Question 2.9.2: Impulse Response of Second-Order Models In Example 2.6.4 we ...

Impulse Response of Second-Order Models

In Example 2.6.4 we obtained the response of the following model to a unit impulse:

\frac{X(s)}{F(s)}=\frac{1}{2s^2+14s+20}

Our analysis showed that if x(0−) = ẋ(0−) = 0, then x(0+) = 0 and ẋ(0+) = 1/2. Use the impulse function to verify these results.

Annotation 2022-10-16 042410

The session is shown here.

≫sys1 = tf(1,[2,14,20]);
≫impulse(sys1)

The plot is shown in Figure 2.9.5. From it we see that x(0+) = 0 and that ẋ(0+) is positive as predicted. We are unable to determine the exact value of ẋ(0+) from this plot, so we multiply the transfer function by s to obtain the transfer function for v = ẋ.

\frac{X(s)}{F(s)}=\frac{1}{2s^2+14s+20}

We now use the impulse function on this transfer function.

≫sys2 = tf([1, 0],[2, 14, 20]);
≫impulse(sys2)

The plot is shown in Figure 2.9.6. From it we see that ẋ(0+) = 0.5 as predicted.

Annotation 2022-10-16 042926

Related Answered Questions

Question: 2.1.6

Verified Answer:

We cannot use Table 2.1.1 for this part because th...
Question: 2.7.7

Verified Answer:

a. The denominator roots are s = −3 ± 5j. To avoid...
Question: 2.7.5

Verified Answer:

There are four repeated roots (s = 0) and one dist...
Question: 2.7.2

Verified Answer:

The denominator roots are s= −12/3= −4, s=0, and s...