Question

1. Calculate the wavelength of a waveform given a period of 17μs.
2. 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 29^oC.
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.

Answer 1

1.

the wavelength:

$\lambda=vT$

where v is wave speed,

T is period

$\lambda=17v$

2) Noise voltage is given by, $V^2=4kT \int_{f1}^{f2} Rdf$

Where V= RMS input noise voltage, $f_2-f_1=band \space width=3.33KHz=3.33\times 10^3Hz$

k= Boltzmann constant$=1.38\times 10^{-23}J/K$

T=temperature in kelvin =29+273=302K

$R=Resistance=42\Omega$

Now, $V=(4kTR(f_2-f_1))^{0.5}=(4\times1.38\times 10^{-23}\times 302\times42\times 3.33\times 1000)^{0.5}=482.86\times 10^{-10}V$

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

4) Initial frequency f_i = 73 kHz

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

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

The f_USB of A = 73 kHz + f_m (6 kHz) = 79 kHz

The f_USB of B = 79 kHz + f_m (6 kHz) = 85 kHz

The f_USB of C = 85 kHz + f_m (6 kHz) = 91 kHz

Where: f_USB is the Frequency of the Upper Side-bands, while f_m is 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, f_LSB = 73 kHz and f_USB = 79 kHz;
2. For B, f_LSB = 79 kHz and f_USB = 85 kHz;
3. For C, f_LSB = 85 kHz and f_USB = 91 kHz.

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

5) Root mean square value of the shot noise current i_n is given by the Schotty formula:

I_n=$\sqrt{2\Iota{q}\Delta\Beta}$

Where:

$\Iota$ =Dc current in Amperes

q= charge of an electron in Coulombs

$\Delta$$\Beta$ = the bandwidth in Hertz

Substituting

I_n= $\sqrt{2×0.5×10^{-3}×1.602×10^{-19}×10×10^3}$

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

I_n =1.266$×10^{-9}A$

=1.266nA