Cellular and mobile telephony, using earth-based repeaters, has received wide acceptance and has become an essential means of co
mmunication for business, even for the household. Cellular telephony by satellites is the wave of the future and communication systems are being designed for that purpose. The present allocated frequency band for satellites is at L-band ( 1.6 GHz). Various antennas are being examined for that purpose; one candidate is the microstrip patch antenna. Design a rectangular microstrip patch antenna, based on the dominant mode, that can be mounted on the roof of a car to be used for satellite cellular telephone. The designed center frequency is 1.6 GHz. the dielectric constant of the substrate is 10.2 (i.e., RT/duroid), and the thickness of the substrate is 0.127 cm.
Determine the (a) dimensions of the rectangular patch (in cm) (b) resonant input impedance, assuming no coupling between the two radiating slots
Determine the (a) dimensions of the rectangular patch (in cm) (b) resonant input impedance, assuming no coupling between the two radiating slots
1 answer:
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Answer:
For any string, we use
Explanation:
The pumping lemma says that for any string s in the language, with length greater than the pumping length p, we can write s = xyz with |xy| ≤ p, such that xyi z is also in the language for every i ≥ 0. For the given language, we can take p = 2.
Here are the cases:
- Consider any string a i b j c k in the language. If i = 1 or i > 2, we take
and y = a. If i = 1, we must have j = k and adding any number of a’s still preserves the membership in the language. For i > 2, all strings obtained by pumping y as defined above, have two or more a’s and hence are always in the language.
- For i = 2, we can take and y = aa. Since the strings obtained by pumping in this case always have an even number of a’s, they are all in the language.
- Finally, for the case i = 0, we take
Answer:
minimum current level required = 8975.95 amperes
Explanation:
Given data:
diameter = 5.5 mm
length = 5.0 mm
T = 0.3
unit melting energy = 9.5 j/mm^3
electrical resistance = 140 micro ohms
thickness of each of the two sheets = 3.5mm
Determine the minimum current level required
first we calculate the volume of the weld nugget
v = =
= 118.73 mm^3
next calculate the required melting energy
= volume of weld nugget * unit melting energy
= 118.73 * 9.5 = 1127.94 joules
next find the actual required electric energy
= required melting energy / efficiency
= 1127 .94 / ( 1/3 ) = 3383.84 J
TO DETERMINE THE CURRENT LEVEL REQUIRED use the relation below
electrical energy = I^2 * R * T
3383.84 / R*T = I^2
3383.84 / (( 140 * 10^-6 ) * 0.3 ) = I^2
therefore 8975.95 = I ( current )