Answer to Question #253012 in Electric Circuits for shdfasdfasd

Question #253012
  1. Find the input thermal noise voltage of a receiver with a bandwidth of 3.33 kHz, with an input resistance of 42 Ω, and a temperature of 29oC.
  2. Calculate the wavelength of a waveform given a period of 17μs.
  3. Given the signal power of 2.3 μW and the noise power of 0.7 μW, calculate the signal-to-noise ratio.
  4. Three (3) AM broadcast stations are spaced at 18 kHz, beginning at 73 kHz. Each station is allowed to transmit modulating up to 6 kHz. Compute for the upper and lower sidebands of each station.
  5. Given a dc bias of 0.5mA and a bandwidth of 10 kHz, calculate for the noise current.
1
Expert's answer
2021-10-20T10:39:24-0400

1.

The input thermal noise voltage is



V=4kTBR==4(1.381023)(29+273)333042= =4.83108 V, or 48.3 nV.V=\sqrt{4kTBR}=\\=\sqrt{4(1.38·10^{-23})(29+273)3330·42}=\\\space\\=4.83·10^{-8}\text{ V, or 48.3 nV}.



2.

v=nλ=λTv=n\lambda=\frac{\lambda}{T}

v=c

λ=c×T\lambda=c\times T



λ=3×108×17×106=5100m\lambda=3\times10^8\times17\times10^{-6}=5100m



3.

Signal-to-noise ratio =2.3/0.7=23/7=23:7=2.3/0.7=23/7=23:7


4.

Initial frequency fi = 73 kHz

Final frequency f= (73 + 18) kHz = 91 kHz


Let Station 1, Station 2 and Station 3 be A, B and C respectively.

The fUSB of A = 73 kHz + fm (6 kHz) = 79 kHz

The fUSB of B = 79 kHz + fm (6 kHz) = 85 kHz

The fUSB of C = 85 kHz + fm (6 kHz) = 91 kHz


Where: fUSB is the Frequency of the Upper Side-bands, while fis the modulating frequency at which each station is allowed to transmit.


Therefore, the frequency of the upper and lower side-bands of each station are given by:

  1. For A, fLSB = 73 kHz and fUSB = 79 kHz;
  2. For B, fLSB = 79 kHz and fUSB = 85 kHz;
  3. For C, fLSB = 85 kHz and fUSB = 91 kHz.


Note: fLSB means Frequency of the Lower Side-bands.


5.

Root mean square value of the shot noise current in is given by the Schotty formula:


In=2IqΔB\sqrt{2\Iota{q}\Delta\Beta}



Where:

I\Iota =Dc current in Amperes

q= charge of an electron in Coulombs

Δ\DeltaB\Beta = the bandwidth in Hertz



Substituting


In2×0.5×103×1.602×1019×10×103\sqrt{2×0.5×10^{-3}×1.602×10^{-19}×10×10^3}

2×0.5×103×1.602×1019×10×103\sqrt{2×0.5×10^{-3}×1.602×10^{-19}×10×10^3}



I=1.266×109A×10^{-9}A

=1.266nA


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