as shown in the bottom two lines of screen 5. (5) © 1999 TEXAS INSTRUMENTS INCORPORATED 04steady.doc . Karen Pressnell Revised: 01/25/99 9:25 AM Printed: 01/28/99 8:40 AM Page 34 of 14 CHAPTER 4: STEADY-STATE CIRCUIT ANALYSIS AND FILTER DESIGN 35 Topic 14: Graphing Frequency Response It is easy to find the voltage across the current source as a function of ω using symbolic expressions. Table 1 shows the variations of phasor circuit elements with radian frequency ω. Element Element Equation Phasor Result jωt 8cos10 t v(t)=Re[Ve ] V=8∠0° 10 Ω ZRa = Ra ZRa = 10 Ω 6Ω ZRb = Rb ZRb = = 6 Ω 80 µH ZL = jωL ZL= jω80×10-6 Ω 5 2 µF ZC = 1 jω C ZC = 1 Ω j ω 2 × 10 −6 Table 1: Frequency Dependence of Phasor Circuit Elements The nodal equation then becomes −8∠0o + v v v + + =0 ra rb + zell zc1 Note that ra, rb, and zc1 are used because r1, r2, and zc are TI-89 system variables. 1. Switch back to Rectangular Complex Format mode and enter the equations as shown in screen 6. · 8 « v e ra « v e c rb « zell d « v e zc1 Á 0 § n1 as shown in screen 6. 2. (6) Note: zell is used to avoid confusing zl with z1 (z followed by a 1), a reserved name. Define the element values from the table (screens 7 and 8). For convenience, w ( j w) is used instead of ω ( ¥ c j w). (7) (8) © 1999 TEXAS INSTRUMENTS INCORPORATED 04steady.doc . Karen Pressnell Revised: 01/25/99 9:25 AM Printed: 01/28/99 8:40 AM Page 35 of 14 36 3. ELECTRICAL ENGINEERING APPLICATIONS WITH THE TI-89 With cSolve( ), the solution in screen 9 shows that the voltage varies with frequency. (9) The complete answer is v= −4000000.00 (iw + 75000.00) w − 125000.00iw − 10000000000.00 2 The answer is a bit of a mess. To check it with the previous solution, enter eqn | w=100000 (screen 10). It’s the same answer as Topic 13, screen 2. (10) 4. To view the variation of the voltage magnitude versus frequency, graph v versus w. Since the original problem used w=100,000, graph from w=0 to w=200,000. Press ¥$ and set xmin to 0, xmax to 200000, ymin to 0, and ymax to 50. Press ¥ # and set y1 to graph the magnitude of v (screen 11). 5. Press ¥ % to see the magnitude graph (screen 12). 6. This graph takes a long time to complete because the “with” substitutions are made over and over again for each pixel. One way to speed it up is to do the “withs” once before graphing and save the result in another variable name which is then graphed. (11) (12) To do this, press " and enter the expression: ½ abs( v Í eqn d § eqn2 as shown in screen 13. (13) Then press ¥ #, deselect y1, and enter y2 (screen 14). (14) Note: Deselect equation y1(x) with †. © 1999 TEXAS INSTRUMENTS INCORPORATED 04steady.doc . Karen Pressnell Revised: 01/25/99 9:25 AM Printed: 01/28/99 8:40 AM Page 36 of 14 CHAPTER 4: STEADY-STATE CIRCUIT ANALYSIS AND FILTER DESIGN 7. 37 Press ¥ %. The same result as the previous graph appears much more quickly (screen 15). (15) 8. The phase can be graphed defining the phase angle of the voltage on the Home screen as eqn3 (screen 16). (16) 9. Press ¥ #, deselect y2, and enter eqn3 as plot variable y3 (screen 17). 10. Press ¥ $ and set ymin to -90 and ymax to 0 (since the calculations have been in the degree mode). xmin and xmax can remain the same. (17) 11. Press ¥ % to see the phase graph as shown in screen 18. (18) 12. Usually the magnitude and phase plots are shown together. This can be done using the split screen mode. To do this, press 3 „ B D. Screen 19 presents the Split Screen options. (19) 13. Press ¸. Move down to Split 2 App, and select 4:Graph. Finally, set Number of Graphs to 2 (screen 20). (20) 14. Press ¸ to view the split screen plots (screen 21). The top graph is the phase plot shown before; the bottom graph contains no data yet. (21) © 1999 TEXAS INSTRUMENTS INCORPORATED 04steady.doc . Karen Pressnell Revised: 01/25/99 9:25 AM Printed: 01/28/99 8:40 AM Page 37 of 14 38 ELECTRICAL ENGINEERING APPLICATIONS WITH THE TI-89 15. Convention says the magnitude plot should be on top. To do this, press ¥ # and use † to select y2 and deselect y3. Next, press ¥ $ and set ymin to 0 and ymax to 50. Finally, press ¥% to see the magnitude plot in the upper graph as shown in screen 22. (22) 16. To set up the phase plot in the lower window, change to the other half of the screen by pressing 2a and set up the graph as before. The following operations will give the phase plot in the bottom window (screen 23). ¥ #, select y3, ¥ $, set xmin to 0, xmax to 200000, ymin to -90 and ymax to 0, and finally, ¥ % (23) 17. You can use 2 ‰ 7:Text to add magnitude and phase labels to the graphs. To do this, press 2 ‰ 7:Text and position the cursor where the text should start. The characters will appear below and to the right of the crosshairs. Be careful; once a letter is placed it can’t be erased except by 2 ‰ 2:Eraser. (24) 18. To return to a single screen, press 3 „ and set Split Screen to 1:FULL. Topic 15: Filter Design Overview A class of realizable frequency responses for lowpass filters has the form H(f) = 2 1 1 + ε Ψ 2 (f) 2 where Ψ(f) is a polynomial in f. If f Ψ(f ) = f p n the filter is a Butterworth filter. An alternative is to make Ψ(f) = Cn(f/fp) where Cn is a Chebyshev polynomial, the filter is a Chebyshev filter. The next topic deals with a Butterworth filter, the following topic with a Chebyshev filter. © 1999 TEXAS INSTRUMENTS INCORPORATED 04steady.doc . Karen Pressnell Revised: 01/25/99 9:25 AM Printed: 01/28/99 8:40 AM Page 38 of 14 CHAPTER 4: STEADY-STATE CIRCUIT ANALYSIS AND FILTER DESIGN 39 The design of lowpass filters requires specification of passband and stopband responses often given in dB. |H(f)|2 in dB is calculated as 1 | H ( f )|2 = 10 log dB 1 + ε2Ψ2 (f ) which becomes 1 2 | H ( f )|dB = 10 log 2n f 2 1+ ε f p for Butterworth and 1 2 | H ( f )|dB = 10 log 1 + ε 2C2 f n fP for Chebyshev. Topic 16: Butterworth Filter The performance specifications of a filter are often given in graphical form as shown in Figure 3. The design of a Butterworth filter with these performance specifications is described here. 1k 2k f (Hz) Passband -3dB -40dB |H(f)|2dB Stopband Figure 3. Filter design specifications for a Butterworth filter Suppose a filter with the maximum passband ripple is -3 dB, and the passband edge is at f p=1kHz is to be designed. Additionally, the stopband gain is to be no more than -40 dB with a stopband edge at fs=2kHz. © 1999 TEXAS INSTRUMENTS INCORPORATED 04steady.doc . Karen Pressnell Revised: 01/25/99 9:25 AM Printed: 01/28/99 8:40 AM Page 39 of 14 40 1. ELECTRICAL ENGINEERING APPLICATIONS WITH THE TI-89 From the Home screen, press ¥ 1 2 ˆ 2:NewProb to reset the TI-89 to a known state. Then enter the Butterworth equation as shown in screen 25. 10 p ½ log( 1 e c 1 « eps Z 2 p c f e fp d Z c 2n d d d § butter (25) 2. Press O 9:Numeric Solver exp=butter ¸ (screen 26). (26) 3. 4. Note that the Numeric Solver listed each of the variables for values to be entered. Find the value of eps by entering the data for the passband edge with a -3 dB response at 1000 Hz. n is unknown but at the passband edge all values give the same result, so for now enter 1 for n as shown in screen 27. (27) Place the cursor on the eps line and press „ to solve for eps. After a second or two, the screen shows eps is about 1 (screen 28). (28) 5. (5 Now, find the order of the filter by setting the stopband edge response (exp) to -40 dB and f to 2000 (screen 29). (29) 6. Solve for n by placing the cursor on n and pressing „. After a couple of seconds the solution of n=6.6 is shown, as in screen 30. (30) 7. Since n must be an integer, set n to the next larger integer value of 7 and solve for exp to find the stopband gain for this value of n (screen 31). With a 7th-order Butterworth filter the stopband gain is - 42 dB, a little better than the minimum needed. (31) © 1999 TEXAS INSTRUMENTS INCORPORATED 04steady.doc . Karen Pressnell Revised: 01/25/99 9:25 AM Printed: 01/28/99 8:40 AM Page 40 of 14 CHAPTER 4: STEADY-STATE CIRCUIT ANALYSIS AND FILTER DESIGN 8. 41 Now, plot the Butterworth equation to see the frequency response. To do this, press ¥ # and enter butter with f replaced by x (screen 32). (32) 9. Since the stopband edge of the filter is 2 kHz, plot x from 0 to 3000. The stopband value is -40 dB, so plot y from -45 to 0. Enter these values in the Window Editor (¥ $) as shown in screen 33. (33) 10. Press 3 to be sure the Graph mode is set to FUNCTION. Then press ¥ % and wait a few seconds to see screen 34. (34) The response in the passband looks very flat, which is correct for Butterworth, but are the passband and stopband edges in the right places? These can be checked graphically by pressing 2 ˆ 2:DrawFunc to draw horizontal lines at -3 and -40 dB and 2 ˆ 3:DrawInv to draw vertical lines at 1000 and 2000 Hz (screen 35). (35) 11. Press ¥ % to plot the results (screen 36). The curve passes through the -3 dB point at 1000 Hz and passes below the -40 dB point at 2000 Hz. The filter meets the required specifications. (36) © 1999 TEXAS INSTRUMENTS INCORPORATED 04steady.doc . Karen Pressnell Revised: 01/25/99 9:25 AM Printed: 01/28/99 8:40 AM Page 41 of 14 42 ELECTRICAL ENGINEERING APPLICATIONS WITH THE TI-89 Topic 17: Chebyshev Filter This section shows how to design a Chebyshev lowpass filter with the same specifications as discussed in Topic 16 and shown in Figure 4. 1k 2k f (Hz) Passband -3dB -40dB Stopband |H(f)|2dB Figure 4. Filter design specifications for a Chebyshev filter The Chebyshev equations are f f C n = cos n cos−1 f f p p f f C n = cosh n cosh −1 f f p p f <1 fp where n is the order of the polynomial. f >1 fp Therefore HdB is 2 H ( f ) dB 1 = 10 log f 22 1 + ε Cn f p Follow these steps to enter these three equations. 1. From the Home screen, clear f, fp, eps, and n using DelVar. 2. Note: DelVar can be entered by pressing †4:DelVar. Enter Cn for |f/fp| < 1 as shown in screen 37. 2 X n p ¥ R f e fp d d § cheb1 (37) © 1999 TEXAS INSTRUMENTS INCORPORATED 04steady.doc . Karen Pressnell Revised: 01/25/99 9:25 AM Printed: 01/28/99 8:40 AM Page 42 of 14 CHAPTER 4: STEADY-STATE CIRCUIT ANALYSIS AND FILTER DESIGN 3. 43 Enter Cn for |f/fp|>1 as shown in screen 38. ½ cosh( n p ½ cosh-1( f e fp d d § cheb2 (38) 4. Define cheb to be cheb1 for |f/fp|<1 and cheb2 for |f/fp|>1. This is done using the “when” function as shown in screen 39. ½ when( ½ abs( f e fp d 2 Ã 1 b cheb2 b cheb1 d § cheb (39) 5. Define hdb to be |H(f)|2dB as shown in screen 40. 10 ½ log( 1 e c 1 « eps Z 2 p cheb Z 2 d d § hdb (40) 6. To design the filter, use the Numeric Solver. Press O 9:Numeric Solver exp=hdb ¸ (screen 41). (41) 7. Proceed as with Butterworth. Find eps for -3dB at the passband edge by entering the values as shown in screen 42. The result is the same as with Butterworth. (42) 8. Now find n for -40 dB at 2000 Hz. It takes a few seconds to find the order of the filter (screen 43). (43) 9. To meet the design specifications, n must be 5. Enter 5 and calculate the response. See what the gain is (screen 44). This filter exceeds the design specifications by more than 10 dB with a lower order than the Butterworth. (44) © 1999 TEXAS INSTRUMENTS INCORPORATED 04steady.doc . Karen Pressnell Revised: 01/25/99 9:25 AM Printed: 01/28/99 8:40 AM Page 43 of 14 44 ELECTRICAL ENGINEERING APPLICATIONS WITH THE TI-89 10. In the Y= Editor, enter the expressions for the two filters to compare them (screen 45). Note that n=7 is added to y1 so that the Butterworth filter uses the order computed for it instead of using the value of n=5 used for the Chebyshev filter. (45) 11. Press ¥ % to see the graphs (screen 46). (See Topic 16 for a review of the instructions for a graphical comparison of the filters’ performances.) As expected, the Chebyshev filter has ripples in the passband, but it drops more rapidly in the transition band than the Butterworth. (46) 12. Compare the filters’ performances with the lines which represent the passband and stopband design specifications (screen 47). Press 2 ˆ 2:DrawFunc to draw horizontal lines at -3 and -40 dB. Press 2 ˆ 3:DrawInv to draw vertical lines at 1000 and 2000 Hz. Both filters show the required -3 dB response at 1000 Hz; both filters exceed the specifications since they are below -40 dB at 2000 Hz. (47) Topic 18: Logarithmic Frequency Plots Often frequency responses are plotted on a log frequency scale. Although the TI-89 doesn’t directly support log plots, they are easy to do. 1. Return to the Y= Editor and alter the “with” operation to include the logarithmic relation of frequency with x as shown in screen 48. y1: butter Í f Á 10 Z x ½ and n Á 7 y2: hdb Í f Á 10 Z x (48) The values for x are linearly spaced, but the values of 10x are logarithmically spaced. 2. Press ¥ $ to adjust the range on x. Graph the functions for f=500 and f=3000. To do this, enter log(500) for xmin and log(3000) for xmax as shown in screen 49. (49) 3. Press ¥ % to display screen 50. The logarithmic plots take on a different appearance than the linear plots of screen 46. (50) © 1999 TEXAS INSTRUMENTS INCORPORATED 04steady.doc . Karen Pressnell Revised: 01/25/99 9:25 AM Printed: 01/28/99 8:40 AM Page 44 of 14 CHAPTER 4: STEADY-STATE CIRCUIT ANALYSIS AND FILTER DESIGN 4. 45 The passband and stopband lines are added by using log() of the f values in DrawInv (screen 51). Press 2 ˆ 2:DrawFunc for -3 and -40 dB. Press 2 ˆ 3:DrawInv for log(1000) and log(2000) Hz. (51) As expected, the plots in screen 52 show that both filters meet the design specifications. (52) Tips and Generalizations Topic 13 show how to enter, solve, and display equations with complex numbers. Multiple equations with multiple complex unknowns can be solved. Also the matrix approach of Topic 3 can be used with complex numbers. The Numeric Solver works nicely for filter design but can easily solve any equation for an unknown value. Once a steady-state response is known, the power dissipated by the various elements can be found. Chapter 5 explores this topic further. © 1999 TEXAS INSTRUMENTS INCORPORATED 04steady.doc . Karen Pressnell Revised: 01/25/99 9:25 AM Printed: 01/28/99 8:40 AM Page 45 of 14 ...
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