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Circuit Analysis Demysti ed
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Fig 8-12 The transfer function for the high-pass lter in Example 8-4
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This condition is known as full-width at half power So the condition we need to solve is
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2 2 = 1 + c R 2 C 2
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From which we nd that c = 1 RC
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So by tuning the values of the resistor and capacitor, we can construct a highpass lter with the desired cutoff frequency A plot of the transfer function is shown in Fig 8-12 Band-pass or band-stop lters can be constructed using RLC circuits In a series RL circuit, the damping parameter is = A lter is underdamped if < 0 It is critically damped if = 0 (821) (820) R 2L (819)
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Frequency Response
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And it is overdamped if > 0 The bandwidth of the lter is = 2 = R L (823) (822)
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This gives the total frequency range about the critical frequency that is allowed to pass That is, the lter will pass frequencies that lie within the range = c 2 (824)
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The quality or Q-factor for the circuit is given by Q= L 0 = 0 R (825)
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EXAMPLE 8-5 An RL lter has R = 200 and L = 10H What is the bandwidth If the resonant frequency is 100 Hz, what is the quality factor SOLUTION Using (823) the bandwidth is = The Q-factor is Q= L 0 = R 10 (2 )(100) = 3142 200 R = 20 rad/s L
Summary
For a series LC circuit, the differential equation describing the voltage across the capacitor is LC d 2vC + vC = 0 dt 2
Circuit Analysis Demysti ed
The natural frequency of the circuit is given by 0 = 1 LC
A resonant frequency for an input is one where = 0 Resonant frequencies can cause the system to blow up The impedance of a circuit is given by Z= R 2 + X 2 tan 1 X R
We can use this to write Ohm s law as V = ZI The susceptance is Y= 1 Z
Given an excitation E we can describe the behavior of a circuit in terms of the transfer function H R = HE where R is the response of the circuit
Quiz
1 A load has a voltage V = 20 0 and current I = 2 20 Find the impedance and determine a series circuit that will model the load Is the circuit inductive or capacitive Assume that = 100 rad/s
4 1/10 + 8 6 I
Fig 8-13 Circuit diagram for Problem 2
Frequency Response
2 Consider the circuit shown in Fig 8-13 The response of the circuit is the current owing through the 6 resistor Determine the resonant frequency if v s (t) = 10 cos t 3 Consider the circuit in Fig 8-6 Suppose that the voltage source is replaced by a current source i s (t) = I0 cos t and the positions of the inductor and capacitor are switched What is the resonant frequency 4 Reverse the positions of the capacitor and resistor in Fig 8-11 Does the circuit still function as a lter 5 An RLC lter has R = 100 and L = 4 H What is the bandwidth If the resonant frequency is 100 Hz, what is the quality factor
Operational Ampli ers
An operational ampli er or op amp is a circuit that takes an input voltage and ampli es it The symbol used to represent an op amp in a circuit diagram is shown in Fig 9-1 An op amp is de ned by two simple equations The rst thing to note is that the voltage across the input terminals is zero Hence Va = Vb (91)
The second relation that is essential for analyzing op amp circuits is that the currents drawn at a and b in Fig 9-1 are zero Ia = Ib = 0 (92)
Despite this, we will see that the op amp will result in voltage gains at the output terminal c How does this work Two voltages are input to terminals a and b Their difference is then ampli ed and output at c, which is taken with referenc to ground Although we won t worry about the internal construction of an op amp, note that it consists of a set of resistors and dependent voltage
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