EECE 290 Homework 1 with Solution, Spring 2011

1 Homework 1 P10.1.5 Determine the response ) ( /) ( ω ω j V j V SRC O , both in magnitude and phase, the passband gain, and the corner frequency in Figure P10.1.5. Solution P10.1.5 The impedance of the 8 nF capacitor is jω 0.125 kΩ, where ω is in Mrad/s. Hence, ( ) ω ω ω ω ω j j j V V j j H SRC O .0 25/ 10 .0 125/ ) ( ) ( + = = ω 40 1 5.0 = + j . It follows that ( ) 2) (40 1 .5 0 ω ω + = H j , tan (40 ) ) ( 1 ω ω − = − ∠ H j . The response is lowpass; the passband gain, when ω → 0 is 0.5; the corner frequency is when 40ωcl = 1, or ωcl = 25 krad/s. – + + – 10 kΩ 8 nF 8 nF VO(jω) VSRC (jω) Figure P10.1.5 – + 4 nF + – 10 kΩ 4 nF 8 nF VO(jω) VSRC (jω) 2 P10.1.7 Determine the response ) ( /) ( ω ω j I j V SRC O , both in magnitude and phase, the passband gain, and the corner frequency in Figure P10.1.7. Solution P10.1.7 The impedance of the capacitor is 3ω 100 j kΩ, where ω is in krad/s. The dependent source of voltage ) ( 103 IO jω , where ( ω) IO j , is the current through the source, is equivalent to a resistance of 1 ) ( ) ( 103 ≡ ω ω j I j I O O kΩ. Hence, ( ) = = ) ( ) ( ω ω ω j I j V j H SRC O = + × × ω ω 100 / 3 3 2 3 10 100 3 j j 0.09ω 1 10 2 3 + j × V/A. It follows that ( ) 2 3 (0.09 ) 1 10 2 ω ω + × = H j V/A, (0.09 ) tan ) ( 1 ω ω − = − ∠ H j . The response is lowpass; the passband gain, when ω → 0 is 2×103; the corner frequency is when 0.09ωcl = 1, or ωcl = 100/9 krad/s. Figure P10.1.7 30 nF IO ( jω) 103IO ( jω) V – + 2 kΩ ISRC ( jω) + – VO ( jω) 3 P10.1.10 Determine the response ) ( /) ( ω ω j V j V SRC O , both in magnitude and phase, the passband gain, and the corner frequency in Figure P10.1.10. Solution P10.1.10 Let the current in the capacitor be IO(jω). Then VO(jω) = IO(jω)/jωC. The current in the 2 kΩ resistor is 2000 ) ( 0.5 VO jω , and the total current is IO(jω) + 2000 ) ( 0.5 VO jω = [ 1/ 4000 ]) ) ( j C + VO j ω ω . It is seen that a 4 kΩ resistor appears in parallel with the capacitor. TEC as seen by the capacitor will have ) ( 0.8 ) ( 5 4 ) ( ω ω ω j V j V j V SRC SRC Th = = ; RTh = 0.8 5 4 = kΩ. TEC could have been derived directly. Thus, on open circuit, the current is 2 ) ( 0.5 ) ( ω ω j V j V O O − mA, so that 4 ) ( ) ( ) ( ω ω ω j V j V j V O O SRC + = , so that ) ( 0.8 ) ( 5 4 ) ( ω ω ω j V j V j V SRC SRC Th = = . On short circuit, 1 ) ( ) ( ω ω j V j I SRC sc = mA. Hence, RTh 0.8 ) ( = = sc Th I j V ω kΩ. The impedance of the capacitor is ω j 20 kΩ, where ω is in krad/s. Hence, ω ω ω ω j j j V j V SRC O 20 / 0.8 20 / ) ( 0.8 ) ( + = , or ω ω ω ω j j V j V j H SRC O 0.04 1 0.8 ) ( ) ( ) ( + = = . It follows that ( ) 2) (0.04 1 .8 0 ω ω + = H j . tan (0.04 ) ) ( 1 ω ω − = − ∠ H j . The response is lowpass; the passband gain, when ω → 0 is 0.8. The corner frequency is when 0.04ωcl = 1, or ωcl = 25 krad/s. Figure P10.1.10 – + 1 kΩ + – 50 nF 2 kΩ – + VO(jω) 0.5VO(jω) VSRC (jω) 4 P10.1.12 Determine the response ) ( /) ( ω ω j V j V SRC O , both in magnitude and phase, the passband gain, and the corner frequency in Figure P10.1.12. Solution P10.1.12 The impedance of the capacitor is 10 9 100 1 − × jω jω ≡ 10 kΩ, where ω is in krad/s. When the source and 2 kΩ resistor are reflected to the secondary side, they become a source of 2VSRC(jω) in series with 8 kΩ. Hence, ω ω ω j j V j V SRC O 10 / 18 10 ) ( 2 ) ( + = , or ω ω ω ω ω ω 1.8 1 2 10 / 18 20 ) ( ) ( ) ( j j j j V j V j H SRC O + = + = = . It follows that ( ) 2) (1.8 1 2 ω ω ω + = H j , tan (1.8 ) 90 ) ( 1 ω ω − − = ∠ o H j . The response is highpass; the passband gain, when ω → ∞ is 10/9 ≡ 0.915 dB; the corner frequency is ωcl = 9 5 krad/s ≡ 88.4 Hz 10 kΩ 2 kΩ – + 100 nF 1:2 + – VO ( jω) VSRC ( jω) 10 kΩ Figure P10.1.12 2 kΩ – + 100 nF 1:2 + – VO(jω) VSRC (jω) 5 P10.2.2 Determine R , L , and C in Figure P10.2.2 such that: (a) the maximum response of VO(s) is 1 V; (b) the bandwidth is 4 krad/s; and (c) the center frequency is 100 krad/s. P10.2.2 Maximum response, which occurs at ω0, is RI; hence R = 10 kΩ. Q = Solution ω0 /BW = 100/4 = 25; it follows that ≡ × = = 4 5 0 10 10 25 R Q C ω 25 nF, and L = 4 25 10 10 5 4 0 ≡ × Q = R ω mH. P10.2.3 Given the transfer function 125 500 5 100 1000 4 ) ( 2 2 + + + + = s s s s H s . Determine the maximum response in dB. Solution P10.2.3. ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + + + + + + ⎥ = ⎦ ⎤ ⎢ ⎣ ⎡ + + + + = 25 100 250 25 100 25 0.8 25 100 25 250 5 4 ) ( 2 2 2 2 2 s s s s s s s s s s H s . The first term is a bandstop response of magnitude 0.8 at low and high frequencies and zero at ω = 5 rad/ s. The second term is a bandpass response of zero at low and high frequencies and magnitude of 2 at ω = 5 rad/ s. The sum will have a maximum magnitude of 2, or 20log102 = 6 dB, at ω = 5 rad/s. Figure P10.2.2 C L + – VO(s) 0.1 mA R