Final answer:
In this case, the convolution integral for y(t) is y(t) = (t-1) * u(t-2) + 2(t-1) * u(t-1)
Step-by-step explanation:
To determine the convolution of the given signals x(t) and h(t), we can use the convolution integral:
y(t) = ∫[x(τ) * h(t-τ)] dτ
where * denotes the convolution operation.
Let's calculate the convolution step by step:
1. Determine the limits of integration:
Since the given signals x(t) and h(t) have non-zero values only for a limited range of t, we need to determine the limits of integration accordingly. The range of t for which x(t) and h(t) are non-zero is from 0 to 10.
2. Rewrite the signals with respect to the integration variable τ:
x(τ) = τ + 1
h(t-τ) = δ((t-τ)+2) + 2δ((t-τ)+1)
3. Determine the non-zero ranges for the signals in terms of τ:
x(τ) is non-zero for 0 ≤ τ ≤ 10
h(t-τ) is non-zero for -2 ≤ t-τ ≤ -1
4. Rewrite the convolution integral based on the non-zero ranges:
y(t) = ∫[(τ+1) * [δ((t-τ)+2) + 2δ((t-τ)+1)]] dτ
= ∫[(τ+1) * δ(t-τ-2) + 2(τ+1) * δ(t-τ-1)] dτ
5. Evaluate the integral:
To evaluate the integral, we need to consider the cases when δ(t-τ-2) and δ(t-τ-1) are non-zero. These occur when the following conditions are satisfied:
-2 ≤ t-τ ≤ -1 => t-1 ≤ τ ≤ t-2
Case 1: When t-1 ≤ τ ≤ t-2
In this case, the integral becomes:
∫[(τ+1) * δ(t-τ-2)] dτ
Case 2: When 0 ≤ τ ≤ t-1
In this case, the integral becomes:
∫[(τ+1) * δ(t-τ-2) + 2(τ+1) * δ(t-τ-1)] dτ
6. Evaluate each case separately:
Case 1:
∫[(τ+1) * δ(t-τ-2)] dτ = (t-2+1) = t-1, since δ(t-τ-2) = 1 when τ = t-2.
Case 2:
∫[(τ+1) * δ(t-τ-2) + 2(τ+1) * δ(t-τ-1)] dτ
= [(t-1+1) * δ(t-t+1-2) + 2(t-1+1) * δ(t-t+1-1)]
= [(t-1+1) * δ(t-1) + 2(t-1+1) * δ(t)].
7. Combine the results from each case:
The convolution integral for y(t) becomes:
y(t) = (t-1) * u(t-2) + 2(t-1) * u(t-1)
Note: u(t) represents the unit step function.