EECE 290 Homework 2, Spring 2015

30 nF IO ( j) 103IO ( j) V – + 2 k ISRC ( j) + – VO ( j) Figure P14.1.15 Homework 2 P14.1.4 Given a lowpass RL filter having R = 1 k and L = 20 mH. Determine: (a) the 3- dB cutoff frequency in krad/s; (b) the lowpass transfer function. Short solution: (a) c = R/L = 1000/(2010-3) rad/s  1000/20 = 50 krad/s. (b) For the LP response, . If frequency is expressed as  in krad/s, then H(j) = . P14.1.14 (a) Determine NEC seen by the resistor in Figure P14.1.14; (b) IO(j); (c) the 3-dB cutoff frequency. Solution: It is convenient to determine the open- circuit voltage and ZTh. On open circuit, each branch has an impedance of j1.5, so that the 6 A divides equally between the two branches. It follows that Vac = j3; Vbc = j1.5; Vba = VTh = Vbc – Vac = j1.5 – j3 = -j1.5. When the current source is replaced by an open circuit, the impedance between terminals ‘ab’ is j1.5 in parallel with j1.5, which is j0.75. TEC will be as shown. Hence, NEC consists of a current source IN = Iba = -j1.5/(j0.75) = -2 A in parallel with j0.75, as shown. The current source - 2 A is directed from terminal ‘b’ to terminal ‘a’ in accordance with the assumed polarity of VTh. (b) It follows that = . (c) ; ; , or, c =4000/3 rad/s  4/3 krad/s. P14.1.15 Determine, for the filter in Figure P14.1.15, (a) the transfer function VO(j)/ISRC(j); (b) the 3-dB cutoff frequency in krad/s; (c) the magnitude of the gain 1 H 1 H 0.5 H 0.5 H 1 k 6 A IO(j) Figure P14.1.14  j 6 A  j j0.5 a b 3 j0.5 3 c j0.75 – + -j1.5 b a j0.75 -2 A b a  1 k IO(j) 1/6 F 6  1/30 F 30  – + VSRC ( j) Figure P14.1.20 1 F 30 nF 103IO(j) V + – + – – + VO(j) IO(j) IT(j) VT(j) + – – +  3 100 j k 3 k VO(j) 2103ISRC(j) and the phase angle as a function of frequency; (d) the maximum gain in the passband. Solution: Since the circuit must reduce to an RC circuit, the equivalent circuit consisting of the dependent source and capacitor should be determined. By applying a test source, VT(j) = 103(j) + , or, , where VT(j) is in V, IO(j) is in A, and  is in krad/s, or k. This is equivalent to a 1 k resistor in series with a capacitor of impedance 100/j3 k. Alternatively, from the source absorption theorem, the dependent source is equivalent to a resistance . The next step is to derive TEC seen by the capacitor. VTh(j) = 2103ISRC(j) in series with 3 k. (a) It follows that ,  is in krad/s. (b) , or c = 1/0.09 = 100/9 krad/s. (c) Magnitude of the gain is , phase angle is -tan-1(0.09). (d) Maximum gain is 2103 as   0. P14.1.20 Reduce the circuit of Figure P14.1.20 to a first- order circuit and specify the values of the circuit elements. Short Solution: Let the impedance of the lower branch be Z, where Z = 6 + = 6 + ; the impedance of the upper branch is 30 + = 30 + = 5Z; the impedance of the two branches in parallel is 5Z/6 = 5 + = 5 + , which is equivalent to 5  in series with 1/5 F. This capacitance in series with 1 F is F. hence the first order circuit is 5  in series with 1/6 F. 2 k ISRC + – VO  1 k  3 100 j k 1/jC R2 R1 jL – + Figure P14.1.22 VSRC ( j) I2 ( j) I1 ( j) Figure P14.2.8 – + VSRC ( j) R 0.5 F L + – VO (j) Figure P14.2.9 – + VSRC ( j) 0.5 F 2 H + – VO (j)  60  30  5  P14.1.22 Determine the transfer functions I1(j)/VSRC(j) and I2(j)/VSRC(j). (Note that each is a first-order transfer function that is independent of the parameters of the other subcircuit.) Short Solution: A/V A/V P14.2.8 Determine R and L in Figure P14.2.8 so that the resonant frequency is 4 kHz, and Q = 5. Solution: ; H 3.17 mH. ; . P14.2.9 Determine in Figure P14.2.9: (a) the minimum value of the transfer function |VO(j)/VSRC(j)|; (b) Q and BW. Solution: The minimum value of VO(j) occurs when the LCR branch has a minimum impedance. This occurs when L and C in series have a zero impedance, that is, at = 103 rad/s. At this frequency, 30||5 = 30/7 . (a) From voltage division, the minimum value of the transfer function . (b) TEC seen by the LCR branch has RTh = (60)||(30) = 20 . For the purpose of determining Q, VTh can be set to zero, the circuit reducing to that shown. This is a series LCR circuit with R = 25 . Hence, 80; BW = 12.5 rad/s. P14.2.15 Determine the frequency at which VO(j) = 0 in Figure P14.2.15 Solution: VO(j) = 0 when the impedance of the LC branch is zero. The impedance of L in parallel with C is . This impedance in series with -j/C = 0, that is, , which gives , or , or rad/s  Mrad/s. Figure P14.2.15 1 H + – VO (j) ISRC(j) 1 F 1 F 10 k