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complex division
- Thread starter c4l3b
- Start date
Hello, c4l3b!
Did you look at your post?
We're still trying to figure out what you wrote . . .
Just guessing . . .
\(\displaystyle \text{We are given: }\:Z_1 \,=\,\frac{1}{a + bi},\quad Z_2 \,=\,\frac{1}{bi}\)
. . \(\displaystyle \text{And we want: }\:\frac{Z_2}{Z_1}\)
\(\displaystyle \text{We have: }\;\frac{Z_2}{Z_1} \:=\:\frac{\dfrac{1}{bi}}{\dfrac{1}{a + bi}} \:=\:\frac{a+bi}{bi}\)
\(\displaystyle \text{Rationalize: }\;\frac{-bi}{-bi}\cdot\frac{a+bi}{bi} \;=\;\frac{-abi - b^2i^2}{-b^2i^2} \:=\:\frac{-abi - b^2(-1)}{-b^2(-1)} \;=\;\frac{b^2-abi}{b^2}\)
\(\displaystyle \text{Reduce: }\;\frac{b(b - ai)}{b^2} \;=\;\frac{b-ai}{b}\)
Did you look at your post?
We're still trying to figure out what you wrote . . .
Complex number division is the calculation below correct?
Z1= 1/j?C+R Z2= 1/j?C
Z2/Z1 = (1/j?C)(1/j?C+R) = ( 1+j?RC )^2 . What??
. . . . You multiplied?
Just guessing . . .
\(\displaystyle \text{We are given: }\:Z_1 \,=\,\frac{1}{a + bi},\quad Z_2 \,=\,\frac{1}{bi}\)
. . \(\displaystyle \text{And we want: }\:\frac{Z_2}{Z_1}\)
\(\displaystyle \text{We have: }\;\frac{Z_2}{Z_1} \:=\:\frac{\dfrac{1}{bi}}{\dfrac{1}{a + bi}} \:=\:\frac{a+bi}{bi}\)
\(\displaystyle \text{Rationalize: }\;\frac{-bi}{-bi}\cdot\frac{a+bi}{bi} \;=\;\frac{-abi - b^2i^2}{-b^2i^2} \:=\:\frac{-abi - b^2(-1)}{-b^2(-1)} \;=\;\frac{b^2-abi}{b^2}\)
\(\displaystyle \text{Reduce: }\;\frac{b(b - ai)}{b^2} \;=\;\frac{b-ai}{b}\)
D
Deleted member 4993
Guest
soroban said:Hello, c4l3b!
Did you look at your post?
We're still trying to figure out what you wrote . . . <<< Second that bewilderment!!!!
Complex number division is the calculation below correct?
Z1= 1/j?C+R Z2= 1/j?C
Z2/Z1 = (1/j?C)(1/j?C+R) = ( 1+j?RC )^2 . What??
. . . . You multiplied?
Just guessing . . .
\(\displaystyle \text{We are given: }\:Z_1 \,=\,\frac{1}{a + bi},\quad Z_2 \,=\,\frac{1}{bi}\)
. . \(\displaystyle \text{And we want: }\:\frac{Z_2}{Z_1}\)
\(\displaystyle \text{We have: }\;\frac{Z_2}{Z_1} \:=\:\frac{\dfrac{1}{bi}}{\dfrac{1}{a + bi}} \:=\:\frac{a+bi}{bi}\)
\(\displaystyle \:\frac{a+bi}{bi} \: = \: \frac{a}{bi} +1 \: = \: -i\frac{a}{b} +1..........because \: \frac{1}{i} \: = \: -i\)
Its a ciruit with that is essentially a voltage divider.Z1 is a capacitor & resistor in series and Z2 has serial connection capacitor in series with Z1
\(\displaystyle Z1= \frac{1}{jwC}+R\)
\(\displaystyle Z2= \frac{1}{jwC}\)
\(\displaystyle v=\frac{Z2}{Z1+Z2}\)
\(\displaystyle \frac{Z1}{Z2}=\frac{1/jwC+R}{1/jwC} = \frac{(1/jwC)(1/jwC-R)}{(1/JwC+R)(1/JwC-R)}= (1+JwRC)^2\)
Is that correct?
This calculation is abuild up to derive a differential equation relating the output & input signal.
\(\displaystyle Z1= \frac{1}{jwC}+R\)
\(\displaystyle Z2= \frac{1}{jwC}\)
\(\displaystyle v=\frac{Z2}{Z1+Z2}\)
\(\displaystyle \frac{Z1}{Z2}=\frac{1/jwC+R}{1/jwC} = \frac{(1/jwC)(1/jwC-R)}{(1/JwC+R)(1/JwC-R)}= (1+JwRC)^2\)
Is that correct?
This calculation is abuild up to derive a differential equation relating the output & input signal.
D
Deleted member 4993
Guest
c4l3b said:Its a ciruit with that is essentially a voltage divider.Z1 is a capacitor & resistor in series and Z2 has serial connection capacitor in series with Z1
\(\displaystyle Z1= \frac{1}{jwC}+R\)
\(\displaystyle Z2= \frac{1}{jwC}\)
\(\displaystyle v=\frac{Z2}{Z1+Z2}\)
\(\displaystyle \frac{Z1}{Z2}=\frac{1/jwC+R}{1/jwC} = \frac{(1/jwC)(1/jwC-R)}{(1/JwC+R)(1/JwC-R)}= (1+JwRC)^2\)
Is that correct?
This calculation is abuild up to derive a differential equation relating the output & input signal.
We understand that - but your work is incorrect.
The real term of (1 + jwRC)[sup:edasq57m]2[/sup:edasq57m] = Re[(1 + jwRC)[sup:edasq57m]2[/sup:edasq57m]] = 1- w[sup:edasq57m]2[/sup:edasq57m]R[sup:edasq57m]2[/sup:edasq57m]C[sup:edasq57m]2[/sup:edasq57m]
That is not the real term (Z[sub:edasq57m]1[/sub:edasq57m]/Z[sub:edasq57m]2[/sub:edasq57m])
Re(Z[sub:edasq57m]1[/sub:edasq57m]/Z[sub:edasq57m]2[/sub:edasq57m]) = 1