8/9/2019 2014_Problema Aliniere US VEm
1/16
Let consider a super-heterodyne radio receiver
designed for SW frequency band 11 ... 19 MHz.
The radio receiver is tuned by means of a gang
variable capacitor with two sections,
CV50 ... 315 pF.
The maximum accepted attenuation due to a
frequency mismatch is amax = 2 dB; 50 % of this
mismatch may come from the alignment error;
19.01.2015 11:091RADIO COMMUNICATIONS: Systems and Equipment
8/9/2019 2014_Problema Aliniere US VEm
2/16
The next parameters are known:
The RF building block consists only in the RFTC.
The maximum frequency of the modulating signal,
fmmax= 4,5 kHz
The quality factor of the tuned circuits, Q = 60.
The intermediate frequency, fi= 455 kHz.
19.01.2015 11:092RADIO COMMUNICATIONS: Systems and Equipment
8/9/2019 2014_Problema Aliniere US VEm
3/16
We are required to:
1. Calculate the maximum allowable alignment error;2. Calculate the central frequencies of the zero
alignment error signals;
3. Calculate the values of the tuned circuits
components considering a = 3 % frequency bandextension; This extension is necessary to ensure the
frequency band coverage in spite of the electronic
components tolerances;
4. Check if the maximum realized alignment error
fulfils the requirements from point 1.
19.01.2015 11:093RADIO COMMUNICATIONS: Systems and Equipment
8/9/2019 2014_Problema Aliniere US VEm
4/16
1. The 3 dB bandwidth of the RF circuits is:
As the 3dB bandwidth of the IFA is
BI = 9 kHz, we obtain:
kHz ,183,5=
Q
fB
sRF
min
,120,37=kHz9kHz183,5=
BB
FI
RF>>
It results that the condition BRF>>BAFI is
fulfilled; It can be concluded that themismatch remained after the alignment
procedure will affect mainly the RF circuits.
19.01.2015 11:094RADIO COMMUNICATIONS: Systems and Equipment
8/9/2019 2014_Problema Aliniere US VEm
5/16
Using the expression of the attenuation introduced
by an RF building block which includes n tuned
circuits :
We can derive the relation for the maximum value of
the variablex:
the maximum allowed useful signal attenuation is:
.dB1=a
2
1a admmaxmax
,)x+1(lgn10=a2
.0,5088=1-10|x| n0a
adm
1
max
19.01.2015 11:09 5RADIO COMMUNICATIONS: Systems and Equipment
.kHz46,644=Q2
fx
Q2
fx=f s
r
a
minmax
8/9/2019 2014_Problema Aliniere US VEm
6/16
The calculations will be preceded by a comparison of
the coverage factors (for the signal frequency range kS
and for the variable capacitor range, kV ); we cancontinue only if kV> kS);
.1,727=f
f=k>2,51=
C
C=k
s
s
s
V
V
v
min
max
min
max
2. The correctly received (aligned) frequencies will be
calculated by means of the empirical relations
proposed by Mohrmann:
f1=fS min. kS0,147= 11,9202 MHz ,
f2=fS min. kS0,852= 17,5236 MHz .
19.01.2015 11:09 6RADIO COMMUNICATIONS: Systems and Equipment
8/9/2019 2014_Problema Aliniere US VEm
7/16
3. Lets consider the schematic diagrams of a pair of
tuned circuits:
19.01.2015 11:09 7RADIO COMMUNICATIONS: Systems and Equipment
8/9/2019 2014_Problema Aliniere US VEm
8/16
To be sure that the required frequency band is
received in spite of the tolerances associated to the
components, the first part of the analysis will be
done considering an extended band;
The extended frequency band will be
f (1 + )
(f=fsmax-fsmin), and the parameters of this band
are:
MHz ,19,12=2
f+f=f
MHz10,88=
2
f-f=f
ss
ss
max
/
max
min
/
min
.1,75=ff=k
s
ss/
min
/
max
19.01.2015 11:09 8RADIO COMMUNICATIONS: Systems and Equipment
8/9/2019 2014_Problema Aliniere US VEm
9/16
The values of the electronic components belonging
to the RF selective circuits can be calculated using
the next expressions:
.)C+C(L2
1=f
)C+C(L2
1=f
TVss
TVss
S
S
max
/
min
min
/
max
.nH546,0209=)C+C(f4
1=L
,pF76,8981=
1-k
Ck-C=C
Tvs
2S
2
s
V2
SV
ST
S
max
2/
min
minmax
19.01.2015 11:09 9RADIO COMMUNICATIONS: Systems and Equipment
8/9/2019 2014_Problema Aliniere US VEm
10/16
In order to calculate the values of the components
constituting the tuned circuit of the local
oscillator, firstly we determine the values of thevariable capacitor for the two zero alignment error
frequencies obtained above:
,2,1=junde,C-Lf4
1=C TS
2j
2V Sj
It results:
CV 1= 249,5856 pFCV 2= 74,1734 pF.
19.01.2015 11:09 10RADIO COMMUNICATIONS: Systems and Equipment
8/9/2019 2014_Problema Aliniere US VEm
11/16
For these signal frequencies we calculate the
frequencies that have to be generated by the localoscillator :
fh j= fj+ fi, cu j = 1 ... 2,
It results:
fh1= 12,3752 MHz
fh2= 17,9786 MHz.
19.01.2015 11:09 11RADIO COMMUNICATIONS: Systems and Equipment
8/9/2019 2014_Problema Aliniere US VEm
12/16
The local oscillator (LO) will generate the required
frequency values if:
.)C+C(L2
1
=f
)C+C(L2
1=f
THVHh
THVH
h
2
2
1
1
Lets use the notation:
kh= fh 2/fh1= 1,45 ,
19.01.2015 11:09 12RADIO COMMUNICATIONS: Systems and Equipment
8/9/2019 2014_Problema Aliniere US VEm
13/16
Using the same procedure as in the case of the RF
circuit one obtains:
.nH496,1650=)C+C(f4
1=L
,pF83,7697=1-k
Ck-C
=C
THV
2
h
2H
2h
V2hV
TH
11
21
19.01.2015 11:09 13RADIO COMMUNICATIONS: Systems and Equipment
8/9/2019 2014_Problema Aliniere US VEm
14/16
4. By definition the alignment error, fa, is given by:
.f-)C+C(L2
1-
)C+C(L2
1
=f-f-ff-ff
i
TVSTVH
irshid
a
SH
=
=
=
19.01.2015 11:09 14RADIO COMMUNICATIONS: Systems and Equipment
8/9/2019 2014_Problema Aliniere US VEm
15/16
Having in mind that the plot of alignment error
versus received signal frequency is:
It follows that large values are obtained for the nextfrequencies:
fS min, fS max , fex= ( f1+ f2) / 2 .
19.01.2015 11:09 15RADIO COMMUNICATIONS: Systems and Equipment
8/9/2019 2014_Problema Aliniere US VEm
16/16
;kHz20,223-=kHz455-kHz10880,000-kHz11314,776=
=f-)C+C(L2
1-
)C+C(L2
1
=|f
i
TVMAXSTVMAXH
f=fa
SH
S
=
'min
.kHz39,337-=kHz455-kHz19120,000-kHz19535,662=
=f-)C+C(L2
1-
)C+C(L2
1
=|f
i
TSVSTHVH
f=fa SMAX
minmin
'
=
19.01.2015 11:09 16RADIO COMMUNICATIONS: Systems and Equipment
Top Related