Chapter 41: Attenuators

Example 1, page no. 763

In [1]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
 #ratio  of  output  power  to  input  power
rp1  =  2;
rp2  =  25;  
rp3  =  1000;
rp4  =  0.01;

#calculation:
 #power  ratio  in  decibels
rpd1  =  10*(1/2.303)*math.log(rp1)
rpd2  =  10*(1/2.303)*math.log(rp2)
rpd3  =  10*(1/2.303)*math.log(rp3)
rpd4  =  10*(1/2.303)*math.log(rp4)
 #power  ratio  in  nepers
rpn1  =  (math.log(rp1))/2
rpn2  =  (math.log(rp2))/2
rpn3  =  (math.log(rp3))/2
rpn4  =  (math.log(rp4))/2


#Results
print  "\n\n  Result  \n\n"
print  "\n  power  ratio  in  decibels  are  (a)",round(rpd1,2),"  dB  (b)",round(rpd2,2),"  dB "
print  "(c)  ",round(rpd3,2),"  dB  and  (d)  ",round(rpd4,2),"  dB"
print  "\n  power  ratio  in  nepers  are  (a)",round(rpn1,2),"  Np  (b)",round(rpn2,2),"  Np"
print   "(c)  ",round(rpn3,2),"  Np  and  (d)  ",round(rpn4,2),"  Np"


  Result  



  power  ratio  in  decibels  are  (a) 3.01   dB  (b) 13.98   dB 
(c)   29.99   dB  and  (d)   -20.0   dB

  power  ratio  in  nepers  are  (a) 0.35   Np  (b) 1.61   Np
(c)   3.45   Np  and  (d)   -2.3   Np

Example 2, page no. 763

In [3]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
rp  =  0.05;#  power  ratio  P2/P1

 #calculation:
 #power  ratio  in  decibels
rpd  =  10*(1/2.303)*math.log(rp)


#Results
print  "\n\n  Result  \n\n"
print  "\nthe  attenuation  is  ",round(abs(rpd),2),"  dB"


  Result  



the  attenuation  is   13.01   dB

Example 3, page no. 764

In [4]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
gain  =  1.5;#  in  dB
Pi  =  0.012;#  in  Watt

#calculation:
 #output  power
Po  =  Pi*10**gain


#Results
print  "\n\n  Result  \n\n"
print  "\noutput  power  is  ",round(Po,2),"  W"


  Result  



output  power  is   0.38   W

Example 4, page no. 764

In [5]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
I2  =  0.05;#  in  Amperes
rin  =  1.32;#  in  Np

 #calculation:
 #current  input,  I1
I1  =  I2*math.e**(rin)
 #current  ratio  in  decibels
rid  =  20*(1/2.303)*math.log(I2/I1)


#Results
print  "\n\n  Result  \n\n"
print  "\ncurrent  input,  I1  is  ",round(I1,2),"  A"
print  "\ncurrent  ratio  in  decibels  is  ",round(rid,2),"  dB"


  Result  



current  input,  I1  is   0.19   A

current  ratio  in  decibels  is   -11.46   dB

Example 5, page no. 769

In [3]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
Ra1 = 8; # in ohms
Ra2 = 8; # in ohms
Ra3 = 21; # in ohms
Rb1 = 10; # in ohms
Rb2 = 10; # in ohms
Rb3 = 15; # in ohms
Rc1 = 200; # in ohms
Rc2 = 200; # in ohms
Rc3 = 56.25; # in ohms

#calculation:
R01 = (Ra1**2 + 2*Ra2*Ra3)**0.5
R02 = (Rb1**2 + 2*Rb2*Rb3)**0.5
R03 = (Rc1**2 + 2*Rc2*Rc3)**0.5

#Results
print  "\n\n  Result  \n\n"
print  "\n(a) the characteristic impedance, R0  is  ",R01,"  ohm"
print  "\n(b) the characteristic impedance, R0  is  ",R02,"  ohm"
print  "\n(c) the characteristic impedance, R0  is  ",R03,"  ohm"


  Result  



(a) the characteristic impedance, R0  is   20.0   ohm

(b) the characteristic impedance, R0  is   20.0   ohm

(c) the characteristic impedance, R0  is   250.0   ohm

Example 6, page no. 769

In [6]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
R1  =  500;#  in  ohm
R2  =  1000;#  in  ohm
I1  =  1;#  in  ampere  (lets  say)

#calculation:
 #  for  symmetrical  pi-attenuator  section
 #characteristic  impedance,  R0
R0  =  (R1*(R2**2)/(R1  +  2*R2))**0.5
 #current  Ix
Ix  =  (R2/(R2  +  R1  +  (R2*R0/(R2  +  R0))))*I1
 #current  I2
I2  =  (R2/(R2  +  R0))*Ix
ri  =  I1/I2;#  retio  of  currents
 #attenuation
attn  =  20*(1/2.303)*math.log(ri)


#Results
print  "\n\n  Result  \n\n"
print  "\n  the  characteristic  impedance  is  ",round(R0,2),"  ohm"
print  "\n  attenuation  is  ",round(attn,2),"  dB"


  Result  



  the  characteristic  impedance  is   447.21   ohm

  attenuation  is   8.36   dB

Example 7, page no. 770

In [2]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
Ra1  =  15;#  in  ohm
Ra2  =  15;#  in  ohm
Ra3  =  10;#  in  ohm
Rb1  =  15;#  in  ohm
Rb2  =  5;#  in  ohm
Rb3  =  5;#  in  ohm

#calculation:
Roc1 = Ra1 + Ra3
Rsc1 = Ra1 + Ra2*Ra3/(Ra2+Ra3)
R01 = (Roc1*Rsc1)**0.5

Roc2 = Rb2* (Rb1 + Rb3)/(Rb1 + Rb2 + Rb3)
Rsc2 = Rb2*Rb1/(Rb2+Rb1)
R02 = (Roc2*Rsc2)**0.5

#Results
print  "\n\n  Result  \n\n"
print  "\n  (a) the input resistance when the output port is open-circuited  is", Roc1," ohm for T-Network"
print   "and ",Roc2," ohm for pi-Network "
print  "\n  (b) the input resistance when the output port is short-circuited  is,", Rsc1," ohm for T-Network"
print   "and ",Rsc2," ohm for pi-Network "
print  "\n  (c) the characteristic impedance.  is,",round(R01,1)," ohm for T-Network and ,",round(R02,2)," ohm for pi-Network "


  Result  



  (a) the input resistance when the output port is open-circuited  is 25  ohm for T-Network
and  4.0  ohm for pi-Network 

  (b) the input resistance when the output port is short-circuited  is, 21.0  ohm for T-Network
and  3.75  ohm for pi-Network 

  (c) the characteristic impedance.  is, 22.9  ohm for T-Network and , 3.87  ohm for pi-Network

Example 8, page no. 770

In [3]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
Vat = 20; # in db
R0  =  600;#  in  ohm

#calculation:
N = math.e**(Vat*2.3/20)
R1 = R0*(N-1)/(N+1)
R2 = R0*2*N/(N**2 - 1)

#Results
print  "\n\n  Result  \n\n"
print  "\n For a T-section symmetrical attenuator pad, Resistance R1 is",round(R1,0)," ohm and Resistance R2 is",round(R2,1)," ohm"


  Result  



 For a T-section symmetrical attenuator pad, Resistance R1 is 491.0  ohm and Resistance R2 is 121.5  ohm

Example 9, page no. 771

In [4]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
Vat = 20; # in db
R0  =  600;#  in  ohm

#calculation:
N = math.e**(Vat*2.303/20)
R1 = R0*(N**2 - 1)/(2*N)
R2 = R0*(N+1)/(N-1)

#Results
print  "\n\n  Result  \n\n"
print  "\n For a pi-section symmetrical attenuator pad, Resistance R1 is",round(R1/1000,2)," Kohm"
print   "and Resistance R2 is",round(R2,0)," ohm"


  Result  



 For a pi-section symmetrical attenuator pad, Resistance R1 is 2.97  Kohm
and Resistance R2 is 733.0  ohm

Example 10, page no. 772

In [7]:
from __future__ import division
import math
#initializing  the  variables:
R1  =  300;#  in  ohm
R2  =  450;#  in  ohm
I1  =  1;#  in  ampere  (lets  say)

#calculation:
 #the  characteristic  impedance  of  a  symmetric  T-pad  attenuator  is  given  by
R0  =  (R1**2  +  2*R1*R2)**0.5
 #By  current  division
 #current  I2
I2  =  (R2/(R2  +  R1+  R0))*I1
ri  =  I1/I2;#  ratio  of  currents
 #insertion  loss
il  =  20*(1/2.303)*math.log(ri)


#Results
print  "\n\n  Result  \n\n"
print  "\n  the  characteristic  impedance  is  ",R0,"  ohm"
print  "\n  insertion  loss  is  ",round(il,2),"  dB"


  Result  



  the  characteristic  impedance  is   600.0   ohm

  insertion  loss  is   9.54   dB

Example 11, page no. 773

In [8]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
r  =  500;#  in  ohm
Rhm  =  3000;#  in  ohm
RL  =  2000;#  in  ohm
r1  =  2000;#  in  ohm
r2  =  1000;#  in  ohm
E  =  1;#  in  volts  (lets  say)

 #calculation:
 #Without  the  rheostat  in  the  circuit  the  voltage  across  the  2  kohm  load,  VL
VL  =  (RL/(RL  +  r))*E
 #voltage  V2  with  2kohm  tapping
V2  =  ((RL*r1/(r1  +  RL))/((RL*r1/(r1  +  RL))  +  Rhm  -  r1  +  r))*E
rv1  =  VL/V2;#  ratio  of  currents
 #insertion  loss  
il1  =  20*(1/2.303)*math.log(rv1)
 #voltage  V1  with  1kohm  tapping
V1  =  ((RL*r2/(r2  +  RL))/((RL*r2/(r2  +  RL))  +  Rhm  -  r2  +  r))*E
rv2  =  VL/V1;#  ratio  of  currents
 #insertion  loss  
il2  =  20*(1/2.303)*math.log(rv2)


#Results
print  "\n\n  Result  \n\n"
print  "\n  insertion  loss  for  2kohm  tap  is  ",round(il1,2),"  dB"
print  "\n  insertion  loss  for  1kohm  tap  is  ",round(il2,2),"  dB"


  Result  



  insertion  loss  for  2kohm  tap  is   6.02   dB

  insertion  loss  for  1kohm  tap  is   11.59   dB

Example 12, page no. 774

In [9]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
R1  =  1000;#  in  ohm
R2  =  500;#  in  ohm
I1  =  1;#  in  amperes  (lets  say)

#calculation:
 #characteristic  impedance  of  a  symmetrical  attenuator
R0  =  (R1*(R2**2)/(R1  +  2*R2))**0.5
 #current  Ix
Ix  =  (R2/(R2  +  R1  +  (R2*R0/(R2  +  R0))))*I1
 #current  I2
I2  =  (R2/(R2  +  R0))*Ix
ri  =  I1/I2;#  retio  of  currents
 #insertion  loss  
il  =  20*(1/2.303)*math.log(ri)


#Results
print  "\n\n  Result  \n\n"
print  "\n  characteristic  impedance  is  ",round(R0,2),"  ohm"
print  "\n  insertion  loss  is  ",round(il,2),"  dB"


  Result  



  characteristic  impedance  is   353.55   ohm

  insertion  loss  is   15.31   dB

Example 13, page no. 776

In [10]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
R1  =  100;#  in  ohm
R2  =  200;#  in  ohm
R3  =  300;#  in  ohm
I1  =  1;#  in  amperes  (lets  say)

#calculation:
 #image  impedance  Roa
Roa  =  ((R1  +  R2)*(R2  +  (R1*R3/(R1  +  R3))))**0.5
 #image  impedance  Rob
Rob  =  ((R1  +  R3)*(R3  +  (R1*R2/(R1  +  R2))))**0.5
 #The  iterative  impedance  at  port  1
Ri1  =  (-1*R1  +  (R1**2  -  (-1*4*((R2*(R1  +  R3))  +  (R3*R1))))**0.5)/2
 #The  iterative  impedance  at  port  2
Ri2  =  (R1  +  (R1**2  -  (-1*4*((R3*(R1  +  R2))  +  (R2*R1))))**0.5)/2


#Results
print  "\n\n  Result  \n\n"
print  "\n  image  impedance  are  ",round(Roa,2),"  ohm  and  ",round(Rob,2),"  ohm  "
print  "\n  iterative  impedances  are  ",round(Ri1,2),"  ohm  and  ",round(Ri2,2),"  ohm  "


  Result  



  image  impedance  are   287.23   ohm  and   382.97   ohm  

  iterative  impedances  are   285.41   ohm  and   385.41   ohm

Example 14, page no. 777

In [11]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
R1  =  1000;#  in  ohm
R2  =  2000;#  in  ohm
R3  =  3000;#  in  ohm
I1  =  1;#  in  amperes  (lets  say)

#calculation:
 #image  impedance  Roa
Roa  =  (((R3  +  R2)*R1/(R1  +  R2  +  R3))*(R1*R3/(R1  +  R3)))**0.5
 #image  impedance  Rob
Rob  =  (((R3  +  R1)*R2/(R1  +  R2  +  R3))*(R2*R3/(R2  +  R3)))**0.5
 #The  iterative  impedance  at  port  1
Ri1  =  (-1*R1  +  ((R1**2)  -  (-1*4*2*R2*R1))**0.5)/(2*2)
 #The  iterative  impedance  at  port  2
Ri2  =  (R1  +  ((-1*R1)**2  -  (-1*4*2*R2*R1))**0.5)/(2*2)


#Results
print  "\n\n  Result  \n\n"
print  "\n  image  impedance  are  ",round(Roa,2),"  ohm  and  ",round(Rob,2),"  ohm  "
print  "\n  iterative  impedances  are  ",round(Ri1,2),"  ohm  and  ",round(Ri2,2),"  ohm  "


  Result  



  image  impedance  are   790.57   ohm  and   1264.91   ohm  

  iterative  impedances  are   780.78   ohm  and   1280.78   ohm

Example 15, page no. 780

In [12]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
r  =  500;#  in  ohm
RL  =  100;#  in  ohm
E  =  1;#  in  volts  (lets  say)

#calculation:
 #res.
R1  =  (r*(r  -  RL))**0.5
R2  =  (r*RL**2/(r  -  RL))**0.5
 #current  I1
I1  =  E/(r  +  R1  +  R2*RL/(RL  +  R2))
 #current  I2
I2  =  (R2/(R2  +  RL))*I1
 #input  power
P1  =  r*I1**2
 #output  power
P2  =  RL*I2**2
 #attenuation
attn  =  10*(1/2.303)*math.log(P1/P2)
 #Load  current,  IL
IL  =  E/(r  +  RL)
 #voltage,  VL
VL  =  IL*RL
 #voltage,  V1
V1  =  E  -  I1*r
 #voltage,  V2
V2  =  V1  -  I1*R1
 #insertion  loss
il  =  20*(1/2.303)*math.log(VL/V2)


#Results
print  "\n\n  Result  \n\n"
print  "\n  R1  =  ",round(R1,2),"  ohm  and  R2  =  ",round(R2,2),"  ohm  "
print  "\n  attenuation  is  ",round(attn,2),"  dB  "
print  "\n  In  decibels,  the  insertion  loss  is  ",round(il,2),"  dB  "


  Result  



  R1  =   447.21   ohm  and  R2  =   111.8   ohm  

  attenuation  is   12.54   dB  

  In  decibels,  the  insertion  loss  is   9.98   dB

Example 16, page no. 783

In [1]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
attnO  =  70;#  in  dB
n  =  5;#  numbers  of  identical  atteneurs
V1  =  0.02;#  in  Volts

#calculation:
 #attenuation  of  each  section
attn  =  attnO/n
 #output  of  the  final  stage
Vo  =  V1/(10**(attnO/20))
 #voltage  output  of  the  third  stage
V3  =  V1/(10**(3*attn/20))


#Results
print  "\n\n  Result  \n\n"
print  "\n  attenuation  of  each  section  =  ",round(attn,2),"  dB  "
print  "\n  output  of  the  final  stage  is  ",round(Vo*1E6,2),"uV  "
print  "\n  voltage  output  of  the  third  stage  is  ",round(V3*1E3,3),"mV  "


  Result  



  attenuation  of  each  section  =   14.0   dB  

  output  of  the  final  stage  is   6.32 uV  

  voltage  output  of  the  third  stage  is   0.159 mV

Example 17, page no. 784

In [5]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
r = 450; # in ohm
R0 = 450; # in ohms
x = 1/8

#calculation:
N = 1/x
R1 = R0*(N-1)/(N+1)
R2 = R0*2*N/(N**2 - 1)

Io = x*x

attn  =  20*math.log10(N)
attnO = 4*attn

#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)For a T-section symmetrical attenuator pad, Resistance R1 is",round(R1,2)," ohm "
print   "and Resistance R2 is",round(R2,0)," ohm"
print  "\n  (b)current flows in the load =  ",round(Io,2),"of the original current."
print  "\n  (c)overall attenuation is  ",round(attnO,2),"dB  "


  Result  



  (a)For a T-section symmetrical attenuator pad, Resistance R1 is 350.0  ohm 
and Resistance R2 is 114.0  ohm

  (b)current flows in the load =   0.02 of the original current.

  (c)overall attenuation is   72.25 dB