Chapter 11 Cathode ray oscilloscope

Ex 11_1 PG-11.16

In [1]:
from math import sqrt, pi
print "Refer to the figure-11.11 showwn"
print "from the figure we can see that the screen is divided such"
print "that one part of the half wave is sub-divided into 5 units"
print "1subdivision=(1/5)units=0.2 units"
print "amplitude of the positive peak signal=2division+3subdivision "
print "ie amplitude of the positive peak signal=2+3*0.2=2.6"
Vp=2+3*0.2##amplitude of the peak voltage in terms of dividion
Vpp=2*Vp##peak to peak voltage in terms of division
VA=2##Vertical attenuation in (mV/div)
V=Vpp*VA##required peak to peak output voltage in volts
Vm=V/2##amplitude of the output voltage in volts
print "\n Therefore amplitude of the output voltage in volts=%.1f mV \n"%(Vm)
Vrms=Vm/sqrt(2)##rms value of the output voltage
print "\n Therefore rms value of the output voltage in volts=%.4f mV \n"%(Vrms)

Refer to the figure-11.11 showwn
from the figure we can see that the screen is divided such
that one part of the half wave is sub-divided into 5 units
1subdivision=(1/5)units=0.2 units
amplitude of the positive peak signal=2division+3subdivision 
ie amplitude of the positive peak signal=2+3*0.2=2.6

 Therefore amplitude of the output voltage in volts=5.2 mV 


 Therefore rms value of the output voltage in volts=3.6770 mV

Ex 11_2 PG-11.19

In [2]:
from math import sqrt, pi
Vd=2##voltage per division in(V/div)
Td=2e-3##time base division in (s/div)
Vdiv=3##vertical occupancy in division as shown in the screen
Vpp=Vd*Vdiv##peak to peak voltage 
Vm=Vpp/2##peak voltage
Vrms=Vm/sqrt(2)##rms value of the output voltage
print "\n Therefore rms value of the output voltage in volts=%.4f V \n"%(Vrms)
Hdiv=2##horizontal occupancy in division as shown in the screen
T=Hdiv*Td##time period of the waveform
f=1/T##frequency
print "\n Therefore frequency f=%.0f Hz \n "%(f)

 Therefore rms value of the output voltage in volts=2.1213 V 


 Therefore frequency f=250 Hz