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Frequency selective networks

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There are man frequency selective networks, the most common two networks are the series and parallel resonant circuits

1

- In this kind of circuits, at the resonant frequency the inductor and capacitors has equal impedances with opposite signs
- At the resonant frequency vO will be in phase with vi
- The quality factor for this circuit is given by

2

- The quality factor gives an indication of the bandwidth of the circuit
- Narrower bandwidth means that the circuit has a larger quality factor as shown in the Figure

3-dB

f2

3

f1

- The quality factor is defined according to the following equation
- The bandwidth is defined as the difference between f1 and f2
- At these two particular frequencies the signal amplitude is less by 3 dB as indicated by the figure in slide 3

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Example: Design a filter to couple a voltage source, with negligible source impedance, to a 50 ohm load resistance. The specifications are that the filter center frequency be 5 MHz and the bandwidth be 100 kHz

Solution

From the quality factor we find that

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Solution

The capacitance value can be found from

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- The circuit gain at any frequency can be determined from
- The attenuation at any harmonic can be found from

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Example: A series tuned circuit is to be used to filter out harmonics of a waveform. What must be the minimum Q for the amplitude of the fifth harmonic to be 40 dB below the amplitude of the fundamental frequency

Solution

40 dB corresponds to a voltage ratio of 100:1, n=5

This means that Q=20.83

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- In parallel resonant circuit two susceptances are added in parallel, so that the admittance, instead of impedance is a minimum at fO

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- In practical life the inductor usually has a finite resistance, therefore a more accurate model for the parallel resonant circuit is shown below

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- Reactive networks can be used to match impedances over a narrow frequency range
- These networks can also be used to filter out some harmonics

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- The input impedance of the two networks shown in the next slide are equal provided
- If we take the real and imaginary parts and equate them we get the following

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- The previous equations can be rewritten as

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Example: The input impedance of a transistor amplifier is equal to 10 Ω in series with 0.2 μH. Design a matching network so that the input impedance is 50 Ω at 20 MHz

Solution:

At 20 MHz the inductive reactance of a 0.2 μH inductor is 25.1 Ω. The combined impedance of the circuit became 10+j25.1 Ω.

To do the impedance transformation we can use the equation

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Solution:

Or

in order to convert the real part of the impedance to 50 Ω, the imaginary part became Xs=20 Ω.

In order to convert the 25.1 Ω to 20 Ω a series reactance of -j5.1Ω must be added to the circuit.

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Solution:

Now Xp can be found from

if a reactance of –j25 is added in parallel with the circuit as shown in the figure of the next slide then input impedance became exactly 50 Ω

The –j25 reactance results from the impedance of a capacitor 318 pF

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Example: Design a lossless matching network to couple the impedance shown in the fig. below to a 50 Ω source impedance at 20 MHz

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solution: Since the series to parallel transformation always results in a parallel resistance larger than 50 Ω, there is no series reactance that will directly transform the 100 Ω resistor to a parallel equivalent of 50 Ω

The matching circuit can be solved based on the following equation

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Solution:

The input impedance of the circuit is

This can be brought to 50 Ω only by adding a –j50Ω reactance in series with the above mentioned circuit.

The completed solution will similar to the one shown in the next circuit

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