PHOTOGRAPHIC COMMUNICATION J 316
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Fundamentals of Electrical Engineering 1st Edition Giorgio Rizzoni Errata Corrige for first printing of 1 Edition Revision 1 August 315 2008 Rev 1 August 31 2008 Note from the Author Dear students or instructors First thank you for using this book This document contains a list of typographical errors misprints and the occasional technical error I am grateful to Dr Ralph Tanner of Western Michigan University for uncovering the majority of these errata andl include a note prepared by him becausel especially liked his suggestion directed at the students I would be grateful if any user who thinks that they might have uncovered an error would be kind enough to send a message with a detailed description of the suggested correction to me at rizzonil osuedu Again thank you for adopting this textbook Giorgio Rizzoni Columbus august 2008 Note by Dr Ralph Tanner Western Michigan University There are a dozen pages of errata contained here At first blush this might be considered excessive However the errors are generally minor and are of the type that the knowledgeable reader would tend to self correct when reading the material But since this is a text intended for the student these errors can cause selfdoubt and can get in the way of learning the material The student may tend to believe that the error is in their understanding rather than in the printing of the book For this reason this errata file has been very picky I would recommend that the student adopt one of two strategies with this errata file 1 Go through the entire errata file and mark the changes in the book 2 Go through the entire errata file and put a prominent red dot on each page where an erratum occurs I believe the first strategy is the best This allows the student to have the correct information at the point when it is needed However the second strategy will alert the student to a possible point of confusion If a topic on the page in question causes confusion the student can make the change at that time If the topic causing confusion was in error the student will have the corrected material If the topic causing confusion was not the one in error the student will know to keep working to understand the area of confusion Whatl do NOT recommend is the filing of the errata then going to it only when confusion arises In my classes I teach the analysis of circuits 1 also teach the engineering methods used to reduce the chance of errors Engineering is a field where errors may cause the loss of life Another is medicine Because of this grave responsibility we need to learn how to minimize errors One of the methods used to minimize errors is multiple cumulative review This errata file represents the start of this process I am certain that T have not found all of the errors in this text Rev 1 August 31 2008 2 Chapter 1 No errata Rev 1 August 31 2008 Chapter 2 Fundamentals of Electric Circuits p 29 The last sentence of the first paragraph should read An example of this dual behavior is exhibited by the photodiode which can act either in a passive mode as a light sensor or in an active mode as a solar cell p 44 CHECK YOUR UNDERSTANDING The first sentence should read Compute the fullscale ie largest output voltage for the forcemeasuring apparatus of Example 216 p 77 Figure 318 VI in the figure for analysis for mesh 2 does not enter into the analysis for mesh 2 so it should not be on that figure delete v1 7 Rev 1 August 31 2008 4 Chapter 3 Resistive Network Analysis p 81 Example 310SolutionArm1ysir We follow the Focus on Measurement should read We follow the Focus on Methodology p 83 on the second line transistors are introduced in Chapter 9 should read transistors are introduced in Chapter 10 p 84 Figure 325 should be replaced by the one shown below Figure 325 p 84 a subscript is missing 21 14 CHECK YOUR UNDERSTANDING Answer should be v 3 V v3 H V p119 Figure P320 The resistor with the missing Value should have a Value of4Q Rev 1 August 31 2008 Chapter 4 AC Network Analysis p 133 Example 41 Solution Find Charge separation at nominal voltage should read Find Charge stored at nominal voltage p 133 Example 41 Solution The line that reads Since we know that the discharge current is 25 A and the available charge separation is 250 F should read Since we know that the discharge current is 25 A and the available charge separation is 250 C p 134 CHECK YOUR UNDERSTANDING Should read Compare the charge storage achieved in this ultracapacitor with a similarly sized electrolytic capacitor used in power electronics applications by calculating the charge storage for a 2000 pF electrolytic capacitor rated at 400 V p 138 CHECK YOUR UNDERSTANDING Should read Compare the energy stored in this ultracapacitor with a similarly sized electrolytic capacitor used in power electronics applications by calculating the energy stored in a 2000 pF electrolytic capacitor rated at 400 V p 146 Under Why Sinusoids regarding the analysis of electric power circuits Note that the methods should read regarding the analysis of electric power circuits The more ambitious reader may wish to explore Fourier Analysis on the web to obtain a more comprehensive explanation of the importance of sinusoidal signals Note that the methods p 148 Equation 427 should read T T W TPAV Tpl I pl39dl39 I Rig t dt le R 0 0 p 149 Equation 428 should read 1 T 17 1512rdz Rev 1 August 31 2008 6 Chapter 5 Transient Analysis p 182 last line on the page By analogy with equation 58 we call should read By analogy with equation 57 we call p 184 Section 53 last line of introductory paragraph using the principle of continuity of inductor voltage and current should read using the principle of continuity of capacitor voltage and inductor current p 191 FOCUS ON METHODOLOGY Step 2 vC vC 7 should read VC 0 vC 7 p 195 Example 57 Solution Step 4 with reference to equation 522 should read with reference to equation 524 p 197 Figure 519 The vertical scale on the graph should be labeled Ratio of capacitor voltage to source voltage vC VB and the horizontal scale should be labeled Time constants RC p 197 Figure 520 The direction of iL should be counterclockwise The vertical scale on the graph should be labeled Ratio of Inductor current to source current 139 L IB and the horizontal scale should be labeled Time constants L4 p 202 two paragraphs above the beginning of Example 510 with reference to equation 525 should read with reference to equation 524 p 204 Example 510 Solution Step 2 equation 525 should read equation 522 p 204 Example 510 Solution Step 4 with reference to equation 522 should read with reference to equation 524 p 210 below Equation 549 Rev 1 August 31 2008 7 in equation 548 should read in equation 549 p 211 five lines under Figure 539 in the differential equation 548 should read in the differential equation 549 p 214 Example 513 Solution Analysis This equation is in the form of equation 550 with KS 1 mi 1LC should re ad This equation is in the form of equation 550 with 7750 1LC p 214 Example 513 Solution Analysis by inspection that K 1 a 11IL should re ad by inspection that 77a 11LC p 214 Example 513 Solution Analysis we have 77312 7125ij3162 should read we have 77312 7126ij3160 The second line of the equation immediately below this should then be D e l 6J316 02 0626912 67316 02 p 216 Five lines below Equation 559 the exponential decay term 20416441 should read the exponential decay 2oqe 4 d t Rev 1 August 31 2008 p 218 FOCUS ON METHODOLOGY Step 2 iLoiLoi should read iL0iLO39 p 218 FOCUS ON METHODOLOGY Step 3 equation 59 or 549 should read equation 513 or 549 p 219 Step 2 continuity of inductor voltage and capacitor currents should read continuity of inductor current and capacitor voltage p 221 Example 515 Solution Find the differential equation in 139 L I describing should re ad the differential equation in vC ldescribing p221 Example 515 Solution Schematics 39 R 500 Q should read R RS 1000 Q p 222 Example 515 Solution Step 2 continuity of inductor voltage and capacitor currents should read continuity of inductor current and capacitor voltage p 224 Example 516 Solution Step 2 continuity of inductor voltage and capacitor currents should re ad continuity of inductor current and capacitor voltage Rev 1 August 31 2008 p 229 The text neglects to divide the voltage drop across the resistor by the inductance This causes errors in 11 and 1 at the bottom of the page and in the voltage across the spark plug on the following page The student should trace through these errors to obtain the correct solution 8192 V Note that there is also one additional sign error in the final computation by the text of the initial voltage across the spark plug 1 have not finished correcting this last error I Will have to retype several equations and recalculate a lot of things Rev 1 August 31 2008 10 Chapter 6 Frequency Response and System Concepts p 254 The equation at the bottom of the page is missing the equal sign between H j39wand the fraction p 255 around the frequency of 300 rads the magnitude should read around the frequency of 800 rads the magnitude Rev 1 August 31 2008 Chapter 7 AC Power p 280 three lines below Equation 71 with phase angle 1 7r6 77and should re ad with phase angle 771 0 and p 290 The equation at the bottom of the page 866j50 W should read 866j50 VA p 291 The equation at the bottom of the page 11927139316 W should read 11927139316 VA p 293 Example 76 Solution Analysis part 1 The equation that reads 77503j839 W should read 77503j839 VA p 297 The equation at the bottom of the page 684j1186 W should re ad 684 j1186 VA Rev 1 August 31 2008 12 p 302 Example 71 Solution Analysis middle of the page where we have selected the positive value of arccos pfl because pfl is lagging and the negative value of arccos I pr should re ad where we have selected the negative value of arccos pr because pfl is lagging and the positive value of arccos 2 p 302 Example 71 Solution Analysis seven lines above the last equation on the page where once again should read where now p 303 Section 73 introductory paragraph last line discussed in Chapter 14 should read discussed in Chapter 13 p 304 seven lines below equation 732 as explained in Chapter 14 should read as explained in Chapter 13 p 306 The equation at the end of the example that reads N 7 500 W 1Wmy m 15625 A should read 7777f M 15625 A 39 primary 4 800 V p 306 CHECK YOUR UNDERSTANDING The answer should be n1 6000 p 313 Section 74 second line of second paragraph as will be explained in Chapter 15 should read as explained in Chapter 14 Rev 1 August 31 2008 13 p 316 CHECK YOUR UNDERSTANDING The second answer should be 7 SL 6912 kW j2074 kVA p 318 Equation 764 1 y should be iny p 326 Conclusion 2 that for which the user is charged should be changed to that which does work for the user A sentence should be added after this sentence which would read However the user is charged for all of the power supplied by the utility company both real and reactive p 326 Conclusion 2 and it can be minimized by should be replaced by and the reactive power can minimized by Rev 1 August 31 2008 14 Chapter 8 Operational Amplifiers p 356 five lines above Figure 814 which is discussed in Section 86 should read which is discussed in Section 85 p 361 third line from the top The second amplifier and inverting should read The second amplifier an inverting p 362 twelve lines below Figure 8 l 8 ranging from 1 to 50 should read ranging from 1 to 10 p 363 Practical OpAmp Design Considerations box 1 see Table 21 should read see Table 22 p 363 Practical OpAmp Design Considerations box 1 inspection of Table 21 reveals should read inspection of Table 22 reveals p 363 Practical OpAmp Design Considerations box 2 as explained in Section 86 should read as explained in Section 85 p 367 line immediately above Figure 822 filter response for frequencies is significantly higher should read filter response for frequencies significantly higher p 370 The last equation on the page which begins Rev 1 August 31 2008 15 R1 ij R11jagt Rlij R1 we should read ZS R1 ij R11jagt R11 R1 me p 371 Figure 828 The dashed lines in the two diagrams refer to the circuit in Figure 820 The solid lines in the two diagrams refer to the circuit in Figure 827 Example 87 p 371 CHECK YOUR UNDERSTANDING for the filter of Example 86 at the cutoff should read for the filter of Example 87 at the cutoff p 372 the analog computer which is discussed in Section 85 Example 88 illustrates should read the analog computer Example 88 illustrates Rev 1 August 31 2008 16 Chapter 9 Semiconductors and Diodes p 408 Learning Objectives 4 Section 94 should read Sections 94 95 p 426 Example 95 Solution Analysis the equation R7 R1 R2 R3 HR4 should read RT R3 R4 R1HR2 p 433 second line below Figure 942 shown in Figure 943a for the case should read shown in Figure 943ab for the case p 433 third line below Figure 942 Figure 943b depicts should read Figure 943c depicts p 433 fourth line below Figure 942 waveform ofFigure 943b is not should read waveform ofFigure 943c is not p 434 Example 98 Problem similar to that in Figure 925 is used should read similar to that in Figure 929 is used p 436 CHECK YOUR UNDERSTANDING rectifier ofFigure 940 assuming should read rectifier ofFigure 94l assuming p 442 Example 912 Solution Analysis the equation rZ HR Rs should re ad Rev 1 August 31 2008 17 VZ RS HRL RSHRLVZ p 448 Problem 936 that ofFigure 925 in the text should re ad that of Figure 929 in the text p 448 Problem 937 that ofFigure 925 in the text should re ad that of Figure 929 in the text Rev 1 August 31 2008 Chapter 10 Bipolar Junction Transistors p 460 third equation from the top PVC V3 8 V should read VCV34V This changes the following equations Vccchil2i8 Vccchil2i4 77 IC 7 4 mA should read quotIC 7 8 mA RC 1000 RC 1000 a 1c 10 1c 1c s ou rea 80 80 80 h 1d d 7777 160 18 IE 18 IE and VCE VC 7 V5 8713 67 Vquot should read 7quotch VC 7 V5 4713 27 Vquot to values which now match Figures 108 and 109 p 465 fourth line from top Thus the diode operating at a higher temperature should read Thus the transistor operating at a higher temperature p 472 Example 107 Solution AnalysisComments VCEQ 7 v ICQ 22 mA IBQ 150 yA should re ad 7ch 6 v ICQ 25 mA IBQ 150 uA Rev 1 August 31 2008 19 This lead to changing 7ch VCEQ gtlt ICQ 154 mw to 7ch VCEQ gtlt ICQ 150 mW p 478 CHECK YOUR UNDERSTANDING 1 kQ 3100 and VCC 14 V should read 1 kQ 100 and VCC 14 V p 479 Eighth line from top if both v7 and va are 0 should read if both v7 and v1 are 0 p 481 Example 1010 Solution Analysis 2 that diode D1 is still forwardbiased but diode D2 is now should read that diode D2 is still forwardbiased but diode D1 is now p 481 Example 1010 Solution Analysis 4 77V 0 m VCC ilcR3 should read V 0 u VCEsat 02 V p 487 Figure P1036 There should be a dot at the intersection of the circuit paths joining R1 R2 C1 and the transistor to indicate that the two crossing paths are connecte p 487 Figure P1037 The transistor figure is incorrectly drawn the emitter symbol should be shifted to a lower position Rev 1 August 31 2008 20 Chapter 11 Field Effect Transistors p 499 Example 113 Solution Analysis iDQ KVGS 7 VT2 48524 714 485 mA should re ad mDQ Kvssi VT2 485gtlt10 324714 485 mA p 504 Equation 1111 The second equation of 1111 is missing a term However the whole second equation may be eliminated because it is duplicated with the missing term present in the first equation of 1112 p 506 4 lines above Figure 1114 VDD 7 VB lt VG 7 V7 so actually should read so actually p 512 Example 118 Solution Analysis VGSQ 0 V should read VGSQ 25 V p 513 Example 119 Solution Analysis The analysis needs substantial revision When an input is off low voltage the corresponding NMOS will be off and the corresponding PMOS will be on When an input is high the corresponding NMOS will be on and the corresponding PMOS will be off Therefore the analysis for case 1 v1 v2 0 should be that M1 and M2 are off that and M3 and M are on as is indicated in Figure 1121 This will result in v0 5 V The analysis for case 2 v1 5 V and v2 0 V should be that v1 turns M1 on and M3 off Since v2 still is 0 M2 remains off and M4 remains on This will result in v0 0 V There is no figure that corresponds to this configuration shown The analysis for case 3 v1 0 V and v2 5 V should be that v1 leaves M1 off and M3 on as it did in case 1 Since v is now 5 V M2 will be on and M will be off This will result in v0 0 V There is no figure that corresponds to this configuration shown The analysis for case 4 is correct and corresponds to Figure 1122c Figures 1122a and 1122b can not be created using the inputs available to Figure 1120 In the table when v1 v2 0 V M3 will be Off39 when v1 0 V and v2 5 V M will be On Otherwise the table is correct Note this will have to be essentially rewritten if the above comments are correct For now 1 am moving forward Rev 1 August 31 2008 21 Chapter 12 Digital Logic Circuits p 522 Learning Objectives 7 Section 127 should read Section 126 p 535 Example 123 Solution Analysis last line on the page Rules 2 and 6 should read Rules 2 6 and 14 p 546 three lines below Figure 1231 fourvariable map of Figure 1229 are shown in Figure 1232 should read fourvariable map of Figure 1231 are shown in Figure 1232 p 547 The second line of the only equation on the page W27YZWXYZW XYZW firZ should read W27i7ZWXi7ZW Xi7ZW 47472 P 555 Example 1219 Problem Find a minimum productofsums realization should re ad Find a minimum sumof products realization p 556 Example 1220 Solution Analysis By appropriately selecting three of the four don t care entries should re ad By appropriately selecting four of the five don t care entries p 564 last line on the page whenever the enable input is high the flipflop is set should re ad Rev 1 August 31 2008 22 whenever the enable input is high the ip op is set to the value of 5quot p 565 Figure 1264b This is not an error but the diagram could be improved to provide a more effective explanation the Preset and Clear are on for most of the timing diagram Therefore the timing of the enable is not effectively shown p 565 Figure 1264c This is the schematic for a FE quad latch latch is another name for a ipflop rather than the RS latch described in parts a and b p 569 Example 1233 Solution Find Output of RS flipflop should read Output of JK flipflop p 574 Example 1225 Analysis The analysis section assumes without explicitly stating that the Preset and Clear are only effective on a clock 21215 Although this doesn t correlate to statements earlier in the text it is correct for many commercially available Note we will need to review earlier statements in the text to find the claim ed inconsistency Rev 1 August 31 2008 23 Chapter 13 Principles of Electromechanics p 597 line below equation 1325 In equation 1324 I represents the length of the coil wire39 should read In equation 1324 I represents the length of the core39 p 604 Figure 1319 The length of the gap should be 00025 m p 605 Example 133 Solution Analysis 3 Calculation of reluctance In the calculation of Rm the final value should be 255 instead of 398 In the calculation of Kq the final value should be 257 instead of 4 In the calculation of 5 the final value should be 392 instead of 251 p 620 Example 139 Solution Analysis The force we must overcome is mg 98 N should re ad The force we must overcome is mg 49 N p 620 Example 139 Solution Analysis Z u0A 2x l1lzj 2 A A 14me ON0 65gtlt104A i255A should read 2 04 2x 11 014 do 4 N2 p 620 Example 139 Solution Analysis 14me 52gtlt105A i721A Rev 1 August 31 2008 24 11EA112 2 2 A 2 fmw 21056x10 3 A N x 0 0459 A shouldread 1 1 2 44H 2 1119 es 1 fmwTA 21x10 A x 0 0649 A p 622 Example 13 10 501mm Aha1ysrs The ha1 answer should be 178 Artums rhstead of 56 4 Artums P 625 Practical facts ahnut snhsruuds texthextto Frgure 13 38 bypassmg the resrstor through the NC swrteh connecting the resrstor 1h sehes shouldread L V L nu Nu m resrstor 1h sehes p 625 Frgure 13 3915notcorrect Please refer to the following gure ustt1muhn Marl p 626 Frgure 1342 In order the reqmred eurreht wru be approxrrhate1y 200 mA Note that 1h the example the value x 0 05 rh 5 cm1s used p 630 text to the rrght of Frgure 13 44 The 1311 law just r11ustrated quot should read The 13111 lzwjust r11ustrated p 630 Equauor 13 62 Rev 1 August 31 2008 25 PM ifmu Bliu should read PM fmu Bliu p 641 Figure P1341 The electrical portion of this figure is incorrect See the correction to Figure 1339 above Rev 1 August 31 2008 Chapter 14 Introduction to Electric Machines p 652 Example 141 Solution Analysis The equation SRMX 1800 shouldread SRMX 1760 p 658 Figure 1411 N and S are reversed p 658 in the paragraph starting with Often The north and south poles indicated in the figure are a consequence of the fact that the flux exits the bottom part of the structure thus the north pole indicated in the figure and enters the top half of the structure thus the south pole should read The north and south poles indicated in the figure are a consequence of the fact that the flux exits the top part of the structure thus the north pole indicated in the figure and enters the bottom half of the structure thus the south pole p 679 Example 149 Solution Analysis dwltrgt 7K I I J TPM a4r dt 1 501 Tm I should read AKTYPMIH IJd6TEIbwI 410010 p 679 The equation 717pr Ia s SJ bQs T10 s should read 717pr Ia s SJ bQs 7T105ds p 679 The equation BL R KHYPM 1 2 VLSJ i KTfM S 17 Tload S should re ad Rev 1 August 31 2008 27 SLH R 10pr 1 I s QM 717va sJb p 679 The equation d rL R KTPM e SL0 Ra 7K T PM Qm s should re ad dawn R Q S KTPM m SL0 Ra de 7K TPM p 680 the equation at the top of the page should have a minus sign in front of Tloads Kn PM sJb s Kn PM sJb p 695 Example 1414 Problem Figures 1439 to 1441 should read Figures 1438 to 1440 Rev 1 August 31 2008