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Hertz's Observations

Power Spectra...

Question

Power Spectral Density (PSD) for a random variable X(t) is given below for positive frequency. The variance of $Y (t) = X (t) c o s 2 π f_{0} t . (f_{0} = 5 k H z)$ is

5600

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4800

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8000

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2000

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Solution

The correct option is C 8000
$R_{γ} (τ) = \frac{1}{T} \int_{0}^{τ} x (t) . c o s (2 π f_{0} τ) . x (t + τ) . c o s (2 π f_{0} (t + τ)) d t$

$= \frac{1}{2} R_{X} (τ) . c o s (2 π f_{0} τ)$

$S_{γ} (f) = \frac{S_{X} (f - f_{0}) + S_{X} (f + f_{0})}{4}$

$E [Y^{2} (t)] = \int_{- \infty}^{\infty} S_{γ} (f) d f = 2 \times 10^{3} \times 4 = 8000$

E[Y(t)] = 0

$∴ σ_{γ}^{2} = 8000$

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Power Spectral Density (PSD) for a random variable X(t) is given below for positive frequency. The variance of Y(t)=X(t) cos 2π f0 t.(f0=5 kHz) is

The correct option is C 8000Rγ(τ)=1T∫τ0x(t).cos(2πf0τ).x(t+τ).cos(2πf0(t+τ))dt =12RX(τ).cos(2πf0τ) Sγ(f)=SX(f−f0)+SX(f+f0)4 E[Y2(t)]=∫∞−∞Sγ(f)df=2×103×4=8000 E[Y(t)] = 0 ∴ σ2γ=8000

Power Spectral Density (PSD) for a random variable X(t) is given below for positive frequency. The variance of $Y (t) = X (t) c o s 2 π f_{0} t . (f_{0} = 5 k H z)$ is