Understand the calculations in an RC circuit - voltage division to polar form

Question

I'm missing some calculation steps in an AC RC-example, and I can't understand how to get to the next step. If someone could show me, I'm sure to understand the rest.

Example: A standard RC-circuit.

----R------
       |
Vin    C      Vout
       |
-----------

Vin=1 sin(wt) -> Vin = 1<0deg (polar form)

1. Vout is given by voltage division

Vout=Zc/(R+Zc)=...=(1/(1+jwRC))(Vin)

Now, apply R,C and w

R=1600ohm

C=0.1 mikrofarad

w=2PIf

f=100

The example then shows:

2. f=100Hz gives Vout=(0.99-j0.10)1<0=0.995<-5.7deg

Given 1), applying numbers would give 1/(1+j0,100530965)

But then I'm stuck.

How do I solve the equatation to get to 2) the above from the voltage division? I will use it to calculate further different frequncies, i.e for 1000Hz and so on, but need to understand what is happening in between 1) and 2). I understand basic complex numbers, but reading right now upon converting to polar form. I understand -j(1/wc)=1/jwc. but have a problem getting to (0,99-j0.10)Vin from 1/(1+j0,100530965)Vin

Answer 1

DK has told you the right viewpoint. You write that:

I understand basic complex numbers

But it appears you don't. One of the first things you are taught is how to put a fraction into standard form of \$a+b i\$ when presented with something of the form \$\frac{a+b i}{c+d i}\$. It's probably late chapter 1 or maybe chapter 2. Not sure. But it is very early. You multiply by \$1=\frac{c-d i}{c-d i}\$, which is 1 made up by a fraction composed of the complex conjugate of the denominator in both the numerator and denominator. Remember?

So,

$$\begin{align*} \frac{V_{_\text{OUT}}}{V_{_\text{IN}}}&=\frac1{1+j \omega R C} \\\\ &=\frac1{1+j \omega R C}\cdot \frac{1-j \omega R C}{1-j \omega R C} \\\\ &=\frac{1-j \omega R C}{1+\left(\omega R C\right)^2} \\\\ &=\frac{1}{1+\left(\omega R C\right)^2}-j\:\frac{\omega R C}{1+\left(\omega R C\right)^2} \end{align*}$$

So, \$a=\frac{1}{1+\left(\omega R C\right)^2}\$ and \$b=-\frac{\omega R C}{1+\left(\omega R C\right)^2}\$. And now the complex number is back in standard Cartesian form.

If you plug in the values from your problem, I think you will see where your #(2) example result comes from.

Answer 2

\$\frac{1}{A-jB}=\frac{1}{A-jB}\times\frac{A+jB}{A+jB}=\frac{A+jB}{A^2+B^2}=\frac{A}{A^2+B^2}+\frac{jB}{A^2+B^2}=\frac{A}{A^2+B^2}+j\frac{B}{A^2+B^2}\$