Is the source impedance of 100ohms really not considered in the calculation? i.e. added to the characteristic impedance. Also, why is the voltage to be absorb by the load base on Z0 = 2/3 and not ZL = 1/3? Thanks for any answers
When firstly consider the source voltage that gonna transmit, it need to be considered as "equivalent voltage", which need to be calculated as the load is open, which is infinite resistance. (thevenin theorem). The transmit medium is consider as "transmission line", so during the calculation, bear in mind the values are like "equivalent", "characteristic" etc. If the transmit medium were considered as just a "perfect wire, real static resistance", which means the signals transmit from source to load is considered "instant", then it would be the normal DC voltage dividing as you implying.
Most coaxial cables and connectors in common use have a 50Ω characteristic impedance, with an exception being the 75 Ω cable used in television systems. The reasoning behind these choices is that an air-filled coaxial line has minimum attenuation for a characteristic impedance of about 77Ω (Problem 2.27), while maximum power capacity occurs for a characteristic impedance of about 30Ω (Problem 3.28). A 50Ω characteristic impedance thus represents a compromise between minimum attenuation and maximum power capacity. - From Pozar, Microwave Engineering pg 134
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Very clear.Confident speaking.
your explanation is very clear , thanks Dr
Great Lectures................
Very useful and very clear thanks so much
Is the source impedance of 100ohms really not considered in the calculation? i.e. added to the characteristic impedance. Also, why is the voltage to be absorb by the load base on Z0 = 2/3 and not ZL = 1/3? Thanks for any answers
very useful
excellent
Why the voltage at the source is 2/3V? Do not consider the load when dividing the voltage? thank you for your answer
When firstly consider the source voltage that gonna transmit, it need to be considered as "equivalent voltage", which need to be calculated as the load is open, which is infinite resistance. (thevenin theorem). The transmit medium is consider as "transmission line", so during the calculation, bear in mind the values are like "equivalent", "characteristic" etc. If the transmit medium were considered as just a "perfect wire, real static resistance", which means the signals transmit from source to load is considered "instant", then it would be the normal DC voltage dividing as you implying.
Thank you
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Why 50 ohm is consider for char. Impedance . plz CLR more
Most coaxial cables and connectors in common use have a 50Ω characteristic impedance,
with an exception being the 75 Ω cable used in television systems. The reasoning behind these
choices is that an air-filled coaxial line has minimum attenuation for a characteristic impedance
of about 77Ω (Problem 2.27), while maximum power capacity occurs for a characteristic
impedance of about 30Ω (Problem 3.28). A 50Ω characteristic impedance thus represents a
compromise between minimum attenuation and maximum power capacity. - From Pozar, Microwave Engineering pg 134