gjfelix2001
Junior Member level 2
Hi, i've tried to solve this problem, but i have only ideas. Can you tell me if this is the way to solve this problem?. Thanks in advance!
Problem:
Resistance \[R_1\] has a mean value of 330 ohms with a tolerance of 1% corresponding to 3 standard deviations. [tex:66f7d338cb]R_1[/tex:66f7d338cb] has an gaussian distribution. If
\[V_2 = \frac{R_2}{R_1+R_2}V\] and
\[V_1=\frac{R_1}{R_1+R_2}V\]
where, V is a 9 Volts constant voltage and \[R_2\] has a value of 330 ohms.
1. Calulate the mean and variance of \[V_2\]
2. Calculate the pdf of \[V_2\]
My ideas:
Only \[V_2\] and \[R_1\] are random variables, and \[V_2\] is a function of \[R_1\], thus \[V_2 = g(R_1)\]. So, for the mean of \[V_2[/tex:66f7d338cb]:
\[E[V_2]=\int{g(r_1)f_{R_1}(r_1)dr_1}\]
right??
Or, should i calculate first the cdf with
\[F_{V_2}(v_2)=\int{f_{R_1}(r_1)}dr_1\]
or, directly the pdf with
\[f_{V_2}(v_2)=f_{R_1}(r_1)|\frac{dx}{dy}|\]
Can you give me some tips and hints??
Thank you all!!
Problem:
Resistance \[R_1\] has a mean value of 330 ohms with a tolerance of 1% corresponding to 3 standard deviations. [tex:66f7d338cb]R_1[/tex:66f7d338cb] has an gaussian distribution. If
\[V_2 = \frac{R_2}{R_1+R_2}V\] and
\[V_1=\frac{R_1}{R_1+R_2}V\]
where, V is a 9 Volts constant voltage and \[R_2\] has a value of 330 ohms.
1. Calulate the mean and variance of \[V_2\]
2. Calculate the pdf of \[V_2\]
My ideas:
Only \[V_2\] and \[R_1\] are random variables, and \[V_2\] is a function of \[R_1\], thus \[V_2 = g(R_1)\]. So, for the mean of \[V_2[/tex:66f7d338cb]:
\[E[V_2]=\int{g(r_1)f_{R_1}(r_1)dr_1}\]
right??
Or, should i calculate first the cdf with
\[F_{V_2}(v_2)=\int{f_{R_1}(r_1)}dr_1\]
or, directly the pdf with
\[f_{V_2}(v_2)=f_{R_1}(r_1)|\frac{dx}{dy}|\]
Can you give me some tips and hints??
Thank you all!!
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