Chapter 2 Linear Waveshaping: High-pass Circuits
|
|
- Teresa Nash
- 6 years ago
- Views:
Transcription
1 Puls and Digital Circuits nkata Ra K., Rama Sudha K. and Manmadha Ra G. Chaptr 2 Linar Wavshaping: High-pass Circuits. A ramp shwn in Fig.2p. is applid t a high-pass circuit. Draw t scal th utput wavfrm fr th cass: (i) T =, (ii) T = 0.2, (iii) T = 5. Fig.2p. A ramp as input Slutin: Frm Eq. (2.64): / v = t T v = t T Th pak f th utput will ccur at t = T. W knw: v (pak) = T (i) Whn T = : = T v (pak) = () = (ii) Whn T = 0.2 : T = 0.2, = 5 T 0.2 v (pak) = (5) = (iii) Whn T = 5 : / / T 200 Drling Kindrsly India Pvt. Ltd
2 Puls and Digital Circuits nkata Ra K., Rama Sudha K. and Manmadha Ra G. T = 5 = 0.2 T 5 v (pak) = (0.2) = Fig. Rspns f th high-pass circuit fr ramp input 2. A wavfrm shwn in Fig.2p.2 is applid as input t an high-pass circuit whs tim cnstant is 250 ps. If th maximum utput vltag acrss th rsistr is 50, what is th pak valu f th input wavfrm? Fig.2p.2 Input t th high-pass circuit Slutin: Fr a ramp input v () t ( ) t At t t s v i vi v (max) 000 i Drling Kindrsly India Pvt. Ltd 2
3 Puls and Digital Circuits nkata Ra K., Rama Sudha K. and Manmadha Ra G. Fr t t v 50 t v v A limitd ramp shwn in Fig.2p.3 is applid t an high-pass circuit f Fig.2.2 (a). Th tim cnstant f th circuit is 2 ms. Calculat th maximum valu f utput vltag and th utput at th nd f th input wavfrm. Fig.2p.3 Input t th high-pass circuit Slutin: Fr a ramp input t v ( t) ( ) 3 At t = t = 0.40 s v ( t) 20 ( ) Th pak valu ccurs nly at t = t = 0.40 v (max) Fr t > t t 3 20 v Th vltag at t = 0.4 ms is: 3 s 200 Drling Kindrsly India Pvt. Ltd 3
4 Puls and Digital Circuits nkata Ra K., Rama Sudha K. and Manmadha Ra G. v v 0.06 Th vltag at t = 0.4 ms is Th utput wavfrm is shwn in Fig.3. Fig. 3 Output f th high-pass circuit fr th givn input 4. Th pridic wavfrm shwn in Fig.2p.4 is applid t an diffrntiating circuit whs tim cnstant is 0 µs. Sktch th utput and calculat th maximum and minimum valus f th utput vltag with rspct t th grund. Fig.2p.4 Pridic squar wav as an input t th high-pass circuit Slutin: Givn T =00 s, T 2 = s, τ = 0 s Th stady-stat utput wavfrm is drawn by calculating,, 2 and 2. At t 0, v 2 andatt 0, v t 0 t T, v Fr At t = T Fr At 200 Drling Kindrsly India Pvt. Ltd 4
5 Puls and Digital Circuits nkata Ra K., Rama Sudha K. and Manmadha Ra G. T 00 0 At t T, v 0 Fr T t ( T T ), v 2 2 T2 T At t T, v Pak-t-pak input is = 00 2 = = T T Fig. 4 Output f th high-pass circuit fr th spcifid input 5. Th pridic ramp vltag as shwn in Fig.2p.5 is applid t a high-pass circuit. Find quatins frm which t dtrmin th stady-stat utput wavfrm whn T =T 2 =. Fig.5 A pridic ramp as input Slutin: / v = t () 200 Drling Kindrsly India Pvt. Ltd 5
6 Puls and Digital Circuits nkata Ra K., Rama Sudha K. and Manmadha Ra G. If thr is an initial vltag f n C, Eq.() gts mdifid as fllws: Fr th ramp input, th slp = t / t v =. T Th capacitr chargs frm t 2 in tim T. At t = T +, th capacitr ds nt rspnd fr suddn changs. Hnc, th utput changs t ( 2 ). During T 2, th capacitr blcks th dc. S th capacitr dischargs frm 3 t. Givn T = T 2 = At t = T v () t = 2 Using () T ( ) (2) = T v T ) = 2 ( T2 r ( ) ( ) ( 0.34) = v( T)= 2 = = Drling Kindrsly India Pvt. Ltd 6
7 Puls and Digital Circuits nkata Ra K., Rama Sudha K. and Manmadha Ra G. Fig. 5 Th changs in vltag with tim 6. A squar wav f puls width 2 ms and pak amplitud f 2 as shwn in Fig.2p.6 is applid t high-pass circuit with tim cnstant 4 ms. Plt th first fur cycls f th utput wavfrm. T/2 = 2 ms Fig.2p.6 Symmtric squar wav as an input Slutin: Givn T = T 2 = 0.2 ms, τ = 4 ms (i) Fr t <0, vi = 0, and hnc v = 0 At t = 0, v i jumps t 2 As th vltag acrss capacitr cannt chang instantanusly, v is als qual t 2. At t = 0, v = a = 2. (ii) During th prid 0 < t < 2.0 ms, as th input is cnstant th utput dcays. v = a t t 2 At t = 2.0 ms, v = b = a = 2 4 = 7.27 At t = 2 ms, th input falls by 2. Th utput als falls by 2. = 2 = = 4.73 c b (iii) Fr 2.0 < t < 4.0 v = c ( tt / 2) At t = T=4 ms, v = d = c 4.73 = 2.86 At t = 4 ms, th input riss by 2. Th utput als riss by 2. = d + 00 = = 9.4 (iv) During th prid T < t < 3T/2, that is btwn 4 t 6 ms, th utput dcays. 2 4 At t = 6 ms v = f = = 9.4(0.606) = 5.53 At 6 ms, th input falls by 2. Hnc g = f 2 = 6.47 (v) During 3T/2 < t < 2T, that is, during 6 t 8 ms, th utput dcays At t = 2T=8 ms, v = h = g = 6.47 = j = h + 00 = = Drling Kindrsly India Pvt. Ltd 7
8 Puls and Digital Circuits nkata Ra K., Rama Sudha K. and Manmadha Ra G. Fig.6 Th utput wavfrm 7. A 20-Hz symmtric squar wav rfrncd t 0 vlts and, with a pak-t-pak amplitud f 0, is fd t an amplifir thrugh th cupling ntwrk shwn in Fig. 2p.7. Calculat and plt th utput wavfrm whn th lwr 3-dB frquncy is: (i) 0.6 Hz, (ii) 6 Hz and (iii) 60 Hz. Fig.2p.7 Th givn cupling ntwrk Slutin: Givn = 0 (i) f = 0.6 Hz = = s 2 f 2 (0.6) T = = = 0.05 s. f 20 T = s. 2 = T / 2 0 = / = = T = = 5.25(0.9) = Drling Kindrsly India Pvt. Ltd 8
9 Puls and Digital Circuits nkata Ra K., Rama Sudha K. and Manmadha Ra G. = 2 and = 2 = 5.25 = 2 = 4.8 = 2 Th utput in this cas is plttd in Fig.7.. (ii) f = 6 Hz Fig.7.Output whn f = 0.6 Hz = = s 2 f 2 (6) = T / 2 0 = / = 7.20 = T = = 7.20(0.389) = 2.8. = 2 and = 2 =7.20 = 2 = 2.8 = 2 Th utput fr this cnditin is plttd in Fig Drling Kindrsly India Pvt. Ltd 9
10 Puls and Digital Circuits nkata Ra K., Rama Sudha K. and Manmadha Ra G. (iii) f = 60 Hz Fig.7.2 Output whn f = 6 Hz = = s 2 f 2 (60) = T / 2 = / =0.0 = T 2 =0( ) = = 2 and = 2 =0.00 = 2 = = 2 Th utput fr this cas is plttd in Fig Drling Kindrsly India Pvt. Ltd 0
11 Puls and Digital Circuits nkata Ra K., Rama Sudha K. and Manmadha Ra G. Fig.7.3 Output whn f = 60 Hz 8. A squar wav is applid as input t an amplifir thrugh a cupling cndnsr f 0 F. Th amplifir has input rsistanc f 0 kω. Dtrmin th lwst frquncy if th tilt is nt t xcd 0 pr cnt. Slutin: W hav P = 0., R = 0 kω and C = 0 F T Pr cnt tilt, P = 00 pr cnt 2 f 50 Hz P f P = 00 pr cnt f f Pf Hz 9. A puls f 0 amplitud and duratin ms is applid t a high-pass circuit with R = 20 k and C = 0.5 F. Plt th utput wavfrm t scal and calculat th pr cnt tilt in th utput. Slutin: 200 Drling Kindrsly India Pvt. Ltd
12 Puls and Digital Circuits nkata Ra K., Rama Sudha K. and Manmadha Ra G. τ = = 0 ms Fr 0< t < t p v i =0 v =0 t At t = t p, v = At t = t p+, v = 2 = 3 t = = 0.95 Fr t > t p, v = pr cnt tilt pr cnt 0 Fig. 9 Th utput wavfrm 0. Th input t th high-pass circuit in Fig. 2p.0 is th wavfrm shwn in Fig. 2p.0. Calculat and plt th utput wavfrm t scal, givn that = τ = 0. ms.. Fig.2p.0 Input t th high-pass circuit Slutin: 200 Drling Kindrsly India Pvt. Ltd 2
13 Puls and Digital Circuits nkata Ra K., Rama Sudha K. and Manmadha Ra G. Fr t < 0. ms v = 0, v = 0 (i) At t = 0. ms, th input suddnly falls t 5, and th utput als changs by th sam amunt as th capacitr acts as a shrt circuit. Fr 0. < t <0.2, v i rmains cnstant at 5. Thrf r, v dcays xpnntially with th tim cnstant 0. ms. (ii) At t = 0.2 ms, i v At t = 0.2 ms, th input suddnly riss by 5, v als riss by th sam amunt. v (t = 0.2 ms) = = 3.6 Fr 0.2 ms < t < 0.3 ms, v i rmains at 0. Hnc v dcays xpnntially with th tim cnstant 0. ms (iii) At t = 0.3 ms v At t = 0.3 ms, input suddnly falls by 20. Th utput als changs by th sam amunt. v (t = 0.3 ms) = = 5.6 Fr 0.3 ms < t < 0.4 ms, v i rmains cn stant at 0. Hnc, v will dcay xpnntially with th tim cnstant 0. ms. (iv) At t = 0.4 ms, v Drling Kindrsly India Pvt. Ltd 3
14 Puls and Digital Circuits nkata Ra K., Rama Sudha K. and Manmadha Ra G. Fig.0 Th utput wavfrm. A puls f 0- amplitud with a puls width f 0.5 ms, as shwn in Fig.2p.9, is applid t a high-pass circuit f Fig. 2.(a), having tim cnstant 0 ms. Sktch th utput wavfrm and dtrmin th pr cnt tilt in th utput. Slutin: 0 ms Fr, 0< t < t p v i =0 =0 t 00 v At t = t p-, v = At t = t p+, v = 2 = = = t Fr t > t p, v = pr cnt tilt pr cnt 0 Th utput is als shwn in Fig. 2p.9. Fig.2p.9 Input and utput f th high-pass circuit 2. A high-pass circuit is dsird t pass a 3-ms swp (ramp input) with lss than 0.4 pr cnt transmissin rrr. Calculat th highst pssibl valu f th lwr 3-dB frquncy. Slutin: Cnsidr th circuit in Fig. 2.(a). T = s pr cnt t(max) =0.4 pr cnt r t(max) = Drling Kindrsly India Pvt. Ltd 4
15 Puls and Digital Circuits nkata Ra K., Rama Sudha K. and Manmadha Ra G. T t ft f A symmtric squar wav with f = 500 khz shwn f Hz in Fig.2p.3 is fd t an high-pass ntwrk f 3 30 Fig.2.(a). Calculat and plt th transint and th stady-stat rspns if: (i) τ = 5T and (ii) τ = T/20. Fig.2p.3 Input t th cupling ntwrk Slutin: Givn f = 500 Hz, hnc T = 2.0 ms. Cas : Givn, τ = 5T =0 ms. Whn τ is larg, th capacitr chargs and dischargs vry slwly. Th utput has a small tilt. Th vltags ar calculatd t plt th transint rspns. i. Fr t < 0, v = 0, and hnc v = 0 At t = 0, jumps t 50. As th vltag acrss capacitr cann t chang instantanusly, v is als qual t 50. At t = 0 v = a = 50. ii. During th prid 0 < t <.0 ms, as th input is cnstant th utput dcays. v = a v i t i t At t =.0 ms, v = b = a = 50 0 = At t =.0 ms, th input falls by 00. Th utput als falls by 00. = 00 = = c b iii. Fr.0 < t < 2.0, v = c ( tt / 2) At t = T = 2 ms, v = d = c = At t = 2 ms, th input riss by 00. Th utput als riss by 00. = d + 00 = = Drling Kindrsly India Pvt. Ltd 5
16 Puls and Digital Circuits nkata Ra K., Rama Sudha K. and Manmadha Ra G. iv. During th prid T < t < 3T/2, that is, btwn 2 t 3 ms, th utput dcays..0 0 At t = 3 ms v = f At 3 ms, th input falls by 00. Hnc g = 00 = 9.73 f = = (0.9048) = v. During 3T /2 < t < 2T, that is, d uring 3 t 4 ms, th utput dcays At t = 2T= 4 ms, v = h = g = 9.73 =7.85. j = h + 00 = =7.85. In a fw cycls, th utput rachs th stady stat. Stady-stat rspns: Undr stady stat, th utput is symmtrical with rspct t zr vlts, sinc th capacitr blcks dc. Thrfr, th dc cmpnnt in th utput is zr. Lt b th vltag at t = 0 t Fr 0 < t < T/2, v = 0. At t = T/2= ms, v = = = = (3) As th input abruptly falls, utput als falls by th sam amunt t 2. (tt / 2) Fr T/2 < t < T v = 2 0. At t = T, v = 2 = 2 = = (4) Fr symmtrical wav = 2 and = 2 (5) 2 = 00 and 2 = 00 (6) Frm (6), w hav 2 = 00 (7) And frm (3), w hav = 2 (8) Substituting (8) in (7), w hav + = 00 (9) Frm (3), w hav = Substituting in (9) = = 00. =52.49 and = = (0.905)(52.49)= Frm (5) as = 2 an d = 2 2 = = W can nw plt th stady-stat rspns as w knw = = Drling Kindrsly India Pvt. Ltd 6
17 Puls and Digital Circuits nkata Ra K., Rama Sudha K. and Manmadha Ra G. 2 = = Th transint and stady-stat rspnss ar plttd in Figs.3. and 3.2. Fi g.3. Transint rspns Fig.3.2 Stady-stat rspns Cas 2: Fr vry lw tim cnstant, i.. whn τ = T/20 = 0. ms. Sinc th tim cnstant is vry small, th capacitr chargs and dischargs vry fast. Th input and utput ar shwn in Fig Drling Kindrsly India Pvt. Ltd 7
18 Puls and Digital Circuits nkata Ra K., Rama Sudha K. and Manmadha Ra G. Fig.3.3 Output fr th givn input whn tim cnstant is vry small 4. A currnt puls f amplitud 5 A in Fig.2p. is applid t a paralll cmbinatin shwn in Fig.2p.2. Plt t scal th wavfrms f th currnt flwing thrugh capacitr fr th cass: (i) t p =0., (ii) t p =, (iii) t p = 5 Fig.2p.Th givn input t th circuit Fig. 2p.2 Th givn circuit Slutin: Till t = t p, using Laplac transfrms, th circuit can b drawn as in Fig Drling Kindrsly India Pvt. Ltd 8
19 Puls and Digital Circuits nkata Ra K., Rama Sudha K. and Manmadha Ra G. Fig.4. Circuit in trms f Laplac transfrms Applying KCL, w hav 5 5s 5 IC () s R s s s R s Cs Taking t Laplac invrs, th charging currnt is ic ( t) 5 At t = t p, th currnt suddnly falls frm 5 A t 0. Th vltag acrss th capacitr at t = t p is [ 5 ic ( t p )] R Thrfr fr t t p, [5 ic( tp)] R [5 ic( tp)] IC () s sr ( ) ( s ) Cs Taking Laplacinvrs i ( t) [5 i ( t )] C C p -t-t ( p ) Th circuit that rprsnts th discharg f th cndnsr is prsntd in Fig.4.2. Cas : Fr 0 < t < t i C p ( t) 5 i C dcays xpnntially, Fig.4.2 Circuit that indicats th discharg f th cndnsr t 0. at t t, i ( t ) A p C p Fr t > t p, i C riss xpnntially as 200 Drling Kindrsly India Pvt. Ltd 9
20 Puls and Digital Circuits nkata Ra K., Rama Sudha K. and Manmadha Ra G. i C = 5 i C ( t )] [ p ( t t p ) = t t p Cas 2: Fr 0 < t < t p i i C dcays xpnntially, t Fr t > t p i C dcays xpnntially as C ( t) 5 at t tp, Ic( tp) A [5 I ( t )] C p t t p t = 3.6 t p Cas 3: Fr 0 < t < t p i C dcays xpnntially, i a C ( t) 5 t 5 5 t t tp, Ic( tp) A Fr t > t p i C riss xpnntially as t t p t t p [5 i C ( t p )] = Th input and utputs ar plttd in Fig Drling Kindrsly India Pvt. Ltd 20
21 Puls and Digital Circuits nkata Ra K., Rama Sudha K. and Manmadha Ra G. Fig.4.3 Input and utputs fr th givn circuit 5. Draw th utput wavfrm if th wavfrm shwn in Fig.2p.5(a) is applid at th input f th circuit shwn in Fig.2p.5(b). 200 Drling Kindrsly India Pvt. Ltd 2
22 Puls and Digital Circuits nkata Ra K., Rama Sudha K. and Manmadha Ra G. Fig.2p.5(a) Th input t th high-pass circuit in Fig.2p.5(a) Slutin: Fig.2p.5(b) Th givn high-pass circuit Tim cnstatnt s Tim prid f input wavfrm is T = 4 ms Sinc <<T, th circuit acts as a gd diffrntiatr. dvi 9 dvi Th xprssin fr utput is v 0000 dt dt 00 Fr 0 t 2 ms, vi t, d v 0000 t m 3 3 dt v rmains at 50 m. At t =2 ms, v i falls by 00. Sinc capacitr acts as a shrt circuit, v als falls by th sam amunt. (t = 2 ms) = = v 200 Drling Kindrsly India Pvt. Ltd 22
23 Puls and Digital Circuits nkata Ra K., Rama Sudha K. and Manmadha Ra G Fr 2 < t < 4 ms, v i ( t 2 0 ) d 00 3 v 0000 ( t 2 0 ) 3 dt 20 9 ( 00) m 3 20 Th utput wavfrm is shwn in Fig. 5.. Fig.5 Output f th high-pass circuit 200 Drling Kindrsly India Pvt. Ltd 23
Topic 5: Discrete-Time Fourier Transform (DTFT)
ELEC36: Signals And Systms Tpic 5: Discrt-Tim Furir Transfrm (DTFT) Dr. Aishy Amr Cncrdia Univrsity Elctrical and Cmputr Enginring DT Furir Transfrm Ovrviw f Furir mthds DT Furir Transfrm f Pridic Signals
More informationSensors and Actuators Introduction to sensors
Snsrs and Actuatrs Intrductin t snsrs Sandr Stuijk (s.stuijk@tu.nl) Dpartmnt f Elctrical Enginring Elctrnic Systms APAITIVE IUITS (haptr., 7., 9., 0.6,.,.) apaciti snsr capacitanc dpnds n physical prprtis
More informationLECTURE 5 Guassian Wave Packet
LECTURE 5 Guassian Wav Pact 1.5 Eampl f a guassian shap fr dscribing a wav pact Elctrn Pact ψ Guassian Assumptin Apprimatin ψ As w hav sn in QM th wav functin is ftn rprsntd as a Furir transfrm r sris.
More informationLecture 26: Quadrature (90º) Hybrid.
Whits, EE 48/58 Lctur 26 Pag f Lctur 26: Quadratur (9º) Hybrid. Back in Lctur 23, w bgan ur discussin f dividrs and cuplrs by cnsidring imprtant gnral prprtis f thrand fur-prt ntwrks. This was fllwd by
More information. This is made to keep the kinetic energy at outlet a minimum.
Runnr Francis Turbin Th shap th blads a Francis runnr is cmplx. Th xact shap dpnds n its spciic spd. It is bvius rm th quatin spciic spd (Eq.5.8) that highr spciic spd mans lwr had. This rquirs that th
More informationModern Physics. Unit 5: Schrödinger s Equation and the Hydrogen Atom Lecture 5.6: Energy Eigenvalues of Schrödinger s Equation for the Hydrogen Atom
Mdrn Physics Unit 5: Schrödingr s Equatin and th Hydrgn Atm Lctur 5.6: Enrgy Eignvalus f Schrödingr s Equatin fr th Hydrgn Atm Rn Rifnbrgr Prfssr f Physics Purdu Univrsity 1 Th allwd nrgis E cm frm th
More informationExam 1. It is important that you clearly show your work and mark the final answer clearly, closed book, closed notes, no calculator.
Exam N a m : _ S O L U T I O N P U I D : I n s t r u c t i o n s : It is important that you clarly show your work and mark th final answr clarly, closd book, closd nots, no calculator. T i m : h o u r
More informationFrequency Response. Lecture #12 Chapter 10. BME 310 Biomedical Computing - J.Schesser
Frquncy Rspns Lcur # Chapr BME 3 Bimdical Cmpuing - J.Schssr 99 Idal Filrs W wan sudy Hω funcins which prvid frquncy slciviy such as: Lw Pass High Pass Band Pass Hwvr, w will lk a idal filring, ha is,
More informationA Unified Theory of rf Plasma Heating. J.e. Sprott. July 1968
A Unifid Thry f rf Plasma Hating by J.. Sprtt July 968 PLP 3 Plasma Studis Univrsity f iscnsin INTRODUCfION In this papr, th majr rsults f PLP's 86 and 07 will b drivd in a mr cncis and rigrus way, and
More informationEven/Odd Mode Analysis of the Wilkinson Divider
//9 Wilkinn Dividr Evn and Odd Md Analyi.dc / Evn/Odd Md Analyi f th Wilkinn Dividr Cnidr a matchd Wilkinn pwr dividr, with a urc at prt : Prt Prt Prt T implify thi chmatic, w rmv th grund plan, which
More informationSinusoidal Response Notes
ECE 30 Sinusoidal Rspons Nots For BIBO Systms AStolp /29/3 Th sinusoidal rspons of a systm is th output whn th input is a sinusoidal (which starts at tim 0) Systm Sinusoidal Rspons stp input H( s) output
More informationLecture 27: The 180º Hybrid.
Whits, EE 48/58 Lctur 7 Pag f 0 Lctur 7: Th 80º Hybrid. Th scnd rciprcal dirctinal cuplr w will discuss is th 80º hybrid. As th nam implis, th utputs frm such a dvic can b 80º ut f phas. Thr ar tw primary
More informationN J of oscillators in the three lowest quantum
. a) Calculat th fractinal numbr f scillatrs in th thr lwst quantum stats (j,,,) fr fr and Sl: ( ) ( ) ( ) ( ) ( ).6.98. fr usth sam apprach fr fr j fr frm q. b) .) a) Fr a systm f lcalizd distinguishabl
More informationZVS Boost Converter. (a) (b) Fig 6.29 (a) Quasi-resonant boost converter with M-type switch. (b) Equivalent circuit.
EEL6246 Pwer Electrnics II Chapter 6 Lecture 6 Dr. Sam Abdel-Rahman ZVS Bst Cnverter The quasi-resnant bst cnverter by using the M-type switch as shwn in Fig. 6.29(a) with its simplified circuit shwn in
More informationELEC 372 LECTURE NOTES, WEEK 11 Dr. Amir G. Aghdam Concordia University
ELEC 37 LECTURE NOTES, WEE Dr Amir Aghdam Cncrdia Univrity Part f th nt ar adaptd frm th matrial in th fllwing rfrnc: Mdrn Cntrl Sytm by Richard C Drf and Rbrt H Bihp, Prntic Hall Fdback Cntrl f Dynamic
More informationAnother Explanation of the Cosmological Redshift. April 6, 2010.
Anthr Explanatin f th Csmlgical Rdshift April 6, 010. Jsé Francisc García Juliá C/ Dr. Marc Mrncian, 65, 5. 4605 Valncia (Spain) E-mail: js.garcia@dival.s h lss f nrgy f th phtn with th tim by missin f
More informationECE 2100 Circuit Analysis
ECE 00 Circuit Analysis Lessn 6 Chapter 4 Sec 4., 4.5, 4.7 Series LC Circuit C Lw Pass Filter Daniel M. Litynski, Ph.D. http://hmepages.wmich.edu/~dlitynsk/ ECE 00 Circuit Analysis Lessn 5 Chapter 9 &
More informationMHT-CET 5 (PHYSICS) PHYSICS CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW / NASHIK /PUNE /NAGPUR / BOKARO / DUBAI # 1
1. (D) Givn, mass f th rckts, m = 5000 kg; Exhaust spd, v = 800 m/s Acclratin, a = 0 m/s m Lt is amunt f gas pr scnd, t Frc = m (a + g) mu m a g t m 800 m a g t 5000 10 0 5000 0 m 5000 0 187.5 kg sc t
More informationChapter 33 Gauss s Law
Chaptr 33 Gauss s Law 33 Gauss s Law Whn askd t find th lctric flux thrugh a clsd surfac du t a spcifid nn-trivial charg distributin, flks all t ftn try th immnsly cmplicatd apprach f finding th lctric
More information6. Negative Feedback in Single- Transistor Circuits
Lctur 8: Intrductin t lctrnic analg circuit 36--366 6. Ngativ Fdback in Singl- Tranitr ircuit ugn Paprn, 2008 Our aim i t tudy t ffct f ngativ fdback n t mall-ignal gain and t mall-ignal input and utput
More informationAdditional Math (4047) Paper 2 (100 marks) y x. 2 d. d d
Aitional Math (07) Prpar b Mr Ang, Nov 07 Fin th valu of th constant k for which is a solution of th quation k. [7] Givn that, Givn that k, Thrfor, k Topic : Papr (00 marks) Tim : hours 0 mins Nam : Aitional
More informationBASIC DIRECT-CURRENT MEASUREMENTS
Brwn University Physics 0040 Intrductin BASIC DIRECT-CURRENT MEASUREMENTS The measurements described here illustrate the peratin f resistrs and capacitrs in electric circuits, and the use f sme standard
More information:2;$-$(01*%<*=,-./-*=0;"%/;"-*
!"#$%'()%"*#%*+,-./-*+01.2(.*3+456789*!"#$%"'()'*+,-."/0.%+1'23"45'46'7.89:89'/' ;8-,"$4351415,8:+#9' Dr. Ptr T. Gallaghr Astrphyscs Rsarch Grup Trnty Cllg Dubln :2;$-$(01*%
More informationA Brief and Elementary Note on Redshift. May 26, 2010.
A Brif and Elmntary Nt n Rdshift May 26, 2010. Jsé Francisc García Juliá C/ Dr. Marc Mrncian, 65, 5. 46025 Valncia (Spain) E-mail: js.garcia@dival.s Abstract A rasnabl xplanatin f bth rdshifts: csmlgical
More information5 Curl-free fields and electrostatic potential
5 Curl-fr filds and lctrstatic tntial Mathmaticall, w can gnrat a curl-fr vctr fild E(,, ) as E = ( V, V, V ), b taking th gradint f an scalar functin V (r) =V (,, ). Th gradint f V (,, ) is dfind t b
More informationSection 11.6: Directional Derivatives and the Gradient Vector
Sction.6: Dirctional Drivativs and th Gradint Vctor Practic HW rom Stwart Ttbook not to hand in p. 778 # -4 p. 799 # 4-5 7 9 9 35 37 odd Th Dirctional Drivativ Rcall that a b Slop o th tangnt lin to th
More informationDefinition1: The ratio of the radiation intensity in a given direction from the antenna to the radiation intensity averaged over all directions.
Dirctivity or Dirctiv Gain. 1 Dfinition1: Dirctivity Th ratio of th radiation intnsity in a givn dirction from th antnna to th radiation intnsity avragd ovr all dirctions. Dfinition2: Th avg U is obtaind
More informationCHAPTER 5. Solutions for Exercises
HAPTE 5 Slutins fr Exercises E5. (a We are given v ( t 50 cs(00π t 30. The angular frequency is the cefficient f t s we have ω 00π radian/s. Then f ω / π 00 Hz T / f 0 ms m / 50 / 06. Furthermre, v(t attains
More informationBicycle Generator Dump Load Control Circuit: An Op Amp Comparator with Hysteresis
Bicycle Generatr Dump Lad Cntrl Circuit: An Op Amp Cmparatr with Hysteresis Sustainable Technlgy Educatin Prject University f Waterl http://www.step.uwaterl.ca December 1, 2009 1 Summary This dcument describes
More informationNARAYANA I I T / P M T A C A D E M Y. C o m m o n P r a c t i c e T e s t 1 6 XII STD BATCHES [CF] Date: PHYSIS HEMISTRY MTHEMTIS
. (D). (A). (D). (D) 5. (B) 6. (A) 7. (A) 8. (A) 9. (B). (A). (D). (B). (B). (C) 5. (D) NARAYANA I I T / P M T A C A D E M Y C o m m o n P r a c t i c T s t 6 XII STD BATCHES [CF] Dat: 8.8.6 ANSWER PHYSIS
More informationWhere k is either given or determined from the data and c is an arbitrary constant.
Exponntial growth and dcay applications W wish to solv an quation that has a drivativ. dy ky k > dx This quation says that th rat of chang of th function is proportional to th function. Th solution is
More informationECE 2100 Circuit Analysis
ECE 2100 Circuit Analysis Lessn 25 Chapter 9 & App B: Passive circuit elements in the phasr representatin Daniel M. Litynski, Ph.D. http://hmepages.wmich.edu/~dlitynsk/ ECE 2100 Circuit Analysis Lessn
More information1973 AP Calculus AB: Section I
97 AP Calculus AB: Sction I 9 Minuts No Calculator Not: In this amination, ln dnots th natural logarithm of (that is, logarithm to th bas ).. ( ) d= + C 6 + C + C + C + C. If f ( ) = + + + and ( ), g=
More informationGUC (Dr. Hany Hammad) 4/20/2016
GU (r. Hay Hamma) 4/0/06 Lctur # 0 Filtr sig y Th srti Lss Mth sig Stps Lw-pass prttyp sig. () Scalig a cvrsi. () mplmtati. Usig Stus. Usig High-Lw mpac Sctis. Thry f priic structurs. mag impacs a Trasfr
More informationEE 119 Homework 6 Solution
EE 9 Hmwrk 6 Slutin Prr: J Bkr TA: Xi Lu Slutin: (a) Th angular magniicatin a tlcp i m / th cal lngth th bjctiv ln i m 4 45 80cm (b) Th clar aprtur th xit pupil i 35 mm Th ditanc btwn th bjctiv ln and
More informationA. Limits and Horizontal Asymptotes ( ) f x f x. f x. x "±# ( ).
A. Limits and Horizontal Asymptots What you ar finding: You can b askd to find lim x "a H.A.) problm is asking you find lim x "# and lim x "$#. or lim x "±#. Typically, a horizontal asymptot algbraically,
More informationLecture Outline. Skin Depth Power Flow 8/7/2018. EE 4347 Applied Electromagnetics. Topic 3e
8/7/018 Cours Instructor Dr. Raymond C. Rumpf Offic: A 337 Phon: (915) 747 6958 E Mail: rcrumpf@utp.du EE 4347 Applid Elctromagntics Topic 3 Skin Dpth & Powr Flow Skin Dpth Ths & Powr nots Flow may contain
More information(1) Then we could wave our hands over this and it would become:
MAT* K285 Spring 28 Anthony Bnoit 4/17/28 Wk 12: Laplac Tranform Rading: Kohlr & Johnon, Chaptr 5 to p. 35 HW: 5.1: 3, 7, 1*, 19 5.2: 1, 5*, 13*, 19, 45* 5.3: 1, 11*, 19 * Pla writ-up th problm natly and
More informationAnswer Homework 5 PHA5127 Fall 1999 Jeff Stark
Answr omwork 5 PA527 Fall 999 Jff Stark A patint is bing tratd with Drug X in a clinical stting. Upon admiion, an IV bolus dos of 000mg was givn which yildd an initial concntration of 5.56 µg/ml. A fw
More informationMicrowave Engineering
Micrwav Enginring hng-hsing Hsu Dpartmnt f Elctrical Enginring Natinal Unitd Univrsity Outlin. Transmissin Lin Thry. Transmissin Lins and Wavguids Gnral lutins fr TEM, TE, and TM wavs ; Paralll Plat wavguid
More informationA Propagating Wave Packet Group Velocity Dispersion
Lctur 8 Phys 375 A Propagating Wav Packt Group Vlocity Disprsion Ovrviw and Motivation: In th last lctur w lookd at a localizd solution t) to th 1D fr-particl Schrödingr quation (SE) that corrsponds to
More informationPHYSICS Unit 3 Trial Examination
STAV Publishing Pty Ltd 005 PHYSICS Unit 3 Trial Examinatin SOLUTIONS BOOK Published by STAV Publishing Pty Ltd. STAV Huse, 5 Munr Street, Cburg VIC 3058 Australia. Phne: 6 + 3 9385 3999 Fax: 6 + 3 9386
More informationMathematics. Complex Number rectangular form. Quadratic equation. Quadratic equation. Complex number Functions: sinusoids. Differentiation Integration
Mathmatics Compl numbr Functions: sinusoids Sin function, cosin function Diffrntiation Intgration Quadratic quation Quadratic quations: a b c 0 Solution: b b 4ac a Eampl: 1 0 a= b=- c=1 4 1 1or 1 1 Quadratic
More informationMATHEMATICS FOR MANAGEMENT BBMP1103
Objctivs: TOPIC : EXPONENTIAL AND LOGARITHM FUNCTIONS. Idntif pnntils nd lgrithmic functins. Idntif th grph f n pnntil nd lgrithmic functins. Clcult qutins using prprtis f pnntils. Clcult qutins using
More informationSignals and Systems View Point
Signals and Sstms Viw Pint Inpt signal Ozt Mdical Imaging Sstm LOzt Otpt signal Izt r Iz r I A signalssstms apprach twards imaging allws s as Enginrs t Gain a bttr ndrstanding f hw th imags frm and what
More informationThree charges, all with a charge of 10 C are situated as shown (each grid line is separated by 1 meter).
Three charges, all with a charge f 0 are situated as shwn (each grid line is separated by meter). ) What is the net wrk needed t assemble this charge distributin? a) +0.5 J b) +0.8 J c) 0 J d) -0.8 J e)
More informationIntroduction to Three-phase Circuits. Balanced 3-phase systems Unbalanced 3-phase systems
Intrductin t Three-hase Circuits Balanced 3-hase systems Unbalanced 3-hase systems 1 Intrductin t 3-hase systems Single-hase tw-wire system: Single surce cnnected t a lad using tw-wire system Single-hase
More information9.5 Complex variables
9.5 Cmpl varabls. Cnsdr th funtn u v f( ) whr ( ) ( ), f( ), fr ths funtn tw statmnts ar as fllws: Statmnt : f( ) satsf Cauh mann quatn at th rgn. Statmnt : f ( ) ds nt st Th rrt statmnt ar (A) nl (B)
More informationINDUCTANCE Self Inductance
DUCTCE 3. Sef nductance Cnsider the circuit shwn in the Figure. S R When the switch is csed the current, and s the magnetic fied, thrugh the circuit increases frm zer t a specific vaue. The increasing
More informationMath 34A. Final Review
Math A Final Rviw 1) Us th graph of y10 to find approimat valus: a) 50 0. b) y (0.65) solution for part a) first writ an quation: 50 0. now tak th logarithm of both sids: log() log(50 0. ) pand th right
More informationSeries and Parallel Resonances
Series and Parallel esnances Series esnance Cnsider the series circuit shwn in the frequency dmain. The input impedance is Z Vs jl jl I jc C H s esnance ccurs when the imaginary part f the transfer functin
More informationAppendices on the Accompanying CD
APPENDIX 4B Andis n th Amanyg CD TANSFE FUNCTIONS IN CONTINUOUS CONDUCTION MODE (CCM In this st, w will driv th transfr funt v / d fr th thr nvrtrs ratg CCM 4B- Buk Cnvrtrs Frm Fig. 4-7, th small signal
More informationECE602 Exam 1 April 5, You must show ALL of your work for full credit.
ECE62 Exam April 5, 27 Nam: Solution Scor: / This xam is closd-book. You must show ALL of your work for full crdit. Plas rad th qustions carfully. Plas chck your answrs carfully. Calculators may NOT b
More information10. The Discrete-Time Fourier Transform (DTFT)
Th Discrt-Tim Fourir Transform (DTFT Dfinition of th discrt-tim Fourir transform Th Fourir rprsntation of signals plays an important rol in both continuous and discrt signal procssing In this sction w
More informationLab 11 LRC Circuits, Damped Forced Harmonic Motion
Physics 6 ab ab 11 ircuits, Damped Frced Harmnic Mtin What Yu Need T Knw: The Physics OK this is basically a recap f what yu ve dne s far with circuits and circuits. Nw we get t put everything tgether
More informationQ1. A string of length L is fixed at both ends. Which one of the following is NOT a possible wavelength for standing waves on this string?
Term: 111 Thursday, January 05, 2012 Page: 1 Q1. A string f length L is fixed at bth ends. Which ne f the fllwing is NOT a pssible wavelength fr standing waves n this string? Q2. λ n = 2L n = A) 4L B)
More informationDifferential Equations
UNIT I Diffrntial Equations.0 INTRODUCTION W li in a world of intrrlatd changing ntitis. Th locit of a falling bod changs with distanc, th position of th arth changs with tim, th ara of a circl changs
More informationDesign Guidelines for Quartz Crystal Oscillators. R 1 Motional Resistance L 1 Motional Inductance C 1 Motional Capacitance C 0 Shunt Capacitance
TECHNICAL NTE 30 Dsign Guidlins for Quartz Crystal scillators Introduction A CMS Pirc oscillator circuit is wll known and is widly usd for its xcllnt frquncy stability and th wid rang of frquncis ovr which
More information2F1120 Spektrala transformer för Media Solutions to Steiglitz, Chapter 1
F110 Spktrala transformr för Mdia Solutions to Stiglitz, Chaptr 1 Prfac This documnt contains solutions to slctd problms from Kn Stiglitz s book: A Digital Signal Procssing Primr publishd by Addison-Wsly.
More informationEE 221 Practice Problems for the Final Exam
EE 1 Practce Prblems fr the Fnal Exam 1. The netwrk functn f a crcut s 1.5 H. ω 1+ j 500 Ths table recrds frequency respnse data fr ths crcut. Fll n the blanks n the table:. The netwrk functn f a crcut
More informationRelationships Between Frequency, Capacitance, Inductance and Reactance.
P Physics Relatinships between f,, and. Relatinships Between Frequency, apacitance, nductance and Reactance. Purpse: T experimentally verify the relatinships between f, and. The data cllected will lead
More information4.2 Design of Sections for Flexure
4. Dsign of Sctions for Flxur This sction covrs th following topics Prliminary Dsign Final Dsign for Typ 1 Mmbrs Spcial Cas Calculation of Momnt Dmand For simply supportd prstrssd bams, th maximum momnt
More informationLectur 22. RF and Microwave Circuit Design Γ-Plane and Smith Chart Analysis. ECE 303 Fall 2005 Farhan Rana Cornell University
ctur RF ad Micrwav Circuit Dig -Pla ad Smith Chart Aalyi I thi lctur yu will lar: -pla ad Smith Chart Stub tuig Quartr-Wav trafrmr ECE 33 Fall 5 Farha Raa Crll Uivrity V V Impdac Trafrmati i Tramii i ω
More informationTypes of Communication
Tps f Cmmunicatin Analg: cntinuus ariabls with nis {rrr 0} 0 (imprfct) Digital: dcisins, discrt chics, quantizd, nis {rrr} 0 (usuall prfct) Mssag S, S, r S M Mdulatr (t) channl; adds nis and distrtin M-ar
More informationChapter 30. Inductance
Chapter 30 nductance 30. Self-nductance Cnsider a lp f wire at rest. f we establish a current arund the lp, it will prduce a magnetic field. Sme f the magnetic field lines pass thrugh the lp. et! be the
More informationChapter 16. Capacitance. Capacitance, cont. Parallel-Plate Capacitor, Example 1/20/2011. Electric Energy and Capacitance
summary C = ε A / d = πε L / ln( b / a ) ab C = 4πε 4πε a b a b >> a Chapter 16 Electric Energy and Capacitance Capacitance Q=CV Parallel plates, caxial cables, Earth Series and parallel 1 1 1 = + +..
More informationEdexcel GCSE Physics
Edexcel GCSE Physics Tpic 10: Electricity and circuits Ntes (Cntent in bld is fr Higher Tier nly) www.pmt.educatin The Structure f the Atm Psitively charged nucleus surrunded by negatively charged electrns
More informationMATHEMATICS (B) 2 log (D) ( 1) = where z =
MATHEMATICS SECTION- I STRAIGHT OBJECTIVE TYPE This sction contains 9 multipl choic qustions numbrd to 9. Each qustion has choic (A), (B), (C) and (D), out of which ONLY-ONE is corrct. Lt I d + +, J +
More informationPotential and Capacitance
Ptential and apacitance Electric Ptential Electric ptential (V) = Electric ptential energy (U e ) per unit charge () Define: ptential energy U e = 0 at infinity (r = ) lim U 0 r e Nte the similarity f
More informationThomas Whitham Sixth Form
Thomas Whitham Sith Form Pur Mathmatics Unit C Algbra Trigonomtr Gomtr Calculus Vctor gomtr Pag Algbra Molus functions graphs, quations an inqualitis Graph of f () Draw f () an rflct an part of th curv
More informationBackground: We have discussed the PIB, HO, and the energy of the RR model. In this chapter, the H-atom, and atomic orbitals.
Chaptr 7 Th Hydrogn Atom Background: W hav discussd th PIB HO and th nrgy of th RR modl. In this chaptr th H-atom and atomic orbitals. * A singl particl moving undr a cntral forc adoptd from Scott Kirby
More informationSections 15.1 to 15.12, 16.1 and 16.2 of the textbook (Robbins-Miller) cover the materials required for this topic.
Tpic : AC Fundamentals, Sinusidal Wavefrm, and Phasrs Sectins 5. t 5., 6. and 6. f the textbk (Rbbins-Miller) cver the materials required fr this tpic.. Wavefrms in electrical systems are current r vltage
More informationAs the matrix of operator B is Hermitian so its eigenvalues must be real. It only remains to diagonalize the minor M 11 of matrix B.
7636S ADVANCED QUANTUM MECHANICS Solutions Spring. Considr a thr dimnsional kt spac. If a crtain st of orthonormal kts, say, and 3 ar usd as th bas kts, thn th oprators A and B ar rprsntd by a b A a and
More informationSFDMB3638F. Specifications and Applications Information. orce LED Driver. Mass: 7 grams typ. 10/15/08 Preliminary. Package Configuration
Specificatins and Applicatins Infrmatin 1/1/8 Prelimary Smart Fr rce LED Driver The ERG Smart Frce Series f LED Drivers are specifically designed fr applicatins which require high efficiency, small ftprt
More informationSolution: APPM 1360 Final (150 pts) Spring (60 pts total) The following parts are not related, justify your answers:
APPM 6 Final 5 pts) Spring 4. 6 pts total) Th following parts ar not rlatd, justify your answrs: a) Considr th curv rprsntd by th paramtric quations, t and y t + for t. i) 6 pts) Writ down th corrsponding
More informationUnit 6: Solving Exponential Equations and More
Habrman MTH 111 Sction II: Eonntial and Logarithmic Functions Unit 6: Solving Eonntial Equations and Mor EXAMPLE: Solv th quation 10 100 for. Obtain an act solution. This quation is so asy to solv that
More information22/ Breakdown of the Born-Oppenheimer approximation. Selection rules for rotational-vibrational transitions. P, R branches.
Subjct Chmistry Papr No and Titl Modul No and Titl Modul Tag 8/ Physical Spctroscopy / Brakdown of th Born-Oppnhimr approximation. Slction ruls for rotational-vibrational transitions. P, R branchs. CHE_P8_M
More informationRevision: August 19, E Main Suite D Pullman, WA (509) Voice and Fax
.7.4: Direct frequency dmain circuit analysis Revisin: August 9, 00 5 E Main Suite D Pullman, WA 9963 (509) 334 6306 ice and Fax Overview n chapter.7., we determined the steadystate respnse f electrical
More information2. Find i, v, and the power dissipated in the 6-Ω resistor in the following figure.
CSC Class exercise DC Circuit analysis. Fr the ladder netwrk in the fllwing figure, find I and R eq. Slutin Req 4 ( 6 ) 5Ω 0 0 I Re q 5 A. Find i, v, and the pwer dissipated in the 6-Ω resistr in the fllwing
More information6. The Interaction of Light and Matter
6. Th Intraction of Light and Mattr - Th intraction of light and mattr is what maks lif intrsting. - Light causs mattr to vibrat. Mattr in turn mits light, which intrfrs with th original light. - Excitd
More informationFourier Transforms and the Wave Equation. Key Mathematics: More Fourier transform theory, especially as applied to solving the wave equation.
Lur 7 Fourir Transforms and th Wav Euation Ovrviw and Motivation: W first discuss a fw faturs of th Fourir transform (FT), and thn w solv th initial-valu problm for th wav uation using th Fourir transform
More informationPitch vs. Frequency:
Pitch vs. Frequency: Pitch = human ear s perceptin f frequency f a sund vibratin Lw pitch lw frequency f vibratin/scillatin High pitch high frequency f vibratin/scillatin Q: Is the relatin between {perceived}
More informationSAFE HANDS & IIT-ian's PACE EDT-15 (JEE) SOLUTIONS
It is not possibl to find flu through biggr loop dirctly So w will find cofficint of mutual inductanc btwn two loops and thn find th flu through biggr loop Also rmmbr M = M ( ) ( ) EDT- (JEE) SOLUTIONS
More informationGENERAL FORMULAS FOR FLAT-TOPPED WAVEFORMS. J.e. Sprott. Plasma Studies. University of Wisconsin
GENERAL FORMULAS FOR FLAT-TOPPED WAVEFORMS J.e. Sprtt PLP 924 September 1984 Plasma Studies University f Wiscnsin These PLP Reprts are infrmal and preliminary and as such may cntain errrs nt yet eliminated.
More informationJAZAN University. Department: Electrical Engineering. Names & ID: Electronics LAB - 1/ / Electronics LAB - EngE : 314 G:...
Electrnics LAB - EngE : 314 G:... Electrnics LAB - 1/2-201. /201. Electrnics LAB - EngE : 314 G:... Electrnics LAB - 2/2-201. /201. Silicn JAZAN University Experiment 1: characteristics Electrnics LAB
More informationPlan o o. I(t) Divide problem into sub-problems Modify schematic and coordinate system (if needed) Write general equations
STAPLE Physics 201 Name Final Exam May 14, 2013 This is a clsed bk examinatin but during the exam yu may refer t a 5 x7 nte card with wrds f wisdm yu have written n it. There is extra scratch paper available.
More information2/12/2013. Overview. 12-Power Transmission Text: Conservation of Complex Power. Introduction. Power Transmission-Short Line
//03 Ovrviw -owr Transmission Txt: 4.6-4.0 ECEGR 45 owr ystms Consrvation of Complx owr hort in owr Transmission owr Transmission isualization Radial in Mdium and ong in owr Transmission oltag Collaps
More informationG D S. Drain-Source Voltage 30 V Gate-Source Voltage. at T =100 C Continuous Drain Current 3
N-channl Enhancmnt-mod Powr MOSFET Simpl Driv Rquirmnt D Fast Switching Charactristics Low Gat Charg R DS(ON) 25mΩ G RoHS-compliant, halogn-fr I D 28A S BV DSS 30V Dscription Advancd Powr MOSFETs from
More informationDSP-First, 2/e. LECTURE # CH2-3 Complex Exponentials & Complex Numbers TLH MODIFIED. Aug , JH McClellan & RW Schafer
DSP-First, / TLH MODIFIED LECTURE # CH-3 Complx Exponntials & Complx Numbrs Aug 016 1 READING ASSIGNMENTS This Lctur: Chaptr, Scts. -3 to -5 Appndix A: Complx Numbrs Complx Exponntials Aug 016 LECTURE
More informationCHAPTER 3 INEQUALITIES. Copyright -The Institute of Chartered Accountants of India
CHAPTER 3 INEQUALITIES Cpyright -The Institute f Chartered Accuntants f India INEQUALITIES LEARNING OBJECTIVES One f the widely used decisin making prblems, nwadays, is t decide n the ptimal mix f scarce
More informationMore on FT. Lecture 10 4CT.5 3CT.3-5,7,8. BME 333 Biomedical Signals and Systems - J.Schesser
Mr n FT Lcur 4CT.5 3CT.3-5,7,8 BME 333 Bimdicl Signls nd Sysms - J.Schssr 43 Highr Ordr Diffrniin d y d x, m b Y b X N n M m N M n n n m m n m n d m d n m Y n d f n [ n ] F d M m bm m X N n n n n n m p
More informationOperating parameters for representative BWR and PWR designs are given below. For the PWR hot channel and the BWR average channel compute and plot:
Opratin paramtrs r rprsntativ BWR an PWR sins ar ivn blw. Fr t PWR t cannl an t BWR avra cannl cmput an plt: 1) t vi an quality istributins ) Dtrmin t iniviual cmpnnts an t ttal prssur rp Cmpar t rsults
More informationApplying Kirchoff s law on the primary circuit. V = - e1 V+ e1 = 0 V.D. e.m.f. From the secondary circuit e2 = v2. K e. Equivalent circuit :
TRANSFORMERS Definitin : Transfrmers can be defined as a static electric machine which cnverts electric energy frm ne ptential t anther at the same frequency. It can als be defined as cnsists f tw electric
More informationActivity Guide Loops and Random Numbers
Unit 3 Lessn 7 Name(s) Perid Date Activity Guide Lps and Randm Numbers CS Cntent Lps are a relatively straightfrward idea in prgramming - yu want a certain chunk f cde t run repeatedly - but it takes a
More informationChapter 3 Lecture 14 Longitudinal stick free static stability and control 3 Topics
Chaptr 3 Lctur 14 Longitudinal stick fr static stability and control 3 Topics 3.4.4 Rquirmnt for propr stick forc variation 3.4.5 Fl of th stability lvl by th pilot Exampl 3.3 3.5 Dtrmination of stick-fr
More informationImpedance Transformation and Parameter Relations
8/1/18 Cours nstructor Dr. Raymond C. Rumpf Offic: A 337 Phon: (915) 747 6958 E Mail: rcrumpf@utp.du EE 4347 Applid Elctromagntics Topic 4 mpdanc Transformation and Paramtr Rlations mpdanc Ths Transformation
More informationFirst derivative analysis
Robrto s Nots on Dirntial Calculus Chaptr 8: Graphical analysis Sction First drivativ analysis What you nd to know alrady: How to us drivativs to idntiy th critical valus o a unction and its trm points
More informationElectrochemical Energy Systems Spring 2014 MIT, M. Z. Bazant. Midterm Exam
10.66 Elctrochmical Enrgy Systms Spring 014 MIT, M. Z. Bazant Midtrm Exam Instructions. This is a tak-hom, opn-book xam du in Lctur. Lat xams will not b accptd. You may consult any books, handouts, or
More informationECE 546 Lecture 02 Review of Electromagnetics
C 546 Lecture 0 Review f lectrmagnetics Spring 018 Jse. Schutt-Aine lectrical & Cmputer ngineering University f Illinis jesa@illinis.edu C 546 Jse Schutt Aine 1 Printed Circuit Bard C 546 Jse Schutt Aine
More informationALL INDIA TEST SERIES
Frm Classrm/Integrated Schl Prgrams 7 in Tp, in Tp, 5 in Tp, 6 in Tp 5 ll India Ranks & Students frm Classrm /Integrated Schl Prgrams & 7 Students frm ll Prgrams have been warded a Rank in JEE (dvanced),
More information