Laplace Transforms and use in Automatic Control

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1 Laplace Transforms and use in Automatic Control P.S. Gandhi Mechanical Engineering IIT Bombay Acknowledgements: P.Santosh Krishna, SYSCON

2 Recap Fourier series Fourier transform: aperiodic Convolution integral Frequency response

3 Convolution Integral Linear, time-invariant (LTI) systems can be completely characterized by their impulse response Consider an LTI system whose impulse response is h(t), the response of the system to arbitrary input u(t) is given by: y(t)=h(t)*u(t) yt () = ht ( τ ) u( τ) dτ

4 Convolution Integral Fundamentals Convolution as sum of inverted and shifted impulse responses at various input points

5 Frequency Response e jωt LTI? H(jω) e jωt Frequency response of the system = fourier transform of the impulse response

6 Complex Exponentials and LTI systems e jωt LTI? H(jω) e jωt What will be the response of the system to a sum of complex exponentials?

7 Why Laplace transforms?? Integration and differentiation can be replaced by algebraic operations in complex plane. It converts complex functions such as sinusoidal, exponential into algebraic functions of complex variable s These algebraic equations can be easily solved in s for dependent variable whose time domain response can be found with inverse Laplace transforms MA 203 Q: cant we do these by using Fourier transform? A: Not always. We will consider some cases later

8 Definition of L.T. f () t ( ) 0 If is a function of time, f t = for t < 0 and s is a complex variable and F(s) is the laplace transform of function f ( t) Then the laplace transform of f t is given by, 0 ( ) () () st F s = L f t = f(). t e dt What if s = jw????? Fourier transform formula

9 Inverse Laplace Transforms If F(s) is the laplace transform of a function, then its time domain response is given by ( ) st L F s = f() t = F() s e ds 2π j 1 1 c+ j c j for t > 0 This is called inverse Laplace transform of F(s)

10 Advantages of L.T over F.T We have already seen Fourier transform which is a function of the complex variable jw. Laplace trans- form is a function of the complex variable s denoting ' σ + jw' in which if σ = 0, then Laplace transforms equals Fourier transforms. What does this mean physically??? Laplace transforms are introduced to fill the gaps which Fourier transform does not; We see these two cases with examples.

11 Advantages of L.T over F.T Case 1: Unstable systems Stable system is one which gives bounded output for bounded input. Consider the unstable system y() t = t. u() t whose output diverge for a unit step input. Applying Fourier transforms on both sides, we get = jwt Y jw tu t e dt ( ) ().

12 Advantages of L.T over F.T But Dirichlets first condition states that [t.u(t)] should be absolutely integrable for the Fourier transform to exist. That is, tu( t) which is violated in this case. Thus, it is clear that we can t t find Fourier transform for unstable systems. Now, taking Laplace transforms, dt <

13 Advantages of L.T over F.T st Y s tu t e dt = ( ) (). st te. = s 0 1 = for Re{s}> 0 2 s e s st st = te. dt 2 0 So, it is clear that for unstable systems, even though Fourier transform does not exist, Laplace transform exits which helps to investigate the stability of a system

14 Advantages of L.T over F.T Case 2: 2 Let us take the input signal as below instead of unit step at x() t = e. u() t Since we need to find F.T of both the system and the input signal in order to obtain the output response, we need to find F.T of this input signal given above jwt X jw e u t e dt = ( ) at ().

15 Advantages of L.T over F.T at jwt X jw e dt ( ) = 0 The right side part is tending to infinite, so we can t find F.T for this input signal. We can conclude this without any calculations directly from Dirichlet first condition as in case-1. Taking Laplace transforms, st X s e u t e dt = ( ) at ().

16 Advantages of L.T over F.T = X s e dt ( ) t( s a) 0 = 1 for s a Re{ s} So, for some values of Re{s}, we are able to find L.T for the given signal. This is called Region of Convergenc (ROC). Here, it is the area in s-plane s whose real part is greater than a.. In case-1, ROC in the right half of the S-plane. > a

17 How to find laplace transform?? Consider a sinusoidal function, f ( t ) = 0, for t < 0 = Asin wt, for t >= 0 But we know that, 1 jwt jwt sin wt = ( e e ) 2 j

18 contd. Hence A [ sin ] jwt jwt st L A wt = ( e e ) e dt 2 j 0 A 1 A 1 =.. 2j s jw 2j s+ jw = s Aw + w 2 2

19 Laplace transforms for some commonly used functions Function Transform Unit impulse 1 Unit step n n! t n 1 s + at 1/ s e 1 s+ a w sin wt 2 2 s + w

20 contd. Function Transform s cos wt 2 2 sinh wt 2 2 s s + w w w s 2 2 cosh wt s w

21 Some other functions Observe how affects these functions, e at Function Transform e e at at t n sin wt n! ( s + a ) n + 1 w ( s+ a) + w 2 2 e at cos wt s+ a ( s+ a) + w 2 2

22 Contd. Function e e at at sinh cosh wt wt Transform w ( s+ a) w 2 2 s+ a ( s+ a) w 2 2 i.e, at. () at st L e f t = e. f ( t). e dt = F( s+ a) 0

23 L.T of a translated function let F(s) be the L.T of f(t) and l(t) is a function such that f( t α). l( t α) = 0 for t < α

24 Contd. then L.T of f( t α). u( t α) is given by, st L[ f ( t α). l( t α)] = f ( t α) l( t α). e dt 0 = α s ( t + α ) f ( t) l( t). e dt = 0 ( ). st α t f t e. e dt = e α s. F( s)

25 Change of time scale If t is changed into t / α, where α is a positive constant. Then function f () t has changed into t then L.T of t is, f, α t t st L f = f. e dt α α 0 α f α. () α st = f t. e dt = α F ( αs) 0

26 Laplace transform theorems Real differentiation theorem d L f () t = sf( s) f (0) dt Real integration theorem 1 Fs () f (0) L f() t dt = + s s Complex differentiation theorem d L t. f () t = F( s) ds IMP for LTI

27 Initial value theorem From this theorem, we can find the value of f(t) at t=0+ directly from the L.T of f(t). This theorem will not give the value of f(t) exactly at t=0 but slightly greater than zero df () t Def: : If f(t) and are laplce transformable dt and if lim sf( s) exists then, s f (0 + ) = lim sf ( s) s

28 Final value theorem This theorem gives the value of f(t) when t This theorem applies if and only if lim f ( t) exits t All the poles of sf() s should lie in the left half of s-plane for lim f ( t) to exist. t df () t Def: : If f(t) and are laplce transformable dt and if lim f ( t) exists then, t lim f ( t) = lim sf( s) t s 0

29 Convolution Integral t f ( t τ ) f ( τ) dτ 1 2 The integral can also be 0 written as f1() t f2() t which is known as convolution integral. If F () s & F () s are L.T s s of f () t & f () t respectively, then L.T of convolution integral is given by, t L f1( t τ) f1( t) dτ = F1( s). F2( s) 0 Blocks can be multiplicative in Laplace domain

30 L.T of product of two time functions If F(s) and G(s) are the laplace transforms of f(t) and g(t) respectively, then the laplace transform of f() t g() t is given by, c+ j 1 L[ f () t g() t ] = F( p). G( s p) dp 2π j c j

31 Example Problem We have already seen block diagram representation to a system with integrators, differentiators and multiplier blocks. But this block diagram representation becomes much simpler with this Laplace transforms as shown. Let us consider the system given in exam,... yt () + 5 yt () + 2 yt () = 10. xt () First let us find the system response applying L.T s 2 s Y s + sy s + Y s = X s () 5. () 2 () 10. ()

32 Ex. Problem (Contd ) Y() s 10 = X s s + s+ 2 () 5 2 This ratio of output to input is called Transfer Function After finding the partial fractions and applying Inverse Laplace transforms to above transfer function with unit Step as input, we get the following output, 0.439t 4.562t yt ( ) = 2.425( e e ) + 5 ut ( ) Solve and prove in class

33 Ex. Problem (Contd ) The block diagram representation for the above system with integrator and multiplier blocks is,

34 Ex. Problem (Contd ) For designing this small system, we needed many integrator, gain and summation blocks leading to big block diagram. Instead, we can form it in S-domain S with only a single transfer function block which is obtained in previous slide.

35 Ex. Problem (Contd ) This reduces the complexity of the block diagram very much and thus representation of much bigger plants also became simpler with just these transfer function blocks and multiplier blocks. The output response of the system will be same in both the block diagram representations with MATLAB. These output responses are shown in next slide.

36 Ex. Problem (Contd ) Response in time domain block diagram

37 Ex. Problem (Contd ) Response in transfer function block diagram

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