−=−, and then, with

(1)

Band pass filter (πfilter)

a) Ignoring resistance r of inductance and assuming the filter on its characteristic impedance Z^C show that the circuit below is a band pass filter with the low and high cutoff frequencies fb_and f_h_.

b) Show that the characteristic impedance can be written:

2 2

2 ,

C

b

a h

Z C

ω ω ω ω ω

= −

−

Calculate its value R^C_for

ω

²−

ω

_b² =

ω

_h²−

ω

², and then, with R_C =75Ω, the values of

a, b

C C _andL_b_{to have} f_b=150KhZ_andf_h =400KhZ.

c) The inductance Lb has the resistance r=0.44Ω ; plot the Bode diagram for f _ranging from 15kHz to 4MHz. Determine the crossover frequencies at −3dB, f^'^b_and f^'h_.

d) The crossover frequencies f^'^b_and f_b_, f^'_h_and f_h are related by

'b b

f b

f = _and

'h h

f h

f = _.

Change the value of Lb using the average value of b and h to closer f^'^b_and f^'h_from

fb_and f_h. Edit r in the same proportion. Determine the new crossover frequencies f^''^b and f^''^h_.

e) Find the pulses for which the asymptotic of expressions

T

apply 1.

Correction a)

4 4

1 1

with and 1

^b

1

^b

.

b b h h b

a a

b b b b

C C

L C L C

ω

≥

ω ω

=

ω

≤

ω

= ⋅ + ⇒

ω

=

ω

+

ZC

Cb

i1

i2

Ca 2

Lb , r

ue u_s

Ca 2

(2)

2 2

2 4 2 4

, filter in π, , in case of pures réactances.

4 4

C C

a b a b

Z Z X X

Z Z

Z Z X X

 

= = −

 

 + + 

 

And a π filter has a bandwidth set, in the case of pure reactance, by:

4

^b

0

a

X

− ≤

X

≤ _and:

(

²

)

4 0 4 1 0

1 1 4

et 1 .

b a

b b

a b

b

b h

b b b b a

X C

X L C C

C

L C L C C

ω

ω ω ω ω

− ≤ ≤ ⇔ − ≤ − ≤

⇒ ≥ = ≤ + =

b)

2 2 2

C

2

2 2 2

4 2 1

Z .

4 4 1

1 2

4

C

b b a b

a b a b

b b a

b b

a a b

b

a b

X X L C

Z X X C C

C L C

L C

Z C

C C

L C C

ω

ω ω

ω

ω ω

−

= − ⇒ =

+ + −

 

 − 

 

=

 + 

 − 

 

2 2

2 2 2 2 2

2

2 2 2 2

2 2 1

, , ,

2

2 2 1

14nF, 1 =21.4 nF , 52.6 μH.

4

C C

C

h b

h b a

a

a h

a b b

h b b b b

Z R C

C R

C C C L

R C

ω ω

ω ω ω ω ω

ω ω

ω

ω ω ω ω

− = − ⇒ = + = = =

 

= + = =  −  = =

2 2

2 ,

C

b

a h

Z C

ω ω ω ω ω

−

=

−

c) Calculate u_e then us and finally T from the equations of meshes :

1 2

2

2 2

,

2 1 2

0 ,

1 2

2 .

1 2

C C

e

a a

b a

a b a

s a

i i

u jC jC

R i

jL r i

jC jC R C jC

u R i

R C

ω ω

ω ω ω

ω



 = −



  

  

 = + + + + −

  

  + 

 



 =

 +



(3)

( ) ( ) ( ) ( )

( ) ( )

2 4 2 3

2

2 3 2

2 4 2

2 3 2

1 2

2 2 2 2

2 ,

1 2

2 2 2

C C

C

C C

C

C C

C

C C

a a

a b b b

a a a a

b b b b b

a

a b a

a b b a b

a a a a

b b b b b

e

R L C C rR C C

C C j R C R C C jL C C r R C C j

i jC i

jC jC jR C

R L C C rR C C

C C j R C R C C jL C C r R C C j

u

ω ω

ω ω ω ω

ω

ω ω ω

ω ω ω ω ω

   

+ − + − − + + −

   

 

 

=    

  +  

   

 

 

+ − +  − − + + −

 

=

( ) ( )

2 3

2 2

2

2 ,

1 2

C

a

a a

a b

i i

R C jC

jC jC j

ω

ω ω







 

  

     −

   +  

    

  



( ) ( )

( )

2 4 2 3

2

2 3 2 2

3 2

2

2 2

2 2 2

1 2 2 2

C C

C

C C C

s C

C C

C

C C C

a a

a b b b

a a a a a

b b b b b b b

e

a b

a a

a b b b

b b b b b

e

b

R L C C rR C C

C C j R C R C C jL C C r R C C j R C C jC

u u

jC jC R

R L C C rR C C

j R C R C L C j r R C j jR C

u jC

ω ω

ω ω ω ω ω ω

ω ω

ω ω ω ω

ω

   

+ − + − − + + − + −

   

 

 

= 

 

 

  

+ + − + + − − −

  

 

=





s ,

C

u R







 

 

 

 



( )

3

1 2 1

2 2 2

C

C C

s C

a a a

b b b

b b

b e

b

L rC r C L C R C

R C j R C

R R C

u u

jR C

ω ω ω

ω

     

− + + + + −

     

     

= 

 

 

2 2

2 3

,

1 1

2 2 2

C

C C C

C C

b

a b a a

b b b b

b

T R C

C L r C C

r R C R C L R C

R R C

ω

ω ω ω

=

  

   

− + + + +  − 

   

 

 

    

The maximum of

T

_is_1.79dB_to_{231.3 kHz} and in the points of -3dB this maximum correspond to f^'b=122kHz_andf^'h =324.8kHz_.( ⇒ b = 0.81 and h = 0.81) d)

   

   

   

2 2

' '

b b b

b h

h

b + h b + h

L = L et r = r , What according to the literal expressions of f

2 2

and f should multiply f and f by 2 = 1.23.

b + h

'b 34.8μH

L = _andr^' =0.29Ω. The maximum is now 2.93dB to 231.3 kHz with

''b 183.4kHz

f = _and f^''h =412.3kHz_.

(4)

e) Asymptotic expressions of

T

take the value 1 for

1

C b

ω

=

R C

_and ²

b a

ω

= L C _{, it} seems then possible to adapt f^''^b_and f^''h by multiplying Cb with the ratio 183.4

150 and Ca _with

412.3 2

400

 

 

  , this translating asymptotes in the desired direction f^''^b_and

''h

f . The maximum is then 2.97dB for 298.7 kHz and points to -3dB are at 167.2 kHz and 396.2 kHz .

ω→0 ,

T

→

R C

^C b

ω

_;

ω

→ ∞, 2

2

b a

T

^→

L C ω

^.

The three curves of |T|Db