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help udrestanding the values in this table

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antenna_beginner

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Hello, anyone has an idea regarding how did they got these values?

thanks
Capture.JPG
 

what kind of trial an error was done here

Hello , i am having trouble understanding what was done in the table in the attahed photo
?

thanks

Capture (1).JPG
 
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Re: what kind of trial an error was done here

The equation seems "interlaced" so I think the author has used iteration technique to find the coefficients.
 

Re: what kind of trial an error was done here

we want a 20dB couping with 1dB ripple, so Cmax is -19 and Cmin is -21

what kind of itterations?, why they pick each time a different C_max?
C max is -19 dB which is 0.112202
C max and Cmin are given to us from the question

Thanks
 

Re: what kind of trial an error was done here

They are trying different values for c1 and c2 to achieve the intended cmax and cmin. What's particularly unclear about the method?

It's appropriate for problems where the solution can't be found analytically. Many Chebyshev (equal ripple) solutions are found in a similar way.
 

Re: what kind of trial an error was done here

ok they pick combinations of C1 and C2 , for what purpose?

in the solution they say
that at minimum point:

theta=pi/2
C_min=-21dB=0.089125
C=C_1*sin(3/2*pi)+(C_2-C_1)sin(pi/2)=-C_1+C_2-C_1=C_2-2*C_1

eq1.
0.089125=C_2-2*C_1

C_max=-19dB=0.1122


there are lots of combinations for C_1 and C_2 to match eq1. how do i test a combination to see that it fits?
what is the addional criteria for C1 and C2 to fit?
Thanks
 
Last edited:

It is stated that Cmax is -19dB so in the righthand side of the table, the numbers are given in such a way that we can obtain -19dB. When you use 0.035=C1, you are getting the closest number to -19dB. Normally we use 10^(-C(dB)/20) , you can see this calculation in the beginning of some questions. If we implement it in reverse way, its calculation is like the following:

log(C1) = log(0.1125) = -0.948847478

-0.948847478*20 = -18.9769496 dB which is close enough to -19dB. It must be the reason of chosing 0.035 to obtain 0.1125. The others are not as close as this.
 

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