Date post: | 14-Feb-2018 |
Category: |
Documents |
Upload: | raghu-raman |
View: | 222 times |
Download: | 0 times |
of 23
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
7/23/2019 Frequency Selective Networks
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
7/23/2019 Frequency Selective Networks
3/23
3
Series resonant circuit
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
7/23/2019 Frequency Selective Networks
4/23
4
Series resonant circuit
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
=
7/23/2019 Frequency Selective Networks
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 =
==
7/23/2019 Frequency Selective Networks
6/23
6
Series resonant circuitexample
Solution
The capacitance value can be found from
LCf
2
10=
( ) ( ) ( ) FLfC =
== 7.12106.791052
1
2
1
625220
2
7/23/2019 Frequency Selective Networks
7/237
Series resonant circuit
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
7/23/2019 Frequency Selective Networks
8/238
Series resonant circuitexample
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
7/23/2019 Frequency Selective Networks
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
7/23/2019 Frequency Selective Networks
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
7/23/2019 Frequency Selective Networks
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
7/23/2019 Frequency Selective Networks
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 ++=
==
7/23/2019 Frequency Selective Networks
13/23
13
Impedance matching
The previous equations can be rewrittenas
7/23/2019 Frequency Selective Networks
14/23
14
Impedance matching
7/23/2019 Frequency Selective Networks
15/23
15
Impedance matching andharmonic fltering using
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+
=
7/23/2019 Frequency Selective Networks
16/23
I d t hi d
7/23/2019 Frequency Selective Networks
17/23
17
Impedance matching andharmonic fltering using
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
7/23/2019 Frequency Selective Networks
18/23
18
Impedance matching andharmonic fltering using
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
7/23/2019 Frequency Selective Networks
19/23
I d t hi d
7/23/2019 Frequency Selective Networks
20/23
20
Impedance matching andharmonic fltering using
reactive networksExample. #esign a lossless matchingnetwork to couple the impedance shownin the fig& below to a $% ; sourceimpedance at 0% '()
I d t hi d
7/23/2019 Frequency Selective Networks
21/23
21
Impedance matching andharmonic fltering using
reactive networkssolution.
7/23/2019 Frequency Selective Networks
22/23
22
Impedance matching andharmonic fltering using
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
7/23/2019 Frequency Selective Networks
23/23
23
Impedance matching andharmonic fltering using
reactive networks