ECE 2100 Circuit Analysis

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1 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.

2 ECE 00 Circuit Analysis Lessn 5 Chapter 9 & App B: Passive circuit elements in the phasr representatin Daniel M. Litynski, Ph.D.

3 Sinusids and Phasr Chapter 9 9. Mtivatin 9. Sinusids features 9.3 Phasrs 9.4 Phasr relatinships fr circuit elements 9.5 Impedance and admittance 9.6 Kirchhff s laws in the frequency dmain 9.7 Impedance cmbinatins 3

4 9.5 Impedance and Admittance ( The impedance Z f a circuit is the rati f the phasr vltage V t the phasr current I, measured in hms Ω. Z V I where = e, Z is the resistance and X = Im, Z is the reactance. Psitive X is fr L and negative X is fr C. jx The admittance Y is the reciprcal f impedance, measured in siemens (S. Y Z I V 4

5 9.5 Impedance and Admittance ( Impedances and admittances f passive elements Element Impedance Admittance Z Y L Z j L Y j L C Z j C Y j C 5

6 9.5 Impedance and Admittance (3 0; Z 0 Z j L ; Z 0; Z Z j C ; Z 0 6

7 9.5 Impedance and Admittance (4 After we knw hw t cnvert LC cmpnents frm time t phasr dmain, we can transfrm a time dmain circuit int a phasr/frequency dmain circuit. Hence, we can apply the KCL laws and ther therems t directly set up phasr equatins invlving ur target variable(s fr slving. 7

8 9.5 Impedance and Admittance (5 Example 8 efer t Figure belw, determine v(t and i(t. v s 5cs(0t Answers: i(t =.8cs(0t 6.56 A; v(t =.36cs(0t V 8

9 9.6 Kirchhff s Laws in the Frequency Dmain ( Bth KVL and KCL hld in the phasr dmain r mre cmmnly called frequency dmain. Mrever, the variables t be handled are phasrs, which are cmplex numbers. All the mathematical peratins invlved are nw in the cmplex dmain. 9

10 9.7 Impedance Cmbinatins ( The fllwing principles used fr DC circuit analysis all apply t AC circuit. Fr example: a. vltage divisin b. current divisin c. circuit reductin d. impedance equivalence e. Y-Δ transfrmatin 0

11 9.7 Impedance Cmbinatins ( Example 9 Determine the input impedance f the circuit in figure belw at ω =0 rad/s. Answer: Z in = 3.38 j73.76

12 Frequency espnse Chapter 4 4. Intrductin 4.5 Series esnance 4.7 Passive Filters

13 4. Intrductin ( What is Frequency espnse f a Circuit? It is the variatin in a circuit s behavir with change in signal frequency and may als be cnsidered as the variatin f the gain and phase with frequency. 3

14 4. Transfer Functin ( The transfer functin H(ω f a circuit is the frequency-dependent rati f a phasr utput Y(ω (an element vltage r current t a phasr input X(ω (surce vltage r current. H( Y( X( H( 4

15 4. Transfer Functin ( Fur pssible transfer functins: H( Vltage gain V ( V ( i H( Transfer Impedance V ( I ( i H( Y( X( H( H( Current gain I( I ( i H( Transfer Admittance I( V ( i 5

16 4. Transfer Functin (3 Example Fr the C circuit shwn belw, btain the transfer functin V/Vs and its frequency respnse. Let v s = V m csωt. 6

17 4. Transfer Functin (4 Slutin: The transfer functin is H( V V s j C / j C j C The magnitude is H(, ( / The phase is tan /C Lw Pass Filter 7

18 4. Transfer Functin (5 Example Obtain the transfer functin V/Vs f the L circuit shwn belw, assuming v s = V m csωt. Sketch its frequency respnse. 8

19 4. Transfer Functin (6 Slutin: The transfer functin is H( V V s j L j L j L High Pass Filter H( The magnitude is, ( The phase is 90 tan /L 9

20 4.3 Series esnance ( esnance is a cnditin in an LC circuit in which the capacitive and inductive reactance are equal in magnitude, thereby resulting in purely resistive impedance. esnance frequency: Z j( L C f rad/s LC Hz LC r 0

21 4.3 Series esnance ( The features f series resnance: The impedance is purely resistive, Z = ; The supply vltage Vs and the current I are in phase, s cs = ; The magnitude f the transfer functin H(ω = Z(ω is minimum; The inductr vltage and capacitr vltage can be much mre than the surce vltage. Z j( L C

22 4.3 Series esnance (3 Bandwidth B The frequency respnse f the resnance circuit current is Z j( L I I V m ( L / C C The average pwer absrbed by the LC circuit is P( I The highest pwer dissipated ccurs at resnance: P( V m

23 4 3 Series esnance (4 Half-pwer frequencies ω and ω are frequencies at which the dissipated pwer is half the maximum value: P( P( (V m / Vm 4 The half-pwer frequencies can be btained by setting Z equal t. L ( L LC L ( L LC Bandwidth B B 3

24 4.3 Series esnance (5 Quality factr, Q Peak energy stred in the circuit Energy dissipated by thecircuit in neperid at resnance L C The relatinship between the B, Q and ω : B L Q C The quality factr is the rati f its resnant frequency t its bandwidth. If the bandwidth is narrw, the quality factr f the resnant circuit must be high. If the band f frequencies is wide, the quality factr must be lw. 4

25 4.3 Series esnance (6 Example 3 A series-cnnected circuit has = 4 Ω and L = 5 mh. a. Calculate the value f C that will prduce a quality factr f 50. b. Find ω and ω, and B. c. Determine the average pwer dissipated at ω = ω, ω, ω. Take V m = 00V. 5

26 4.4 Parallel esnance ( It ccurs when imaginary part f Y is zer Y j( C L esnance frequency: LC rad/s r f LC Hz 6

27 4.4 Parallel esnance ( Summary f series and parallel resnance circuits: characteristic Series circuit Parallel circuit ω LC LC Q B ωl r Q ω C L r Q C ω, ω ( Q Q ( Q Q Q 0, ω, ω B B 7

28 4.4 Parallel esnance (3 Example 4 Calculate the resnant frequency f the circuit in the figure shwn belw. Answer: 9.79 rad/s 8

29 4.5 Passive Filters ( A filter is a circuit that is designed t pass signals with desired frequencies and reject r attenuate thers. Passive filter cnsists f nly passive element, L and C. There are fur types f filters. Lw Pass High Pass Band Pass Band Stp 9

30 4.5 Passive Filters ( Example 5 Fr the circuit in the figure belw, btain the transfer functin V(ω/Vi(ω. Identify the type f filter the circuit represents and determine the crner frequency. Take =00 = and L =mh. Answer: 5 krad/s 30

ECE 2100 Circuit Analysis

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