Question

The frequency response of an LTI system is H(ω) = 1 + ω^2/2. The impulse response of the system is h(t). The input of the system is x(t), and the output is y(t).

a) What is ∫₋[infinity]⁺[infinity] h(t)dt?

Answer 1

The value of this integral depends on the specific form of the impulse response
`\(h(t)\)` obtained from the inverse Fourier transform of the given frequency response
`\(H(\omega) = 1 + (\omega^2)/(2)\)`. To find the exact value, you'll need to calculate the inverse Fourier transform and evaluate the resulting expression for
`\(h(t)\).`

`\[ \int_(-\infty)^(\infty) h(t) dt \]`

Step-by-step explanation:

The question asks for the integral over all time of the impulse response of a linear time-invariant (LTI) system whose frequency response is given by H(ω) = 1 + ω² : 2. To find this integral, which is representatively ∫−∞⁺∞ h(t)dt, we can use the fact that the integral of the impulse response of an LTI system over all time is equal to the system's frequency response evaluated at ω = 0. From the given frequency response, H(ω) = 1 + ω² : 2, we can see that H(0) = 1. Therefore, the integral of the impulse response h(t) over all time is 1.

The impulse response of a system, denoted as
`\(h(t)\)`, is related to the system's frequency response
`\(H(\omega)\)` through the inverse Fourier transform. Specifically, the impulse response
`\(h(t)\)` is the inverse Fourier transform of the frequency response
`\(H(\omega)\)`.

The inverse Fourier transform is given by:

`\[ h(t) = (1)/(2\pi) \int_(-\infty)^(\infty) H(\omega) e^(j\omega t) d\omega \]`

Given that
`\(H(\omega) = 1 + (\omega^2)/(2)\)`, let's find the inverse Fourier transform
`\(h(t)\):\[ h(t) = (1)/(2\pi) \int_(-\infty)^(\infty) \left(1 + (\omega^2)/(2)\right) e^(j\omega t) d\omega \]`

Now, integrate this expression to find \(h(t)\). After finding \(h(t)\), you can then evaluate the integral:

`\[ \int_(-\infty)^(\infty) h(t) dt \]`

This integral represents the total area under the impulse response
`(h(t)\)`curve. The result of this integral will depend on the specific form of
`\(h(t)\)`obtained from the inverse Fourier transform.