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ECE3050 Assignment 181. Sketc h and label the Bode magnitude and phase plots for the transfer functions given. Use log-log scales for the magnitude plots and linear-log scales for the phase plots. On the magnitudeplots, label the slopes of all asymptotes in dec/dec, label the break frequencies in rad/s, labelthe gain magnitude on all zero-slope asymptotes, and label the approximate gain magnitudefor the actual plot at the break frequencies. You can check your plots with a convenientcomputer program, e.g. PSpice, Mathcad, MatLab, etc., to obtain computer generated plots.An example PSpice deck is given for one of the transfer functions. The output voltage isV(2), i.e. the voltage at node 2. It is obtained by the LAPLACE statement, which multipliesthe voltage V(1) by the transfer function after the word LAPLACE.They-axisinthePSpiceProbegraphicsroutinemustbechangedtoalogscaletoseethecorrectslopesontheBodemagnitude plot. The number of decades displayed on the y-axis should be c hanged to nomore than 3 or 4 to get the best looking plot.T (s)=1011+s/100T (s)=10s/1001+s/100T (s)=10(1 + s/100) (1 + s/1000)T (s)=10(s/1000)2(1 + s/100) (1 + s/10000)T (s)=10(s/100)2+2(s/100) + 1T (s)=10(s/100)2(s/100)2+2(s/100) + 1T (s) = 1002(s/100)(s/100)2+2(s/100) + 1T (s)=10(s/100)2+1(s/100)2+2(s/100) + 1T (s)=101(s/100)2+√2(s/100) + 1T (s)=10(s/100)2(s/100)2+√2(s/100) + 1T (s) = 100√2(s/100)(s/100)2+√2(s/100) + 1T (s)=10(s/100)2+1(s/100)2+√2(s/100) + 1T (s)=101(s/100)2+0.5(s/100) + 1T (s)=10(s/100)2(s/100)2+0.5(s/100) + 1T (s) = 1000.5(s/100)(s/100)2+0.5(s/100) + 1T (s)=10(s/100)2+1(s/100)2+0.5(s/100) + 1T (s)=10s/10 − 1s/10 + 1T (s)=10(s/100)2−√2(s/100) + 1(s/100)2+√2(s/100) + 1EXAMPLE TRANSFER FUNCTION BODE PLOTVS10AC1E1 2 0 LAPLACE {V(1)}={10*PWR(S/100,2)/(PWR(S/100,2)+SQRT(2)*S/100+1)}.AC DEC 50 1 100K.PROBE.END2. The figure shows an RLC circuit. Show that the voltage gain transfer function is of the formT (s)=(s/ω0)2(s/ω0)2+ s/ (Qω0)+11where you must give the equations for ω0and Q.ForQ =0.5, show that the transfer functionbecomesT (s)=(s/ω0)2(s/ω0+1)2For Q<0.5, show that the transfer function becomesT (s)=s/ω1s/ω1+1s/ω2s/ω2+1whereω1,2= ω012Q±sµ12Q¶2− 1Sketc h the Bode magnitude and phase plots as a function of ω for the cases Q<0.5, Q =0.5,and Q>0.5. Label the slopes in dec/dec on the magnitude plot and label all break frequen-cies in the asymptotes.3. Repeat problem 2 for the circuit given. Show that the transfer function is given byT (s)=s/ (Qω0)(s/ω0)2+ s/ (Qω0)+1For Q =0.5, show that the transfer function becomesT (s)=s/ω0(s/ω0+1)2For Q<0.5, show that the transfer function becomesT (s)=rω1ω2s/ω1s/ω1+11s/ω2+1whereω1,2= ω012Q±sµ12Q¶2− 1Sketc h the Bode magnitude and phase plots as a function of ω for the cases Q<0.5, Q =0.5,and Q>0.5. Label the slopes in dec/dec on the magnitude plot and label all break frequen-cies in the asymptotes.24. Show that the voltage gain transfer function for the circuit is given byT (s)=R2k (R3+ R4)R1+ R2k (R3+ R4)1+R3kR4Cs1+(R1kR2+ R4) kR3CsShow that the input impedance is given byZin=[R1+ R2k (R3+ R4)]1+[R3k (R1kR2+ R4)] Cs1+[R3k (R2+ R4)] Cs5. Show that the voltage gain transfer function for the circuit is given byT (s)=R4R1k (R2+ R3)+R41+(R1+ R2) kR3Cs1+(R1kR4+ R2) kR3CsShow that the input impedance is given byZin=[R1k (R2+ R3)+R4]1+[R3k (R1kR4+ R2)] Cs1+[R3k (R1+ R2)] Cs6. Solve for the transfer function for Vo/Vifor the circuit below. Th e short cut method wecovered in class does not work with this circuit. However, the short cut method can be usedto solve for Va/Vi. Once this is obtained, superposition of Viand VacanbeusedtosolveforVo.This eliminates writing node equations, but there is some algebra involved in combining termsto put the transfer function into the ratio of two polynomials. Sketch the Bode magnitudeplot, label the break frequencies, and label the gain on the zero-slope asymptotes. Answer:VoVi=1+[R2R3/ (R1+ R2+ R3)] Cs1+[(R1+ R2) kR3] CsTheelementvaluesaretobechosensothat the high-frequency asymptotic gain is 0.05 andthe high-frequency asymptotic output resistance (with Vi=0)is100 Ω. The frequency of the3zero in the transfer function is to be 100 Hz. If C = 220 µF, specify the element values in thecircuit and calculate the pole frequency. Answers: R1=2kΩ, R2=105.26 Ω, R3= 155.36 Ω,fp=5Hz.7. Solve for the transfer function for Vo/Vifor the circuit below. Sketch the Bode plot, label thebreak frequencies, and label the gain on the zero-slope asymptotes. Answ er:VoVi= −ZFR1= −R2R11+R3Cs1+(R2+ R3) CsThe circuit is to be designed as a lag-lead compensator for a motor control system. The spec-ifications are low-frequency asymptotic gain: −2, input resistance: 10 kΩ, pole frequency:1Hz, zero frequency: 10 Hz. Specify the element values. Answers: R1=10kΩ, R2=20kΩ,R3= 2222.2 Ω,andC =7.1620 µF.8. Solve for Vo/Vifor the circuits below. Sketch and label the Bode magnitude plots.Answers: (a) The transfer function is a low-pass shelving function with a dc gain ofKdc=R2+ R3R1+ R2+ R34and a high-frequency gain ofK∞=R2+ R3kR4R1+ R2+ R3kR4The transfer function isVoVi=R2+ R3R1+ R2+ R31+(R2kR3+ R4) Cs1+[(R1+ R2) kR3+ R4] Cs(b) The transfer function is a high-pass shelving function. The zero-frequen cy gain isKdc=1+R1+ R2R3The high-frequency gain isK∞=1+R2R3The transfer function is given byVoVi=µVfVo¶−1=µ1+R1+ R2R3¶1+[(R2+ R3) kR1] Cs1+R1Cs9. It is desired to design a circuit that realizes the following impedance transfer function:Z = 1000(1 + s/2πf2)(1+s/2πf4)(1 + s/2πf1)(1+s/2πf3)where f1=10Hz, f2= 100 Hz, f3=1kHz,andf4=10kHz.(a) Sketch the Bode magnitude plot for Z. Note that the impedance starts at 1000 Ω,shelvesat 100 Ω, then shelves again at 10 Ω. Label the break frequencies and label the magnitudeof the impedance on each zero-slope asymptote.(b) A possible circuit realization is in the figure below.At low frequencies, the impedance starts at the value R + R1+ R2. As frequency is in-creased, suppose that C1becomes a short circuit well before C2becomes a short circuit.When C1becomes a short, the impedance shelves at the value R + R2.Therefore,C1causes both a pole and a zero. As frequency is increased further, C2becomes a shortand the impedance shelves at the value R.ThusC2also causes a pole and a zero. Withthis information show that the input impedance is approximately given byZin(s) ' (R + R1+ R2)1+R1k (R + R2) C1s1+R1C1s1+(R2kR)


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