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 option is δ[n - 2] + δ[n - 1] + δ[n] + δ[n + 1].

To find the impulse response of two linear time-invariant (LTI) systems connected in cascade, we need to perform the convolution of their individual impulse responses, $h_1\left[n\right]$ and $h_2\left[n\right]$.

The overall impulse response $h\left[n\right]$ of the cascaded system is given by:
$h\left[n\right]=h_1\left[n\right]*h_2\left[n\right]$
where $*$ denotes the discrete-time convolution operation.

We are given:
$h_1\left[n\right]=\delta [n-1]+\delta [n+1]$
$h_2\left[n\right]=\delta \left[n\right]+\delta [n-1]$
Substituting these into the convolution formula:
$h\left[n\right]=\left(\delta \right[n-1]+\delta [n+1\left]\right)*\left(\delta \right[n]+\delta [n-1\left]\right)$
Using the distributive property of convolution, we can expand this expression:
$h\left[n\right]=\left(\delta \right[n-1]*\delta [n\left]\right)+\left(\delta \right[n-1]*\delta [n-1\left]\right)+\left(\delta \right[n+1]*\delta [n\left]\right)+\left(\delta \right[n+1]*\delta [n-1\left]\right)$
Recall the property of convolution with shifted impulse functions:
$\delta [n-n_1]*\delta [n-n_2]=\delta [n-(n_1+n_2\left)\right]$
Applying this property to each term individually:
1) $\delta [n-1]*\delta \left[n\right]=\delta [n-1]$
2) $\delta [n-1]*\delta [n-1]=\delta [n-2]$
3) $\delta [n+1]*\delta \left[n\right]=\delta [n+1]$
4) $\delta [n+1]*\delta [n-1]=\delta [n-0]=\delta \left[n\right]$
Combining all the simplified terms, we get:
$h\left[n\right]=\delta [n-1]+\delta [n-2]+\delta [n+1]+\delta \left[n\right]$
Rearranging the terms in ascending order of the time shift:
$h\left[n\right]=\delta [n-2]+\delta [n-1]+\delta \left[n\right]+\delta [n+1]$