18Spring_hw07_solns.pdf - University of Illinois Spring...

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University of Illinois Spring 2018 ECE210 / ECE211 - Homework 07 Solutions 1. Consider the circuit drawn below,where R = 1 Ω , L = 0 . 2 H, C = 0 . 05 F and the frequency response of the circuit is: H ( ω ) = V I . (a) What is the resonant frequency of this circuit? (b) Plot | H ( ω ) | , and label the resonant frequency on this plot; (c) Plot Re { H ( ω ) } and Im { H ( ω ) } . (d) Explain why this circuit might be called a “bandpass” filter. (e) Repeat (a) and (b) for resistor values of 10 Ω and 0 . 1 Ω . (f) Based on (d), how does the resistor value relate to the passband of the filter (e.g., does a larger value for the resistor give a narrower or wider passband)? Solution The phasor equivalent circuit has an inductance impedance jωL and a capacitor impedance 1 jωC . Applying KCL yields I = I R + I L + I C = V 1 R + 1 jωL + jωC Therefore, the frequency response of the system is H ( ω ) = V I = 1 1 R + 1 jωL + jωC and the magnitude response | H ( ω ) | = 1 q 1 R 2 + ( ωC - 1 ωL ) 2 . In terms of physical meaning, H ( ω ) is the reciprocal of impedence. For this standard RLC circuit, When ω = 0 , H ( ω ) 0 . Intuitively, it is because capacitor performs as a short circuit for DC. When ω = ω resonant , H ( ω ) = 1 R = 1 Z min . H ( ω ) achieves its maximum value. When ω → ∞ , H ( ω ) 0 . Intuitively, it is because inductor performs as a short circuit. (a) To find the resonant frequency of this circuit we need to find a frequency ω that maximizes | H ( ω ) | , which is the same as finding the ω that minimizes the denominator. The square root is a monotonically increasing function, so minimizing q 1 R 2 + ( ωC - 1 ωL ) 2 is the same as minimizing its argument G ( ω ) = 1 R 2 + ωC - 1 ωL 2 = R 2 + C 2 ω 2 ω 2 - 1 LC 2 . The function G ( ω ) has a minimum ( | H ( ω ) | has a maximum) at ω o = q 1 LC rad / s . Plugging in the values we find that the resonant frequency is ω o = r 1 LC = r 1 0 . 2 × 0 . 05 = r 1 0 . 01 = 10 rad / s .
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