如何计算给定输入信号的系统响应？附具体系统及输入定义

Let's break this down step by step so you can follow exactly how to compute the output ( y(n) ). We'll cover two approaches: direct substitution into the system equation (the simplest way here) and using the impulse response with convolution (to tie back to your question about impulse responses).

First, Let's Formalize the Input ( x(n) ) #

Your input signal has non-zero values only at specific indices:

• ( x(-2) = -3 )
• ( x(0) = 2 ), ( x(1) = 2 )
• For all other ( n ), ( x(n) = 0 )

Approach 1: Direct Substitution into the System Equation #

The system is defined by:

y(n) = 3x(n+2) + 2x(n) - 4x(n-2)

This is a linear combination of shifted versions of the input. Since ( x(n) ) only has non-zero values at 3 indices, we only need to calculate ( y(n) ) for indices where at least one of the terms ( x(n+2) ), ( x(n) ), or ( x(n-2) ) is non-zero.

Let's compute ( y(n) ) for every relevant ( n ):

• n = -4: Only ( 3x(n+2) = 3x(-2) = 3*(-3) = -9 ) is non-zero. Others are 0. So ( y(-4) = -9 )
• n = -3: All shifted input terms point to indices where ( x(n) = 0 ). ( y(-3) = 0 )
• n = -2: ( 3x(0) = 6 ), ( 2x(-2) = -6 ), ( -4x(-4) = 0 ). Sum: ( 6 + (-6) + 0 = 0 ). ( y(-2) = 0 )
• n = -1: ( 3x(1) = 6 ), others are 0. ( y(-1) = 6 )
• n = 0: ( 2x(0) = 4 ), ( -4x(-2) = 12 ), ( 3x(2) = 0 ). Sum: ( 4 + 12 = 16 ). ( y(0) = 16 )
• n = 1: ( 2x(1) = 4 ), others are 0. ( y(1) = 4 )
• n = 2: ( -4x(0) = -8 ), others are 0. ( y(2) = -8 )
• n = 3: ( -4x(1) = -8 ), others are 0. ( y(3) = -8 )
• All other n: Every shifted input term points to 0, so ( y(n) = 0 )

Approach 2: Using Impulse Response + Convolution #

To connect this to impulse responses, first find the system's impulse response ( h(n) )—this is the output when the input is the unit impulse ( \delta(n) ) (where ( \delta(0)=1 ), ( \delta(n)=0 ) for ( n≠0 )).

Substitute ( x(n) = \delta(n) ) into the system equation:

h(n) = 3δ(n+2) + 2δ(n) - 4δ(n-2)

So ( h(n) ) has non-zero values at:

• ( h(-2) = 3 )
• ( h(0) = 2 )
• ( h(2) = -4 )

The system response ( y(n) ) is the convolution of ( x(n) ) and ( h(n) ):

y(n) = x(n) * h(n) = Σₖ x(k) * h(n - k)

Calculating this convolution will give the exact same results as direct substitution. For example:

• For ( n=0 ): ( Σₖ x(k)h(0-k) = x(-2)h(2) + x(0)h(0) + x(1)h(-1) = (-3)(-4) + 22 + 2*0 = 12 +4 =16 ), which matches our earlier result.

This makes sense because the system equation is already a convolution-like sum—LTI systems' outputs are always the convolution of input and impulse response.

内容的提问来源于stack exchange，提问作者zorro