Chapter9:MICROWAVE LINEAR-BEAM TUBES(O TYPE)

Eg9.2.1:pg-377

In [4]:
import math
#(a) Calculate the input gap voltage to give maximum voltage V2

#For maximum V2, J1(X) must be maximum.This means J1(X)=0.582 at X=1.841.  
X=1.841   #bunching parameter
J1X=0.582 
V0=10**3  #dc voltage in Volt
v0=0.593*(10**6)*(sqrt(V0)) #The electron velocity just leaving the cathode
f=3*(10**9)                 #operating frequency in Hertz
d=1*(10**-3)                #Gap spacing in either cavity in meter
w=(2*math.pi*f)             #angular frequency in Hertz
Og=(w*d)/v0                 #calculation of gap transit angle in radians
Bi=(math.sin(Og/2))/(Og/2)  #value of Bi is wrong in book because of calculation mistake
Bo=Bi                       #value of Bo is wrong in book
L=4*(10**-2)                #Spacing between the two cavities in meter
O0=(w*L)/v0                 #DC transit angle between the cavaties in radians
V1max=(2*V0*X)/(round(Bi,3)*int(O0)) 
print"The maximum input voltage V1 (in Volts) is =",round(V1max,1),"V" #value of Bi used in book for calculation of V1 is wrong so answer is wrong in book  

#(b) Calculate the voltage gain
R0=40*(10**3)             #in Ohms
Rsh=30*(10**3)            #Effective shunt impedance excluding beam loading in Ohms
Av=((round(Bo,3)**2)*int(O0)*J1X*Rsh)/(R0*X) #voltage gain
print"The voltage gain,neglecting the beam loading in the output cavity is =",round(Av,3) #answer is wrong in book as the value of Bo used is wrong       

#(c)Calculate the efficiency of the  amplifier
I0=25*(10**-3)    #in ampere
I2=2*I0*J1X 
V2=round(Bo,3)*I2*Rsh 
efficiency=(round(Bo,3)*I2*int(V2))/(2*I0*V0) 
efficiency=100*efficiency 
print"The efficiency of the  amplifier,neglecting beam loading =",round(efficiency,1),"%" #calculation mistake in book    

#(d)Calculate the beam loading conductance
G0=25*(10**-6)    #G0=I0/V0 is the DC beam conductance in mho
Og=round(Og)
GB=(G0/2)*((round(Bo,3)**2)-(round(Bo,3)*math.cos(Og/2))) 
print"The beam loading conductance GB (in mho) is =","{:.1e}".format(GB),"mho" #answer is wrong in book as the value of Bo in book is wrong
RB=1/GB 
print"Then the beam loading resistance RB (ohm)is =","{:.2e}".format(RB),"ohms"
print('In comparasion with RL and Rsho or the effective shunt resistance Rsh,the beam loading resistance is like an open circuit and thus can be neglected in the preceding calculations') 
The maximum input voltage V1 (in Volts) is = 96.1 V
The voltage gain,neglecting the beam loading in the output cavity is = 8.704
The efficiency of the  amplifier,neglecting beam loading = 46.6 %
The beam loading conductance GB (in mho) is = 9.6e-07 mho
Then the beam loading resistance RB (ohm)is = 1.04e+06 ohms
In comparasion with RL and Rsho or the effective shunt resistance Rsh,the beam loading resistance is like an open circuit and thus can be neglected in the preceding calculations

Eg9.3.1:pg-385

In [6]:
import math
#(a) Calculate the dc electron velocity
V0=14.5*(10**3)            #beam voltage in Volt
v0=0.593*(10**6)*sqrt(V0)  #dc electron velocity 
print"The dc electron velocity(in m/s)is =","{:.2e}".format(v0),"m/s"

#(b) Calculate the dc phase constant
f=(10*(10**9))          #operating frequency in Hertz
Be=(2*math.pi*f)/v0     #angular frequency in Hertz    
print"The dc phase constant(in rads/m) is =","{:.2e}".format(Be),"rads/m"

#(c)Calculate the plasma frequency
po=1*(10**-6)          #dc electron charge density in C/m**3
wp=sqrt((1.759*(10**11)*po)/(8.854*(10**-12))) 
print"The plasma frequency(in rad/s)is =","{:.2e}".format(wp),"rad/s"

#(d) Calculate the reduced plasma frequency for R=0.4
R=0.4 
wq=R*wp 
print"The reduced plasma frequency for R=0.4 (in rad/s) is =","{:.2e}".format(wq),"rad/s" 

#(e)Calculate the dc beam current density
J0=po*v0 
print"The dc beam current density(in A/m2) is =",round(J0,1),"A/m2" 

#(f) Calculate the instantaneous beam current density
p=1*(10**-8)       #RF charge density in C/m**3
v=1*(10**5)        #velocity perturbation in m/s
J=(p*v0)-(po*v) 
print"The instantaneous beam current density (in A/m2) is =",round(J,3),"A/m2"
The dc electron velocity(in m/s)is = 7.14e+07 m/s
The dc phase constant(in rads/m) is = 8.80e+02 rads/m
The plasma frequency(in rad/s)is = 1.41e+08 rad/s
The reduced plasma frequency for R=0.4 (in rad/s) is = 5.64e+07 rad/s
The dc beam current density(in A/m2) is = 71.4 A/m2
The instantaneous beam current density (in A/m2) is = 0.614 A/m2

Eg9.3.2:pg-386

In [7]:
import math
#(a) Calculate the dc electron velocity
V0=18*(10**3)              #beam voltage in Volt
v0=0.593*(10**6)*sqrt(V0) 
print"The dc electron velocity(in m/s)is =","{:.2e}".format(v0),"m/s" 

#(b) Calculate the dc electron phase constant
f=(10*(10**9))            #Operating frequency in Hertz
w=2*math.pi*f             #angular frequency in Hertz
Be=w/v0 
print"The dc electron phase constant(in rads/m) is =","{:.3e}".format(Be),"rads/m" 

#(c) Calculate the plasma frequency
po=1*(10**-8)            #dc electron beam current density in C/m**3
wp=sqrt((1.759*(10**11)*po)/(8.854*(10**-12))) 
print"The plasma frequency(in rad/s) is =","{:.2e}".format(wp),"rad/s"

#(d) Calculate the reduced plasma frequency for R=0.5
R=0.5 
wq=R*wp 
print"The reduced plasma frequency for R=0.5 (in rad/s)is =","{:.2e}".format(wq),"rad/s" 

#(e) Calculate the reduced plasma phase constant
Bq=wq/v0 
print"The reduced plasma phase constant(in rad/m) is =",round(Bq,3),"rad/m"

#(f) Calculate the transit time across the input gap
d=1*(10**-2)            #gap distance in m
t=d/v0 
t=t*(10**9) 
print"The transit time across the input gap(in ns) is =",round(t,4),"ns" 

#(g) Calculate the electron velocity leaving the input gap
V1=10              #signal voltage in Volt
Bi=1.0             #beam coupling coefficient
Vt1=v0*(1+(((Bi*V1)/(2*V0))*math.sin(w*round(t,4)*(10**-9)))) 
print"the electron velocity leaving the input gap(in m/s)is =","{:.2e}".format(v0),"+","{:.2e}".format(v0*(((Bi*V1)/(2*V0))*math.sin(w*round(t,4)*(10**-9)))),"m/s =","{:.2e}".format(Vt1),"m/s"                                
The dc electron velocity(in m/s)is = 7.96e+07 m/s
The dc electron phase constant(in rads/m) is = 7.897e+02 rads/m
The plasma frequency(in rad/s) is = 1.41e+07 rad/s
The reduced plasma frequency for R=0.5 (in rad/s)is = 7.05e+06 rad/s
The reduced plasma phase constant(in rad/m) is = 0.089 rad/m
The transit time across the input gap(in ns) is = 0.1257 ns
the electron velocity leaving the input gap(in m/s)is = 7.96e+07 + 2.21e+04 m/s = 7.96e+07 m/s

Eg9.3.3:pg-388

In [8]:
import math
#(a)Calculate the plasma frequency
po=1*(10**-6)               #dc electron beam current density in C/m**3
e0=8.854*(10**-12)          #permittivity of free space in F/m
wp=sqrt((1.759*(10**11)*po)/e0) 
print"The plasma frequency(in rad/s)is =","{:.2e}".format(wp),"rad/s"

#(b) Calculate the reduced plasma frequency for R=0.5
R=0.5 
f=8*(10**9)          #operating frequency in Hertz
w=2*math.pi*f        #angular frequency in Hertz
wq=R*wp 
print"The reduced plasma frequency for R=0.5(in rad/s)is =","{:.2e}".format(wq),"rad/s" 

#(c) Calculate the induced current in the output cavity
V0=20*(10**3)        #beam voltage in Volt
I0=2                 #beam current in ampere
V1=10                #Signal voltage in Volt
Bo=1                 #Beam coupling coefficient
I2=(I0*w*(Bo**2)*V1)/(2*V0*wq) 
print"The induced current in the output cavity(in Ampere)is =",round(I2,4),"A" 

#(d) Calculate the induced voltage in the output cavity
Rshl=30*(10**3)      #total shunt resistance including load in Volt
V2=round(I2,3)*Rshl 
V2=V2/1000           #in KV
print"The induced voltage in the output cavity(in KV)is =",V2,"KV"

#(e) Calculate the output power delivered to the load
Rsh=10*(10**3)       #shunt resistance of the cavity
Rshl=30*(10**3)      #total shunt resistance including load
Pout=(round(I2,3)**2)*Rshl 
Pout=Pout/1000 
print"The output power delivered to the load(in KW)is =",round(Pout,2),"KW"

#(f) Calculate the power gain
powergain=(((I0*w)**2)*(Bo**4)*Rsh*Rshl)/(4*((V0*wq)**2)) 
powergain=10*math.log10(powergain)      #powergain in dB
print"The power gain is =",round(powergain,1),"dB"

#(g) Calculate the electronic efficiency
n=(Pout*1000)/(I0*V0) 
n=n*100 
print"The electronic efficiency (in %)is =",round(n,1),"%" 
The plasma frequency(in rad/s)is = 1.41e+08 rad/s
The reduced plasma frequency for R=0.5(in rad/s)is = 7.05e+07 rad/s
The induced current in the output cavity(in Ampere)is = 0.3566 A
The induced voltage in the output cavity(in KV)is = 10.71 KV
The output power delivered to the load(in KW)is = 3.82 KW
The power gain is = 55.8 dB
The electronic efficiency (in %)is = 9.6 %

Eg9.3.4:pg-390

In [9]:
import math
#(a) Calculate the plasma frequency
po=5*(10**-5)          #dc electron beam current density in C/m**3
wp=sqrt((1.759*(10**11)*po)/(8.854*(10**-12))) 
print"The plasma frequency(in rad/s)is =","{:.2e}".format(wp),"rad/s" 

#(b) Calculate the reduced plasma frequency for R=0.6
R=0.6 
f=4*(10**9)        #operating frequency in Hertz
w=2*math.pi*f      #angular frequency in Hertz
wq=R*wp 
print"The reduced plasma frequency for R=0.6(in rad/s)is =","{:.2e}".format(wq),"rad/s" 

#(c) Calculate the induced current in the output cavity
Rsh=10*(10**3)    #shunt resistance of the cavity in Ohms
Rshl=5*(10**3)    #total shunt resistance including load in Ohms
V0=10*(10**3)     #beam voltage in volt
I0=0.7            #beam current in ampere
V1=2              #Signal voltage in volt
Bo=1              #Beam coupling coefficient
I4=(((I0*w)**3)*(Bo**6)*V1*(Rsh**2))/(8*((V0*wq)**3)) 
print"The induced current in the output cavity(in Ampere)is =",round(I4,4),"A"

#(d) Calculate the induced voltage in the output cavity
V4=I4*Rshl 
V4=V4/1000      #in KV
print"The induced voltage in the output cavity(in KV)is =",round(V4,2),"KV"

#(e) Calculate the output power delivered to the load
Pout=(I4**2)*Rshl 
Pout=Pout/1000    #in KW
print"The output power delivered to the load(in KW)is =",round(Pout,2),"KW" 
The plasma frequency(in rad/s)is = 9.97e+08 rad/s
The reduced plasma frequency for R=0.6(in rad/s)is = 5.98e+08 rad/s
The induced current in the output cavity(in Ampere)is = 0.6366 A
The induced voltage in the output cavity(in KV)is = 3.18 KV
The output power delivered to the load(in KW)is = 2.03 KW

Eg9.4.1:pg-399

In [10]:
import math
#(a) Calculate the value of the repeller voltage
V0=600           #beam voltage in volt
n=2              #mode=2
fr=9*(10**9)     #operating frequency in Hertz
w=2*math.pi*fr   #angular frequency in Hertz
L=1*(10**-3)     #in meter
em=1.759*(10**11)   #em=e/m is the charge to mass ratio of electron
x=((em)*(((2*math.pi*n)-(math.pi/2))**2))/(8*(w**2)*(L**2)) #x=V0/(V0+Vr)**2
y=V0/x           #y=(V0+Vr)**2
z=sqrt(y)        #z=V0+Vr
Vr=z-V0 
print"The value of the repeller voltage(volts)is =",int(round(Vr)),"V" 

#(b)Calculate the direct current necessary to give a microwave gap voltage of 200V
        #Assume that Bo=1
        #V2 = I2*Rsh = 2*I0*J1(X)*Rsh   
V2=200    #gap voltage in volt
Rsh=15*(10**3) #shunt resistance of te cavity in ohms
X=1.841         #bunching parameter
J1X=0.582 
I0 = V2/(2*J1X*Rsh) 
I0=I0*1000   #in mA
print"The direct current necessary to give a microwave gap voltage of 200V(in mA)is =",round(I0,2),"mA" 

#(c) Calculate the electronic efficiency 
efficiency=(2*X*J1X)/((2*math.pi*n)-(math.pi/2)) 
efficiency=efficiency*100 
print"The electronic efficiency(in %)is =",round(efficiency,2),"%"
The value of the repeller voltage(volts)is = 250 V
The direct current necessary to give a microwave gap voltage of 200V(in mA)is = 11.45 mA
The electronic efficiency(in %)is = 19.49 %

Eg9.5.1:pg-415

In [15]:
import math
#(a) Calculate the gain parameter
I0=30*(10**-3)     #Beam current in ampere
V0=3*(10**3)       #Beam voltage in volt
Z0=10              #characteristic impedance of the helix in ohms
C=(((I0*Z0)/(4*V0))**(1/3.0)) 
print"The gain parameter is =","{:.2e}".format(C) 

#(b) Calculate the output power gain in dB
N=50             #Crcular length
Ap=-9.54+(47.3*N*round(C,4)) 
print"The output power gain (in dB) is =",round(Ap,2),"dB"

#(c) Calculate the four propagation constants
f=10*(10**9)     #frequency in Hertz
V0=3*(10**3)     #beam voltage in volt
w=2*(math.pi)*f  #nagular frequency in Hertz
v0=0.593*(10**6)*sqrt(V0) 
Be=w/v0 
Be=int(Be/10)
Be=Be*10
r1=(-1*Be*round(C,4)*(round((sqrt(3)/2),2)))+1j*Be*(1+(round(C,4)/2)) 
X=round(r1.real,2)
Y=int(r1.imag)
r1=X+1j*Y
print"The four propagation constants are:"
print"\nThe first propagtaion constant is =",r1    #value of imaginary part is wrong in book
r2=(Be*round(C,4)*(round((sqrt(3)/2),2)))+1j*Be*(1+(round(C,4)/2))
X=round(r2.real,2)
Y=int(r2.imag)
r2=X+1j*Y
print"The second propagtaion constant is =",r2   #value of imaginary part is wrong in book
r3=1j*Be*(1-round(C,4)) 
print"The third propagtaion constant is =",r3    #answer is wrong in book
r4=(-1*1j*Be*(1-((C**3)/4)))
r4=0+1j*round(r4.imag)
print"The fourth propagtaion constant is =",r4
The gain parameter is = 2.92e-02
The output power gain (in dB) is = 59.52 dB
The four propagation constants are:

The first propagtaion constant is = (-49.03+1958j)
The second propagtaion constant is = (49.03+1958j)
The third propagtaion constant is = 1873.644j
The fourth propagtaion constant is = -1930j

Eg9.7.1:pg-427

In [16]:
#(a) Calculate the number of electrons returned per second
Ir=0.85         #returned current in ampere
q=1.6*(10**-19) #electronic charge in C
Nr=Ir/q 
print"The number of electrons returned (per second)  is =","{:.2e}".format(Nr),"electrons/s" 

#(b)Calculate the Energy associated with these returning electrons in 20ms
V=-11*(10**3)   #overdepreesion collector voltage in volt
V=-1*V
t=20*(10**-3)   #in seconds
W=V*Nr*t 
print"The Energy associated with these returning electrons in 20ms(in eV) is =","{:.3e}".format(W),"eV"

#(c) Calculate the Power for returning electrons
P=V*Ir 
P=P/1000   #in KW
print"The Power for returning electrons(in KW)is =",P,"KW" 

#(d) Calculate the Heat in calories associated with the returning electrons(a factor for converting joules to calories is 0.238)
t=20*(10**-3)  #in seconds
H=0.238*P*1000*t 
print"The Heat associated with the returning electrons(in calories)is =",round(H,2),"calories"

#(e) Calculate the temperature
mass=250*(10**-3)    #mass of heated iron pole piece in gram
specific_heat=0.108 
T=round(H,2)/(mass*specific_heat) 
print"The temperature(in degree Celsius)is =",round(T,2),"degree celsius"

#(f) Calculate whether the output iron pole piece is melted
print"The output iron pole piece is melted"
The number of electrons returned (per second)  is = 5.31e+18 electrons/s
The Energy associated with these returning electrons in 20ms(in eV) is = 1.169e+21 eV
The Power for returning electrons(in KW)is = 9.35 KW
The Heat associated with the returning electrons(in calories)is = 44.51 calories
The temperature(in degree Celsius)is = 1648.52 degree celsius
The output iron pole piece is melted