2014_Problema Aliniere US VEm

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


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.



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.



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.



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


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 .



7/16

3. Lets consider the schematic diagrams of a pair of

tuned circuits:



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



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



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.



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.



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 ,



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



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

=

=

=



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 .



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

'

=


2014_Problema Aliniere US VEm

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