Presentation is loading. Please wait.

Presentation is loading. Please wait.

1 Section 2 SECTION 2 Partial Fractions. 2 We need to split the following into separate terms: Roots of the denominator D(s): Case I – unrepeated factor.

Similar presentations


Presentation on theme: "1 Section 2 SECTION 2 Partial Fractions. 2 We need to split the following into separate terms: Roots of the denominator D(s): Case I – unrepeated factor."— Presentation transcript:

1 1 Section 2 SECTION 2 Partial Fractions

2 2 We need to split the following into separate terms: Roots of the denominator D(s): Case I – unrepeated factor Case II – repeated factor Case III – complex factors Case IV – repeated complex factors Section 2

3 3 Case I – unrepeated factor example Section 2

4 4 Use the "cover up" rule …

5 5 Section 2 Inverse Laplace Transform:

6 6 Section 2 example Inverse Laplace Transform:

7 7 Section 2 Why does the "cover up" rule work ? alternative method solve these

8 8 Question: Section 2 Obtain the Partial Fractions to

9 9 Section 2 Case II – repeated factor example unrepeated factor repeated factor

10 10 Section 2 Use the repeated factor rule …

11 11 Section 2 Inverse Laplace Transform:

12 12 Section 2 Case III – complex factors example complex conjugates Inverse Laplace Transform:

13 13 Section 2 example Inverse Laplace Transform: where

14 14 Section 2 (1)Attacking Polynomials Directly (rather than using the complex formulae) Other Topics not Covered Inverse Laplace Transform: using

15 15 Section 2 (2) To solve these, you just use the methods of cases II and III, only with complex number calculations Case IV – repeated complex factors


Download ppt "1 Section 2 SECTION 2 Partial Fractions. 2 We need to split the following into separate terms: Roots of the denominator D(s): Case I – unrepeated factor."

Similar presentations


Ads by Google