Complex Pole Network Analysis: 1/2 --> Root2/2?

mbferguson

Joined Apr 23, 2017
94
I'm working on an inverse laplace transform that involves complex poles.

The question is (4s^2 + 7s + 13)/((s+2)(s^2+2s+5))

I can succesfully reduce it via partial fractions, but there is a step where he has (1/2)(1+j) and converts this to (Root2)/2 x (Root2/2 + Root2/2 x j)

Also, he then converts that to Root2/2 Phasor(45)

Could anyone explain the 1/2 to Root2/2 conversion, as well as how phasor conversion work? Thank you.

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Papabravo

Joined Feb 24, 2006
17,292
What is the problem? You want to convert from Cartesian Coordinates to Polar Coordinates. To that you compute the magnitude via the Pythagorean formula, and you get the angle by taking the arctangent of the imaginary part over the real part. So the magnitude of 1 + j is:

$$\sqrt{1^2 + 1^2}\;=\;\sqrt{2}$$
multiply that by
$$\frac{1}{2}$$
and you get
$$\frac{\sqrt{2}}{2}$$
To get the angle you do
$$tan^{-1}(\frac{1}{1})\;=\;45^\circ$$
The other bit is just unnecessary algebra to amaze and confuse you, and demonstrate what a high opinion the author has of himself.

Last edited:

mbferguson

Joined Apr 23, 2017
94
What is the problem? You want to convert from Cartesian Coordinates to Polar Coordinates. To that you compute the magnitude via the Pythagorean formula, and you get the angle by taking the arctangent of the imaginary part over the real part. So the magnitude of 1 + j is:

$$\sqrt{1^2 + 1^2}\;=\;\sqrt{2}$$
multiply that by
$$\frac{1}{2}$$
and you get
$$\frac{\sqrt{2}}{2}$$
To get the angle you do
$$tan^{-1}(\frac{1}{1})\;=\;45^\circ$$
The other bit is just unnecessary algebra to amaze and confuse you, and demonstrate what a high opinion the author has of himself.
Thank you!!

He thinks VERY highly of himself.

Papabravo

Joined Feb 24, 2006
17,292
There was a bit hand waving in my previous post. Actually the magnitude of a complex number is defined as the square root of the product of a complex numbe z and it's conjugate z*. I should have written:

$$\sqrt{(1+j)(1-j)}\;=\;\sqrt{1^2-j^2}\;=\;\sqrt{1^2+1^2}$$