Unit 2: Modeling in the Frequency Domain Part 2: The Laplace Transform. The Laplace Transform. The need for Laplace

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1 Unit : Modeling in the Frequency Domain Part : Engineering 81: Control Systems I Faculty of Engineering & Applied Science Memorial University of Newfoundland January 1, Pair Table Unit, Part : Unit, Part : Pair Table In the previous section we saw that the responses to a series RL circuit (which could be any other linear system) was composed of constants, decaying exponentials, and sinusoids. The complex frequency representation can handle all of these by utilizing the following representation with different values for s: x(t) R{Xe st } However, if the input is not of this form it is more difficult to solve for the system s response. The Laplace transform allows almost any realistic input to be described as an infinite sum of complex exponentials. The output will also be of this form, and it becomes much easier to describe how the system converts from input to output. Unit, Part : is defined as, L{f (t)} F (s) f (t)e st dt 0 where f (t) is the time-domain signal that we wish to describe in terms of an infinite sum of complex exponentials. In other words, we want to transform f (t) into the frequency-domain. The LT exists if the integral converges. The lower-limit is 0 means that if there is a discontinuity at 0, we start before the discontinuity. This allows impulse functions to be included. Notice that we are only integrating for positive time (including 0). We will be assuming that our time-domain functions are 0 for negative time.

2 is defined as, L 1 {F (s)} 1 σ+j F (s)e st ds f (t)u(t) πj σ j The ILT can be viewed as reconstructing the original time-domain signal f (t) from an infinite sum of complex exponentials. Notice the use of u(t) (the unit step function). If f (t) ever had a negative-time part, it would not have been captured by the LT. Therefore, the ILT can t bring it back. Also notice the limits of integration. The constant σ must be chosen so that the path of integration stays clear of any singular points of F (s). Example e.g. Find the LT of f (t) Ae at u(t). The function does not contain an impulse, so integration can begin at 0. F (s) f (t)e st dt 0 Ae at e st dt 0 A e (s+a)t dt 0 A s + a e (s+a)t t0 A s + a Visualizing the LT Pair Table 1 s+ : Real and Imaginary Components What does the frequency-domain signal F (s) look like for a time-domain signal f (t)? e.g. f (t) u(t)e t The result on the previous slide tells us that F (s) 1 s+. For every point s σ + jω on the complex plane we get a complex number F (s). We will visualize these complex numbers in two ways... Unit, Part :

3 1 s+ : Magnitude and Phase Pair Table We don t often utilize the Laplace integral directly. The transforms for a number of important functions appear below: f (t) F (s) 1. δ(t) 1. u(t) 1 s 3. tu(t) 1 s 4. t n u(t) n! s n+1. e at u(t) 1 s+a 6. (sin ωt)u(t) ω s +ω 7. (cos ωt)u(t) s s +ω It is clear that F (s) near s. This is a pole of this function. (Typo in book for item 4.) s s +9 : Magnitude and Phase e.g. f (t) u(t) cos 3t According to the table F (s) s s +9. There are now two poles, corresponding to the roots of the denominator. Pair Table Theorem Name 1. L{f (t)} F (s) 0 f (t)e st dt Definition. L{kf (t)} kf (s) Linearity theorem 3. L{f1(t) + f(t)} F1(s) + F(s) Linearity theorem 4. L{e at f (t)} F (s + a) Frequency shift theorem. L{f (t T )} e st F (s) Time shift theorem 6. L{f (at)} 1 a F ( s a ) Scaling theorem Unit, Part :

4 Example Pair Table e.g. Find the inverse Laplace transform of F (s) 1 (s+3). This F (s) does not appear directly in the LT table. However, we see that it is a shifted version of 1 which corresponds to tu(t) s (the ramp function). The frequency shift theorem is, L{e at f (t)} F (s + a) Therefore, we can conclude that f (t)u(t) e 3t tu(t). Note: Unless otherwise specified, we will assume that the inputs to the systems we are studying do not begin until t 0. Hence, we will leave off u(t) from our time-domain responses. Therefore, for the example above the answer is f (t) e 3t t. Unit, Part : Pair Table Theorem 7. L{ df dt } sf (s) f (0 ) Differentiation 8. L{ d f dt } s F (s) sf (0 ) f (0 ) Name Differentiation 9. L{ dn f dt n } sn F (s) n k1 sn k f (k 1) (0 ) Differentiation 10. L{ t F (s) 0 f (τ)dτ} s Integration theorem 11. f ( ) lims 0 sf (s) Final value theorem 1. f (0+) lims sf (s) Initial value theorem See the textbook for special conditions on theorems 11 and 1. (Typos in book for items 8 and 10.) Unit, Part : Example Pair Table e.g. What is the inverse Laplace transform of s? Assume initial conditions are zero. Use the first differentiation theorem (theorem 7): L{ df } sf (s) f (0 ) dt with F (s) 1. The ILT of 1 is δ(t). Therefore, f (t) dδ(t) dt Unit, Part : If F (s) is complicated it can be difficult to find the ILT. We will often see rational functions which have the form, F (s) N(s) D(s) Where N(s) and D(s) are polynomials. If the order of N(s) is less than the order of D(s) then we can apply a partial-fraction expansion. Consider the following function, F (s) s3 + 3s + s s + 3 If we want a lower-order numerator we can actually carry out polynomial division until the remainder has this property, or we can find other ways to simplify. For the example above we can factor out s from the first three terms of the numerator and get, F (s) s s + 3

5 F (s) s s + 3 We can further factor the quadratic term, F (s) s (s + 1)() Functions like the second term can be expanded as follows, (s + 1)() K1 s K In general, there are three cases for partial fraction expansion. We will use the current example to illustrate case 1... Case 1: Real and Distinct Roots (s + 1)() K1 s K We must solve for K1 and K. Multiply the equation by (s + 1), K(s + 1) K1 + This should be valid for all s. Let s approach -1 to eliminate everything else on the R.H.S.. We get K1. Apply the same strategy to obtain K. Returning to our full F (s) we have, F (s) s s We can now apply known Laplace transforms and theorems, f (t) dδ(t) e t + e t dt Case : Real and Repeated Roots e.g. This can be expanded as follows, F (s) (s + 1)() (s + 1)() K1 s K () + K3 We can solve for K1 using the previously described method. To find K we multiply by () to isolate K, () + K + ()K3 s + 1 s + 1 Letting s we get K. To find K3 we differentiate the equation above... () + + ()K3 s + 1 s + 1 Differentiating w.r.t. s, s() (s + 1) (s + 1) + K3 Letting s we find K3. Thus, the whole expansion is, (s + 1)() s + 1 ()

6 Case 3: Complex Roots Same procedure as for case 1, except we have to deal with complex roots. (Book presents another alternative). e.g. 3 F (s) s(s + s + ) 3 s(s j)(s + 1 j) K1 s + K s j + K3 s + 1 j Using the same procedure as before we get K1 3. Likewise we can solve for K and K3 only now we get complex numbers, (workings not shown) K ( + j), K3 0 ( j). ( ) F (s) 3 s 3 + j 0 s j + j s + 1 j Applying the ILT we obtain, ) f (t) (( j)e ( 1 j)t + ( j)e ( 1+j)t 0 Assuming that f (t) should be purely real, we can utilize Euler s formula to capture the complex numbers. Finally we arrive at, ( ) f (t) 3 3 e t cos t + 1 sin t