17
Enter frequency @ attn2
f2
.
100010
6
w1
. .
2
f1
F1 in radians/sec
w2
. .
2
f2
F2 in radians/sec
Attenuation factor for f1 in nepers/meter
Attenuation factor for f2 in nepers/meter
alpha1
.
1
328.1
10
attn1
20
alpha2
.
1
328.1
10
attn2
20
= alpha1 0.0042559231
= alpha2 0.0108141844
* r and g are both functions of frequency, and are computed using the method
* described in "Transmission Lines" by Robert Chipman, McGraw-Hill, 1968,
* pp 65-66. r is assumed to increase in proportion to the square root
* of frequency, while g varies in direct proportion to frequency. A high
* frequency relationship for the attenuation factor is:
*
* alpha = ((r / z0) + (g * z0)) / 2,
*
* and r and g can be found by selecting values of alpha at two frequencies
* (100 MHz and 1 GHz are used here) and solving two simultaneous equations:
*
* alpha1 = (.5 / z0) * r1 + (.5 * z0) * g1
* alpha2 = (.5 / z0) * sqrt(w2 / w1) * r1 + (.5 * z0) * (w2 / w1) * g1
*
* The alpha's are converted to units of nepers per meter, and the frequencies
* (w1 and w2) are in units of radians per second. Kramer's rule givs:
r1
. .
alpha1
w2
w1
alpha2
z0
2
w2
w1
w2
w1
4
g1
alpha2
.
alpha1
w2
w1
( )
.
2 z0
w2
w1
w2
w1
4
= r1 0.6963951914
= g1 0.0000103123
* Then the frequency-dependent expressions for r and g are:
