给定块信号的自相关确定及计算方法技术问询

Hey there! Let's break down how to calculate the autocorrelation of a block signal, with clear, actionable methods for both continuous-time and discrete-time cases. First, a quick refresher: a block signal is one that's non-zero only over a finite time interval (e.g., from ( t=0 ) to ( t=T ) for continuous time, or ( n=0 ) to ( n=N-1 ) for discrete time) and zero everywhere else.

Autocorrelation measures how similar a signal is to a shifted version of itself. The formal definitions vary slightly for continuous vs discrete signals:

• Continuous-time: The autocorrelation function ( R_x(\tau) ) is given by:
  [
  R_x(\tau) = \int_{-\infty}^{\infty} x(t) \cdot x(t+\tau) , dt
  ]
  Since ( x(t) ) is a block signal, we can narrow the integral to where both ( x(t) ) and ( x(t+\tau) ) are non-zero (their overlapping interval).
• Discrete-time: The autocorrelation sequence ( r_x[k] ) is:
  [
  r_x[k] = \sum_{n=-\infty}^{\infty} x[n] \cdot x[n+k]
  ]
  Again, we only sum over indices where both ( x[n] ) and ( x[n+k] ) are non-zero.

Let's use a common example: a rectangular block signal ( x(t) = A ) for ( 0 \leq t \leq T ), and ( x(t)=0 ) otherwise. Here's how to compute its autocorrelation:

• Step 1: Identify non-zero intervals
  ( x(t) ) is non-zero on ( [0, T] ); ( x(t+\tau) ) is non-zero when ( 0 \leq t+\tau \leq T ), or ( -\tau \leq t \leq T-\tau ).
• Step 2: Find overlapping intervals for different ( \tau )
  • Case 1: ( \tau \geq T ) or ( \tau \leq -T ): No overlap between ( [0, T] ) and ( [-\tau, T-\tau] ). So ( R_x(\tau) = 0 ).
  • Case 2: ( 0 \leq \tau < T ): Overlap is ( [0, T-\tau] ). Integrate ( A \cdot A ) over this interval:
    [
    R_x(\tau) = \int_{0}^{T-\tau} A^2 dt = A^2(T - \tau)
    ]
  • Case 3: ( -T < \tau < 0 ): Let ( \tau' = -\tau ) (positive), overlap is ( [-\tau, T] ). Integrate to get:
    [
    R_x(\tau) = \int_{-\tau}^{T} A^2 dt = A^2(T + \tau)
    ]
• Step 3: Combine cases
  The final autocorrelation is a triangular pulse centered at ( \tau=0 ), peaking at ( A^2T ) and tapering to 0 at ( \tau=\pm T ).

Take a discrete rectangular block: ( x[n] = A ) for ( 0 \leq n \leq N-1 ), ( x[n]=0 ) otherwise.

• Step 1: Identify non-zero indices
  ( x[n] ) is non-zero for ( n \in [0, N-1] ); ( x[n+k] ) is non-zero when ( 0 \leq n+k \leq N-1 ), or ( -k \leq n \leq N-1 -k ).
• Step 2: Sum over overlapping indices
  • Case 1: ( |k| \geq N ): No overlap, so ( r_x[k] = 0 ).
  • Case 2: ( 0 \leq k < N ): Overlap has ( N - k ) terms. Sum gives:
    [
    r_x[k] = A^2(N - k)
    ]
  • Case 3: ( -N < k < 0 ): Let ( k' = -k ), overlap has ( N + k ) terms. Sum gives:
    [
    r_x[k] = A^2(N + k)
    ]
• Step 3: Combine cases
  The result is a discrete triangular sequence, peaking at ( A^2N ) when ( k=0 ).

For non-rectangular block signals (e.g., a Gaussian pulse limited to ( [t_1, t_2] )):

• Step 1: Define the signal's non-zero interval
  Note exactly where ( x(t) ) (or ( x[n] )) is non-zero: ( t \in [t_a, t_b] ) (continuous) or ( n \in [n_a, n_b] ) (discrete).
• Step 2: Calculate the overlap region
  For continuous time, find ( t ) values where both ( t \in [t_a, t_b] ) and ( t+\tau \in [t_a, t_b] ). This simplifies to ( t \in [\max(t_a, t_a - \tau), \min(t_b, t_b - \tau)] ).
• Step 3: Integrate/Sum over the overlap
  Compute ( \int_{\text{overlap}} x(t)x(t+\tau) dt ) (continuous) or ( \sum_{\text{overlap}} x[n]x[n+k] ) (discrete) to get the autocorrelation for that shift value.

The key takeaway: autocorrelation of block signals depends entirely on the overlap between the original signal and its shifted version—more overlap means higher autocorrelation, zero overlap means zero autocorrelation.

内容的提问来源于stack exchange，提问作者user463102