Lab11_AC Circuits - FREQUENCY RESPONSE AND PASSIVE FILTERS...

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FREQUENCY RESPONSE AND PASSIVE FILTERS LABORATORY In this experiment we will analytically determine and measure the frequency response of networks containing resistors, AC source/sources, and energy storage elements (inductors and capacitors). Given an input sinusoidal voltage, we will analyze the circuit using the frequency-domain method to determine the phasor of output voltage in the ac steady state. The response function is defined as the ratio of the output and input voltage phasors. It is a function of the input frequency and the values of the circuit elements (resistors, inductors, capacitors). We start with examples of a few filter circuits to illustrate the concept. RC Low-Pass Filter : Consider the series combination of the resistor R and the capacitor C, connected to an input signal represented by AC voltage source of frequency . v in (t) = V s cos( t + I ) (1) Figure 11.1 Suppose we are interested in monitoring the voltage across the capacitor. We designate this voltage as the output voltage. We know that it will be a sinusoid of frequency . Thus, v out (t) = V o cos( t + o ) (2) We will now determine expressions for the amplitude V o and the phase angle o . First we convert the network to frequency domain. It is shown in Figure 11.2.
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Figure 11.2 In the above circuit, the voltage source is represented by its phasor and the resistor and capacitor by their impedance. We wish to evaluate the phasor V out for the output sinusoid. Since the three elements are in series, the voltage divider formula can be used and we obtain: R Z Z V V C C in OUT (3) where V in is the phasor of the input voltage. It is given by: V in = V s e j I (4) Z c = 1/j C (5) Manipulation of Equation (3) gives the frequency response as: RC j V V j H in out 1 1 ) ( (6) The product RC has units of the inverse of angular frequency. We define o = 1/RC as a characteristic frequency of the network and write the frequency response as: H(j ) = 1/(1 + j / o ) (7) In other words, we are measuring frequency in units of o . The sinusoid corresponding to the output voltage can be written as v out (t) = Re{ V out e j t } = Re{ H(j )Vin e j t } = Re{V s e j I e j t /(1+j / o )} (8) v out (t) = {V s /[1+( / o ) 2 ] 1/2 }cos( t + I tan 1 ( / o ) ) (9)
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Returning to the frequency response, H(j ) is a complex number. It has a magnitude and phase. Both depend on the frequency, R and C. Thus, H(j ) = H exp(j H ) (10) The magnitude (absolute value) of H is a measure of the ratio of the amplitudes of the output and input voltages. It is given by: H = | H(j )| = V o / V s = 1/[1+( / o ) 2 ] 1/2 (11) On the other hand, the phase angle of H measures the difference in the output and input phase angles. It is given by: o - I = H = tan 1 ( / o) (12) The frequency dependence of the magnitude H is sketched in Figure 11.3
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Lab11_AC Circuits - FREQUENCY RESPONSE AND PASSIVE FILTERS...

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