9.5 The Transfer Function

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1 Lecture Notes on Control Systems/D. Ghose/ The Transfer Function Consider the n-th order linear, time-invariant dynamical system. dy a 0 y + a 1 dt + a d 2 y 2 dt + + a d n y 2 n dt b du 0u + b n 1 dt + b d 2 u 2 dt + + b d m u 2 m dt m with zero initial conditions on all derivatives. Taking the Laplace transform on both sides, we get, a 0 Y (s)+a 1 sy (s)+a 2 s 2 Y (s)+ + a n s n Y (s) b 0 U(s)+b 1 su(s)+b 2 s 2 U(s)+ + b m s m U(s) From which, Y (s) U(s) N(s) D(s) b 0 + b 1 s + b 2 s b m s m a 0 + a 1 s + a 2 s a n s n mk1 b k s k nj1 a j s j Where, G(s) is called the transfer function. It is defined from differential equations for which the initial condition is zero. G(s) issaidtobeproperifm n. G(s) is said to be strictly proper if m<n. Note: We will understand these terms in a better way a little later. However, in most cases we will be dealing with strictly proper systems. Let us do some more algebraic manipulations. Let, Then, K b m a n bk b k b m, for k 0,...,m 1 ā j a j a n, for j 0,...,n 1 b ms m + b m 1 s m b 1 s + b 0 a n s n + a n 1 s n a 1 s + a 0

2 Lecture Notes on Control Systems/D. Ghose/ where, m n b ( ) m s m + b m 1 b m s m b 1 b m s + b 0 b ( m ) a n sn + a n 1 a n s n a 1 a n s + a 0 a n b ( m s m + b m 1 s m b 1 s + b ) 0 a n s n +ā n 1 s n 1 + +ā 1 s +ā 0 ( s m + K b m 1 s m b 1 s + b ) 0 s n +ā n 1 s n 1 + +ā 1 s +ā 0 K (s z 1)(s z 2 ) (s z m ) (s p 1 )(s p 2 ) (s p n ) 1. The numerator roots z 1,,z m are called system zeros. 2. The denominator roots p 1,,p n are called the system poles. Figure 9.9: A sketch of the poles and zeros along the real line. The denominator polynomial is called the characteristic polynomial. 4. The transfer function is the Laplace transform of the impulse response function 4. An Example. An example of why m>n(or, a system which is not proper) is a bad idea. Newton s law says f m v. Let us identify the input and the output.

3 Lecture Notes on Control Systems/D. Ghose/ Figure 9.10: Force applied on a mass Figure 9.11: Case 1 Case 1: Let us say that input is v and output is f. Also, let v be a step input. Looks like a bad idea!!! Case 2: Let the input be f and the output be v, and let f be the unit force. Figure 9.12: Case 2 Looks OK. Now, look at the transfer function. 4 To see this, put u(t) δ(t), then U(s) 1. So, Y (s) G(s) andy(t) L 1 [G(s)]

4 Lecture Notes on Control Systems/D. Ghose/2012 f(t) m v F (s) msv (s) (assuming zero initial condition) For Case 1: F (s) msv (s) F (s) V (s) ms (which is an improper transfer function) Note that V (s) e st 0 F (s) me st 0. Thus, a delayed step function is caused by a s delayed impulse signal. For Case 2: V (s) 1 V (s) F (s) 1 ms F (s) ms (which is strictly proper). So, F (s) e st 0 V (s) me st 0 1 me st 0. This may be interpreted as the integral of s s 2 s s a delayed step function, giving rise to a ramp function. 9.6 Initial and Final Value Theorems Initial Value Theorem Since [ ] df L sf (s) f(0) dt and since, s e st 0, we have, So, Therefore, Final Value Theorem [ ] df lim L s 0 dt Therefore, and hence, [ df lim L s dt ] df lim s 0 dt e st dt 0 lim [sf (s) f(0)] 0 s f(0) lim s sf (s) df lim s 0 0 dt e st dt 0 lim[sf (s) f(0)] f( ) f(0) s 0 df dt f( ) f(0) dt lim sf (s) f( ) lim f(t) s 0 t However, the final value theorem is meaningful only if the following conditions are met.

5 Lecture Notes on Control Systems/D. Ghose/ The Laplace transform of f(t) and df exists. dt 2. lim t f(t) exists.. All poles of F (s) are on the left half plane except for one which may be at the origin. 4. No poles on the imaginary axis. 9.7 Partial Fraction Expansion The reason we introduced the Laplace transform is to devise an easy way to find the system response. Figure 9.1: Input-output representation One way to do this would be by using the convolution integral, y(t) t 0 g(t τ)u(τ)dτ But if the Laplace transforms are known for g and u, then Y (s) G(s)U(s) and then y(t) is obtained by finding the inverse Laplace transform. To achieve this, we need to break up Y (s) into pieces for which the inverse Laplace transforms are available, and then use the Laplace transform tables to find the inverse Laplace transform of the complete Y (s). Let, N(s) D(s) b ms m + b m 1 s m b 1 s + b 0 a n s n + a n 1 s n a 1 s + a 0 mk1 (s z k ) K nj1 (s p j ) Note that G(s) is not necessarily the transfer function of the system. It could be the output Y (s) or any other function in the s-domain having a numerator polynomial and a denominator polynomial of appropriate order.

6 Lecture Notes on Control Systems/D. Ghose/ Case 1: Distinct Poles: p i p j,i j Then, the partial fraction expansion of G(s) is n k i s p i where, k i are called the residues. How do we find k i? To find k i : Multiply G(s) by(s p i ) and let s p i. i1 Example. Let n k j k i s pi lim (s p i ) s pi lim (s p i ) j1 (s p j ) It is easy to find the poles of G(s), s s 2 +s +2 s (s +1)(s +2) Since the poles are distinct (p 1 1,p 2 2), we may expand G(s) into partial fractions as, To find the residues, So, k 1 lim (s +1) lim s 1 s 1 k 2 lim (s +2) lim s 2 s 2 k 1 s +1 + k 2 s +2 s s +2 s s +1 1 s s Verify that the above is indeed the same as the original G(s). Case 2: A pole of multiple order or repeated pole. N(s) (s p 1 )(s p 2 ) (s p i 1 )(s p i ) l (s p i+1 ) (s p n )

7 Lecture Notes on Control Systems/D. Ghose/ In the above, the pole p i has order l>1. All other poles have order 1. Then the partial fraction expansion is given by, where, k 1 s p 1 + k 2 s p k i+1 s p i A 1 s p i + k n k i 1 s p i 1 Simple poles Simple poles s p n A 2 (s p i ) + + A l Repeated poles 2 (s p i ) l k j (s p j )G(s) spj For simple poles and for the repeated pole, A l (s p i ) l G(s) spi [ d { A l 1 (s pi ) l G(s) }] ds sp i A l 2 1 [ d 2 { (s pi ) l G(s) } ] 2! ds 2. A 2 A 1 [ 1 d (l 2) (l 2)! 1 (l 1)! sp i { (s pi ) l G(s) } ] ds (l 2) [ d (l 1) { (s pi ) l G(s) } ds (l 1) sp i ] sp i Example. Consider a system which has a response given by the differential equation, ẍ +2ẋ + x 0, x(0) a, ẋ(0) b Taking Laplace transform on both sides, s 2 X(s) sx(0) ẋ(0) + 2(sX(s) x(0)) + X(s) 0 s 2 X(s) sa b +2(sX(s) a)+x(s) 0 From which, X(s) as +(2a + b) s 2 +2s +1 The partial fraction expansion is then, as +(2a + b) (s +1) 2 X(s) A 1 s +1 + A 2 (s +1) 2

8 Lecture Notes on Control Systems/D. Ghose/ where, A 2 (s +1) 2 X(s) as +(2a + b) s 1 s 1 a +2a + b a + b [ ] d A 1 ds {(s +1)2 X(s) a s 1 So, X(s) a s +1 + a + b (s +1) 2 Verify that the above is indeed the same as the original X(s). Case : Complex poles TRANSFER FUNCTION HAVING DISTINCT POLES + TRANSFER FUNCTION HAVING REPEATED POLES + TRANSFER FUNCTION HAVING COMPLEX POLES The complex roots are expressed as, When b>0, k 1 s + k 2 (s + a) 2 + b 2 (s + a) 2 + b 2 (s + a) 2 (jb) 2 (s + a + jb)(s + a jb) Since the complex roots are distinct, one can use the method of distinct roots as given earlier to obtain the residues. But in that case the residues will also be complex. On further manipulations we can get back the real numbers. Finally we can use the following inverse Laplace transforms, [ ] L 1 b e at sin bt (s + a) 2 + b 2 [ ] L 1 s + a e at cos bt (s + a) 2 + b 2 Example.

9 Lecture Notes on Control Systems/D. Ghose/ s +s 2 +6s +4 s + s 2 +2s 2 +2s +4s +4 s 2 (s +1)+2s(s +1)+4(s +1) (s 2 +2s +4)(s +1) (s +1)[(s +1) 2 +( ) 2 ) So, the poles are, p 1 1, p 2 1 j,p 1+j Since all the poles are distinct, by partial fraction expansion, k 1 s +1 + k 2 s +1+j + k s +1 j and, the residues are computed as, k 1 (s +1)G(s) s 1 (s +1) s 1 Similarly, k 2 (s +1+j )G(s) (s +1)(s +1 j ) 1 j + ( j )( j2 ) 2 j j 6 k (s +1 j )G(s) s 1 j s 1 j s 1+j

10 Lecture Notes on Control Systems/D. Ghose/ (s +1)(s +1+j ) 1+j + (j )(j2 ) 2+j 6 1 j 6 Substituting these values, (1/) + j( /6) + s +1+j s 1+j 2/ + (1/) j( /6) s +1 s +1 j 2/ s +1 +( 2/) s +1 (s +1) 2 + +(1/ ) (s +1) 2 + Applying the inverse transform we get, g(t) 2 e t 2 e t cos t + 1 e t sin t An alternative way to solve the same problem is by equating the coefficients. This avoids the complication of using imaginary numbers. Let, So, (s +1)[(s +1) 2 +] (s +1)[(s +1) 2 +] k 1 s +1 + k 2s + k (s +1) 2 + k 1[(s +1) 2 +]+(s +1)(k 2 s + k ) (s +1)[(s +1) 2 +] Since the denominators are the same, the numerators must also be the same. Thus, (k 1 + k 2 )s 2 +(2k 1 + k 2 + k )s +4k 1 + k Comparing the coefficients of the powers of s, Solving, k 1 + k 2 0 2k 1 + k 2 + k 1 4k 1 + k k 1 2 k 2 2 k 1

11 Lecture Notes on Control Systems/D. Ghose/ The rest follows in the same way as before. Question. How would you handle repeated complex roots?

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