Overcoupling always tends to cause oscillations, in a manner similar to when a microphone is held too close to a speaker and overcoupling in a RADAR receiver is no exception, so Answer D. is the right answer.
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The Intermediate Frequency (I.F.) amplifier bandwidth is wide for short ranges and narrow for long ranges in the context of radio frequency (RF) communication systems.
In RF communication systems, the I.F. amplifier is a crucial component that processes the intermediate frequency signal. The purpose of the I.F. amplifier is to amplify the desired signal while rejecting unwanted noise and interfering signals.
For short ranges, such as in local communication systems or within a limited geographical area, the I.F. amplifier's bandwidth is typically wide. This wide bandwidth allows for a larger range of frequencies to pass through, capturing a broader spectrum of signals. In short-range applications, there is less concern about unwanted interference from distant sources, so a wider bandwidth is beneficial for capturing and amplifying the desired signal.
On the other hand, for long ranges, such as in long-distance communication systems or broadcasting, the I.F. amplifier's bandwidth is usually narrow. This narrower bandwidth is necessary to reject a larger portion of the spectrum, including unwanted noise and interfering signals from distant sources. In long-range applications, there is a higher likelihood of encountering more interfering signals, so a narrower bandwidth helps in filtering out unwanted signals and improving the signal-to-noise ratio.
Mnemonic: "Wide for Short, Narrow for Long".
Remember, noise level is proportional to bandwidth, so weaker signals, as in distan signals, require narrower bandwidth for the signal to be successfully decoded.
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Dynamic range is the ability of a receiver to receive both very strong and very weak signals.
A high dynamic range is a very desirable quality in a receiver.
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\(dB = 10 \times \log{( \frac{P_{out}}{P_{in}})}\)
\(dB = 10 \times \log{( \frac{V_{out}}{V_{in}})}^2\)
\(dB = 20 \times \log{( \frac{V_{out}}{V_{in}})}\)
\(dB = 20 \times \log{( \frac{2V}{2 \times 10^{-6}})}\)
\(dB = 20 \times \log (10^6)\)
\(dB = 20 \times 6\)
\(dB = 120\)
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