Laplace Transform Solution Manual
Math 344, Maple Lab Manual Chapter 3: Laplace Transforms, 2 Inverse Transform, Dirac Delta. Solution h t = 1 4 2 eKt sin 2 t. The mass first passes through equlibrium. May 24, 2012 - Using the Laplace transform find the solution for the following equation. Without the Laplace transform we can obtain this general solution. Laplace Transform Solution Manual. 1 minute ago 0 views. Laplace Transform Solution Manual. Report this video. Select an issue. Sexual content.
Example: Distinct Real Roots Find the inverse Laplace Transform of the function F(s). Solution: We can find two of the unknown coefficients using the.
We find the other term using: Equating like powers of 's' gives us: power of 's' Equation s 2 s 1 s 0 We could have used these relationships to determine A 1, A 2, and A 3. But A 1 and A 3 were easily found using the ' method. The top relationship tells us that A 2=-0.25, so and (where, again, it is implicit that f(t)=0 when t. Example: Complex Conjugate Roots (Method 1) Using the complex (first order) roots Simplify the function F(s) so that it can be looked up in the.
Solution: If we use complex roots, we can expand the fraction as we did before. This is not typically the way you want to proceed if you are working by hand, but may be easier for computer solutions (where complex numbers are handled as easily as real numbers). To perform the expansion, continue as before. Where Note that A2 and A3 must be complex conjugates of each other since they are equivalent except for the sign on the imaginary part.
Laplace Transform Solution Of Differential Equations A Programmed Text
Performing the required calculations: so The inverse Laplace Transform is given. Example: Complex Conjugate Roots (Method 2) Method 2 - Using the second order polynomial Simplify the function F(s) so that it can be looked up in the. Solution: Another way to expand the fraction without resorting to complex numbers is to perform the expansion as follows.
Note that the numerator of the second term is no longer a constant, but is instead a first order polynomial. From above (or using the cover-up method) we know that A=-0.2. We can find the quantities B and C from cross-multiplication. If we equate like powers of 's' we get order of coefficient left side coefficient right side coefficient 2 nd (s 2) 0 A+B 1 st (s 1) 1 4A+5B+C 0 th (s 0) 3 5A+5C Since we already know that A=-0.2, the first expression (0=A+B) tells us that B=0.2, and the last expression (3=5A+5C) tells us that C=0.8.
We can use the middle expression (1=4A+5B+C) to check our calculations. Finally, we get The inverse Laplace Transform is given. Some for handling complex roots The two previous examples have demonstrated two techniques for performing a partial fraction expansion of a term with complex roots.
The first technique was a simple extension of the rule for dealing with distinct real roots. It is conceptually simple, but can be difficult when working by hand because of the need for using complex numbers; it is easily done by computer. The second technique is easy to do by hand, but is conceptually a bit more difficult.
It is easy to show that the two resulting partial fraction representations are equivalent to each other. Let's first examine the result from Method 1 (using two techniques). We start with Method 1 with no particular simplifications. Method 1 - brute force technique (The last line used ) We now repeat this calculation, but in the process we develop a general technique (that proves to be useful when using MATLAB to help with the partial fraction expansion. We know that F(s) can be represented as a partial fraction expansion as shown below: We know that A 2 and A 3 are complex conjugates of each other: Let We can now find the inverse transform of the complex conjugate terms by treating them as simple first order terms (with complex roots). In this expression M=2K.
The frequency (ω) and decay coefficient (σ) are determined from the root of the denominator of A 2 (in this case the root of the term is at s=-2+j; this is where the term is equal to zero). The frequency is the imaginary part of the root (in this case, ω=1), and the decay coefficient is the real part of the root (in this case, σ=-2). Using the cover-up method (or, more likely, a computer program) we get and This yields It is easy to show that the final result is equivalent to that previously found, i.e., While this method is somewhat difficult to do by hand, it is very convenient to do by computer. This is the approach used on the page that shows. Finally we present Method 2, a technique that is easier to work with when solving problems for hand (for homework or on exams) but is less useful when using MATLAB. Completing the square. (The last line used the entry ' from ) Thus it has been shown that the two methods yield the same result.
Use Method 1 with MATLAB and use Method 2 when solving problems with pencil and paper. Example - Combining multiple expansion methods Find the inverse Laplace Transform of Solution: The fraction shown has a second order term in the denominator that cannot be reduced to first order real terms.
As discussed in the page describing, we'll use two techniques. The first technique involves expanding the fraction while retaining the second order term with complex roots in the denominator. The second technique entails '.' Since we have a repeated root, let's cross-multiply to get Then equating like powers of s. Previous Example, Using Brute Force We repeat the previous example, but use a brute force technique. You will see that this is harder to do when solving a problem manually, but is the technique used by MATLAB. It is important to be able to interpret the MATLAB solution.
Find the inverse Laplace Transform of Solution: We can express this as four terms, including two complex terms (with A 3=A 4.) Cross-multiplying we get (using the fact that (s+1-2j)(s+1+2j)=(s 2+2s+5)) Then equating like powers of s. Power of s Equation s 3 0=A 1+A 3+A 4 s 2 5=2A 1+A 2+(1+2j)A 3+(1-2j)A 4 s 1 8=5A 1+2A 2 s 0 -5=5A 2 We could solve by hand, or use MATLAB: A=1 0 1 1; 2 1 1+2j 1-2j; 5 2 0 0; 0 5 0 0; b=0 5 8 -5'; inv(A).b ans = 2.0000 -1.0000 -1.0000 - 1.0000i -1.0000 + 1.0000i So, and We will use the notation derived above. The root of the denominator of the A 3 term in the partial fraction expansion is at s=-1+2j (i.e., the denominator goes to 0 when s=-1+2j), the magnitude of A 3 is √2, and the angle of A3 is 225°. So, M=2√2, φ=225°, ω=2, and σ=-1. Solving for f(t) we get This expression is equivalent to the one obtained in the previous example.
When the Laplace Domain Function is not strictly proper (i.e., the order of the numerator is different than that of the denominator) we can not immediatley apply the techniques described above. Example: Order of Numerator Equals Order of Denominator Find the inverse Laplace Transform of the function F(s). Solution: For the fraction shown below, the order of the numerator polynomial is not less than that of the denominator polynomial, therefore we first perform long division Now we can express the fraction as a constant plus a ratio of polynomials. Using the cover up method to get A 1 and A 2 we get so The last case we will consider is that of exponentials in the numerator of the function. Example: Exponentials in the numerator Find the inverse Laplace Transform of the function F(s).
Solution: The exponential terms indicate a time delay (see the). The first thing we need to do is collect terms that have the same time delay. We now perform a partial fraction expansion for each time delay term (in this case we only need to perform the expansion for the term with the 1.5 second delay), but in general you must do a complete expansion for each term. Now we can do the inverse Laplace Transform of each term (with the appropriate time delays).
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Is there any way to get a printable version of the solution to a particular Practice Problem? As we saw in the last computing Laplace transforms directly can be fairly complicated.
Usually we just use a when actually computing Laplace transforms. The table that is provided here is not an inclusive table, but does include most of the commonly used Laplace transforms and most of the commonly needed formulas pertaining to Laplace transforms. Before doing a couple of examples to illustrate the use of the table let’s get a quick fact out of the way. Fact Given f(t) and g(t) then, for any constants a and b.
In other words, we don’t worry about constants and we don’t worry about sums or differences of functions in taking Laplace transforms. All that we need to do is take the transform of the individual functions, then put any constants back in and add or subtract the results back up. So, let’s do a couple of quick examples. Example 1 Find the Laplace transforms of the given functions.
(a) (b) (c) (d) Solution Okay, there’s not really a whole lot to do here other than go to the, transform the individual functions up, put any constants back in and then add or subtract the results. We’ll do these examples in a little more detail than is typically used since this is the first time we’re using the tables. (a) (b) (c) (d). Make sure that you pay attention to the difference between a “normal” trig function and hyperbolic functions. The only difference between them is the “+ a 2” for the “normal” trig functions becomes a “- a 2” in the hyperbolic function!
It’s very easy to get in a hurry and not pay attention and grab the wrong formula. If you don’t recall the definition of the hyperbolic functions see the notes for the. Let’s do one final set of examples.
Example 2 Find the transform of each of the following functions. (a) (b) (c) (d) (e) Solution (a) This function is not in the table of Laplace transforms. However we can use in the table to compute its transform.
This will correspond to #30 if we take n=1. So, we then have, Using #30 we then have(b) This part will also use in the table. In fact we could use #30 in one of two ways. We could use it with. Or we could use it with. Since it’s less work to do one derivative, let’s do it the first way. So using we have, The transform is then(c) This part can be done using either (with ) or (along with ).
We will use #32 so we can see an example of this. In order to use #32 we’ll need to notice that Now, using #5, we get the following. This is what we would have gotten had we used #6. (d) For this part we will use along with the answer from the previous part. To see this note that if then Therefore, the transform is.
(e) This final part will again use from the table as well as. Remember that g(0) is just a constant so when we differentiate it we will get zero!