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Rahul Maheshwari

This document provides an introduction to resonant circuits, discussing both series and parallel resonance. It defines resonance as occurring when the input voltage and current are in phase for a passive RLC circuit. For series resonance, the input impedance is completely real at resonance. Key equations for the resonant frequency, bandwidth, and quality factor Q are derived. Matlab programs are included to simulate the frequency response for different Q values. Duality between series and parallel circuits is also explained, where equations can be transformed between the two by substituting component values. Examples are worked through to calculate resonant components and bandwidth from given parameters.

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1 Introduction To Resonant Circuits University of Tennessee, Knoxville ECE Department wlg
2 Resonance In Electric Circuits Any passive electric circuit will resonate if it has an inductor and capacitor. Resonance is characterized by the input voltage and current being in phase. The driving point impedance (or admittance) is completely real when this condition exists. In this presentation we will consider (a) series resonance, and (b) parallel resonance.
3 Series Resonance Consider the series RLC circuit shown below. R L C + _ IV V = VM 0 The input impedance is given by: 1 ( )Z R j wL wC = + The magnitude of the circuit current is; 2 2 | | 1 ( ) mV I I R wL wC = = +
4 Series Resonance Resonance occurs when, 1 wL wC = At resonance we designate w as wo and write; 1 ow LC = This is an important equation to remember. It applies to both series And parallel resonant circuits.
5 Series Resonance The magnitude of the current response for the series resonance circuit is as shown below. mV R 2 mV R w |I| wow1 w2 Bandwidth: BW = wBW = w2 w1 Half power point
6 Series Resonance The peak power delivered to the circuit is; 2 mV P R = The so-called half-power is given when 2 mV I R = . We find the frequencies, w1 and w2, at which this half-power occurs by using; 2 21 2 ( )R R wL wC = +
7 Series Resonance After some insightful algebra one will find two frequencies at which the previous equation is satisfied, they are: 2 1 1 2 2 R R w L L LC 錚 錚 = + +錚 錚 錚 錚 and 2 2 1 2 2 R R w L L LC 錚 錚 = + +錚 錚 錚 錚 The two half-power frequencies are related to the resonant frequency by 1 2ow w w=
8 Series Resonance The bandwidth of the series resonant circuit is given by; 2 1b R BW w w w L = = = We define the Q (quality factor) of the circuit as; 1 1o o w L L Q R w RC R C 錚 錚 = = = 錚 錚 錚 錚 Using Q, we can write the bandwidth as; ow BW Q = These are all important relationships.
9 Series Resonance An Observation: If Q > 10, one can safely use the approximation; 1 2 2 2 o o BW BW w w and w w= = + These are useful approximations.
10 Series Resonance An Observation: By using Q = woL/R in the equations for w1and w2 we have; 2 2 1 1 1 2 2 ow w Q Q 錚 錚刻 錚駈 錚= + +錚 錚 錚 錚削 錚醐０ 錚 2 1 1 1 1 2 2 ow w Q Q 錚 錚刻 錚金錚 錚= + +錚 錚 錚 錚削 錚醐０ 錚 and
11 Series Resonance In order to get some feel for how the numerical value of Q influences the resonant and also get a better appreciation of the s-plane, we consider the following example. It is easy to show the following for the series RLC circuit. 2 1 ( ) 1 1( ) ( ) s I s L RV s Z s s s L LC = = + + In the following example, three cases for the about transfer function will be considered. We will keep wo the same for all three cases. The numerator gain,k, will (a) first be set k to 2 for the three cases, then (b) the value of k will be set so that each response is 1 at resonance.
12 Series Resonance An Example Illustrating Resonance: The 3 transfer functions considered are: Case 1: Case 2: Case 3: 2 2 400 ks s s+ + 2 5 400 ks s s+ + 2 10 400 ks s s+ +
13 Series Resonance An Example Illustrating Resonance: The poles for the three cases are given below. Case 1: Case 2: Case 3: 2 2 400 ( 1 19.97)( 1 19.97)s s s j s j+ + = + + + 2 5 400 ( 2.5 19.84)( 2.5 19.84)s s s j s j+ + = + + + 2 10 400 ( 5 19.36)( 5 19.36)s s s j s j+ + = + + +
14 Series Resonance Comments: Observe the denominator of the CE equation. 2 1R s s L LC + + Compare to actual characteristic equation for Case 1: 2 2 400s s+ + 2 400ow = 20w = 2 R BW L = = 10ow Q BW = = rad/sec rad/sec
15 Series Resonance Poles and Zeros In the s-plane: s-plane jw axis axis 0 0 20 -20 xx x x x x ( 3) (2) (1) ( 3) (2) (1) -5 -2.5 -1 Note the location of the poles for the three cases. Also note there is a zero at the origin.
16 Series Resonance Comments: The frequency response starts at the origin in the s-plane. At the origin the transfer function is zero because there is a zero at the origin. As you get closer and closer to the complex pole, which has a j parts in the neighborhood of 20, the response starts to increase. The response continues to increase until we reach w = 20. From there on the response decreases. We should be able to reason through why the response has the above characteristics, using a graphical approach.
17 Series Resonance Matlab Program For The Study: % name of program is freqtest.m % written for 202 S2002, wlg %CASE ONE DATA: K = 2; num1 = [K 0]; den1 = [1 2 400]; num2 = [K 0]; den2 = [1 5 400]; num3 = [K 0]; den3 = [1 10 400]; w = .1:.1:60; grid H1 = bode(num1,den1,w); magH1=abs(H1); H2 = bode(num2,den2,w); magH2=abs(H2); H3 = bode(num3,den3,w); magH3=abs(H3); plot(w,magH1, w, magH2, w,magH3) grid xlabel('w(rad/sec)') ylabel('Amplitude') gtext('Q = 10, 4, 2')
18 0 10 20 30 40 50 60 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 w(rad/sec) Amplitude Q = 10, 4, 2 Series Resonance Program Output
19 Series Resonance Comments: cont. From earlier work: 2 1 2 1 1 , 1 2 2 ow w w Q Q 錚 錚刻 錚饗縁 錚= + +錚 錚 錚 錚削 錚醐０ 錚 With Q = 10, this gives; w1= 19.51 rad/sec, w2 = 20.51 rad/sec Compare this to the approximation: w1 = w0 BW = 20 1 = 19 rad/sec, w2 = 21 rad/sec So basically we can find all the series resonant parameters if we are given the numerical form of the CE of the transfer function.
20 Series Resonance Next Case: Normalize all responses to 1 at wo 0 10 20 30 40 50 60 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 w(rad/sec) Amplitude Q = 10, 4, 2
21 Series Resonance Three dB Calculations: Now we use the analytical expressions to calculate w1 and w2. We will then compare these values to what we find from the Matlab simulation. Using the following equations with Q = 2, 錚 錚 錚 錚 錚 錚 錚 錚 +錚 錚 錚 錚 錚 錚 += 1 2 1 2 1 , 2 21 Q w Q www oo we find, w1 = 15.62 rad/sec w2 = 21.62 rad/sec
22 Series Resonance Checking w1 and w2 15.3000 0.6779 15.4000 0.6871 15.5000 0.6964 15.6000 0.7057 15.7000 0.7150 15.8000 0.7244 w1 25.3000 0.7254 25.4000 0.7195 25.5000 0.7137 25.6000 0.7080 25.7000 0.7023 25.8000 0.6967 25.9000 0.6912 w2 This verifies the previous calculations. Now we shall look at Parallel Resonance. (cut-outs from the simulation)
23 Parallel Resonance Background Consider the circuits shown below: I R L C V I R L CV 錚 錚 錚 錚 錚 錚 ++= jwL jwC R VI 11 錚 錚 錚 錚 錚 錚 ++= jwC jwLRIV 1
24 錚 錚 錚 錚 錚 錚 ++= jwL jwC R VI 11 錚 錚 錚 錚 錚 錚 ++= jwC jwLRIV 1 We notice the above equations are the same provided: VI R R 1 CL If we make the inner-change, then one equation becomes the same as the other. For such case, we say the one circuit is the dual of the other. Series Resonance Duality If we make the inner-change, then one equation becomes the same as the other. For such case, we say the one circuit is the dual of the other.
25 Parallel Resonance Background What this means is that for all the equations we have derived for the parallel resonant circuit, we can use for the series resonant circuit provided we make the substitutions: R bereplacedR 1 LbyreplacedC CbyreplacedL What this means is that for all the equations we have derived for the parallel resonant circuit, we can use for the series resonant circuit provided we make the substitutions: What this means is that for all the equations we have derived for the parallel resonant circuit, we can use for the series resonant circuit provided we make the substitutions:
26 Parallel Resonance Parallel Resonance Series Resonance R Lw Q O = LC wO 1 = LC wO 1 = RCwQ o = L R wwwBW BW === )( 12 RC wBW BW 1 == w 21 ,ww 錚 錚 錚 錚 錚 錚 錚 錚 +錚 錚 錚 錚 錚 錚 += LCL R L R ww 1 22 , 2 21 錚 錚 錚 錚 錚 錚 錚 錚 +錚 錚 錚 錚 錚 錚 += LCRCRC ww 1 2 1 2 1 , 2 21 21 ,ww 錚 錚 錚 錚 錚 錚 錚 錚 +錚 錚 錚 錚 錚 錚 += 1 2 1 2 1 , 2 21 QQ www o 錚 錚 錚 錚 錚 錚 錚 錚 +錚 錚 錚 錚 錚 錚 += 1 2 1 2 1 , 2 21 QQ www o
27 Resonance Example 1: Determine the resonant frequency for the circuit below. jwRCLCw jwLLRCw jwC jwLR jwC RjwL Z NI + + = ++ + = )1( )( 1 ) 1 ( 2 2 At resonance, the phase angle of Z must be equal to zero.
28 jwRCLCw jwLLRCw + + )1( )( 2 2 Resonance Analysis For zero phase; LCw wRC LCRw wL 22 1()( = This gives; 12222 = CRwLCw or )( 1 22 CRLC wo =
29 Parallel Resonance Example 2: A parallel RLC resonant circuit has a resonant frequency admittance of 2x10-2 S(mohs). The Q of the circuit is 50, and the resonant frequency is 10,000 rad/sec. Calculate the values of R, L, and C. Find the half-power frequencies and the bandwidth. First, R = 1/G = 1/(0.02) = 50 ohms. Second, from R Lw Q O = , we solve for L, knowing Q, R, and wo to find L = 0.25 H. Third, we can use F xRw Q C O 袖100 50000,10 50 === A parallel RLC resonant circuit has a resonant frequency admittance of 2x10-2 S(mohs). The Q of the circuit is 50, and the resonant frequency is 10,000 rad/sec. Calculate the values of R, L, and C. Find the half-power frequencies and the bandwidth. A series RLC resonant circuit has a resonant frequency admittance of 2x10-2 S(mohs). The Q of the circuit is 50, and the resonant frequency is 10,000 rad/sec. Calculate the values of R, L, and C. Find the half-power frequencies and the bandwidth.
30 Parallel Resonance Example 2: (continued) Fourth: We can use sec/200 50 101 4 rad x Q w w o BW === and Fifth: Use the approximations; w1 = wo - 0.5wBW = 10,000 100 = 9,900 rad/sec w2 = wo - 0.5wBW = 10,000 + 100 = 10,100 rad/sec
31 Extension of Series Resonance Peak Voltages and Resonance: VS R L C + _ I + + + _ _ _ VR VL VC We know the following: When w = wo = 1 LC , VS and I are in phase, the driving point impedance is purely real and equal to R. A plot of |I| shows that it is maximum at w = wo. We know the standard equations for series resonance applies: Q, wBW, etc.
32 Extension of Series Resonance Reflection: A question that arises is what is the nature of VR, VL, and VC? A little reflection shows that VR is a peak value at wo. But we are not sure about the other two voltages. We know that at resonance they are equal and they have a magnitude of QxVS. max 2 1 1 2 ow w Q = The above being true, we might ask, what is the frequency at which the voltage across the inductor is a maximum? We answer this question by simulation Irwin shows that the frequency at which the voltage across the capacitor is a maximum is given by; Irwin shows that the frequency at which the voltage across the capacitor is a maximum is given by; Irwin shows that the frequency at which the voltage across the capacitor is a maximum is given by;
33 Extension of Series Resonance Series RLC Transfer Functions: The following transfer functions apply to the series RLC circuit. 2 1 ( ) 1( ) C S V s LC RV s s s L LC = + + 2 2 ( ) 1( ) L S V s s RV s s s L LC = + + 2 ( ) 1( ) R S R s V s L RV s s s L LC = + +
34 Extension of Series Resonance Parameter Selection: We select values of R, L. and C for this first case so that Q = 2 and wo = 2000 rad/sec. Appropriate values are; R = 50 ohms, L = .05 H, C = 5袖F. The transfer functions become as follows: 6 2 6 4 10 1000 4 10 C S V x V s s x = + + 2 2 6 1000 4 10 L S V s V s s x = + + 2 6 1000 1000 4 10 R S V s V s s x = + +
35 Extension of Series Resonance Matlab Simulation: % program is freqcompare.m % written for 202 S2002, wlg numC = 4e+6; denC = [1 1000 4e+6]; numL = [1 0 0]; denL = [1 1000 4e+6]; numR = [1000 0]; denR = [1 1000 4e+6]; w = 200:1:4000; grid HC = bode(numC,denC,w); magHC = abs(HC); grid HC = bode(numC,denC,w); magHC = abs(HC); HL = bode(numL,denL,w); magHL = abs(HL); HR = bode(numR,denR,w); magHR = abs(HR); plot(w,magHC,'k-', w, magHL,'k--', w, magHR, 'k:') grid xlabel('w(rad/sec)') ylabel('Amplitude') title(' Rsesponse for RLC series circuit, Q =2') gtext('VC') gtext('VL') gtext(' VR')
36 0 500 1000 1500 2000 2500 3000 3500 4000 0 0.5 1 1.5 2 2.5 w(rad/sec) Amplitude Rsesponse for RLC series circuit, Q =2 VC VL VR Exnsion of Series Resonance Simulation Results Q = 2
37 Exnsion of Series Resonance Analysis of the problem: VS R=50 L=5 mH C=5 袖F + _ I + + + _ _ _ VR VL VC Given the previous circuit. Find Q, w0, wmax, |Vc| at wo, and |Vc| at wmax Solution: sec/2000 1051050 11 62 rad xxxLC wO === 2 50 105102 23 === xxx R Lw Q O
38 Exnsion of Series Resonance Problem Solution: oOMAX w Q ww 9354.0 2 1 1 2 == )(212|||| peakvoltsxVQwatV SOR === ( ))066.2 968.0 2 4 1 1 || || 2 peakvolts Q VQx watV S MAXC == = Now check the computer printout.
39 Exnsion of Series Resonance Problem Solution (Simulation): 1.0e+003 * 1.8600000 0.002065141 1.8620000 0.002065292 1.8640000 0.002065411 1.8660000 0.002065501 1.8680000 0.002065560 1.8700000 0.002065588 1.8720000 0.002065585 1.8740000 0.002065552 1.8760000 0.002065487 1.8780000 0.002065392 1.8800000 0.002065265 1.8820000 0.002065107 1.8840000 0.002064917 Maximum
40 Extension of Series Resonance Simulation Results: 0 500 1000 1500 2000 2500 3000 3500 4000 0 2 4 6 8 10 12 w(rad/sec) Amplitude Rsesponse for RLC series circuit, Q =10 VC VL VR Q=10
41 Exnsion of Series Resonance Observations From The Study: The voltage across the capacitor and inductor for a series RLC circuit is not at peak values at resonance for small Q (Q <3). Even for Q<3, the voltages across the capacitor and inductor are equal at resonance and their values will be QxVS. For Q>10, the voltages across the capacitors are for all practical purposes at their peak values and will be QxVS. Regardless of the value of Q, the voltage across the resistor reaches its peak value at w = wo. For high Q, the equations discussed for series RLC resonance can be applied to any voltage in the RLC circuit. For Q<3, this is not true.
42 Extension of Resonant Circuits Given the following circuit: I + _ + _ VC R L We want to find the frequency, wr, at which the transfer function for V/I will resonate. The transfer function will exhibit resonance when the phase angle between V and I are zero.
43 Extension of Resonant Circuits The desired transfer functions is; (1/ )( ) 1/ V sC R sL I R sL sC + = + + This equation can be simplified to; 2 1 V R sL I LCs RCs + = + + With s jw 2 (1 ) V R jwL I w LC jwR + = +
44 Extension of Resonant Circuits Resonant Condition: For the previous transfer function to be at a resonant point, the phase angle of the numerator must be equal to the phase angle of the denominator. num dem慮 慮 = 1 tannum wL R 慮 錚 錚 = 錚 錚 錚 錚 or, 1 2 tan (1 ) den wRC w LC 慮 錚 錚 = 錚 錚 錚 錚 , . Therefore; 2 (1 ) wL wRC R w LC =
45 Extension of Resonant Circuits Resonant Condition Analysis: Canceling the ws in the numerator and cross multiplying gives, 2 2 2 2 2 (1 )L w LC R C or w L C L R C = = This gives, 2 2 1 r R w LC L = Notice that if the ratio of R/L is small compared to 1/LC, we have 1 r ow w LC = =
46 Extension of Resonant Circuits Resonant Condition Analysis: What is the significance of wr and wo in the previous two equations? Clearly wr is a lower frequency of the two. To answer this question, consider the following example. Given the following circuit with the indicated parameters. Write a Matlab program that will determine the frequency response of the transfer function of the voltage to the current as indicated. I + _ + _ VC R L
47 Extension of Resonant Circuits Resonant Condition Analysis: Matlab Simulation: We consider two cases: Case 1: R = 3 ohms C = 6.25x10-5 F L = 0.01 H Case 2: R = 1 ohms C = 6.25x10-5 F L = 0.01 H wr= 2646 rad/sec wr= 3873 rad/sec For both cases, wo = 4000 rad/sec
48 Extension of Resonant Circuits Resonant Condition Analysis: Matlab Simulation: The transfer functions to be simulated are given below. 8 2 7 0.001 3 6.25 10 1.875 10 1 V s I x s x + = + + Case 1: Case 2: 8 2 5 0.001 1 6.25 10 6.25 10 1 V s I x s x + = + +
49 Extension of Resonant Circuits 0 1000 2000 3000 4000 5000 6000 7000 8000 0 2 4 6 8 10 12 14 16 18 w(rad/sec) Amplitude Rsesponse for Resistance in series with L then Parallel with C R= 3 ohms R=1 ohm 2646 rad/sec
50 Extension of Resonant Circuits What can be learned from this example? wr does not seem to have much meaning in this problem. What is wr if R = 3.99 ohms? Just because a circuit is operated at the resonant frequency does not mean it will have a peak in the response at the frequency. For circuits that are fairly complicated and can resonant, It is probably easier to use a simulation program similar to Matlab to find out what is going on in the circuit.
51 End of Lesson Basic Laws of Circuits Resonant Circuits Circuits

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  • 1. 1 Introduction To Resonant Circuits University of Tennessee, Knoxville ECE Department wlg
  • 2. 2 Resonance In Electric Circuits Any passive electric circuit will resonate if it has an inductor and capacitor. Resonance is characterized by the input voltage and current being in phase. The driving point impedance (or admittance) is completely real when this condition exists. In this presentation we will consider (a) series resonance, and (b) parallel resonance.
  • 3. 3 Series Resonance Consider the series RLC circuit shown below. R L C + _ IV V = VM 0 The input impedance is given by: 1 ( )Z R j wL wC = + The magnitude of the circuit current is; 2 2 | | 1 ( ) mV I I R wL wC = = +
  • 4. 4 Series Resonance Resonance occurs when, 1 wL wC = At resonance we designate w as wo and write; 1 ow LC = This is an important equation to remember. It applies to both series And parallel resonant circuits.
  • 5. 5 Series Resonance The magnitude of the current response for the series resonance circuit is as shown below. mV R 2 mV R w |I| wow1 w2 Bandwidth: BW = wBW = w2 w1 Half power point
  • 6. 6 Series Resonance The peak power delivered to the circuit is; 2 mV P R = The so-called half-power is given when 2 mV I R = . We find the frequencies, w1 and w2, at which this half-power occurs by using; 2 21 2 ( )R R wL wC = +
  • 7. 7 Series Resonance After some insightful algebra one will find two frequencies at which the previous equation is satisfied, they are: 2 1 1 2 2 R R w L L LC 錚 錚 = + +錚 錚 錚 錚 and 2 2 1 2 2 R R w L L LC 錚 錚 = + +錚 錚 錚 錚 The two half-power frequencies are related to the resonant frequency by 1 2ow w w=
  • 8. 8 Series Resonance The bandwidth of the series resonant circuit is given by; 2 1b R BW w w w L = = = We define the Q (quality factor) of the circuit as; 1 1o o w L L Q R w RC R C 錚 錚 = = = 錚 錚 錚 錚 Using Q, we can write the bandwidth as; ow BW Q = These are all important relationships.
  • 9. 9 Series Resonance An Observation: If Q > 10, one can safely use the approximation; 1 2 2 2 o o BW BW w w and w w= = + These are useful approximations.
  • 10. 10 Series Resonance An Observation: By using Q = woL/R in the equations for w1and w2 we have; 2 2 1 1 1 2 2 ow w Q Q 錚 錚刻 錚駈 錚= + +錚 錚 錚 錚削 錚醐０ 錚 2 1 1 1 1 2 2 ow w Q Q 錚 錚刻 錚金錚 錚= + +錚 錚 錚 錚削 錚醐０ 錚 and
  • 11. 11 Series Resonance In order to get some feel for how the numerical value of Q influences the resonant and also get a better appreciation of the s-plane, we consider the following example. It is easy to show the following for the series RLC circuit. 2 1 ( ) 1 1( ) ( ) s I s L RV s Z s s s L LC = = + + In the following example, three cases for the about transfer function will be considered. We will keep wo the same for all three cases. The numerator gain,k, will (a) first be set k to 2 for the three cases, then (b) the value of k will be set so that each response is 1 at resonance.
  • 12. 12 Series Resonance An Example Illustrating Resonance: The 3 transfer functions considered are: Case 1: Case 2: Case 3: 2 2 400 ks s s+ + 2 5 400 ks s s+ + 2 10 400 ks s s+ +
  • 13. 13 Series Resonance An Example Illustrating Resonance: The poles for the three cases are given below. Case 1: Case 2: Case 3: 2 2 400 ( 1 19.97)( 1 19.97)s s s j s j+ + = + + + 2 5 400 ( 2.5 19.84)( 2.5 19.84)s s s j s j+ + = + + + 2 10 400 ( 5 19.36)( 5 19.36)s s s j s j+ + = + + +
  • 14. 14 Series Resonance Comments: Observe the denominator of the CE equation. 2 1R s s L LC + + Compare to actual characteristic equation for Case 1: 2 2 400s s+ + 2 400ow = 20w = 2 R BW L = = 10ow Q BW = = rad/sec rad/sec
  • 15. 15 Series Resonance Poles and Zeros In the s-plane: s-plane jw axis axis 0 0 20 -20 xx x x x x ( 3) (2) (1) ( 3) (2) (1) -5 -2.5 -1 Note the location of the poles for the three cases. Also note there is a zero at the origin.
  • 16. 16 Series Resonance Comments: The frequency response starts at the origin in the s-plane. At the origin the transfer function is zero because there is a zero at the origin. As you get closer and closer to the complex pole, which has a j parts in the neighborhood of 20, the response starts to increase. The response continues to increase until we reach w = 20. From there on the response decreases. We should be able to reason through why the response has the above characteristics, using a graphical approach.
  • 17. 17 Series Resonance Matlab Program For The Study: % name of program is freqtest.m % written for 202 S2002, wlg %CASE ONE DATA: K = 2; num1 = [K 0]; den1 = [1 2 400]; num2 = [K 0]; den2 = [1 5 400]; num3 = [K 0]; den3 = [1 10 400]; w = .1:.1:60; grid H1 = bode(num1,den1,w); magH1=abs(H1); H2 = bode(num2,den2,w); magH2=abs(H2); H3 = bode(num3,den3,w); magH3=abs(H3); plot(w,magH1, w, magH2, w,magH3) grid xlabel('w(rad/sec)') ylabel('Amplitude') gtext('Q = 10, 4, 2')
  • 18. 18 0 10 20 30 40 50 60 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 w(rad/sec) Amplitude Q = 10, 4, 2 Series Resonance Program Output
  • 19. 19 Series Resonance Comments: cont. From earlier work: 2 1 2 1 1 , 1 2 2 ow w w Q Q 錚 錚刻 錚饗縁 錚= + +錚 錚 錚 錚削 錚醐０ 錚 With Q = 10, this gives; w1= 19.51 rad/sec, w2 = 20.51 rad/sec Compare this to the approximation: w1 = w0 BW = 20 1 = 19 rad/sec, w2 = 21 rad/sec So basically we can find all the series resonant parameters if we are given the numerical form of the CE of the transfer function.
  • 20. 20 Series Resonance Next Case: Normalize all responses to 1 at wo 0 10 20 30 40 50 60 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 w(rad/sec) Amplitude Q = 10, 4, 2
  • 21. 21 Series Resonance Three dB Calculations: Now we use the analytical expressions to calculate w1 and w2. We will then compare these values to what we find from the Matlab simulation. Using the following equations with Q = 2, 錚 錚 錚 錚 錚 錚 錚 錚 +錚 錚 錚 錚 錚 錚 += 1 2 1 2 1 , 2 21 Q w Q www oo we find, w1 = 15.62 rad/sec w2 = 21.62 rad/sec
  • 22. 22 Series Resonance Checking w1 and w2 15.3000 0.6779 15.4000 0.6871 15.5000 0.6964 15.6000 0.7057 15.7000 0.7150 15.8000 0.7244 w1 25.3000 0.7254 25.4000 0.7195 25.5000 0.7137 25.6000 0.7080 25.7000 0.7023 25.8000 0.6967 25.9000 0.6912 w2 This verifies the previous calculations. Now we shall look at Parallel Resonance. (cut-outs from the simulation)
  • 23. 23 Parallel Resonance Background Consider the circuits shown below: I R L C V I R L CV 錚 錚 錚 錚 錚 錚 ++= jwL jwC R VI 11 錚 錚 錚 錚 錚 錚 ++= jwC jwLRIV 1
  • 24. 24 錚 錚 錚 錚 錚 錚 ++= jwL jwC R VI 11 錚 錚 錚 錚 錚 錚 ++= jwC jwLRIV 1 We notice the above equations are the same provided: VI R R 1 CL If we make the inner-change, then one equation becomes the same as the other. For such case, we say the one circuit is the dual of the other. Series Resonance Duality If we make the inner-change, then one equation becomes the same as the other. For such case, we say the one circuit is the dual of the other.
  • 25. 25 Parallel Resonance Background What this means is that for all the equations we have derived for the parallel resonant circuit, we can use for the series resonant circuit provided we make the substitutions: R bereplacedR 1 LbyreplacedC CbyreplacedL What this means is that for all the equations we have derived for the parallel resonant circuit, we can use for the series resonant circuit provided we make the substitutions: What this means is that for all the equations we have derived for the parallel resonant circuit, we can use for the series resonant circuit provided we make the substitutions:
  • 26. 26 Parallel Resonance Parallel Resonance Series Resonance R Lw Q O = LC wO 1 = LC wO 1 = RCwQ o = L R wwwBW BW === )( 12 RC wBW BW 1 == w 21 ,ww 錚 錚 錚 錚 錚 錚 錚 錚 +錚 錚 錚 錚 錚 錚 += LCL R L R ww 1 22 , 2 21 錚 錚 錚 錚 錚 錚 錚 錚 +錚 錚 錚 錚 錚 錚 += LCRCRC ww 1 2 1 2 1 , 2 21 21 ,ww 錚 錚 錚 錚 錚 錚 錚 錚 +錚 錚 錚 錚 錚 錚 += 1 2 1 2 1 , 2 21 QQ www o 錚 錚 錚 錚 錚 錚 錚 錚 +錚 錚 錚 錚 錚 錚 += 1 2 1 2 1 , 2 21 QQ www o
  • 27. 27 Resonance Example 1: Determine the resonant frequency for the circuit below. jwRCLCw jwLLRCw jwC jwLR jwC RjwL Z NI + + = ++ + = )1( )( 1 ) 1 ( 2 2 At resonance, the phase angle of Z must be equal to zero.
  • 29. 29 Parallel Resonance Example 2: A parallel RLC resonant circuit has a resonant frequency admittance of 2x10-2 S(mohs). The Q of the circuit is 50, and the resonant frequency is 10,000 rad/sec. Calculate the values of R, L, and C. Find the half-power frequencies and the bandwidth. First, R = 1/G = 1/(0.02) = 50 ohms. Second, from R Lw Q O = , we solve for L, knowing Q, R, and wo to find L = 0.25 H. Third, we can use F xRw Q C O 袖100 50000,10 50 === A parallel RLC resonant circuit has a resonant frequency admittance of 2x10-2 S(mohs). The Q of the circuit is 50, and the resonant frequency is 10,000 rad/sec. Calculate the values of R, L, and C. Find the half-power frequencies and the bandwidth. A series RLC resonant circuit has a resonant frequency admittance of 2x10-2 S(mohs). The Q of the circuit is 50, and the resonant frequency is 10,000 rad/sec. Calculate the values of R, L, and C. Find the half-power frequencies and the bandwidth.
  • 30. 30 Parallel Resonance Example 2: (continued) Fourth: We can use sec/200 50 101 4 rad x Q w w o BW === and Fifth: Use the approximations; w1 = wo - 0.5wBW = 10,000 100 = 9,900 rad/sec w2 = wo - 0.5wBW = 10,000 + 100 = 10,100 rad/sec
  • 31. 31 Extension of Series Resonance Peak Voltages and Resonance: VS R L C + _ I + + + _ _ _ VR VL VC We know the following: When w = wo = 1 LC , VS and I are in phase, the driving point impedance is purely real and equal to R. A plot of |I| shows that it is maximum at w = wo. We know the standard equations for series resonance applies: Q, wBW, etc.
  • 32. 32 Extension of Series Resonance Reflection: A question that arises is what is the nature of VR, VL, and VC? A little reflection shows that VR is a peak value at wo. But we are not sure about the other two voltages. We know that at resonance they are equal and they have a magnitude of QxVS. max 2 1 1 2 ow w Q = The above being true, we might ask, what is the frequency at which the voltage across the inductor is a maximum? We answer this question by simulation Irwin shows that the frequency at which the voltage across the capacitor is a maximum is given by; Irwin shows that the frequency at which the voltage across the capacitor is a maximum is given by; Irwin shows that the frequency at which the voltage across the capacitor is a maximum is given by;
  • 33. 33 Extension of Series Resonance Series RLC Transfer Functions: The following transfer functions apply to the series RLC circuit. 2 1 ( ) 1( ) C S V s LC RV s s s L LC = + + 2 2 ( ) 1( ) L S V s s RV s s s L LC = + + 2 ( ) 1( ) R S R s V s L RV s s s L LC = + +
  • 34. 34 Extension of Series Resonance Parameter Selection: We select values of R, L. and C for this first case so that Q = 2 and wo = 2000 rad/sec. Appropriate values are; R = 50 ohms, L = .05 H, C = 5袖F. The transfer functions become as follows: 6 2 6 4 10 1000 4 10 C S V x V s s x = + + 2 2 6 1000 4 10 L S V s V s s x = + + 2 6 1000 1000 4 10 R S V s V s s x = + +
  • 35. 35 Extension of Series Resonance Matlab Simulation: % program is freqcompare.m % written for 202 S2002, wlg numC = 4e+6; denC = [1 1000 4e+6]; numL = [1 0 0]; denL = [1 1000 4e+6]; numR = [1000 0]; denR = [1 1000 4e+6]; w = 200:1:4000; grid HC = bode(numC,denC,w); magHC = abs(HC); grid HC = bode(numC,denC,w); magHC = abs(HC); HL = bode(numL,denL,w); magHL = abs(HL); HR = bode(numR,denR,w); magHR = abs(HR); plot(w,magHC,'k-', w, magHL,'k--', w, magHR, 'k:') grid xlabel('w(rad/sec)') ylabel('Amplitude') title(' Rsesponse for RLC series circuit, Q =2') gtext('VC') gtext('VL') gtext(' VR')
  • 36. 36 0 500 1000 1500 2000 2500 3000 3500 4000 0 0.5 1 1.5 2 2.5 w(rad/sec) Amplitude Rsesponse for RLC series circuit, Q =2 VC VL VR Exnsion of Series Resonance Simulation Results Q = 2
  • 37. 37 Exnsion of Series Resonance Analysis of the problem: VS R=50 L=5 mH C=5 袖F + _ I + + + _ _ _ VR VL VC Given the previous circuit. Find Q, w0, wmax, |Vc| at wo, and |Vc| at wmax Solution: sec/2000 1051050 11 62 rad xxxLC wO === 2 50 105102 23 === xxx R Lw Q O
  • 38. 38 Exnsion of Series Resonance Problem Solution: oOMAX w Q ww 9354.0 2 1 1 2 == )(212|||| peakvoltsxVQwatV SOR === ( ))066.2 968.0 2 4 1 1 || || 2 peakvolts Q VQx watV S MAXC == = Now check the computer printout.
  • 39. 39 Exnsion of Series Resonance Problem Solution (Simulation): 1.0e+003 * 1.8600000 0.002065141 1.8620000 0.002065292 1.8640000 0.002065411 1.8660000 0.002065501 1.8680000 0.002065560 1.8700000 0.002065588 1.8720000 0.002065585 1.8740000 0.002065552 1.8760000 0.002065487 1.8780000 0.002065392 1.8800000 0.002065265 1.8820000 0.002065107 1.8840000 0.002064917 Maximum
  • 40. 40 Extension of Series Resonance Simulation Results: 0 500 1000 1500 2000 2500 3000 3500 4000 0 2 4 6 8 10 12 w(rad/sec) Amplitude Rsesponse for RLC series circuit, Q =10 VC VL VR Q=10
  • 41. 41 Exnsion of Series Resonance Observations From The Study: The voltage across the capacitor and inductor for a series RLC circuit is not at peak values at resonance for small Q (Q <3). Even for Q<3, the voltages across the capacitor and inductor are equal at resonance and their values will be QxVS. For Q>10, the voltages across the capacitors are for all practical purposes at their peak values and will be QxVS. Regardless of the value of Q, the voltage across the resistor reaches its peak value at w = wo. For high Q, the equations discussed for series RLC resonance can be applied to any voltage in the RLC circuit. For Q<3, this is not true.
  • 42. 42 Extension of Resonant Circuits Given the following circuit: I + _ + _ VC R L We want to find the frequency, wr, at which the transfer function for V/I will resonate. The transfer function will exhibit resonance when the phase angle between V and I are zero.
  • 43. 43 Extension of Resonant Circuits The desired transfer functions is; (1/ )( ) 1/ V sC R sL I R sL sC + = + + This equation can be simplified to; 2 1 V R sL I LCs RCs + = + + With s jw 2 (1 ) V R jwL I w LC jwR + = +
  • 44. 44 Extension of Resonant Circuits Resonant Condition: For the previous transfer function to be at a resonant point, the phase angle of the numerator must be equal to the phase angle of the denominator. num dem慮 慮 = 1 tannum wL R 慮 錚 錚 = 錚 錚 錚 錚 or, 1 2 tan (1 ) den wRC w LC 慮 錚 錚 = 錚 錚 錚 錚 , . Therefore; 2 (1 ) wL wRC R w LC =
  • 45. 45 Extension of Resonant Circuits Resonant Condition Analysis: Canceling the ws in the numerator and cross multiplying gives, 2 2 2 2 2 (1 )L w LC R C or w L C L R C = = This gives, 2 2 1 r R w LC L = Notice that if the ratio of R/L is small compared to 1/LC, we have 1 r ow w LC = =
  • 46. 46 Extension of Resonant Circuits Resonant Condition Analysis: What is the significance of wr and wo in the previous two equations? Clearly wr is a lower frequency of the two. To answer this question, consider the following example. Given the following circuit with the indicated parameters. Write a Matlab program that will determine the frequency response of the transfer function of the voltage to the current as indicated. I + _ + _ VC R L
  • 47. 47 Extension of Resonant Circuits Resonant Condition Analysis: Matlab Simulation: We consider two cases: Case 1: R = 3 ohms C = 6.25x10-5 F L = 0.01 H Case 2: R = 1 ohms C = 6.25x10-5 F L = 0.01 H wr= 2646 rad/sec wr= 3873 rad/sec For both cases, wo = 4000 rad/sec
  • 48. 48 Extension of Resonant Circuits Resonant Condition Analysis: Matlab Simulation: The transfer functions to be simulated are given below. 8 2 7 0.001 3 6.25 10 1.875 10 1 V s I x s x + = + + Case 1: Case 2: 8 2 5 0.001 1 6.25 10 6.25 10 1 V s I x s x + = + +
  • 49. 49 Extension of Resonant Circuits 0 1000 2000 3000 4000 5000 6000 7000 8000 0 2 4 6 8 10 12 14 16 18 w(rad/sec) Amplitude Rsesponse for Resistance in series with L then Parallel with C R= 3 ohms R=1 ohm 2646 rad/sec
  • 50. 50 Extension of Resonant Circuits What can be learned from this example? wr does not seem to have much meaning in this problem. What is wr if R = 3.99 ohms? Just because a circuit is operated at the resonant frequency does not mean it will have a peak in the response at the frequency. For circuits that are fairly complicated and can resonant, It is probably easier to use a simulation program similar to Matlab to find out what is going on in the circuit.
  • 51. 51 End of Lesson Basic Laws of Circuits Resonant Circuits Circuits