Answer:
QPSK: 7.5 MHz
64-QAM:2.5 MHz
64-Walsh-Hadamard: 160 MHz
Explanation:
See attached picture.
I can't log in even though it says all ages are accepted.Instead when I TRY logging in it says this. "We're sorry, but we are no
1 answer:
You might be interested in
Answer:
a) Mechanical efficiency ()=63.15% b) Temperature rise= 0.028ºC
Explanation:
For the item a) you have to define the mechanical power introduced (Wmec) to the system and the power transferred to the water (Pw).
The power input (electric motor) is equal to the motor power multiplied by the efficiency. Thus, .
Then, the power transferred (Pw) to the fluid is equal to the flow rate (Q) multiplied by the pressure jump . So
.
The efficiency is defined as the ratio between the output energy and the input energy. Then, the mechanical efficiency is
For the b) item you have to consider that the inefficiency goes to the fluid as heat. So it is necessary to use the equation of the heat capacity but in a "flux" way. Calling <em>H</em> to the heat transfered to the fluid, the specif heat of the water and the density of the water:
[/tex]
Finally, the temperature rise is:
As there are 10 V, for Vp1, that is the peak-voltage of the source:
Then, transformer's theory says that the relation of transformations is:
V1/V2=a
Where V1 is the voltage in the primary and V2 in the secondary.
V1=14.14 V
V2=8.55 V
a=1.65
Then, with the 8.5 V, we find the real peak-voltage, taking in account that in the diodes we have a drop of 0.7 V each, so:
8.5 -1.4=7.1 V
And this will be called VpL
Now we proceed to calculate the mean voltage:
Where Vr is the ripple voltage, we asume that is 1 V
So, Vmean = 6.6 V
Then we have
Vmean/R= I mean
We have that R=1000 Ohm
Imedia=6.6 V/1000 Ohm
Imedia=6.6 mAmps
Finally, we can calculate the capacitor:
C=Q/Vr
C=Imean/(Vr*2f)
Where f is 60Hz
C=6.6mA/(1V*120)
C=5.5 uFarads
Therefore:
C=5.5 uFarads that works at 12 V