homework_6_solutions

homework_6_solutions - EE40 Problem 1 Homework#6 Solutions...

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Unformatted text preview: EE40 Problem 1 Homework #6 Solutions As the frequency increases, the transfer function magnitude decreases, for the f=0, f=1000, f=1500, and f=2000. So it might be a low‐pass filter. HOWEVER, NO ONE WILL BE SURE THAT THIS IS INDEED A LOW‐PASS FILTER UNLESS HE/SHE CAN KNOW THE EXPRESSION FOR THE TRANSFER FUNCTION! Problem 2 1 Rt = 1.5kΩ fB=106Hz H( f ) = 11 21+ j f fB Part (a): So long as the capacitor C and the load resistor RL are connected in parallel, we can switch their positions – in other words, switching the positions of C and RL will not at all change the circuit or the node voltages and branch currents associated with it. 2 Problem 3 The slope of the magnitude plot is 20dB/decade. part (a) For the magnitude response, we see that the higher the frequency, the larger the magnitude of the transfer function. This trend extends from infinitely low frequency well to infinitely high frequency. The correct answer for this part is therefore: “when you take the derivative of a sin wave, the higher the freq. the higher the resulting slope.” part (b) For an integrator, let us first examine its time‐domain “input‐output” characteristic: 3 Vout (t) = ∫ Vin (t)dt + Vout (t = 0) 0 t Differentiating both sides gives dVout (t) = Vin (t) dt If we are in sinusoidal steady‐state, then we can express the above relationship in terms of phasors. Note that differentiation in time domain is essentially a multiplication of jω in the phasor domain. So we have Vout × ( jω) = Vin which means Vout 1 1 = = Vin jω j 2π f Therefore, the Bode plot of an integrator‐like transfer function is similar to Problem 4 The transfer function of the circuit is 4 f j Vout jωRC j1000fC = = = 1000 Vin 1 + jωRC 1 + j1000fC 1 + j f 1000 Applying the “superposition” to this linear circuits, we see that the output response to vin = 5+5cos(wt) is the sum of the output responses to vin1=5 and vin2=5cos(wt) when these two inputs (vin1 and vin2) are applied separately. So we have Vout (t) = 0 × 0 + 0.707 × 5cos(2000π t + 45 ) part (a) vin(t) = 5 + 5cos(200πt) + 5cos(2000πt) H(f = 100) = 0.1 j = 0.1∠(90 − 5.7 ) = 0.1∠84.3 1 + 0.1 j Vout (t) = 0.1cos(200π t + 5.7 ) + 3.536cos(2000π t + 45 ) Problem 5 (b) The Bode magnitude plot follows. 5 A solution with only sketch will also receive full credit. For interested readers, the Matlab code to obtain the above plot is as follows: R1 = 9000 R2 = 1000 C = 1e‐8 logf = 1:0.01:6; f = 10.^logf; w = 2*pi*f; H = R2./(R2+1./(j*w*C + 1/(R1))); semilogx(f,20*log10(abs(H))) (c) At very low frequencies, with the capacitance considered to be an open circuit, we have a two‐resistance voltage divider and which means that, at very low frequency, the magnitude of the transfer function should be roughly ‐20dB, which coincides with our plot in part (b) 6 (d) At very high frequencies, with the capacitance considered to be a short circuit, which directly transmits Vin to Vout, meaning that H(f)=1, the magnitude of which is 0dB. This result also fits the curve in part (b) Problem 6 Note: the current flowing through the three series elementsis i (ω) = Vs R + jωL + 1 jωC = Vs R + j (ωL − So the voltage across the resistor is VR (ω) = RVs 1 R + jωL + jωC = RVs 1 ) R + j(ωL − ωC = 1 ) ωC = 1 Vs R 1 + jQ( ω − ω0 ) ω0 ω ωω 1 + jQ( − 0 ) ω0 ω Vs where Q is the quality factor, and w0 is the resonance frequency in unit of radians/second. 7 The above expressions tells us a lot of information. Firstly, when the angular frequency is equal to the resonance frequency, the phase of the current, and the phase of the voltage across the resistor, are both exactly the same as the phase of Vs. Secondly, if we write | VR (ω) 1 1 |=| |= Vs (ω) 1 + jQ( ω − ω0 ) 2 ω0 ω we have ω1 = ω0 ( 1 + 1 1 + ) 2 4Q 2Q ω2 = ω0 ( 1 + Then we have 1 1 − ) 2 4Q 2Q ω1 − ω2 = ω0 Q ω1 × ω2 = ω0 ω1 + ω2 2 = ω0 1 + 1 4Q2 Therefore, strictly speaking, w0 is the GEOMETRIC MEAN, instead of the ARITHMETIC MEAN, of w1 and w2. However, engineers like approximations which simplify things a lot. We can see from the above expressions that, when Q is greater than 5, the approximation that ω1 + ω2 ω0 2 is very accurate, with less than 0.5 percent error. Under such condition, we can say that 8 ω1,2 ω0 ± ω0 Q What is the qualify factor in this problem? It is 100, which means the above approximation is very accurate. However, if the given Q is, say, 0.5, then we have to be very careful with the above approximations! Problem 7 (c) A rough sketch will also full credit. 9 For interested readers, the Matlab code for obtaining the above plot is R = 1; L = 1e‐3; C = 0.25e‐6; f0 = (1/(2*pi))*sqrt((L ‐ C*R^2)/(C*L^2)); f = 0.95*f0:1:1.05*f0; Y = 1./(R + i*2*pi*f*L) + i*2*pi*f*C; Z = 1./Y; plot(f,abs(Z)) Problem 8 (b) The magnitude (in dB) of the transfer function is 10 (c) At very low frequencies, with the capacitance considered to be an open circuit, no current flows through the resistor, which means the voltage drop across the resistor is zero. Therefore Vout should be equal to Vin in this case; in other words, the magnitude of the transfer function at very low frequencies should be 1, or 0dB, which coincides with the plot in part (b) (d) At very high frequencies with the inductance considered as an open circuit, no current flows through the resistor, which means the voltage drop across the resistor is zero. Therefore Vout should be equal to Vin in this case; in other words, the magnitude of the transfer function at very high frequencies should be 1, or 0dB, which coincides with the plot in part (b) 9, (a) 1 vout 1 1 jwC H= = = = 1 vin R + 1 + jwRC 1 + j 2π fRC jwC wB = 1 = 106 rad / s RC 11 (b) H= vout jwL 1 1 = = = vin R + jwL 1 − j R 1 − j R wL 2π fL R R → = 103 rad / s L L NOTE: THE RATIO OF R TO L IS MEANINGLESS, SINCE THIS RATIO HAS AN UNIT OF RAD/S, AND ITS PHYSICAL MEANING IS NOT SO OBVIOUS AS THAT FOR, SAY, R1/R2. (c) a is low‐pass filter, b is high‐pass filter. (d) The key point is that C is in parallel with the series combination of R2 and L. Then this parallel combination comes series with R1. So the transfer function changes significantly. wB = 1 jwC Z3 = 1 (R2 + jwL) + jwC (R2 + jwL) × H3 = 1 ) jwC 1+ 1 (R2 + jwL) × jwC 1 1 = = 1 R1 ) 1 + jwCR1 + jwCR1 × (R2 + jwL + jwC R2 + jwL 1+ R2 + jwL 1 = R1 1 + j 2π fCR1 + R2 + j 2π fL R1 × (R2 + jwL + vout 3 Z3 1 = = = vin Z 3 + R1 1 + R1 Z3 1 (e) if R2>100R1, then H3(w) = 1 1 + jwCR1 + R1 R2 + jwL 1 = H1(w) 1 + jwCR1 H3( f ) 1 = H1( f ) 1 + j 2π fCR1 12 So long as R2>100R1, H3(f) is approximately equal to H1(f); in this case, if we further choose the values in part (a) for R1 and C, then H3(f) would have the same wB as H1(f) (f) Using the values in part (a) and (b), we have R1=1000 ohm C=1nF R2/L=103 rad/s An additional constraint is that R2 should be greater than 100R1, or, R2>200kohm (g) The overall transfer function is H(w) = 1 Vout Vout Vout _ 3 jwL jwL = × = H3(w) × = × Vin Vout _ 3 Vin R2 + jwL 1 + jwCR1 R2 + jwL 1 j 2π fL × H( f ) = 1 + j 2π fCR1 R2 + j 2π fL (h) 13 ...
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This note was uploaded on 10/24/2009 for the course EE 40 taught by Professor Chang-hasnain during the Fall '07 term at Berkeley.

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