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Second-Order Filters SECTION 3: SECOND-ORDER FILTERS ENGR 202 – Electrical Fundamentals II 2 Introduction K. Webb ENGR 202 Second-Order Circuits 3 Order of a circuit (or system of any kind) Number of independent energy-storage elements Order of the differential equation describing the system Second-order circuits Two energy-storage elements Described by second-order differential equations We will primarily be concerned with second- order RLC circuits Circuits with a resistor, an inductor, and a capacitor K. Webb ENGR 202 Second-Order Circuits 4 In this and the following section of notes, we will look at second-order RLC circuits from two distinct perspectives: Section 3 Second-order filters Frequency-domain behavior Section 4 Second-order transient response Time-domain behavior K. Webb ENGR 202 5 Second-Order Filters K. Webb ENGR 202 Second-Order Filters 6 First-order filters Roll-off rate: 20 dB/decade This roll-off rate determines selectivity Spacing of pass band and stop band Spacing of passed frequencies and stopped or filtered frequencies Second-order filters Roll-off rate: 40 dB/decade In general: Roll-off = 20 / , where is the filter order K. Webb ⋅ ENGR 202 Resonance 7 Resonance Tendency of a system to oscillate at certain frequencies – resonant frequencies – often with larger amplitude than any input Phenomenon that occurs in all types of dynamic systems (mechanical, electrical, fluid, etc.) Examples of resonant mechanical systems: Mass on a spring Pendulum, playground swing Tacoma Narrows Bridge ©Barney Elliot – The Camera Shop K. Webb ENGR 202 Electrical Resonance 8 Electrical resonance Cancellation of reactances, resulting in purely resistive network impedance Occurs at resonant frequencies Second- and higher-order circuits Reactances cancel – sum to zero ohms Inductive reactance is positive Capacitive reactance is negative Voltages/currents in the circuit may be much larger than source voltages/currents We’ll take a look at resonance in two classes of circuits: Series resonant circuits Parallel resonant circuits K. Webb ENGR 202 9 Series Resonant Circuits K. Webb ENGR 202 Series Resonant RLC Circuit 10 Series RLC circuit Second-order – one capacitor, one inductor Circuit will exhibit resonance Impedance of the network: 1 1 = + + = + − At the resonant frequency, or : + = 0 0= 0 1 1 = → = − 2 so 0 → 0 0 = and = 1 1 and 0 0 2 = K. Webb ENGR 202 0 Series RLC Circuit – Quality Factor 11 Quality factor, Ratio of inductive reactance at the resonant frequency to resistance = = 0 20 At resonance, inductive and capacitive reactances (magnitudes) are equal, so 1 1 = = The ratio of voltage magnitude0 across2 the0 inductor or capacitor to the voltage across the whole RLC network at resonance A measure of the sharpness of the resonance K. Webb ENGR 202 Series RLC Circuit – Zin vs. Qs 12 At = = 0 = 0° Q determines sharpness∠ of the resonance Higher Q yields faster transition from capacitive, through resistive, to inductive regions To increase Q: Increase L Reduce R and/or C K. Webb ENGR 202 Series RLC Circuit – Zin 13 Understanding the impedance of a series resonant circuit Capacitor impedance goes up as f Inductor goes down. impedance goes up as f = : goes up. = 0 is real =0° ∠ : : = +90° 0 = 90° ≫ looks 0 ≪ looks ∠inductive ∠capacitive − K. Webb ENGR 202 Series RLC Circuit – Voltages and Currents 14 At , = , so the current phasor is =0 = ∠0° Capacitor voltage at resonance: = = = 90° 90° ∠0° ∠ − Recalling the expression for 0quality 0 factor∠ of a series0 resonant circuit, we have = 90° The voltage across the capacitor ⋅ ∠ is− the source voltage multiplied by the quality factor and phase shifted by 90° K. Webb − ENGR 202 Series RLC Circuit – Voltages and Currents 15 The inductor voltage at resonance: = = ∠0° ⋅ 0∠90° ⋅ 0 = ⋅ 0 Again, substituting in the expression for quality factor gives = + 90° The voltage across the inductor is the source⋅ ∠ voltage multiplied by the quality factor and phase shifted by +90° Capacitor and inductor voltage at resonance: Equal magnitude 180° out of phase – opposite sign – they cancel K. Webb ENGR 202 Series RLC Circuit – Voltages and Currents 16 Now assign component values The resonant frequency is 1 = = 100 0 The quality factor is 100 1 = = = 10 10 0 ⋅ The current phasor at the resonant Ωfrequency is 1 = = = 10 ∠0° 100∠0° K. Webb Ω ENGR 202 Series RLC Circuit – Voltages and Currents 17 The capacitor voltage at the resonant frequency is = = 90° = 90° ∠ − ⋅ ∠ − = 1001 090° = 10 ⋅ ∠90°− The inductor voltage at the resonant∠ − frequency: = = = + 90° ⋅ 0 ⋅ 0 ⋅ ∠ = 10 1 + 90° = 10 ⋅ +∠90° K. Webb ∠ ENGR 202 Series RLC Circuit – Voltages and Currents 18 = 1 = 0° ∠ = = = 10 and are 180° out of phase They cancel KVL is satisfied K. Webb ENGR 202 19 Parallel Resonant Circuits K. Webb ENGR 202 Parallel Resonant RLC Circuit 20 Parallel RLC circuit Second-order – one capacitor, one inductor Circuit will exhibit resonance Impedance of the network: 1 1 1 1 = + + −1 = + −1 − At the resonant frequency, or : + = 0 0= 0 1 1 = → = − 2 so 0 0 0 → = and = and 1 1 0 0 2 = 0 K. Webb ENGR 202 Parallel RLC Circuit – Quality Factor 21 Quality factor, Ratio of inductive susceptance at the resonant frequency to conductance 1/ = = = 1/ 0 At resonance, inductive and capacitive 0susceptances 20 (magnitudes) are equal, so = = 0 0 The ratio of current magnitude through2 the inductor or capacitor to the current through the whole RLC network at resonance A measure of the sharpness of the resonance K. Webb ENGR 202 Parallel RLC Circuit – vs. 22 At = Impedance Magnitude Normalized by R = 1 0 0.8 = 0° |/R 0.6 in |Z Q determines 0.4 ∠ 0.2 sharpness of the 0 0.8 0.85 0.9 0.95 1 1.05 1.1 1.15 1.2 resonance Phase Angle of the Impedance Higher Q yields 100 faster transition Qp = 10 50 from inductive, Qp = 20 Q = 50 through resistive, to 0 p capacitive regions Qp = 100 Phase [deg] -50 To increase Q: -100 0.8 0.85 0.9 0.95 1 1.05 1.1 1.15 1.2 Reduce L Normalized Frequency (f/fo) Increase R and/or C K. Webb ENGR 202 Parallel RLC Circuit – 23 Understanding the impedance of a series resonant circuit = : = Capacitor 0 is real tends toward a short as Inductor tends toward a short 0 as → ∞ →: = +90° 0 : ≪ looks = 90° ∠inductive 0 ≫ looks ∠capacitive − = : is real 0 = 0° K. Webb ENGR 202 ∠ Parllel RLC Circuit – Voltages and Currents 24 Sinusoidal current source, At resonance, = , so the voltage across the network is: = = Current through the capacitor at resonance: ∠0° ⋅ = = Recalling the expression ⋅ for0 quality ⋅ factor0 of the parallel resonant circuit, we have = The current through the ⋅ capacitor∠90° is the source current multiplied by the quality factor and phase shifted by 90° K. Webb ENGR 202 Parallel RLC Circuit – Voltages and Currents 25 The inductor current at resonance: = = 90° 0 0 ∠ − Again, substituting in the expression for quality factor gives = 90° The current through the inductor is the ⋅source∠ − current multiplied by the quality factor and phase shifted by 90° Capacitor and inductor current at resonance:− Equal magnitude 180° out of phase – opposite sign – they cancel K. Webb ENGR 202 Parallel RLC Circuit – Voltages and Currents 26 Now, assign component values The resonant frequency is 1 = = 1 The quality factor is 0 100 = = = 100 1 Ω1 0 ⋅ The phasor for the voltage across the network at the resonant frequency is = = 100 100 = � ⋅ Ω 10∠0° K. Webb ENGR 202 Parallel RLC Circuit – Voltages and Currents 27 The capacitor current at the resonant frequency is = = = ⋅ 0 = 1⋅000100 = 10 ⋅ ∠90° ⋅ ∠90° The inductor current at the resonant frequency: = = 90° = 90° ∠ − ⋅ ∠ − = 1000 1000 90° = 10 ⋅ 90°∠ − ∠ − K. Webb ENGR 202 Parallel RLC Circuit – Voltages and Currents 28 = 1 = 0° ∠ = = = 10 180° and are out of phase They cancel KCL is satisfied K. Webb ENGR 202 29 Example Problems K. Webb ENGR 202 Determine the voltage across the capacitor, , at the resonant frequency. K. Webb ENGR 202 K. Webb ENGR 202 Determine , , , and at the resonant frequency. K. Webb ENGR 202 K. Webb ENGR 202 Determine R, such that = 100 at the resonant frequency. K. Webb ENGR 202 35 Second-Order Filters K. Webb ENGR 202 Second-Order Filters as Voltage Dividers 36 Derive the frequency response functions of second-order filters by treating the circuits as voltage dividers = + 2 1 2 Now, and can be either a single R, L, or C, or a series or parallel combination of any two 1 2 Possible combinations of components for or : 1 2 K. Webb ENGR 202 37 Second-Order Band Pass Filter K. Webb ENGR 202 Second-Order Band Pass Filter 38 One option for a second-order band pass filter: The frequency response function: = + 2 where 1 2 = and = + = 1 −1 so 2 1 2 1+ 1 + = = + 2 + + 1 + 2 2 / = + / + 1/ 2 K. Webb ENGR 202 Second-Order Band Pass Filter 39 Consider the filter’s behavior at three limiting cases for frequency 0: = : : open , cancel short 0 → short || open → ∞open → → shorted to = shorted to ground → → ground → Gain 1 Gain 0 Gain 0 → K. Webb → → ENGR 202 Second-Order Band Pass Filter 40 A second option for a second-order band pass filter: Now, the impedances are: 1 + 1 = + = 2 1 = The 2frequency response function: = = + 1 + + 1 + 2 2 / = + / + 1/ 2 K. Webb ENGR 202 Second-Order Band Pass Filter 41 Consider the filter’s behavior at three limiting cases for frequency 0: = : : open , cancel open 0 → short , short → ∞ short → → Current 0 = Current 0 → → → 0 Gain 1 0 → → Gain 0 Gain 0 → → → K. Webb → → ENGR 202 2nd-Order BPF – General-Form Frequency Response 42 Each of the two BPF variations has the same resonant frequency: 1 = 0 They have different frequency response2 functions and quality factors: 1 = = = = 0 0 0 0 / / = = + / + 1/ + / + 1/ 2 2 Each frequency response function can be expressed in terms of and : 0 = 0 + + 2 0 2 0 K.
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