Answer:
Explanation:
The frequency components in the message signal are
f1 = 100Hz, f2 = 200Hz and f3 = 400Hz
When amplitude modulated with a carrier signal of frequency fc = 100kHz
Generates the following frequency components
Lower side band
Carrier frequency 100kHz
Upper side band
After passing through the SSB filter that filters the lower side band, the transmitted frequency component will be
At the receive these are mixed (superheterodyned) with local ocillator frequency whichh is 100.02KHz, the output frequencies will be
After passing through the SSB filter that filters the higher side band, the transmitted frequency component will be
At the receive these are mixed (superheterodyned) with local oscillator frequency which is 100.02KHz, and then fed to the detector whose output frequencies will be
Is miraculous ladybug one
Answer: the ripple factor is 0.005
Explanation:
Given the data in the question;
we know that expression of ripple factor is;
r = Vr(pp) / Vdc
where Vr(pp) the peak to peak is ripple voltage ( 100mv = 0.1 V)
and Vdc is the dc value of the filter output voltage ( 20 V)
so we substitute our given values;
r = 0.1 / 20
r = 0.005
Therefore; the ripple factor is 0.005
Answer:
The split is given by including spaces in both tabs
Explanation:
The bracket notation can be used to indicate the split. Here is an example:
String [ ] parts = s. split ( "[/]")
Answer: the mass flow rate of concentrated brine out of the process is 46,666.669 kg/hr
Explanation:
F, W and B are the fresh feed, brine and total water obtained
w = 2 x 10^4 L/h
we know that
F = W + B
we substitute
F = 2 x 10^4 + B
F = 20000 + B .................EQUA 1
solute
0.035F = 0.05B
B = 0.035F/0.05
B = 0.7F
now we substitute value of B in equation 1
F = 20000 + 0.7F
0.3F = 20000
F = 20000/0.3
F = 66666.67 kg/hr
B = 0.7F
B = 0.7 * F
B = 0.7 * 66666.67
B = 46,666.669 kg/hr
the mass flow rate of concentrated brine out of the process is 46,666.669 kg/hr
