Evaluate the integral “sin 2t d. (). Determine the convolution flt) = e2 COS What is the corresponding F(8)?’
The Correct Answer and Explanation is:
Problem 1: Evaluate the integral of sin(2t)
We are asked to compute the indefinite integral: ∫sin(2t) dt\int \sin(2t) \, dt∫sin(2t)dt
Solution:
Using the standard integration rule: ∫sin(ax) dx=−1acos(ax)+C\int \sin(ax) \, dx = -\frac{1}{a} \cos(ax) + C∫sin(ax)dx=−a1cos(ax)+C
Apply this to our integral: ∫sin(2t) dt=−12cos(2t)+C\int \sin(2t) \, dt = -\frac{1}{2} \cos(2t) + C∫sin(2t)dt=−21cos(2t)+C
Problem 2: Determine the convolution f(t)=et∗cos(t)f(t) = e^t * \cos(t)f(t)=et∗cos(t)
We are given two functions: f1(t)=etandf2(t)=cos(t)f_1(t) = e^t \quad \text{and} \quad f_2(t) = \cos(t)f1(t)=etandf2(t)=cos(t)
The convolution of two functions is defined as: f(t)=(f1∗f2)(t)=∫0tf1(τ)f2(t−τ) dτf(t) = (f_1 * f_2)(t) = \int_0^t f_1(\tau) f_2(t – \tau) \, d\tauf(t)=(f1∗f2)(t)=∫0tf1(τ)f2(t−τ)dτ
Substitute the functions: f(t)=∫0teτcos(t−τ) dτf(t) = \int_0^t e^\tau \cos(t – \tau) \, d\tauf(t)=∫0teτcos(t−τ)dτ
We simplify this using the convolution theorem in the Laplace domain.
Problem 3: What is the corresponding F(s)F(s)F(s)?
Using the Laplace transform:
- L{et}=1s−1\mathcal{L}\{e^t\} = \frac{1}{s – 1}L{et}=s−11
- L{cost}=ss2+1\mathcal{L}\{\cos t\} = \frac{s}{s^2 + 1}L{cost}=s2+1s
By the Convolution Theorem: L{f(t)}=L{et}⋅L{cost}\mathcal{L}\{f(t)\} = \mathcal{L}\{e^t\} \cdot \mathcal{L}\{\cos t\}L{f(t)}=L{et}⋅L{cost} F(s)=1s−1⋅ss2+1=s(s−1)(s2+1)F(s) = \frac{1}{s – 1} \cdot \frac{s}{s^2 + 1} = \frac{s}{(s – 1)(s^2 + 1)}F(s)=s−11⋅s2+1s=(s−1)(s2+1)s
Textbook-Style Explanation
In integral calculus, evaluating trigonometric functions often follows from standard integration formulas. The function sin(2t)\sin(2t)sin(2t) is a composite trigonometric function with a linear argument. Applying the substitution rule ∫sin(ax) dx=−1acos(ax)+C\int \sin(ax) \, dx = -\frac{1}{a} \cos(ax) + C∫sin(ax)dx=−a1cos(ax)+C, the integral becomes −12cos(2t)+C-\frac{1}{2} \cos(2t) + C−21cos(2t)+C. This antiderivative is foundational in analyzing oscillatory systems and arises frequently in physics and engineering.
Next, we examine the convolution of two functions, ete^tet and cos(t)\cos(t)cos(t). Convolution integrals are essential in systems analysis, particularly in linear time-invariant (LTI) systems where they represent the system’s response to an input. The convolution f(t)=(et∗cost)(t)f(t) = (e^t * \cos t)(t)f(t)=(et∗cost)(t) is defined by the integral ∫0teτcos(t−τ) dτ\int_0^t e^\tau \cos(t – \tau) \, d\tau∫0teτcos(t−τ)dτ, which measures the accumulated effect of eτe^\taueτ scaled by the delayed response cos(t−τ)\cos(t – \tau)cos(t−τ).
To evaluate the convolution efficiently, we use the Laplace transform, which converts convolution in the time domain into multiplication in the frequency domain. The Laplace transform of ete^tet is 1s−1\frac{1}{s – 1}s−11, and for cos(t)\cos(t)cos(t), it is ss2+1\frac{s}{s^2 + 1}s2+1s. Multiplying these gives the transform of the convolution: F(s)=s(s−1)(s2+1)F(s) = \frac{s}{(s – 1)(s^2 + 1)}F(s)=(s−1)(s2+1)s
This expression describes the system behavior in the Laplace domain and facilitates further analysis such as inverse transforms, stability analysis, or control design.
