Frequency Response of Circuits: Series and Parallel Resonance - Prof. Olivera K. Notaros, Study notes of Electrical and Electronics Engineering

Download Frequency Response of Circuits: Series and Parallel Resonance - Prof. Olivera K. Notaros and more Study notes Electrical and Electronics Engineering in PDF only on Docsity! 1 EE221 Circuits II Chapter 14 Frequency Response 2 Frequency Response Chapter 14 14.1 Introduction 14.2 Transfer Function 14.3 Bode Plots 14.4 Series Resonance 14.5 Parallel Resonance 14.6 Passive Filters 14.7 Active filters 5 14.2 Transfer Function Four possible transfer functions: )(V )(V gain Voltage )(H i o ω ωω == )(I )(I gain Current )(H i o ω ωω == )(I )(V ImpedanceTransfer )(H i o ω ωω == )(V )(I AdmittanceTransfer )(H i o ω ωω == φω ω ωω ∠== |)(H| )(X )(Y )(H 6 14.2 Transfer Function Example 1 For the RC circuit shown below, obtain the transfer function Vo/Vs and its frequency response. Let vs = Vmcosωt. 7 14.2 Transfer Function Solution: The transfer function is , The magnitude is 2)/(1 1)(H oωω ω + = The phase is oω ωφ 1tan−−= 1/RC=oω RC j1 1 C j1/ R Cj 1 V V)(H s o ωω ωω + = + == Low Pass Filter 10 14.4 Bode Plots Bode Plots are semilog plots of the magnitude (in dB) and phase (in deg.) of the transfer function versus frequency. HH HeH dB j 10log20= = φ Bode Plot of Gain K HA 20 log io & | Ly 0.1 1 10 100 w b t | 0 | | | | i J. |___j 0.1 | 10 100 w 11 12 Bode Plot of a zero (jω) Faetor Magnitude Phase: ON UBecade Sov" ey" en AY Si decade _ 4 bind (a a, iy, o 4 4 Oy a td ae ee = a Ny [ty = i Ce - Ll Beaton + (furan rl AON dE decode ne Example 1 — 200 jo H(w) a ne = (ja + 2) jo + 10) Solution: We first put H(w) in the standard form by dividing out the poles « zeros. Thus, 10 ja (1 + jw /2)0 + jw/10) 10 | joo| ~ WL + jw/2||1 + jew/10| H(w) = /80° — tan! w/2 — tan! w/ 1 Hence, the magnitude and phase are jw Hyp = 20 logy 10 + 20 logio|je@| — 20 logio 1+ Je 1 jo — 2010 l, + — 810 | 10 a 6 = 90° — tan! 2 — tan! 16 Example 1 «pd gae o° —90° 20 log ,yl0 _ 20 log; pl jaal 1 con es | Ll yd 10 =-20. 100 200 hl ee re 7 | "ae A —— sr OFT + jes 10] 1 1 1 1 —» Sy Dey O20 100 200 @ 20 14.4 Series Resonance The features of series resonance: The impedance is purely resistive, Z = R; • The supply voltage Vs and the current I are in phase, so cos θ = 1; • The magnitude of the transfer function H(ω) = Z(ω) is minimum; • The inductor voltage and capacitor voltage can be much more than the source voltage. ) C 1L ( jRZ ω ω −+= 21 14.4 Series Resonance Bandwidth B The frequency response of the resonance circuit current is )C 1L ( jRZ ω ω −+= 22 m )C /1L (R V|I|I ωω −+ == The average power absorbed by the RLC circuit is RI 2 1)(P 2=ω The highest power dissipated occurs at resonance: R V 2 1)(P 2 m=oω 22 14 4 Series Resonance Half-power frequencies ω1 and ω2 are frequencies at which the dissipated power is half the maximum value: The half-power frequencies can be obtained by setting Z equal to √2 R. 4R V R )2/(V 2 1)(P)(P 2 m 2 m 21 === ωω LC 1) 2L R( 2L R 2 1 ++−=ω LC 1) 2L R( 2L R 2 2 ++=ω 21ωωω =o Bandwidth B 12 B ωω −= 25 14.5 Parallel Resonance Resonance frequency: Hzo LC2 1for rad/s LC 1 o π ω == ) L 1C ( j R 1Y ω ω −+= It occurs when imaginary part of Y is zero 26 Summary of series and parallel resonance circuits: 14.5 Parallel Resonance LC 1 LC 1 RC 1or R L o o ω ω RCor L R o o ω ω Q oω Q oω 2Q ) 2Q 1( 1 2 oo ωω ±+ 2Q ) 2Q 1( 1 2 oo ωω ±+ 2 B ±oω 2 B ±oω characteristic Series circuit Parallel circuit ωo Q B ω1, ω2 Q ≥ 10, ω1, ω2 27 14.5 Resonance Example 4 Calculate the resonant frequency of the circuit in the figure shown below. rad/s2.179 2 19 ==ωAnswer: High Pass Filter (Passive) e joRC a4 Hiw) = = fd + FORC [Fei | hee Ue dl te). tho pe eRe 0.707 be Actual ee 30 Oo % 31 Band Pass Filter (Passive) See Equation 14.33 for corner Frequencies ω1 and ω2 Band Reject Filter (Passive) R pee =] os Ny. Hak — 1f@c) C “T_ ¥ Ha) ee ee ee = i — F : 1 R+ ftwl. — L/w) e,(t) pte) : Le | ror — | | feat ye YEE | H(e)| f I = 0.707 ea iw Ma \ } Actual \ | \ i Ideal \ / Ake Na e a QO de I tig tua aw Band Pass Filter (Active) K ANAS i ' —— fc |! My = Va or 1 } = t+} —- % R ik po AAV 5B = @ — a, : id | R ~ {> | vy ' | ° : v i a | | | | rs B : | { - Qe. ce _ o : ts, = ~ RC Hie) = v2 i] (- joCok Ga) om ee NT jotaR/\ 14+ jee. BFR KR; a, + as J& I — JoOnk dehy i R, 1 + joC Rk 1+ JaCoR RC> Band Reject Filter (Active) R WY @ en | Ro 6 NV Se i + NNN 7] os ~ R R; | * | poe OA | R C5 | | v3 lemcih AAA—t oth | | ee I 14 Ry ( i jojo, > fin > i Nd ae ji! Wy l + J@/@,, Ry (1 + j2w/w, 4 (jw) /ww;) R. (I + jo/ws\(| | joo/wy) Ry Pe a | | OR 2h | , K=— et — R, | Uy | : je z Ry (1 + j2@o/@, + (jw) /w;@,) Alay) =| = —— i 2 R te + Jog / Woh ] + J@o/@,) R, R; @, > we 20 37 Magnitude and Frequency Scaling Example: 4th order low-pass filter Corner Frequency: 1 rad/sec Resistance: 1Ω Corner Frequency: 100π krad/sec Resistance: 10 kΩ