# The value of fa and fb in the Book of Erickson

Joined May 17, 2014
60
Hi the transfer function is

and its phase according to the Power Electronics book of Erickson is

the value of fa and fb is

How have been calculated these two points?

Thanks

Joined May 17, 2014
60
Is there anybody who is expertise both in control and Power?

#### BR-549

Joined Sep 22, 2013
4,928
I am not an expert in either one. Don't care much for math either. But it's easy to see where those values come from......just by relating the lines on the graph.

f0 has a certain slope at 45 degrees. Just extend that slope the the 0 and 90 degree mark......those resulting intersections give you the terms.

Taking the context of the X and Y axis......it should relate to your particular study example.

Joined May 17, 2014
60
thank you so much.
The slop is -0.5 according to -(pi/2)/Ln(e^pi). But, why we should use Ln(x2/x1) rather than Log(x2/x1)

Joined May 17, 2014
60
My means that why the slop of straight line in semi logarithm is (y2-y1)/(Ln(x2/x1))?

#### MrAl

Joined Jun 17, 2014
10,620
Hi the transfer function is
View attachment 136518
and its phase according to the Power Electronics book of Erickson is
View attachment 136519
the value of fa and fb is
View attachment 136520
How have been calculated these two points?

Thanks
Hello,

Here is a workup of the solution to calculate the 90 degree intercept. See attachment. "log()" is the natural log "ln()" not log base 10.

First we transform into a linear domain, then calculate the slope, then calculate the linear intercept, then transform back into the log system. The details to the calculation are shown in the attachment. This is using basic geometry with wo normalized. Also, the y axis is in rads not degrees, and note 0 to -1.6 is approximately 0 to -pi/2 so half of that is -pi/4.
Once we transform into Ph=-atan(10^x) we can then find the slope by differentiating with respect to x. The slope is different in linear coordinates so we get a different value. This is the same as doing a pure graphic solution using a straight edge as shown by the red line.

The fictitious overlay scale shown as x=0 to 400 and y=0 to 160 is just a linear 1:1 scale grid that allows adjusting (and confirming) both the x and y axis to appear as having the same scale factor graphically for better visualization. This allows us to see the approximate slope just by looking at the graph.

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Joined May 17, 2014
60
Thank you so much for your answer. But why you replace Log with Ln?
In my opinion, the exact answer is not e^(pi/2); rather, is 10^(pi/2)