22.11.2014 Views

Sample, Final exam #2

Sample, Final exam #2

Sample, Final exam #2

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

The radiated power is therefore:<br />

d. At 1200 MHz:<br />

P rad = 1.8535 η 0I 0<br />

2<br />

4π = 1.8535×377×4<br />

4π<br />

= 222.43 w<br />

The radiated power is:<br />

λ = c f = 3×10 8<br />

= 0.25 m → L = 4λ<br />

1200×106 P rad = η 0I 0<br />

2<br />

4π ⌠ ⌡<br />

θ=π<br />

θ=0<br />

( ( ) )<br />

cos (4πcosθ − cos(4π) 2<br />

sinθ<br />

dθ = 3.5168η 0I 0<br />

2<br />

4π<br />

= 3.5168×377×4<br />

4π<br />

= 422<br />

w<br />

Note how much more power is radiated for the long antennas compared with the short dipole in<br />

(a).

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!