frequency selective networks

7/23/2019 Frequency Selective Networks

1/23

1

Frequency selectivenetworks

There are man frequency selectivenetworks, the most common twonetworks are the series and parallel

resonant circuits


2/23

2

Series resonant circuit

In this kind ofcircuits, at theresonant frequencythe inductor and

capacitors has equalimpedances withopposite signs

At the resonantfrequency vO will bein phase with vi

The quality factor forthis circuit is givenby


3/23

3


The quality factor gives an indication ofthe bandwidth of the circuit

Narrower bandwidth means that the

circuit has a larger quality factor as shownin the Figure

f1

f2

3-dB


4/23

4


The quality factor is dened according tothe following equation

The bandwidth is dened as the dierencebetween f1and f2

At these two particular frequencies thesignal amplitude is less by ! d" asindicated by the gure in slide !

B

Q O

=


5/23

5

Series resonant circuitexample

Example:#esign a lter to couple a voltagesource, with negligible source impedance,to a $% ohm load resistance& The

specications are that the lter centerfrequency be $ '() and the bandwidth be*%% k()

Solution

From the quality factor we nd thatH

B

RL

6.79

102

505 =

==


6/23

6


Solution

The capacitance value can be found from

LCf

2

10=

( ) ( ) ( ) FLfC =

== 7.12106.791052

1

2

1

625220

2


7/237


The circuit gain at any frequency can bedetermined from

The attenuation at any harmonic can befound from

22 )1(

/

)(O

n

LRn

jnA

O

O

=

)1()(

)(2

=nQ

n

jA

jnA

O

O


8/238


Example: A series tuned circuit is to be used tolter out harmonics of a waveform& +hat mustbe the minimum for the amplitude of the fthharmonic to be -% d" below the amplitude of

the fundamental frequencySolution

-% d" corresponds to a voltage ratio of *%%.*, n/$

This means that /0%&1!)15(

501.0)5(2

==Q

jA O


9/239

Parallel resonant circuit

In parallel resonant circuit twosusceptances are added in parallel, sothat the admittance, instead of impedance

is a minimum at fO


10/2310

Parallel resonant circuit

In practical life the inductor usually has anite resistance, therefore a moreaccurate model for the parallel resonant

circuit is shown below


11/2311

Impedance matching andharmonic fltering using

reactive networks 2eactive networks can be used to matchimpedances over a narrow frequencyrange

These networks can also be used to lterout some harmonics


12/23

12

Impedance matching

The input impedance of the two networksshown in the ne3t slide are equal provided

If we take the real and imaginary partsand equate them we get the following

22

2

22

2

pp

pp

pp

pp

pp

pp

ssiXR

XR

jXR

XR

jXR

jXR

jXRZ ++=

==


13/23

13

Impedance matching

The previous equations can be rewrittenas


14/23

14

Impedance matching


15/23

15


reactive networksExample. The input impedance of a transistoramplifier is equal to *% 4 in series with %&05(& #esign a matching network so that theinput impedance is $% 4 at 0% '()

Solution:

At 0% '() the inductive reactance of a %&0 5(inductor is 0$&* 4& The combined impedanceof the circuit became *%670$&* 4&

To do the impedance transformation we canuse the equation

s

ssp

R

XRR

22+

=


16/23

I d t hi d


17/23

17


reactive networksSolution:8r

in order to convert the real part of theimpedance to $% 4, the imaginary partbecameXs=20 4&

In order to convert the 0$&* 4 to 0% 4 aseries reactance of 9j5.1 4 must beadded to the circuit&

10

1050

22

sX+= 201001050 ==sX

fcj

j2

11.5 = nFc 56.1=

I d t hi d


18/23

18


reactive networksSolution:NowXpcan be found from

if a reactance of j25 is added in parallelwith the circuit as shown in the gure of

the ne3t slide then input impedancebecame e3actly $% 4The :70$ reactance results from the

impedance of a capacitor !*1 pF

s

ss

p X

XRjjX

22+

= =+

= 25

20

2010 22

jjjXp


19/23

I d t hi d


20/23

20


reactive networksExample. #esign a lossless matchingnetwork to couple the impedance shownin the fig& below to a $% ; sourceimpedance at 0% '()

I d t hi d


21/23

21


reactive networkssolution.


22/23

22


reactive networksSolution.

The input impedance of the circuit is

This can be brought to $% ; only by adding a

j50 ; reactance in series with the abovementioned circuit&

The completed solution will similar to the oneshown in the ne3t circuit

5050 jZi +=

I d t hi d


23/23

23


reactive networks

frequency selective networks

Documents