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GATE 2017 EC – Question 61

Communications · Random Processes · 2 marks · Multiple choice

Let $X(t)$ be a wide sense stationary random process with the power spectral density $S_X(f)$ as shown in Figure (a), where $f$ is in Hertz (Hz). The random process $X(t)$ is input to an ideal lowpass filter with the frequency response

$$H(f) = \begin{cases} 1, & |f| \leq \frac{1}{2}\text{ Hz} \\ 0, & |f| > \frac{1}{2}\text{ Hz} \end{cases}$$

as shown in Figure (b). The output of the lowpass filter is $Y(t)$.

[Figure: (a) The power spectral density $S_X(f) = e^{-|f|}$, peaked at $f = 0$. (b) $X(t)$ passes through an ideal lowpass filter with cutoff $\frac{1}{2}$ Hz to give $Y(t)$.]

Let $E$ be the expectation operator and consider the following statements:

I. $E(X(t)) = E(Y(t))$

II. $E(X^2(t)) = E(Y^2(t))$

III. $E(Y^2(t)) = 2$

Select the correct option:

Diagram for GATE 2017 EC question 61
  1. only I is true
  2. only II and III are true
  3. only I and II are true
  4. only I and III are true

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Show answer and explanation

Correct answer: (A) only I is true

Explanation

The power spectral density has no impulse at $f = 0$, so the mean of $X$ is 0, and the filter keeps this, so $E(Y) = 0$ and I is true. The power of $X$ is $\int e^{-|f|}df = 2$. The power of $Y$ is the part within $|f| \leq \frac{1}{2}$, which is $2(1 - e^{-1/2}) = 0.787$. So $E(X^2) \neq E(Y^2)$ and $E(Y^2) \neq 2$, which makes II and III false.