complex division

C

c4l3b

New member
Joined
Jul 17, 2009
Messages
37
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
 
Hello, c4l3b!

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?
Click to expand...

Just guessing . . .

We are given:  Z1 = 1a+bi,Z2 = 1bi\displaystyle \text{We are given: }\:Z_1 \,=\,\frac{1}{a + bi},\quad Z_2 \,=\,\frac{1}{bi}

. . And we want:  Z2Z1\displaystyle \text{And we want: }\:\frac{Z_2}{Z_1}


We have:   Z2Z1 = 1bi1a+bi = a+bibi\displaystyle \text{We have: }\;\frac{Z_2}{Z_1} \:=\:\frac{\dfrac{1}{bi}}{\dfrac{1}{a + bi}} \:=\:\frac{a+bi}{bi}


Rationalize:   −bi−bi⋅a+bibi  =  −abi−b2i2−b2i2 = −abi−b2(−1)−b2(−1)  =  b2−abib2\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}


Reduce:   b(b−ai)b2  =  b−aib\displaystyle \text{Reduce: }\;\frac{b(b - ai)}{b^2} \;=\;\frac{b-ai}{b}

 
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?
Click to expand...

Just guessing . . .

We are given:  Z1 = 1a+bi,Z2 = 1bi\displaystyle \text{We are given: }\:Z_1 \,=\,\frac{1}{a + bi},\quad Z_2 \,=\,\frac{1}{bi}

. . And we want:  Z2Z1\displaystyle \text{And we want: }\:\frac{Z_2}{Z_1}


We have:   Z2Z1 = 1bi1a+bi = a+bibi\displaystyle \text{We have: }\;\frac{Z_2}{Z_1} \:=\:\frac{\dfrac{1}{bi}}{\dfrac{1}{a + bi}} \:=\:\frac{a+bi}{bi}

 a+bibi = abi+1 = −iab+1..........because 1i = −i\displaystyle \:\frac{a+bi}{bi} \: = \: \frac{a}{bi} +1 \: = \: -i\frac{a}{b} +1..........because \: \frac{1}{i} \: = \: -i
Click to expand...
 
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

Z1=1jwC+R\displaystyle Z1= \frac{1}{jwC}+R

Z2=1jwC\displaystyle Z2= \frac{1}{jwC}

v=Z2Z1+Z2\displaystyle v=\frac{Z2}{Z1+Z2}

Z1Z2=1/jwC+R1/jwC=(1/jwC)(1/jwC−R)(1/JwC+R)(1/JwC−R)=(1+JwRC)2\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.
 
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

Z1=1jwC+R\displaystyle Z1= \frac{1}{jwC}+R

Z2=1jwC\displaystyle Z2= \frac{1}{jwC}

v=Z2Z1+Z2\displaystyle v=\frac{Z2}{Z1+Z2}

Z1Z2=1/jwC+R1/jwC=(1/jwC)(1/jwC−R)(1/JwC+R)(1/JwC−R)=(1+JwRC)2\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
Click to expand...
 
You must log in or register to reply here.
Top