Question

2. Find the unit step response of: *(t) = -2 o=[,60 + [] .c) (t)- O y(t) = [2 3]x(t) by two methods (1): transfer function an

I have the first method complete, but I can't figure out the second method Could someone please show how to use the second method?

2. Find the unit step response of:

$$ \begin{aligned} \dot{\overrightarrow{\mathbf{x}}}(t) &=\left[\begin{array}{cc} 0 & 1 \\ -2 & -2 \end{array}\right] \overrightarrow{\mathbf{x}}(t)+\left[\begin{array}{l} 1 \\ 1 \end{array}\right] u(t) \\ y(t) &=\left[\begin{array}{cc} 2 & 3 \end{array}\right] \overrightarrow{\mathbf{x}}(t) \end{aligned} $$

by two methods (1): transfer function and then (2) \(y(t)=\mathbf{C} e^{\mathbf{A} t} \overrightarrow{\mathbf{x}}(0)+\mathbf{C} \int_{0}^{t} e^{\mathbf{A}(t-\tau)} \mathbf{B} u(\tau) d \tau+\mathbf{D} u(t)\). Re-

member that the Laplace transform of the unit step is \(u(s)=\frac{1}{s}\).

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Answer #1

AS given 39.2.2 toge 39 - 13.) ; 8(:) (= [2 3 D=0 ; x (0) = 0 D ie Initial conditions are zero. so as we know that state tranY(t) = s(4-2) 3.2lt?) ult-2) -(2) I lutede. s(t-1)-22 (+- 2 - [-st-2) ett ult-2) 4(e)] ult-t 3e - 9() - (2 de | 2 (2-t) - 2ewhen we will getting the same solve past I of question then also we are answes ie step respense is zeso.

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