If a power meter measures 1% of the total channel power to be -33.9dBm, what is the total channel power?
The key concept here is that dBs enable the use of addition vs. multiplication:
Pt[in Watts] = 100 * Pm[in Watts] ==>
Pt[in dBm's] = 10 log10(100) + Pm[in dBm's] =
20 + -33.9 =
-13.9dBm
Check:
Pm = (10^(-33.9dBm/10))*0.001 = 0.000000407380278 Watts
Pt = (10^(-13.9dBm/10))*0.001 = 0.00004073802778 Watts
Pt[in milliWatts] = 100 * Pm[in milliWatts] =
0.00004073802778 =? 100 * 0.000000407380278 =
0.0000407380278
Related
I have some doubt with these kind of problems, example:
"If we asked 20,000 in a stadium to toss a coin 10 times, what it's the probability of at least one person getting 10 heads?"
I took this example from Practical Statistics for Data Scientist.
So, the probability of at least one person getting 10 heads it's calculated using: 1 - P(of nobody in the stadium getting 10 heads).
So we kind of doing an exclude procedure here, first I get the probability of the contrary event I am trying to measure, not the ACTUAL experiment I want to measure: at least one people getting 10 heads.
Why do we do it this way?
How can I calculate the probability of at least someone getting 10 heads but without passing through the probability of no one getting 10 heads?
As #Robert Dodier mentioned in the comments, the reason is that the calculations are simpler. I will use a stadium of 20 people instead of 20000 as an example:
Method 1:
Probability of not getting 10 heads for one individual
= 1 - probability of getting 10 heads
= 1 - 10!/(10!0!)*0.5^10*(1-0.5)^0
= 0.9990234375
Probability of at least one person in the stadium getting 10 heads
= 1 - P(of nobody in the stadium getting 10 heads)
= 1 - 0.9990234375**20 (because all coin tosses are independent)
= 0.019351109194852834
Method 2:
Probability of getting 10 heads for one individual
= 10!/(10!0!)*0.5^10*(1-0.5)^0
= 0.0009765625
Probability of exactly 1, 2, 3, etc. persons in the stadium getting 10 heads:
p1 = 20!/(1!19!)*0.0009765625^1*(1-0.0009765625)^(20-1) = 0.019172021325613825
p2 = 20!/(2!18!)*0.0009765625^2*(1-0.0009765625)^(20-2) = 0.00017803929872270904
p3 = 20!/(3!17!)*0.0009765625^3*(1-0.0009765625)^(20-3) = 1.0442187608370032e-06
p4 = 20!/(4!16!)*0.0009765625^4*(1-0.0009765625)^(20-4) = 4.338152232216289e-09
p5 = 20!/(5!15!)*0.0009765625^5*(1-0.0009765625)^(20-5) = 1.3569977656981548e-11
p6 = 20!/(6!14!)*0.0009765625^6*(1-0.0009765625)^(20-6) = 3.316221323798032e-14
p7 = 20!/(7!13!)*0.0009765625^7*(1-0.0009765625)^(20-7) = 6.483326146232712e-17
p8 = 20!/(8!12!)*0.0009765625^8*(1-0.0009765625)^(20-8) = 1.029853859983202e-19
p9 = 20!/(9!11!)*0.0009765625^9*(1-0.0009765625)^(20-9) = 1.342266353839299e-22
p10 = 20!/(10!10!)*0.0009765625^10*(1-0.0009765625)^(20-10) = 1.443297154665913e-25
p11 = 20!/(11!9!)*0.0009765625^11*(1-0.0009765625)^(20-11) = 1.2825887804726853e-28
p12 = 20!/(12!8!)*0.0009765625^12*(1-0.0009765625)^(20-12) = 9.403143551852531e-32
p13 = 20!/(13!7!)*0.0009765625^13*(1-0.0009765625)^(20-13) = 5.656451493707817e-35
p14 = 20!/(14!6!)*0.0009765625^14*(1-0.0009765625)^(20-14) = 2.7646390487330485e-38
p15 = 20!/(15!5!)*0.0009765625^15*(1-0.0009765625)^(20-15) = 1.0809927854283668e-41
p16 = 20!/(16!4!)*0.0009765625^16*(1-0.0009765625)^(20-16) = 3.3021529369146104e-45
p17 = 20!/(17!3!)*0.0009765625^17*(1-0.0009765625)^(20-17) = 7.59508466888531e-49
p18 = 20!/(18!2!)*0.0009765625^18*(1-0.0009765625)^(20-18) = 1.2373875315877011e-52
p19 = 20!/(19!1!)*0.0009765625^19*(1-0.0009765625)^(20-19) = 1.2732289258503896e-56
p20 = 20!/(20!0!)*0.0009765625^20*(1-0.0009765625)^(20-20) = 6.223015277861142e-61
Probability of at least one person in the stadium getting 10 heads
= p1 + p2 + p3 + p4 + p5 + p6 + p7 + p8 + p9 + p10 +
p11 + p12 + p13 + p14 + p15 + p16 + p17 + p18 + p19 + p20
= 0.01935110919485281
So the result is the same (the tiny difference is due to floating point precision), but as you can see the first calculation is slightly simpler for 20 people, never mind for 20000 ;)
I have a problem and i need your help!
Here is the code:
kg_lemons = float(input())
kg_sugar = float(input())
water = float(input())
total_lemon_juice = kg_lemons * 980 #in mililiters need to multiply by 1000
total_lemonade = total_lemon_juice + 5 * 1000 + (0.3 * kg_sugar)
cups_made = total_lemonade / 150
money_made = cups_made * 1.20
print(f'All cups sold: {cups_made:.2f}')
print(f'Money earned: {money_made:.2f}')
At then end, after I print it, it must shown the numbers:
All cups sold: 66
Money earned: 79.20
But I got:
All cups sold: 66.01
Money earned: 79.21
So I need to round it up to the second decimal (the lowest number). Should I use math.floor and, if so, how?
I need some help with a quiete simple problem in FiPy. My goal is to simulate a fluid flowing through a concrete block while phase change.
But first of all I tried to do a simple 1D simulation assumed a fluid massflow and a constant wall temperature without any phase change.
from fipy import *
from fipy.meshes import CylindricalGrid2D, Grid1D
import matplotlib.pyplot as plt
import numpy as np
#%%
L = 1.5 #length transfer surface
bS = 0.75 #wide
AV = L * bS #transfer surface
tS0 = 350. #tWall
rhoWF = 880. #density fluid
mWF = 0.036 #mass flow
u = 5e-4 #Fluid speed
hWF = mWF / AV / rhoWF / u #height "fluid block"
nx = 50
VWF = hWF * L * bS/nx #fluid volumen
lambdaWF = 0.6 # thermal conductivity
alpha = 500. #heat transfer coefficient
tWF0 = 371.
mesh = Grid1D(dx=L/nx, nx=nx)
tWF = CellVariable(name="Fluid",
mesh=mesh,
value= tWF0,
hasOld=True)
tS = CellVariable(name="storage",
mesh=mesh,
value=tS0,
hasOld=True)
sourceWF=CellVariable(name="source Fluid", #Variable der Konvektion
mesh=mesh,
value=0.)
cvrho = CellVariable(name = 'cprho',#Fluid
mesh = mesh,
value = rhoWF * 4215.2,
hasOld = True)
tWF.constrain(tWF0, mesh.facesLeft()) #constant inlet temperature
t = 6*3600. #time
timeStepDuration = 1e2
#outflow boundary condition
outlet = mesh.facesRight
ConvCoeff = FaceVariable(mesh,value=u,rank=1)
exteriorCoeff = FaceVariable(mesh,value=0.,rank=1)
exteriorCoeff.setValue(value=ConvCoeff, where=outlet)
ConvCoeff.setValue(0., where=outlet)
residual1 = 1.
elapsedTime = 0.
tWFall = np.zeros(nx)[None,:]
while elapsedTime < t:
tWF.updateOld()
it = 0 #iterations
while residual1> 1e-2:
sourceWF.value = - AV / nx * alpha*(tWF - tS)/ cvrho / VWF #this will be a variable convection source
eq1 = HybridConvectionTerm(coeff=ConvCoeff) + TransientTerm(coeff=1.) == \
+ sourceWF\
- ImplicitSourceTerm(exteriorCoeff.divergence) \
#+ DiffusionTerm(coeff= lambdaWF / cvrho) #not necessary(?)
residual1 = eq1.sweep(dt = timeStepDuration, var = tWF)
print('res1: ' + str(residual1) )
it += 1
if it > 10:
raise ValueError (r'MaxIter reached')
elapsedTime += timeStepDuration ; print('t= ' + str(round(elapsedTime,2)))
residual1 = 1.
tWFall = np.r_[tWFall, tWF.value[None,:]] #value collection
#%% outlet fluid temperature and storage temperature
plt.plot(np.linspace(0,t/3600.,int(t/timeStepDuration)), tWFall[1:,-1], label=r'$\vartheta_{WF}$')
plt.legend()
I would expect a constant fluid outlet temperature because of the constant wall temperature and constant fluid inlet temperature. I have not defined the wall temperature as a boundary condition because some day I would like to analyse heat conduction and variable temperature gradients too. Running my mwe you can see that the fluid temperature at the outlet declines.
Could someone please help at this case?
Thanks in advance!
I changed the script around and it seem to give a constant temperature of 371.0. See this link.
The sourceWF term has been removed. I was unsure what this was for, but I think it would take time for the wall temperature to adjust to this.
The equation declaration has been moved outside the loop. This is the correct way to use FiPy, but shouldn't impact the results in this case.
I tried to sent this to URI but got 10% wrong. For the input 576.43 it gives 2 cents in the end. I did all the math in calculator exactly as my code and got 3 cents in the end. What is wrong?
total = float(input())
bill100 = int((total) / 100)
total = ((total) - (bill100*100))
bill50 = int((total)/50)
total = (total - (bill50*50))
bill20 = int(total/20)
total = (total - (bill20*20))
bill10 = int(total/10)
total = (total - (bill10*10))
bill5 = int(total/5)
total = (total - (bill5*5))
bill2 = int(total/2)
total = (total - (bill2*2))
coin1dolar = int(total)
total = (total - coin1dolar)
coin50 = int(total/0.50)
total = (total - (coin50*0.50))
coin25 = int(total/0.25)
total = (total - (coin25*0.25))
coin10 = int(total/0.10)
total = (total - (coin10*0.10))
coin5 = int(total/0.05)
total = (total - (coin5*0.05))
coin1cent = int(total/0.01)
print("BILLS:")
print(str(bill100) + " bills of R$ 100.00")
print(str(bill50) + " bills of R$ 50.00")
print(str(bill20) + " bills of R$ 20.00")
print(str(bill10) + " bills of R$ 10.00")
print(str(bill5) + " bills of R$ 5.00")
print(str(bill2) + " bills of R$ 2.00")
print("COINS:")
print(str(coin1dolar) + " coins of R$ 1.00")
print(str(coin50) + " coins of R$ 0.50")
print(str(coin25) + " coins of R$ 0.25")
print(str(coin10) + " coins of R$ 0.10")
print(str(coin5) + " coins of R$ 0.05")
print(str(coin1cent) + " coins of R$ 0.01")
Your issue is that floating point math isn't exact. You can read about it in the python docs. You are continuously doing division's with floating point values. You can fix this by doing coin1cent = int(round(total/.01)).
On a separate note, I think you should also look into the mod operator, linked here.
From my text, I read:
Estimated soma area, in μm^2, is from 1073 to 2400 and estimated total
somadendritic area is from 3914 to 11,158 μm^2.
How do I convert μm^2 to meters^2?
1 μm = 10^-6 m
Hence
1 μm = 10^-4 cm
so
1 μm^2 = 10^-8 cm^2