Frequency Response of Networks - Essay Example

Frequency Response of Networks Insert Insert Insert This essay presents the theory behind frequency response in inductive and capacitive reactances, in relation to how they vary with frequency. The essay introduce the plot of frequency response in the RC filter and which regions that these filters operates in. importance in analysis. There is a vivid explanation on how to determine the resonance frequency and the Q factor from the plot of current variation with frequency in a series RLC circuit. The essay evaluates briefly band pass and band stop filters. Introduction The frequency response is a representation of the system’s response to sinusoidal inputs at varying frequencies; it is defined as the magnitude ratio and phase difference between the input and output signals. If the frequency of the source in a circuit is used as a reference, it is possible to have a complete analysis in the frequency domain and time domains. Frequency domain analysis is easier than time domain analysis It is important to obtain the frequency response of a circuit because we can predict its response to any Input signal. . There are four general types of filters:- Low-pass filters (LPF), Band-pass filters, High-pass filter (HPF) and Band-Reject (Stop) (Dhogal, 2009). In this laboratory experiment we will plot the frequency response of a network by analyzing RC passive filters Theory 1) Capacitive Reactance This is the internal resistance that a capacitor offers to the flow of current from an alternating source (Theraja, 2008) ................................ (1) Capacitive reactance derivation from equation Vcos(ω t + p) where V is the amplitude (can be current or potential), ω is the angular frequency, t is time, and Φ is phase shift. current flowing through a capacitor is given by i = C(dv/dt) Naturally, since v = V cos(ωt+ Φ), would equal - ω Vsin(ωt+ Φ). Using trigonometric identity we can rewrite sin as cos and get –ωVcos (ωt+ {Φ -90}) (subtracting 90 degrees from sin to get cosine) by taking the derivative of the voltage and multiplying it by the capacitance we get the current flowing through the capacitor i = - ωCVcos(ωt+ [Φ -90]). This is in the time domain. In order to derive the impedance/reactance, it must be converted to the frequency domain by writing the voltage and current as a phasor. solving using Eulers Identiy where ejx = cosx + jsinx j is the imaginary number . Taking the phasor transform of Vcos(ωt+ [Φ -90]). gives jVp and taking the phasor of i we get I Φ where V Φ and I Φ are used to represent the phasor So far we have I Φ = -jωCVΦ recalling from Ohms law V/I=R. Impedance is much like resistance and is assigned the variable Z. Notice that in the equation Ip= -jωCV Φ solving for VΦ, becomes Vp = IΦ Setting = Z we get VΦ=IΦZ Reactance is the imaginary part of impedance. The impedance of a capacitor is -1/jωC. The reactance is -1/ωC we know that w is angular frequency. w is related to f by = Hence Where capacitive reactance (Ω) r.m.s voltage (v). r.m.s current (A). Capacitive reactance varies with the frequency of the supply fc,. The relationship is linear and the capacitive reactance may be calculated as follows. ...................................... (2) Where capacitive charging and discharging frequency (Hz) 2) Inductive Reactance The opposition to the follow of current in an ac circuit of pure inductance is kwon as inductive resistance. It is calculated by the formula bellow. Inductive reactance derivation equation is based on the Sine function we know ω = 2πƒ The basic V I equation for an inductor is V(t) = L d/dt I(t) Now I(t) = A Sin(ω t) V(t) = L d/dt I(t) V(t) = L A ω Cos(ω t) Z = V rms / I rms Now V rms = ALω /2 I rms = A/2 Therefore XL = V rms / I rms XL= XL =Lω ω = 2πƒ Hence XL = 2πƒ L 2πƒ L 2πƒ L= ω L............. (3) Where XL- inductive resistance (Ω) ƒ is the Frequency of the supply voltage L is the Inductance of the Coil 3) Simple RC filter An RC low-pass filter consists of a resistor and a capacitor in parallel with the source as shown by figure 1 below. The filter exhibits a certain frequency response by allowing only frequencies that are lower than the cut off frequency to pass through the filter as show below. s Its frequency response is shown below. G(jω)= ................................................. (4) The magnitude response is given by │G (jω)│= (5) Or Vout = Vin × the total resistance of the circuit But XC = (Ω) Z =(R2 + X2).................................AC Circuit impedance. Vout = Vin × = Vin .............................................. This is the output voltage for any frequency. and can be expressed in dBs as ............................................ (6) When ωCR > 1, the gain is -20 log10(ωCR ) = -20(log10 CR + log10ω) From the graph above the cut off frequency of the filter is given by fc = . When ω is increased by a factor of 10 db the gain on the other hand reduces by 20 dB. 4) Resonance Resonance frequency is the frequency at which the indicative and capacitive reactances are equal in a RLC circuit. The resonance frequency is given by the following equation. Impedance of series RLC circuit is the sum of the three impedances. i.e. Z = R + jXL -jXc keep note of the signs used for capacitance and inductive At resonance the two reactive elements cancel out. Hence at this freq XL = Xc 2π L = 2πL × = 1 2 = 1/{(2π)2LC} = 1/(2π 8(LC)) ............................................. (7) At the resonance frequency the circuit obeys ohms laws as total reactance is equal to true resistance of the circuit Z = R and I = V/R. A typical sketch of the variation of current with frequency is show in Fig. 4. Quality factor of the system may be determined by the following equation at this frequency. Q factor equation More generally and in the context of reactive component specification (especially inductors), the frequency-dependent definition of Q is used: Q(ω) = ω× Where ω is angular frequency The width (bandwidth) of the resonance is given by Change in f = f0/Q f 0 is the resonant frequency Q=(8) Also .......................................... (9) whereBw is the bandwidth in Hz between the half-power frequencies ƒ2 and ƒ1, i.e., Bw= ƒ2 – ƒ1 Bibliography Dhogal. (2009). BASIC ELECTRICAL ENGINEERING. New Delhi: Tata McGraw-Hill Education. Prasad, R. (2009). Fundamentals Of Electrical Engineering. New Delhi: PHI Learning Pvt. Ltd. Rajput, R. (2007). Basic Electrical and Electronics Engineering. New Delhi: Firewall Media. Theraja, B. L. (2008). A Text Book Of Electrical Technology. New delhi: S. Chand & Company Ltd. Warne, M. A. (2002). Electrical Engineers Reference Book. 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