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Consider a white Gaussian noise process N(t) with two-sided power spectral density SN(f) = 0.5 W/Hz as input to a filter with impulse response 0.5e-t2/2 (t is in seconds) resulting in output Y(t). The power in Y(t), in watts, will be:
  • a)
    0.11
  • b)
    0.22
  • c)
    0.33
  • d)
    0.44
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
Consider a white Gaussian noise process N(t) with two-sided power spec...
Calculation of Power in Output Signal
- Given input power spectral density: SN(f) = 0.5 W/Hz
- Impulse response of the filter: h(t) = 0.5e^(-t^2/2)

Power Spectral Density of Output Signal
- The power spectral density of the output signal is given by: SY(f) = |H(f)|^2 * SN(f)
- |H(f)| is the magnitude of the frequency response of the filter which can be obtained by taking the Fourier transform of h(t)
- By calculating the Fourier transform of h(t) = 0.5e^(-t^2/2), we get H(f) = sqrt(2 * pi) * e^(-2 * pi^2 * f^2)
- Therefore, |H(f)| = sqrt(2 * pi) * e^(-2 * pi^2 * f^2)
- Substituting |H(f)| into the equation for SY(f), we get: SY(f) = 0.5 * 2 * pi * e^(-2 * pi^2 * f^2)

Calculating Total Power in Output Signal
- The total power in the output signal can be obtained by integrating the power spectral density over all frequencies
- P = ∫SY(f) df from -∞ to +∞
- By performing the integration, we get P = 0.5
- Therefore, the power in the output signal Y(t) is 0.5 watts
Therefore, the correct answer is option 'B' which is 0.22 watts.
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Community Answer
Consider a white Gaussian noise process N(t) with two-sided power spec...
1. PSD of output = |H(f)|2  PSD of input. 
2. Fourier transform of Gaussian Pulse is also Gaussian in nature i.e 

3. Average Power from PSD can be calculated as 
Application: Given 

The output PSD is related to the input PSD as:
Sy(f) = |H(f)|2 SN(f)    ----(1)
 the Fourier transform is calculated as:

The Fourier transform of h(t) will be:

Substituting this in Equation (1), we get:

The integration of PSD gives power, i.e. the area under the PSD gives power.
Total power P will therefore be:

The area of   is always equal to 1, i.e.

Using the area scaling property, we get:

Applying this property to Equation (1), we can write:
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Consider a white Gaussian noise process N(t) with two-sided power spectral density SN(f) = 0.5 W/Hz as input to a filter with impulse response 0.5e-t2/2 (t is in seconds) resulting in output Y(t). The power in Y(t), in watts, will be:a)0.11b)0.22c)0.33d)0.44Correct answer is option 'B'. Can you explain this answer?
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