Two discrete-time linear time-invariant systems with impulse responses h₁[n] = δ[n - 1] + δ[n + 1] and h₂[n] = δ[n] + δ[n - 1] are connected in cascade, where δ[n] is the Kronecker delta. The impulse response of the cascaded system is

The correct answer is:
δ[n - 2] + δ[n - 1] + δ[n] + δ[n + 1]

To find the impulse response of two discrete-time linear time-invariant (LTI) systems connected in cascade, we need to perform the discrete-time convolution of their individual impulse responses, denoted as h₁[n] and h₂[n]. Let the impulse response of the cascaded system be h[n]. Therefore, we have:

$h\left[n\right]=h_1\left[n\right]*h_2\left[n\right]$

where $*$ denotes the convolution operation. The given impulse responses are:

$h_1\left[n\right]=\delta \left[n-1\right]+\delta \left[n+1\right]$

$h_2\left[n\right]=\delta \left[n\right]+\delta \left[n-1\right]$

Substituting these expressions into the convolution equation, we get:

$h\left[n\right]=\left[\delta \left[n-1\right]+\delta \left[n+1\right]\right]*\left[\delta \left[n\right]+\delta \left[n-1\right]\right]$

Since convolution is distributive, we can expand this expression term-by-term:

$h\left[n\right]=\left[\delta \left[n-1\right]*\delta \left[n\right]\right]+\left[\delta \left[n-1\right]*\delta \left[n-1\right]\right]+\left[\delta \left[n+1\right]*\delta \left[n\right]\right]+\left[\delta \left[n+1\right]*\delta \left[n-1\right]\right]$

We use the fundamental property of convolution with shifted Kronecker delta functions:
$\delta \left[n-n_1\right]*\delta \left[n-n_2\right]=\delta \left[n-n_1-n_2\right]$

Applying this property to each term in the expanded expression:
1. $\delta \left[n-1\right]*\delta \left[n\right]=\delta \left[n-1\right]$
2. $\delta \left[n-1\right]*\delta \left[n-1\right]=\delta \left[n-2\right]$
3. $\delta \left[n+1\right]*\delta \left[n\right]=\delta \left[n+1\right]$
4. $\delta \left[n+1\right]*\delta \left[n-1\right]=\delta \left[n+1-1\right]=\delta \left[n\right]$

Summing these terms together, we find the overall impulse response:
$h\left[n\right]=\delta \left[n-1\right]+\delta \left[n-2\right]+\delta \left[n+1\right]+\delta \left[n\right]$

Rearranging the terms in order of their delays/advances, we obtain:
$h\left[n\right]=\delta \left[n-2\right]+\delta \left[n-1\right]+\delta \left[n\right]+\delta \left[n+1\right]$