Mettre le schéma des blocs successifs ici
PARTIE SUR L’ANTI-ALIASING
The aim of this filter is to improve the precision of the bandwidth.
To improve the efficency of this filter, its order must be at least 2. Then our anti-aliasing filter will be an order 2 Butterworth filter model. It will respect the following circuit:
Thus, we will use the LM741 because we just need one amplifier.
1.1.1
Cut off frequency
Gabarit:
Amax = - 3db and Amin = -20 db.
Fo trouver un otre gabarit…. Logikement la reponse doit etre plat entre 0 et Wp
We have to set a cut off frequency in order to limit the received frequencies at the next element of the circuit, the MF10.

Filter order:
In order to find the order of the anti-aliasing filter we use the formula:

0


P

 10
A max
10

1


1
2 n
From this formula we can find the order of the filter n
 log
 
10
A max/ 10

1
 
10
A min/ 10

1
 
2 log( wp wa
)
With Wo

Wp
Wa

2

* fo and f
0
= 15 kHz (Commuted-Capacities filter cut-off frequency)
Then

0
= 2
 f
0
= 94247,7796 rad.s
-1
So we have n = 1,6
Consequently, to realize the first block, we will use the Butterworth model with a Sallen-
Key structure with an order 2. This is the cell we will use: http://www.ecircuitcenter.com
1.1.2
Finding the poles
We know that T ( j

)

1

1
 
2 n with =



(the standardized pulsation)
Moreover, the Laplace variable is: p
 j
    p j
For these calculations ω is standardized and so p too in the transfer function.

To get back a normal p, we will make the opposite operation at the end.
So now, let’s replace the ω.
We obtain:
T ( p )

1
1



P j


2 n
=
[ 1

1 p
2 n
] 2 j
2 n
1
=
1

1

(

1 ) n
P
2 n
1 
2
Then we can find the poles solving the following equation:
1

1

(

1 ) n
P
2 n
1 
2 or more easily 1 + (-1) n p
2n
= 0
= 0
In our case n=2, so1+ (-1)² p

1+ p
4
= 0
 p
4
= -1
4
= 1+ p
4
 p
 re j

Let’s find the  k
: n = 2 (par)
  k


2 n
 k
 n
  k


4

Finally, p
 e j (


2 k

4 4
)
= e j (

4
 k

2
) k   k

2
Now we must make the k values vary in order to obtain the different values of p .
Knowing that all the p have modules equal to 1, we can represent them on a circle which has a unit ray.
These are the values of

depending on the k variation: k

0

4
1
3

4
2
5

4
3
7

4
P
3
4
9

4
Poles P
0
P
1
P
2
We observe that after the value 3, we go back to first value of

(2
 modulo).
P
4=
P
0

P
1
=
3

4
P
0
=

4
=
9

4
=P
4
P
2
=
5

4
P
3
=
7

4
In order to have a stable system, we keep only the poles placed on the left part of the graph.
3
 j (
P
1
( e 4
)
) and P
2
( e j (
5

4
)
) are the only poles on the left part.
We validate the results thanks to fact that one complex is never pole without its combined pole. In our case, P
1
( e j (
3

4
)
) and P
2
( e j (
5

4
)
) are combined.
P
2
P
1
 e
 j
5

4 e j
3

4
 cos
3

4
 cos
3



4 j sin j sin
3

4
3


4
 sin 45

  sin j cos 45
45
 j cos 45
Thanks to the last results we can find the transfer function for n = 2, associated to our system…
We use this formula:
T ( p )

( p

1 p
1
)( p
 p
2
)
Or (p – p
1
)*(p - p
2
) = (p-
= (p e
= (p²+
3
 j (
4 cos cos
)
3
)*(p-

 j e j ( sin
2
4
3

 sin
4
5

4
)
3
)

2
4
3

4
) (p -
 p cos
(

2
3

4 cos
3

= p² - 2 p cos +1
4
3


4 j
) sin
)
3

4
)
=>
T ( p )

( p
2 
1
2 p cos
3

4

1 )

1.1.3
Realization
As we said before, we will use this cell in order to realize this block:
Knowing that we are going to use the Millmann theorem, we can write:
V s
V e
=
( Y
1

Y
2
)( Y
3
 k .
Y
1
Y
4
)
.
Y
3

Y
3
( Y
4
 kY
2
)
with Y1=
The fact that we work on a low-pass cell gives us some information usefull for the
Y2=
Y3=
Y4= calculation. Moreover we define that this amplifier is a non-reverser.
Then we have k=1 and so we choose RA=∞ and RB=0
Know thanks to Millmann at the point 2, we can write: V2 =
1
R
1

R
1

V e jV s

C
1
1
R
2
 j

C
2
V a
And: V3 =
After replacing, we obtain:
T (j

) =
V s
V e
=
1
R
2

1

R
2 j

C
2 j

R
1
(Changer Va en V2)
( C
1

C
2
)
 j

R
2
C
2
1
 kj

R
1
C
1

R
1
R
2
C
1
C
2
( j

)²
If we apply the Laplace Transformation, we have finally:
1
T (p) = p ²

1 p (
R
2
C
1
R
1
R
2
C
1
C
2

1
R
1
C
1
)

1
R
1
R
2
C
1
C
2

1.1.4
Quality Factor
We know that an order 2 low-pass filter transfer function follows the formula::
T ( j

)

1

2 j


1

0
 j ²

²

0
²
Then, identifying the components of this formula and the one we found before we obtain:
2 j



0
= 2 p cos
3

4
Moreover, we know the expression of the damping coefficient. It’s:
Q

1
2


0 , 707
 k
  cos(
 k
)
So:
   cos
3
4


0 , 707
Now thanks to this formula, we can calculate the quality factor Q…

1
2

0 , 707
1.1.5
Components value
Cut-off pulsation:
According to the MF 10 datasheet, the cut off frequency of the switched capacitor filters depends on the clock frequency. fc
 f
CLK
50

15 kHz so f = 750 kHz
CLK

In order to respect the shanon conditions we have to set the cut frequency of the antialiasing filter with :
 f c
 f clk
 f c

750 kHz
2 2
Moreover, we have an attenuation of 30 dB/dec.
So we use

0

A

( 10 10

1 )
1
2 n
 f c

375 KHz . with

= 2* ∏ *f
CLK
/2
A = 30dB/dec n = 2 (filter order).
 o = 419 kHz and f
0


0
2

= 66,7 kHz. So respect the Nyquist – Shannon conditions. f must be less than 66,7 kHz in order to
0
1
Moreover we know that: T (p) = p ²

1 p (
R
2
C
1
R
1
R
2
C
1
C
2

1
R
1
C
1
)

1
R
1
R
2
C
1
C
2 and that
T ( j

)

1

2 j


1

0


² j ²

0
²
Then, identifying, we obtain:

0
² =
1
R
1
R
2
C
1
C
2
We also know that: C1 = 2Q and C2= Q/2 (with normalized values). If we don’t have normalized values, there is a formula explaining:
C new
 C old
Ro * Wc
We choose: Ro = R1 = R2 = 10 k

and Wc = 2

Fc = 2

* 66700 = 419088 rd/s
Then: C
1

2 Q
R
0
* W c

337 pF
1
And C
2

2 Q
R
0
* W c
= 84, 3 pF
FINALLY WE HAVE: C1 = 337 pF / C2 = 84, 3 pF / R1 = R2 = 10 k
