Understanding the Impulse Response and Final Value

The question asks for the final value of the impulse response of a system. The system's behavior is described by its transfer function. The impulse response, often denoted as \(h(t)\), is the output of the system when the input is a Dirac delta function (an impulse). The final value refers to the steady-state value of this impulse response as time approaches infinity, i.e., \(\lim_{t \to \infty} h(t)\).

To find the final value of a time-domain function from its Laplace transform, we can use the Final Value Theorem (FVT). This theorem provides a direct way to calculate the steady-state value without needing to perform the inverse Laplace transform, which can often be complex.

Applying the Final Value Theorem

The Final Value Theorem states that if a function \(f(t)\) has a Laplace transform \(F(s)\), and if the limit of \(f(t)\) as \(t \to \infty\) exists, then:

$$ \lim_{t \to \infty} f(t) = \lim_{s \to 0} sF(s) $$

For the impulse response \(h(t)\), its Laplace transform is the system's transfer function \(H(s)\). Therefore, to find the final value of the impulse response, we will apply the FVT to \(H(s)\):

$$ \lim_{t \to \infty} h(t) = \lim_{s \to 0} sH(s) $$

The given transfer function is:

$$ H(s) = \frac{2s + 1}{s^4 + 8s^3 + 16s^2 + s} $$

Step-by-Step Calculation of Final Value

Let's calculate \(sH(s)\) first:

$$ sH(s) = s \times \frac{2s + 1}{s^4 + 8s^3 + 16s^2 + s} $$

Notice that the denominator has 's' as a common factor. We can factor it out:

$$ s^4 + 8s^3 + 16s^2 + s = s(s^3 + 8s^2 + 16s + 1) $$

Now substitute this back into the expression for \(sH(s)\):

$$ sH(s) = s \times \frac{2s + 1}{s(s^3 + 8s^2 + 16s + 1)} $$

We can cancel the 's' term from the numerator and the denominator:

$$ sH(s) = \frac{2s + 1}{s^3 + 8s^2 + 16s + 1} $$

Next, we apply the limit as \(s \to 0\):

$$ \lim_{s \to 0} sH(s) = \lim_{s \to 0} \frac{2s + 1}{s^3 + 8s^2 + 16s + 1} $$

Substitute \(s = 0\) into the expression:

$$ = \frac{2(0) + 1}{(0)^3 + 8(0)^2 + 16(0) + 1} $$

$$ = \frac{0 + 1}{0 + 0 + 0 + 1} $$

$$ = \frac{1}{1} $$

$$ = 1 $$

Applicability of Final Value Theorem

The Final Value Theorem is applicable only if all the poles of \(sH(s)\) lie in the left half of the s-plane. Let's check the poles of \(sH(s)\) in our case. The denominator of \(sH(s)\) is \(P(s) = s^3 + 8s^2 + 16s + 1\). We can use the Routh-Hurwitz criterion to check the stability (i.e., if all roots are in the LHP).

Since all the elements in the first column of the Routh array (\(1, 8, \frac{127}{8}, 1\)) are positive and there are no sign changes, all the roots of \(P(s) = s^3 + 8s^2 + 16s + 1\) are in the left half of the s-plane. Therefore, the Final Value Theorem is indeed applicable.

Conclusion on Final Value

Based on the application of the Final Value Theorem, the final value of the impulse response for the given system is 1.