2 Signal Frequency and Impedances First Order Filter Circuits Resonant and Second Order Filter Circuits... 13

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1 Lecture Notes: 3454 Physics and Electronics Lecture ( nd Half), Year: 7 Physics Department, Faculty of Science, Chulalongkorn University //7 Contents Power in Ac Circuits Signal Frequency and Impedances 9. First Order Filter Circuits esonant and Second Order Filter Circuits Problems 8 Power in Ac Circuits Time Dependent Power Given: V (t) = V m sin ωt I(t) = I m sin(ωt φ) From: P (t) = V (t)i(t) P (t) = V m I m sin ωt sin(ωt φ) Trigonometry: sin A sin B = cos(b A) cos(b A) Power in ac circuit depends on time. P (t) = V mi m [cos φ cos(ωt φ)] Power of Pure esistive Load V (t) = V m sin ωt I(t) = I m sin ωt P (t) = V m I m sin ωt 3454 Physics and Electronics (file: lec.tex) Tianprateep, M.

2 V, I V (t) I(t) t P t Figure Current, voltage, and power versus time for a purely resistive load. V, I V (t) I(t) t P t Figure Current, voltage, and power versus time for a purely inductive load Physics and Electronics (file: lec.tex) Tianprateep, M.

3 V, I V (t) I(t) t P t Figure 3 Current, voltage, and power versus time for a purely capacitive load. Power of Pure Inductive Load V (t) = V m sin ωt I(t) = I m sin(ωt 9 o ) P (t) = V mi m [cos(9 o ) cos(ωt 9 o )] = V mi m sin ωt Power of Pure Capacitive Load V (t) = V m sin ωt I(t) = I m sin(ωt 9 o ) P (t) = V mi m [cos(9 o ) cos(ωt 9 o )] = V mi m sin ωt Active Power or Average Power 3454 Physics and Electronics (file: lec.tex) 3 Tianprateep, M.

4 Active power or average power (unit in Watt,W): the energy delivered to the electrical elements in one cycle divided by the period. P = P av = T T P (t)dt Because T = π ω and cos(θ φ)dθ = sin(θ φ) P = π π { } V [ ] mi m cos φ cos(ωt φ) d(ωt) = V mi m cos φ = V rms I rms cos φ Power Factor and Power Angle Definition: In case: cos φ = power factor (P F ) φ = power angle V (t) = V m sin(ωt φ v ) I(t) = I m sin(ωt φ i ) φ = φ v φ i pure resistive load (φ = ; P F = ) : P = V rmsi rms pure inductive load (φ = 9 o ; P F = ) : P = pure capacitive load (φ = 9 o ; P F = ) : P = Power in Complex Number System In general cases; That is: V = V m φ v = V m cos φ v j V m sin φ v I = I m φ i = I m cos φ i j I m sin φ i V I = (V m cos φ v j V m sin φ v )(I m cos φ i j I m sin φ i ) = V m I m cos(φ v φ i ) j V m I m sin(φ v φ i ) P = e(v I ) Q = Im(V I ) 3454 Physics and Electronics (file: lec.tex) 4 Tianprateep, M.

5 Im Z X φ e Figure 4 The load impedance in a complex plane eactive Power and Apparent Power reactive power, Q (unit in Voltage Amperes eactive, VA): the energy flow back and forth to the energy storage elements (inductances and/or capacitances). Q = Im(V I ) = V rms I rms sin φ P Q = (V rms I rms ) cos φ (V rms I rms ) sin φ = (V rms I rms ) = AP AP = V rms I rms : apparent power (unit in Voltage Amperes, VA) The reactive power effects the power dissipation in the lines and transformers of a power distribution system. Thus, the industrial factory has to pay for this kind of power. That is 5 kw load means P = 5 kw, 5 kva load means AP = 5 kva, and a load absorbs 5 kva means Q = 5 kva. Power Triangle inductive load (I lags V ) φ AP P Q (φ = positive) φ P capacitive load (I leads V ) AP Q (φ = negative) Additional Power elationship Z = Z φ = j X 3454 Physics and Electronics (file: lec.tex) 5 Tianprateep, M.

6 X L I I C V (t) X C I V (t) = 9 I(t) = I =. I C =. 9 X L = j Ω X C = j Ω = Ω Figure 5 Circuit for example X = X = for inductance for capacitance cos φ = Z sin φ = X Z P = V mi m cos φ = V mi m = I m = I rms Q = V mi m sin φ = V mi m = I m X = I rmsx Z X Z P = V rms ; V rms: across resistance Q = V Xrms X ; V Xrms: across reactance Example Physics and Electronics (file: lec.tex) 6 Tianprateep, M.

7 Find power and reactive power φ = φ v φ i = 9 o (35 o ) = 45 o V Srms = V m / = / = 7.7 V I rms = I m / =.44/ =. A P = V Srms I rms cos φ = 7.7. cos(45 o ) =.5 W Q = V Srms I rms sin φ = 7.7. sin(45 o ) =.5 VA Q = Q L = I rmsx L = (.) () =. VA Q C = I CrmsX C = (. ) () =.5 VA Q S = Q L Q C P L = P C = P = I rms = (. ) () =.5 W Example. Find P S, Q, P F, and phasor current I P F A =.5 leading: capacitive (X A =, Q A = ) P F B =.7 lagging: inductive (X B =, Q B = ) P A = V rms I rms P F A = ( 4 )(.5) = 5 kw Q A = APA P A = ( 4 ) (5) = 8.67 kva 3454 Physics and Electronics (file: lec.tex) 7 Tianprateep, M.

8 I V (t) A B I A I B V (t) = 44 3 AP A = kva P F A =.5 leading P B = 5 kw P F B =.7 lagging Figure 6 Circuit for example AP B = P B /P F = 5/.7 = 7.4 kva Q B = APB P B = (74) (5) = 5. kva P S = P A P B = kw Q S = Q A Q B = 3.56 kva φ S = tan Q S P S = 9.6 o P F S = cos φ S =.94 (94.% leading) AP S = P Q =.6 kva V Srms = 44 = kv I rms = AP S V Srms =.6 A I m = I rms = 5 A φ i = φ v φ S = 3 o (9.6 o ) = 49.6 o I = o A Power Factor Correction 3454 Physics and Electronics (file: lec.tex) 8 Tianprateep, M.

9 φ AP P Q Q Figure 7 Power triangle for power factor correction In heavy industry, Q causes higher currents in the power distribution system. The energy rates charged to industry depend on the P F. To decrease P F, the capacitors is placed in parallel with an inductive load to decrease the sum of Q. Example 3. A 5 kw load operates from a 6 Hz kv rms line with a power factor of 6 % lagging. Compute the capacitance that must be placed in parallel with the load to achieve a 9 % lagging power factor. Load power angle (φ L ) = cos (.6) = 53.3 o From power triangle concept; Q L = P L tan φ L = kva After adding the capacitor, P new = P L = 5 kw and new power angle (φ new ): φ new = cos (.9) = 5.84 o From power triangle concept; equired capacitance; C = Q new = P L tan φ new = 4. kva Q C = Q new Q L = 4.45 kva X C = V rms Q C πfx C = = 356 Ω =.6 µf π(6)(356) Signal Frequency and Impedances Frequency Dependent Impedance esistance: = Constant 3454 Physics and Electronics (file: lec.tex) 9 Tianprateep, M.

10 , X X L O X C f Figure 8 Graph shows the frequency effects the impedance value. I C V out Figure 9 A first order lowpass filter. Inductance: Capacitance: X L = ωl = πfl X C = ωc = πfc. First Order Filter Circuits First Order Lowpass Filter From this circuit, first order differential equation can be written as: Thus, it is called first order circuit. C dv C(t) dt V C (t) = V s. I = = Z T /j πfc V out = V C = j πfc I = j πfc /j πfc 3454 Physics and Electronics (file: lec.tex) Tianprateep, M.

11 H(f) H(f) o o f B f B 3f B 4f B f 9 o f B f B 3f B 4f B f Figure Magnitude and phase of the first order lowpass transfer function versus frequency. H(f) = V out = = j (f/f B ) j πfc H(f): transfer function f B : half power frequency In case, f = f B : f B = πc H(f) = / =.77 P = P max / H(f) = (f/fb ) ( ) f H(f) = arctan f B H(f) db = log H(f) = log P out P in Magnitude and Phase Plots of H(f) f increases: H(f) decreases f increases: H(f) decreases to -9 o The low frequency signal can be detected at V out port. On another word, the low frequency signal can be passed this circuit(fig.). (lowpass filter) 3454 Physics and Electronics (file: lec.tex) Tianprateep, M.

12 L V out Figure Another first order lowpass filter. Example 4. Given an input signal = 5 sin(πt)5 sin(πt) is applied to the lowpass C filter. Let = /(π) and C = µ F. Find an expression for the output signal. For the first component of : f B = πc = π(/π)( 6 = Hz ) (t) = 5 sin(πt); = 5 o f = ω π = π = H() = j (f /f B ) = = o j (/) V out = H() = o V out (t) = sin(πt 5.7 o ) The second component of : (t) = 5 sin(πt); = 5 o f = ω π = π = H() = j (f /f B ) = = o j (/) V out = H() = o V out (t) =.4975 sin(πt 84.9 o ) V out = sin(πt 5.7 o ).4975 sin(πt 84.9 o ) Another First Order Lowpass Filter 3454 Physics and Electronics (file: lec.tex) Tianprateep, M.

13 C V out Figure A first order highpass filter. First Order Highpass Filter From this circuit: I = Z T = /j πfc V out = V = I = /j πfc = j πfc j πfc H(f) = V out = j(f/f B) j (f/f B ) f B = πc f/f B H(f) = (f/fb ) ( ) f H(f) = 9 o arctan f B Magnitude and Phase Plots of H(f) f increases: H(f) increases f increases: H(f) decreases to o The high frequency signal can be detected at V out port. On another word, the high frequency signal can be passed this circuit(fig.3). (highpass filter) Another First Order Highpass Filter. esonant and Second Order Filter Circuits Series esonance 3454 Physics and Electronics (file: lec.tex) 3 Tianprateep, M.

14 H(f) H(f) 9 o o f B f B 3f B 4f B f o f B f B 3f B 4f B f Figure 3 Magnitude and phase of the first order highpass transfer function versus frequency. L V out Figure 4 Another first order highpass filter. L I C Figure 5 The series resonant circuit Physics and Electronics (file: lec.tex) 4 Tianprateep, M.

15 From Kirchhoff s Voltage Law: L d Q dt dq dt Q C = V m sin ωt Q = Q m sin(ωt φ) V m Q m = L (ω ω ) ω tan φ = ( ) ωc ωl Compare with the equation of spring motion; natural frequency: ω = LC If the frequency of, ω = ω, Q m (or I m ) is maximum. It is called resonant frequency. On the other hands, resonant frequency: f : the frequency at which the impedance is purely resistive (i.e.,the total reactance is zero). Z T = j (X L X C ) resonant frequency can be found from: Define quality factor current in this circuit: X L X C = πf L = πf C f = π LC Q s = πf L = πf C [ Z T (f) = j Q s ( f f ] f f ) I = = Z T (f) / j Q s (f/f f /f) V = j Q s (f/f f /f) 3454 Physics and Electronics (file: lec.tex) 5 Tianprateep, M.

16 H(f) Q s > Q s > Q s3 > Q s4 H(f).5 Q s4 Q s3 Q s Q s f f.77.5 f Lf f H f Figure 6 Magnitude of the series resonant transfer function versus frequency. C L V out Figure 7 The parallel resonant circuit. Magnitude and Phase Plots of H(f) The middle frequency signal can be detected at V port(fig.3). It can be called second order bandpass filter. Defind bandwidth B = f H f L. At Q s : B = f Q s f H = f B/ f L = f B/ Parallel esonance Z T = / j (πfc /πfl) f : the frequency at which the impedance is purely resistive (i.e.,the total reactance is zero). πf C = πf L f = π LC 3454 Physics and Electronics (file: lec.tex) 6 Tianprateep, M.

17 H(f) Q p > Q p > Q p3 > Q p4 H(f).5 Q p4 Q p3 Q p Q p f f.77.5 f Lf f H f Figure 8 Magnitude of the parallel resonant transfer function versus frequency. L H(f) (db) C V out first order filter 4 second order filter 6 log f f / f f f Figure 9 Second order lowpass filter and Transfer function magnitudes Define quality factor, Q p Q p = πf L = πf C Z T = j Q p (f/f f /f) I V out = j Q p (f/f f /f) Magnitude and Phase Plots of H(f) The middle frequency signal can be detected at V port(fig.8). It can be called second order bandpass filter. Defind bandwidth B = f H f L. At Q p : Second Order Lowpass Filter B = f Q p 3454 Physics and Electronics (file: lec.tex) 7 Tianprateep, M.

18 C H(f) (db) L V out 4 second order filter first order filter 6 f / f f f log f Figure Second order highpass filter and Transfer function magnitudes H(f) (db) L V out C 4 6 f / f f log f Figure Second order band reject (notch) filter and Transfer function magnitudes f = π LC ; and Q s = πf L = πf C H(f) = V out = j Q s (f /f) j Q s (f/f f /f) At Q p : H(f) reaches a high peak at f. Q s = (maximally flat or Butterworth function) is selected for filter design. Second Order Highpass Filter Second Order Band-eject (Notch) Filter 3 Problems. (a) A voltage source V = o delivers 5 kw to a load with a power factor of percent. Find the reactive power and the phasor current. (b) epeat if the power factor is percent lagging Physics and Electronics (file: lec.tex) 8 Tianprateep, M.

19 Figure For Problem 4 Figure 3 For Problem 5 (c) For which power factor would the current ratings of the conductors connecting the source to the load be higher? In which case could the wiring be a lower cost?. A load has an impedance given by Z = j5ω. The current flowing through this load is I = 5 3 o. Is the load inductive or capacitive? Determine the power factor, power, and reactive power delivered to the load. 3. The phasor voltage across a certain load is 3 o, and the phasor current through it is 5 6 o. impedance. Is the power factor leading or lagging? Determine the power factor, power, reactive power, and 4. Determine the power for each source shown in Figure. Also, state whether each source is delivering or absorbing energy. 5. Find the power, reactive power, and apparent power delivered by the source in Figure 3. Find the power factor and state whether it is leading or lagging. 6. Derive and expression for the transfer function H(f) = V out / of the filter shown in Figure. 7. A first order C lowpass filter with a half power frequency of khz is needed. Determine the value of the capacitance if the resistance is kω Physics and Electronics (file: lec.tex) 9 Tianprateep, M.

20 8. A first order lowpass filter has a break frequency of 5 Hz. The input is (t) = 5 3 sin(πt 3 o ) cos( 3 πt) Find an expression for the output voltage. 9. A first order C highpass filter is required to attenuate a 6 Hz input component by 4 db. What value is required for the break frequency of the filter? By how many db is the 6 Hz component attenuated by this filter? If = kω, what is the value of C?. At the resonant frequency f = MHz, a series resonant circuit with = 5Ω has V = V and V L = V. Determine the values of L and C. What is the value of V C?. A series resonant circuit has B = 5 khz, f = 4 khz, and = Ω. Determine the values of L and C.. A parallel resonant circuit has f = MHz and B = khz. The maximum value of Z T is kω. Determine the values of, L, and C Physics and Electronics (file: lec.tex) Tianprateep, M.

REACTANCE. By: Enzo Paterno Date: 03/2013

REACTANCE. By: Enzo Paterno Date: 03/2013 REACTANCE REACTANCE By: Enzo Paterno Date: 03/2013 5/2007 Enzo Paterno 1 RESISTANCE - R i R (t R A resistor for all practical purposes is unaffected by the frequency of the applied sinusoidal voltage or