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In the world of electronics and telecommunication engineering, signals are often not deterministic but random in nature. These random signals arise due to noise, interference, or inherent randomness in the source. Understanding how these signals behave and how to process them is crucial for designing reliable communication systems.
One of the fundamental tools for signal processing is the Linear Time-Invariant (LTI) system. LTI systems are widely used because they are mathematically tractable and model many practical devices like filters, amplifiers, and communication channels.
Filtering random signals through LTI systems is essential to reduce noise, enhance desired signals, and analyze system performance. This section will build from the basics of random signals and LTI systems to the statistical analysis of filtered outputs, equipping you with the knowledge to solve typical ISRO Scientist/Engineer exam problems.
Before diving into filtering, let's revisit the key concepts of LTI systems and random signals.
An LTI system is defined by two properties:
The output \( y(t) \) of an LTI system with input \( x(t) \) is given by the convolution integral:
Here, \( h(t) \) is the impulse response of the system, which completely characterizes the system's behavior.
A random signal is a signal whose amplitude varies in an unpredictable manner over time. Unlike deterministic signals, random signals are described statistically.
Key statistical properties include:
When a random signal \( x(t) \) passes through an LTI system with impulse response \( h(t) \), the output \( y(t) \) is also a random signal. The system modifies the statistical properties of the input.
The output is given by:
Because \( x(t) \) is random, we analyze the output statistically, focusing on mean and autocorrelation functions.
The mean of the output signal \( \mu_y(t) \) is the convolution of the input mean \( \mu_x(t) \) with the impulse response:
If the input is zero-mean, the output mean is zero unless the system adds a bias.
The autocorrelation function \( R_x(\tau) \) of the input signal describes how the signal correlates with itself at different time lags \( \tau \). The output autocorrelation \( R_y(\tau) \) is related to the input autocorrelation and the system impulse response by:
This convolution reflects how the system shapes the correlation properties of the input signal.
To understand random signals in the frequency domain, we use the Power Spectral Density (PSD), which is the Fourier transform of the autocorrelation function.
The autocorrelation function \( R_x(\tau) \) of a stationary random signal \( x(t) \) is defined as:
This function measures how the signal values at two time instants separated by \( \tau \) are related.
The PSD \( S_x(f) \) shows how the power of a signal is distributed over frequency \( f \). It is the Fourier transform of the autocorrelation function:
PSD is especially useful to analyze how filtering affects the frequency content of random signals.
graph TD A[Input Random Signal x(t)] B[Input Autocorrelation R_x(τ)] C[Input PSD S_x(f)] D[LTI System with h(t), H(f)] E[Output Autocorrelation R_y(τ)] F[Output PSD S_y(f)] A --> B B --> C C --> D D --> F F --> E
Flowchart: The input signal's autocorrelation and PSD are transformed by the LTI system to produce the output autocorrelation and PSD.
The output PSD \( S_y(f) \) is related to the input PSD \( S_x(f) \) and the system's frequency response \( H(f) \) by:
This formula is fundamental and allows quick calculation of output spectral characteristics without performing time-domain convolution.
White noise is a special random signal with equal power at all frequencies, making it a common model for noise in communication systems.
Here, \( N_0 \) is the noise power spectral density constant.
| Characteristic | White Noise Input | Filtered Output |
|---|---|---|
| Mean | 0 | 0 |
| Autocorrelation \( R_x(\tau) \) | \( \frac{N_0}{2} \delta(\tau) \) | \( \frac{N_0}{2} (h * h^-)(\tau) \) |
| PSD \( S_x(f) \) | \( \frac{N_0}{2} \) (constant) | \( \frac{N_0}{2} |H(f)|^2 \) |
Filtering white noise shapes its PSD according to the system's frequency response, effectively "coloring" the noise.
Step 1: Identify the given functions:
Step 2: Recall the output autocorrelation formula:
\[ R_y(\tau) = R_x(\tau) * h(\tau) * h(-\tau) \]
Step 3: Compute \( h(-\tau) \):
\[ h(-\tau) = e^{\tau} u(-\tau) \]
Step 4: Perform the convolution \( h(\tau) * h(-\tau) \):
Since \( h(t) \) is causal, the convolution results in a function \( g(\tau) \) which can be found by integration (details omitted for brevity).
Step 5: Convolve \( R_x(\tau) \) with \( g(\tau) \) to get \( R_y(\tau) \).
Answer: The output autocorrelation \( R_y(\tau) \) is the convolution of \( e^{-|\tau|} \) with \( g(\tau) \), reflecting the smoothing effect of the system.
Step 1: Calculate the magnitude squared of \( H(f) \):
\[ |H(f)|^2 = \frac{1}{1 + (2\pi f)^2} \]
Step 2: Use the formula for output PSD:
\[ S_y(f) = |H(f)|^2 S_x(f) = \frac{1}{1 + (2\pi f)^2} \times \frac{1}{1 + f^2} \]
Answer: The output PSD is \( S_y(f) = \frac{1}{(1 + (2\pi f)^2)(1 + f^2)} \).
Step 1: Calculate magnitude squared of \( H(f) \):
\[ |H(f)|^2 = \frac{1}{1 + (2\pi f RC)^2} \]
Step 2: Use the output PSD formula:
\[ S_y(f) = |H(f)|^2 S_x(f) = \frac{N_0}{2} \times \frac{1}{1 + (2\pi f RC)^2} \]
Step 3: Interpretation:
The filter attenuates high-frequency noise components, reducing noise power at frequencies above the cutoff \( \frac{1}{2\pi RC} \).
Answer: The output noise PSD is shaped by the RC filter, effectively low-pass filtering the white noise.
Step 1: Calculate output mean using:
\[ \mu_y = \mu_x \int_{-\infty}^{\infty} h(\tau) d\tau \]
Since \( h(t) = e^{-t} u(t) \),
\[ \int_0^\infty e^{-\tau} d\tau = 1 \]
Thus, \( \mu_y = 2 \times 1 = 2 \).
Step 2: Calculate output autocorrelation:
\[ R_y(\tau) = R_x(\tau) * h(\tau) * h(-\tau) \]
Convolution of exponentials yields \( R_y(0) \) (variance) as:
\[ \sigma_y^2 = R_y(0) - \mu_y^2 \]
After performing convolution (details omitted), \( R_y(0) = 4 \times 0.5 = 2 \).
Step 3: Compute variance:
\[ \sigma_y^2 = 2 - (2)^2 = 2 - 4 = -2 \]
Note: Negative variance is impossible, indicating a misinterpretation. Remember variance is \( \sigma_y^2 = R_y(0) - \mu_y^2 \) only if \( R_y(0) \) is the second moment.
Since \( R_x(0) = E[x^2(t)] = \sigma_x^2 + \mu_x^2 = 4 + 4 = 8 \), the output second moment is:
\[ R_y(0) = 8 \times \int h(\tau) d\tau \times \int h(-\tau) d\tau = 8 \times 1 \times 1 = 8 \]
Therefore, variance is:
\[ \sigma_y^2 = 8 - (2)^2 = 8 - 4 = 4 \]
Answer: Output mean \( \mu_y = 2 \), variance \( \sigma_y^2 = 4 \).
Step 1: Find the system frequency response \( H(f) \):
\[ H(f) = 1 - 0.5 e^{-j 2 \pi f} \]
Step 2: Calculate magnitude squared:
\[ |H(f)|^2 = \left(1 - 0.5 e^{-j 2 \pi f}\right) \left(1 - 0.5 e^{j 2 \pi f}\right) = 1 - 0.5 e^{j 2 \pi f} - 0.5 e^{-j 2 \pi f} + 0.25 \]
Using Euler's formula:
\[ |H(f)|^2 = 1 + 0.25 - 0.5 \times 2 \cos(2 \pi f) = 1.25 - \cos(2 \pi f) \]
Step 3: Find input PSD \( S_x(f) \), Fourier transform of \( R_x(\tau) = e^{-|\tau|} \):
\[ S_x(f) = \int_{-\infty}^\infty e^{-|\tau|} e^{-j 2 \pi f \tau} d\tau = \frac{2}{1 + (2 \pi f)^2} \]
Step 4: Calculate output PSD:
\[ S_y(f) = |H(f)|^2 S_x(f) = \left(1.25 - \cos(2 \pi f)\right) \times \frac{2}{1 + (2 \pi f)^2} \]
Answer: The output PSD is \( S_y(f) = \frac{2 (1.25 - \cos(2 \pi f))}{1 + (2 \pi f)^2} \).
When to use: Quickly determine output spectral characteristics without performing convolution.
When to use: While finding output autocorrelation from input autocorrelation and impulse response.
When to use: When dealing with white noise filtering problems.
When to use: During time-constrained entrance exams like ISRO Scientist/Engineer.
When to use: When solving problems involving random signal statistics.
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