Chapter 9 Laplace transforms Part 1

Chapter 9 Laplace transforms Part 1 - ECSE-303A Signals and...

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Unformatted text preview: ECSE-303A Signals and Systems I Friday, March 12, 2010 LECTURE 20 Chapter 9 The Laplace Transform Introduction to the Laplace transform Chapter 9: The Laplace transform 9.1 The Laplace transform 9.2 The region of convergence of the Laplace transform 9.3 The inverse Laplace transform 9.4 Geometric evaluation of the Fourier transform from the pole-zero plot 9.5 Properties of the Laplace transform 9.7 Analysis and characterization of LTI systems using the Laplace transform 9.8 System function algebra and block diagram representations 9.9 The unilateral Laplace transform NIB Application of Laplace Transforms to circuit analysis NIB Transform circuit diagrams 9.1 The Laplace Transform In Fourier theory, we have learned that a signal can be represented by a sum of sinuses and cosinuse x (t ) = ∞ ∑ k =−∞ x (t ) = ∞ 1 2π ∫ ak e jkω0t X ( jω ) e jωt d ω ∞ FS ↔ ak = ∫ x ( t ) e − jkω0t dt −∞ FT ↔ −∞ X ( jω ) = ∞ ∫ x ( t ) e − jωt dt −∞ Replacing jω by a general complex variable s=jω+σ leads to the Laplace representation, a generalization of the Fourier (frequency) representation. x (t ) = σ + j∞ 1 2π j ∫ σ − j∞ ECSE303 Chap. 9 MR - 3/12/2010 LT X ( s ) e st ds ↔ X (s) = ∞ ∫ x ( t ) e − st dt −∞ 3 Definition of the Laplace Transform The Laplace transform of x(t) is defined as follows: X (s) +∞ ∫ x ( t ) e − st dt {s = σ + jω} −∞ Notice that s=jω (i.e. σ=0) brings back the Fourier transform solution. One way to understand the Laplace transform: the Laplace transform can be interpreted as the Fourier transform of the modified signal x(t)e-σt X (s) = +∞ ∫ −∞ ECSE303 Chap. 9 MR - 3/12/2010 x ( t ) e −( jω +σ )t dt = +∞ ∫ −∞ ⎡ x ( t ) e −σ t ⎤ e − jωt dt ⎣ ⎦ 4 Benefits of the Laplace transform over the Fourier transform The Laplace transform of an unbounded signal or of an unstable impulse response can be defined. The Laplace transform can be used to analyze differential LTI systems with nonzero initial conditions. ECSE303 Chap. 9 MR - 3/12/2010 5 X (s) Example Find the Laplace transform of x(t) = u(t) +∞ ∫ x ( t ) e − st dt −∞ +∞ X ( s) = u (t )e − st dt ∫ −∞ ∞ = ∫ e dt − st 0 1 = , Re{s} > 0 s ECSE303 Chap. 9 MR - 3/12/2010 6 X (s) X (s) = +∞ ∫e − at u ( t ) e dt − st −∞ ∫ x ( t ) e − st dt −∞ Example Find the Laplace transform of x ( t ) = e − at u ( t ) , a real +∞ x (t ) a<0 +∞ = e −( s + a )t dt ∫ 0 a>0 1 , Re {s + a} > 0 s+a 1 = , Re {s} > −a s+a = A specificity of the Laplace transform is the condition applied on s. This condition is often plotted on a complex s plane. ECSE303 Chap. 9 MR - 3/12/2010 7 For x ( t ) = e u ( t ) , a real − at LT ↔ X (s) = 1 , Re {s} > −a s+a the Laplace transform converges only for values of s anywhere in the open half-plane to the right of s = a Im {s Im{s}} = ω This half plane is called Region Of Convergence (ROC) of the Laplace transform. ECSE303 Chap. 9 MR - 3/12/2010 ROC -a Re{s} Re {s} = σ 8 Chapter 9: The Laplace transform 9.1 The Laplace transform 9.2 The region of convergence of the Laplace transform 9.3 The inverse Laplace transform 9.4 Geometric evaluation of the Fourier transform from the pole-zero plot 9.5 Properties of the Laplace transform 9.7 Analysis and characterization of LTI systems using the Laplace transform 9.8 System function algebra and block diagram representations 9.9 The unilateral Laplace transform NIB Application of Laplace Transforms to circuit analysis NIB Transform circuit diagrams 9.2 The ROC Recall that the Fourier transform of x(t ) = e − at u (t ), a ∈ R converges only for a>0 (i.e. decaying exponentials), In contrast, the Laplace transform converges for any value of a as long as we set Re{s}>-a (even for growing exponentials). For example 1 x (t ) = e u (t ) ↔ Re {s} > 2 s−2 The Laplace transform can converge even with unstable signals or systems. It is thus a useful tool ECSE303 Chap. 9 for unstable system analysis. 2t MR - 3/12/2010 LT 10 The convergence of the Laplace integral depends on the value of the real part of the complex Laplace variable, that is σ. As a consequence, the region of convergence in the complex plane is either: • The entire s plane, • a vertical half-plane (left or right), • a vertical strip, • or no convergence possible. ECSE303 Chap. 9 MR - 3/12/2010 11 . Example Find the Laplace transform of x ( t ) = e − at u ( −t ) , a real +∞ X (s) = ∫ e − at u (−t )e − st dt a>0 x (t ) −∞ 0 = ∫ e − ( s + a ) t dt −∞ a<0 1 =− , Re{s + a} < 0 s+a 1 =− , Re{s} < − a s+a ECSE303 Chap. 9 MR - 3/12/2010 12 LT x ( t ) = e u ( −t ) , a real ↔ − at X (s) = − 1 , Re{s} < −a s+a Im{s} This Laplace transform converges only in the ROC which is the open half-plane to the left of s=-a. ROC -a Re{s} Important note: The ROC is an integral part of a Laplace transform. It must be specified. Without it, the corresponding time-domain signal cannot be retrieved. ECSE303 Chap. 9 MR - 3/12/2010 13 If the signal is right-sided in time e.g. x ( t ) = e u ( t ) , a real − at LT ↔ 1 X (s) = , Re {s} > − a s+a then the ROC is to the right of the pole at s=-a If the signal is left-sided in time e.g. LT x ( t ) = e u ( −t ) , a real ↔ − at 1 , Re{s} < −a X ( s) = − s+a then the ROC is to the left of the pole at s=-a ECSE303 Chap. 9 MR - 3/12/2010 14 . Example Find the Laplace transform of x (t ) = e −t x ( t ) = e −t u ( t ) + et u ( −t ) = x1 ( t ) + x2 ( t ) 1 X 1 ( s) = Re {s} > −1 s +1 −1 X 2 (s) = Re {s} < 1 s −1 ECSE303 Chap. 9 MR - 3/12/2010 −2 X (s) = 2 1 > Re {s} > −1 s −1 15 . Example Find the Laplace transform of x ( t ) = u ( t ) − u ( t − t0 ) +∞ X (s) = ∫ −∞ {t ≤ t0 } ⎡u ( t ) − u ( t − t0 ) ⎤ e − st dt ⎣ ⎦ t0 = ∫ e − st dt 0 ( 1 = 1 − e − st0 s ECSE303 Chap. 9 MR - 3/12/2010 ) ∀S 16 . Example Find the Laplace transform of x (t ) = e − at ⎡ u ( t ) − u ( t − t0 ) ⎤ ⎣ ⎦ {t ≤ t0 ,a > 0} t0 X ( s ) = ∫ e − at e − st dt 0 ( 1 − ( s + a ) t0 1− e = s+a ) ∀S Proof that the function still converges when s=-a t0 X ( s ) = ∫ e − at e − st dt = t0 0 ECSE303 Chap. 9 MR - 3/12/2010 17 x (t ) Example Consider the signal x ( t ) = e −2 t u ( t ) + e t u ( − t ) Its Laplace transform is given by +∞ X (s) = ∫ e −2t u (t )e − st dt + −∞ +∞ = ∫ 0 t +∞ et u (−t )e− st dt ∫ −∞ e − ( s + 2)t dt + 0 e − ( s −1)t dt ∫ −∞ 1 1 = − , Re{s} > −2 and Re{s} < 1 s + 2 s −1 −3 =2 , − 2 < Re{s} < 1 s +s−2 ECSE303 Chap. 9 MR - 3/12/2010 18 The ROC is a vertical strip between σ=-2 and σ=1 Im{s} ROC -2 ECSE303 Chap. 9 MR - 3/12/2010 1 Re{s} 19 Example Consider the Laplace transform of the signal x ( t ) = e 2t u ( t ) +e − t u ( −t ) +∞ −∞ X ( s) = +∞ −∞ e 2t u (t )e − st dt + ∫ +∞ = e − ( s − 2)t dt + ∫ 0 0 ∫ ∫ e− t u (−t )e − st dt e− ( s +1)t dt −∞ 1 1 = {Re{s} > 2} − {Re{s} < −1} s−2 s +1 −3 =2 { No ROC} s +s−2 ECSE303 Chap. 9 MR - 3/12/2010 20 The region of convergence in the complex plane is either: • The entire s plane, When the signal has a finite duration • a vertical half-plane (left or right), When the signal is semi-infinite • a vertical strip, When the signal extends from –∞ to ∞ • or no convergence possible When the signal extends from –∞ to ∞, and has infinite energy ECSE303 Chap. 9 MR - 3/12/2010 21 Example Consider the signal x (t ) = 1 Does this function has a ROC? ECSE303 Chap. 9 MR - 3/12/2010 22 Suggested exercises 9.1-9.3, 9.21 (b, d, f), 9.22 (b, d, f) ECSE303 Chap. 9 MR - 3/12/2010 23 ECSE-303A Signals and Systems I Friday, March 12, 2010 LECTURE 21 Section 9.3 The Inverse Laplace Transform Is this yours? ECSE303 Chap. 9 MR - 3/12/2010 25 Example Problem 4.21f: Calculate the Fourier transform of the signal ⎡ sin (π t ) ⎤ ⎡ sin ( 2π ( t − 1) ) ⎤ ⎥ ⎢ ⎥⎢ ⎣ π t ⎦ ⎢ π ( t − 1) ⎥ ⎣ ⎦ ECSE303 Chap. 4 MR - 3/12/2010 26 ωc π x2 ( t ) X 2 ( jω ) FT x2 ( t ) ↔ X 2 ( jω ) 1 ω t −π ωc π ωc −ωc ωc ⎡ sin (π t ) ⎤ ⎡ sin ( 2π ( t − 1) ) ⎤ ⎥ = ⎡ sinc ( t ) ⎤ ⎡ 2 sin c ( 2t ) ∗ δ ( t − 1) ⎤ ⎢ ⎥⎢ ⎦⎣ ⎦ π t ⎦ ⎢ π ( t − 1) ⎥ ⎣ ⎣ ⎣ ⎦ FT ⎧ 1, ω < π δ ( t − 1) ↔ e − jω sinc ( t ) ↔ ⎨ ⎩0 , otherwise FT ⎧ 1, ω < 2π 2 sinc ( 2t ) ↔ ⎨ ⎩0 , otherwise FT ECSE303 Chap. 4 MR - 3/12/2010 27 1 ⎡ ⎤⎡ ⎤ ⎣ sinc ( t ) ⎦ ⎣ 2 sin c ( 2t ) ∗ δ ( t − 1) ⎦ ↔ 2π FT Sq4π ( Ω ) ⎡ e − jΩ , Ω < 2π ⎤ Sq4π ( Ω ) = ⎢ ⎥ 0 , otherwise ⎦ ⎣ 1 -2π -π π 2π Sq2π (ω − Ω ) 1 ω-π ECSE303 Chap. 4 MR - 3/12/2010 ω+π ⎡ 1, ω < π ⎤ ⎡ e − jω , ω < 2π ⎤ ⎥ ⎢ ⎥* ⎢ 0 , otherwise ⎦ ⎣ 0 , otherwise ⎦ ⎣ Ω ⎡1, ω − Ω < π ⎤ Sq2π (ω − Ω ) = ⎢ ⎥ 0 , otherwise ⎦ ⎣ Ω 28 Pattern 1: no overlap Pattern 2: partial overlap ω-π ω+π -2π ω +π ∫π −2 -3π<ω<-π -π π Sq4π ( Ω ) Sq2π (ω − Ω )d Ω = Ω 2π ω +π ∫π e − jΩ d Ω −2 e − j ( ω +π ) − e j 2π = −j ( ) = j −e − jω − 1 ECSE303 Chap. 4 MR - 3/12/2010 29 -π<ω<π Pattern 3: full overlap ω-π -2π ω +π ∫π −2 -π ω+π π 2π Sq4π ( Ω ) Sq2π (ω − Ω )d Ω = ω +π ∫π ω Ω e − jΩ d Ω − e − j ( ω + π ) − e − j ( ω −π ) = −j ( = j −e − jω + e − jω ) =0 ECSE303 Chap. 4 MR - 3/12/2010 30 Pattern 4: partial overlap π<ω<3π ω-π -2π ω +π ∫π −2 -π π Sq4π ( Ω ) Sq2π (ω − Ω )d Ω = ω+π Ω 2π 2π ∫π ω e − jΩ d Ω − e − j ( 2 π ) − e − j ( ω −π ) = −j ( = j 1 + e − jω ECSE303 Chap. 4 MR - 3/12/2010 ) 31 1 ⎡ ⎤⎡ ⎤ ⎣ sinc ( t ) ⎦ ⎣ 2 sin c ( 2t ) ∗ δ ( t − 1) ⎦ ↔ 2π FT ⎡ 1, ω < π ⎤ ⎡ e − jω , ω < 2π ⎤ ⎥ ⎢ ⎥* ⎢ 0 , otherwise ⎦ ⎣ 0 , otherwise ⎦ ⎣ X ( jω) 1 − jω × e 2π 2π -3π ECSE303 Chap. 4 MR - 3/12/2010 -π π 3π ω 32 Chapter 9: The Laplace transform 9.1 The Laplace transform 9.2 The region of convergence of the Laplace transform 9.3 The inverse Laplace transform 9.4 Geometric evaluation of the Fourier transform from the pole-zero plot 9.5 Properties of the Laplace transform 9.7 Analysis and characterization of LTI systems using the Laplace transform 9.8 System function algebra and block diagram representations 9.9 The unilateral Laplace transform NIB Application of Laplace Transforms to circuit analysis NIB Transform circuit diagrams 9.3 Inverse Laplace Transform We have seen that the Laplace transform is equivalent to a Fourier transform of the signal x(t)e-σt X (s) = ∞ ∫ −∞ ⎡ x ( t ) e −σ t ⎤ e − jω t dt ⎣ ⎦ The corresponding inverse Fourier transform is ∞ therefore 1 jω t −σ t x (t ) e = 2π ∫ X (s)e dω −∞ Multiplying both sides of the equation by eσt 1 x (t ) = 2π x (t ) ECSE303 Chap. 9 MR - 3/12/2010 1 ∞ ∫ X ( s ) e(σ + jω )t d ω −∞ σ + j∞ ∫ 2π j σ − j∞ X ( s ) e st ds s = σ + jω ds = jd ω {σ cons tan t} 34 The mathematical definition of the inverse Laplace transform is given by a contour integral x (t ) 1 σ + j∞ ∫ 2π j σ − j∞ X ( s ) e st ds The variable σ is chosen to ensure the convergence of the integral As it is, this integral form is rarely used for our applications and in most cases it is faster to use the tables determined from Laplace transformation of x(t) ECSE303 Chap. 9 MR - 3/12/2010 35 ECSE303 Chap. 9 MR - 3/12/2010 36 Prove them ECSE303 Chap. 9 MR - 3/12/2010 37 Poles and Zeros of Rational Laplace Transforms Most signals in the time domain turn into rational Laplace transforms. That is, they become ratios of polynomials. N (s) A0 + A1s + A2 s 2 + = X (s) = D ( s ) B0 + B1s + B2 s 2 + • The zeros of the numerator N(s) are called the zeros of the Laplace transform. If z1 is a zero, then X(z1)=0. • The zeros of the denominator D(s) are called the poles of the Laplace transform. If p1 is a pole, then X(p1)=∞. ECSE303 Chap. 9 MR - 3/12/2010 38 Example 2s + 1 X (s) = 2 s +s−2 2 ( s + 0. 5 ) 2s + 1 X (s) = 2 = s + s − 2 ( s + 2 )( s − 1) has a zero at s=z1=-0.5 and poles at s=p1=1 and s=p2=-2 ECSE303 Chap. 9 MR - 3/12/2010 39 Partial Fraction Expansion-Standard procedure Assuming m 1. No multiple-order poles in the set of poles { pk }k =1 of the rational transform X(s) 2. The order of the denominator polynomial is greater than the order of the numerator polynomial Then we can expand X(s) in the sum of first order polynom form N A X (s) = ∑ k =1 k s − pk From the ROC of X(s), the ROC of each of the individual terms can be calculated, and then the inverse transform of each of these terms can be determined ECSE303 Chap. 9 MR - 3/12/2010 40 N Ak A1 A2 X (s) = ∑ = + + s − p1 s − p2 k =1 s − pk s∈ROC s∈ROC1 s∈ROC2 AN + s − pN s∈ROC N From this point, a table of Laplace transform pairs leads to the time-domain signal x(t). The ROC of X(s) must at least contain the intersection of all the ROCs of the fractions ROC ⊇ ∩ ROCi i =1,…, N ECSE303 Chap. 9 MR - 3/12/2010 41 If the ROC is to the right of the pole at s=pi, then the inverse transform of this term is Ai e pi t u (t ) i.e., a right-sided signal. If the ROC is to the left of the pole at s=pi, then the inverse transform of this term is − Ai e u ( −t ) pi t i.e., a left-sided signal. ECSE303 Chap. 9 MR - 3/12/2010 42 Example Calculate the inverse of the following Laplace transform: s+3 X (s) = , 0 < Re{s} < 2 s ( s + 1)( s − 2) A3 A1 A2 = + + s +1 s s−2 s∈ROC1 s∈ROC2 s∈ROC3 In order to have ROC ⊇ ROC1 ∩ ROC2 ∩ ROC3 the only possible solution is X ( s) = A1 + A2 + A3 s +1 s s−2 Re{ s }>−1 ECSE303 Chap. 9 MR - 3/12/2010 Re{ s }> 0 Re{ s }< 2 43 A A1 A s+3 = + 2+ 3 s( s + 1)( s − 2) s + 1 s s−2 We calculate the coefficient A1 Re{ s }>−1 Re{ s}> 0 Re{ s}< 2 ⎛ ⎞ ( s + 3) 2 2 A1 = ( s + 1) ⎜ = = ⎟ ⎝ s ( s + 1)( s − 2) ⎠ s =−1 (−1)( −3) 3 Then, we obtain A2 ⎛ ⎞ ( s + 3) 3 3 =− A2 = s ⎜ ⎟= 2 ⎝ s ( s + 1)( s − 2) ⎠ s =0 (1)(−2) and A3 ⎛ ⎞ ( s + 3) 5 5 = = A3 = ( s − 2) ⎜ ⎟ s ( s + 1)( s − 2) ⎠ s = 2 (2)(3) 6 ⎝ ECSE303 Chap. 9 MR - 3/12/2010 44 X ( s) = s+3 , 0 < Re{s} < 2 s ( s + 1)( s − 2) Hence, the Laplace transform can be expanded as 21 31 51 X (s) = − + 3 s +1 2 s 6 s−2 Re{ s }>−1 Re{ s }> 0 Re{ s }< 2 And from 2 −t 3 5 2t x(t ) = e u (t ) − u (t ) − e u (−t ) 3 2 6 ECSE303 Chap. 9 MR - 3/12/2010 45 Partial Fraction Expansion-Second-Order Poles For a problem with a double pole (r=2) X (s) = = ( s − p1 ) N (s) ( s − pm −1 )( s − pm ) 2 ( s − pm +1 ) A1 + s − p1 Am −1 s − pm −1 ( s − pN ) Am Am +1 + + s − pm ( s − pm )2 + Am + r + s − pm +1 + AN s − pN Most coefficients are calculated from residues at poles and Am can be calculated multiplying both sides of the partial fraction equation by s and letting s→∞. This yields Am = [ sX ( s)]s →∞ − ( A1 + A2 + + Am−1 + Am+ 2 + + AN ) ECSE303 Chap. 9 MR - 3/12/2010 46 Example A3 A1 A2 s−2 X ( s) = = + + 2 2 ( s + 1)( s + 3) s + 3 ( s + 3) s +1 . s−2 5 A2 = = s + 1 s =−3 2 s−2 A3 = ( s + 3)2 ECSE303 Chap. 9 MR - 3/12/2010 3 =− 4 s =−1 s−2 = A1 + A3 ( s + 1)( s + 3) s →∞ 0 = A1 − 3 4 3 A1 = + 4 47 Not sure of the answer? Just use a proof A3 A1 A2 s−2 = + + 2 2 ( s + 1)( s + 3) s + 3 ( s + 3) s + 1 31 51 31 = + − 2 4 s + 3 2 ( s + 3) 4 s + 1 s−2 = ( s + 1)( s + 3) 2 ECSE303 Chap. 9 MR - 3/12/2010 48 Partial Fraction Expansion-Multiple-Order Poles Assuming m 1. Multiple-order poles in the set of poles { pk }k =1 of the rational transform X(s) 2. The order of the denominator polynomial is greater than the order of the numerator polynomial Consider X(s) with one multiple pole pm of multiplicity r, that is, X (s) Re{ s}> Re{ pm+1 } ECSE303 Chap. 9 MR - 3/12/2010 = ( s − p1 ) n( s ) ( s − pm −1 )( s − pm ) r ( s − pm +1 ) ( s − pN ) 49 X (s) = ( s − p1 ) n( s ) ( s − pm −1 )( s − pm ) r ( s − pm +1 ) A1 + X (s) = s − p1 Am −1 + s − pm −1 Am Am +1 + + + 2 s − pm ( s − pm ) Am + r + + s − pm +1 ECSE303 Chap. 9 MR - 3/12/2010 ( s − pN ) + Am + r −1 ( s − pm ) r AN + s − pN 50 After multiplication by (s-pm)r on both sides N ( s) ( s − p1 ) ( s − pm −1 )( s − pm ) r ( s − pm +1 ) ( s − pN ) = A1 ( s − pm ) r s − p1 + Am ( s − pm ) + + r −1 Am + r ( s − pm ) s − pm +1 + Am +1 ( s − pm ) r + + r −2 + AN ( s − pm ) + Am + r −1 r s − pN All terms on the right-hand side vanish upon letting s=pm, except the term Am+r-1. The remaining coefficients Am, Am+1,… are calculated by first derivative, second derivative… ECSE303 Chap. 9 with respect to s and evaluation at s=pm MR - 3/12/2010 51 Example X (s) ( s − 1) ( s + 1) = = ( s + 3)3 A B C + + ( s + 3)3 ( s + 3) 2 s + 3 Multiplying both sides by (s+3)3 ( s − 1) ( s + 1) = A + B( s + 3) + C ( s + 3)2 ( −3 − 1)( −3 + 1) = 8 = A Taking the first derivative of X(s)(s+3)3 with respect to s 2s = B + 2C ( s + 3) −6 = B And the second derivative X ( s ) 2 = 2C 1= C ECSE303 Chap. 9 MR - 3/12/2010 ( s − 1) ( s + 1) = ( s + 3)3 8 −6 1 = + + ( s + 3)3 ( s + 3) 2 s + 3 52 Example s2 A B C D X (s) = = + + + 3 3 2 ( s − 2)( s + 3) ( s + 3) ( s + 3) s+3 s−2 Multiplying both sides by (s-2) s2 A( s − 2) B( s − 2) C ( s − 2) = + + +D 3 3 2 s+3 ( s + 3) ( s + 3) ( s + 3) ⎡ s2 ⎤ =D ⎢ 3⎥ ⎣ ( s + 3) ⎦ s =−2 4=D ECSE303 Chap. 9 MR - 3/12/2010 53 s2 A B C D = + + + 3 3 2 s+3 s−2 ( s − 2)( s + 3) ( s + 3) ( s + 3) Multiplying both sides by (s+3)3 and setting s=-3 s2 ( s + 3)3 = A + B( s + 3) + C ( s + 3) 2 + D ( s − 2) ( s − 2) −9 =A 5 Taking the first derivative of X(s)(s+3)3 with respect to s ⎡ ( s + 3) 2 ( s + 3)3 ⎤ 2s s2 − = B + C ( s + 3) + D ⎢ − ⎥ 2 ( s − 2) ( s − 2) ( s − 2) ( s − 2) 2 ⎦ ⎣ 6 9 21 − = =B 5 25 25 And so on for C ECSE303 Chap. 9 MR - 3/12/2010 54 Complex conjugate poles A pair of complex conjugate poles in a partial fraction expansion express a sinus and/or cosinus modulation with an exponential envelope. For example X ( s ) = ( A + Bs ) 1 1 s + (α − jω0 ) s + (α + jω0 ) (where A and B are arbitrary constants) has poles and express s = −α ± jω0 ω0 , Re{s} > −α 2 2 ( s + α ) + ω0 LT s +α −α t e cos(ω0t ) ↔ , Re{s} > −α 2 2 ( s + α ) + ω0 e ECSE303 Chap. 9 MR - 3/12/2010 −α t LT sin(ω0t ) ↔ 55 As a result, a pair of complex conjugate poles in a partial fraction expansion is dealt with by expressing 1 1 X ( s ) = ( A + Bs ) s + (α − jω0 ) s + (α + jω0 ) With A ' ω0 + B '( s + α ) X (s) = ( s + α ) 2 + ω0 2 to enable a straightforward use of the damped or amplified sinusoids in a table of Laplace transforms ω0 e sin(ω0t ) ↔ , Re{s} > −α ( s + α ) 2 + ω0 2 LT s +α −α t e cos(ω0t ) ↔ , Re{s} > −α 2 2 ( s + α ) + ω0 −α t ECSE303 Chap. 9 MR - 3/12/2010 LT 56 Example Compute the inverse Laplace transform of 2 s 2 + 3s − 2 X ( s) = 2 , Re{s} > 0 s + 2s + 4 s ( ) First, we search for the poles of the function s 2 + 2s + 4 = 0 and get thus s1 = −1 + j 3 s2 = −1 − j 3 s 2 + 2 s + 4 = ( s + 1) 2 + ( 3) 2 ECSE303 Chap. 9 MR - 3/12/2010 57 The transform X(s) can be expanded as follows: 2 s 2 + 3s − 2 X ( s) = 2 s + 2s + 4 s ( = ) A3 ( s + 1) 2 +3 Re{ s}>−1 + B ( s + 1) ( s + 1) 2 +3 Re{ s}>−1 C s + Re{ s}> 0 We find C as a residue at pole s=0, after multiplication by s ⎡ 2 s 2 + 3s − 2 ⎤ −1 = =C ⎢2 ⎥ ⎣ s + 2 s + 4 ⎦ s =0 2 Now, set s=-1 to calculate A ECSE303 Chap. 9 MR - 3/12/2010 1 3 −3 1 X ( 1) = = A+ ⇒ A = 2 2 −3 3 58 and multiply both sides of X(s) by s and let s→∞ to get B=5/2 Then we have the following expansion: 3 5 ( 3) ( s + 1) 12 X ( s) = 2 2 +2 2 − ( s + 1) + 3 ( s + 1) + 3 s Re{ s}>−1 Re{ s}>−1 Re{ s}> 0 Taking the inverse Laplace transform ⎡ 3 −t ⎤ 5 −t 1 x(t ) = ⎢ e sin( 3t ) + e cos( 3t ) ⎥ u (t ) − u (t ) 2 2 ⎣2 ⎦ ECSE303 Chap. 9 MR - 3/12/2010 59 Suggested exercises 9.1-9.6 ECSE303 Chap. 9 MR - 3/12/2010 60 Mid term II Friday room 0100 18:00-20:00 Chapters 1-4 See exams from previous years No book or spreadsheet Standard calculators only Office hours Friday 13:30-15:30 (no class) ECSE303 Chap. 9 MR - 3/12/2010 61 ECSE-303A Signals and Systems I Friday, March 12, 2010 LECTURE 23 Section 9.4: Geometric evaluation of the Fourier transform from the pole-zero plot Chapter 9: The Laplace transform 9.1 The Laplace transform 9.2 The region of convergence of the Laplace transform 9.3 The inverse Laplace transform 9.4 Geometric evaluation of the Fourier transform from the pole-zero plot 9.5 Properties of the Laplace transform 9.7 Analysis and characterization of LTI systems using the Laplace transform 9.8 System function algebra and block diagram representations 9.9 The unilateral Laplace transform NIB Application of Laplace Transforms to circuit analysis NIB Transform circuit diagrams 9.4 Geometric evaluation of the Fourier transform from the pole-zero plot The transfer function H(s) can be converted into a frequency response H(jω) if the system is stable, that is, if Re{s}=0 is included in the ROC The poles and zeros of H(s) then convert into specific features of the frequency response In this section, we aim at understanding the frequency response of a system from its pole-zero plot. ECSE303 Chap. 9 MR - 3/12/2010 64 Consider a stable transfer function written in its polezero format H (s) = ( s − zm ) s→ jω H jω = K ( jω − z1 ) ( jω − zm ) ⇒ () ( s − p1 ) ( s − pn ) ( jω − p1 ) ( jω − pn ) K ( s − z1 ) When s=jω, each first-order polynomial represents a vector whose origin is at the zero (z1, zm) or pole (p1, pm) and its endpoint is at jω. For a fixed frequency ω, each of these vectors adds a phase and magnitude contribution to the overall frequency response. ECSE303 Chap. 9 MR - 3/12/2010 65 Example Consider the first-order system with transfer function 1 H (s) = , Re{s} > −2 s+2 s + 2 → jω + 2 It has a frequency response 1 H ( jω ) = jω + 2 In the s-plane, the complex denominator jω+2 taken individually is as a vector-valued function of ω ECSE303 Chap. 9 MR - 3/12/2010 66 as ω is varied from –∞ to 0… …the magnitude of 2+jω (the vector's length) varies from ∞ to 2. …the phase of 2+jω varies from –π/2 to 0. Thus the magnitude of H(jω)=1/(2+jω) varies from 0 to 0.5 while its phase varies from π/2 to 0. ECSE303 Chap. 9 MR - 3/12/2010 67 as ω is varied from 0 to ∞… …the magnitude of 2+jω (the vector's length) varies from 2 to ∞. …the phase of 2+jω varies from 0 to π/2. Thus the magnitude of H(jω)=1/(2+jω) varies from 0.5 to 0 while its phase varies from 0 to -π/2. ECSE303 Chap. 9 MR - 3/12/2010 68 1 H ( jω ) = jω + 2 H ( jω ) = ECSE303 Chap. 9 MR - 3/12/2010 1 ω2 + 4 e ⎛ −ω ⎞ j arctan ⎜ ⎟ ⎝2⎠ 69 Systems of higher order With higher-order transfer functions, the vector representation of each first-order pole and zero factor is taken individually into contribution to the overall frequency response. For this purpose, the function is expressed as a ratio of complex phasors K ( jω − z1 ) ( jω − zm ) H ( jω ) = ( jω − p1 ) ( jω − pn ) jθ1 jθ 2 N1e N 2 e … ⇒ H ( jω ) = K jφ1 jφ2 D1e D2 e … ECSE303 Chap. 9 MR - 3/12/2010 N1 N 2 … H ( jω ) = K D1 D2 … H ( jω ) = θ1 + θ 2 … − (φ1 + φ2 …) 70 N1e jθ1 N 2 e jθ2 … H ( jω ) = K D1e jφ1 D2 e jφ2 … The magnitude at frequency ω is given by the product of the lengths of the vectors originating at zeros divided by the product of the lengths of the vectors originating at poles. N1 N 2 … H ( jω ) = K D1 D2 … The phase at frequency ω is given by the sum of the angles of the vectors originating at zeros minus the sum of the angles of the vectors originating at poles. ∠H ( jω ) = θ1 + θ 2 … − (φ1 + φ2 …) ECSE303 Chap. 9 MR - 3/12/2010 71 Example Plot the frequency response of the complex impulse response h(t) s+2 , Re{s} > −2 H (s) = s + (2 − 2 j) jω + 2 H ( jω ) = jω + ( 2 − 2 j ) ECSE303 Chap. 9 MR - 3/12/2010 72 Im{s} N1e jθ1 jω + 2 = H ( jω ) = jω + ( 2 − 2 j ) D1e jφ1 (-2,2j) ω Re{s} -2 |H(jω)| ∠H(jω) ∞/∞=1 ∞ π/2-(π/2)=0 √8/2=√2 2 π/4-(0)=π/4 1 1 0.46-(-0.46)=0.93 2/√8=1/√2 0 0-(-π/4)=π/4 ∞/∞=1 ECSE303 Chap. 9 MR - 3/12/2010 ω -∞ −π/2-(-π/2)=0 73 N1e jθ1 jω + 2 = H ( jω ) = jω + ( 2 − 2 j ) D1e jφ1 ECSE303 Chap. 9 MR - 3/12/2010 74 Example Consider a typical second-order system with transfer function H ( s) = s +α (s +α ) +ω 2 2 0 , Re{s} > −α It has a frequency response H ( jω ) = ECSE303 Chap. 9 MR - 3/12/2010 jω + α ( jω + α ) + ω 2 2 0 (−α,ω0) −α Im{s} ω Re{s} (−α,−ω0) 75 H ( jω ) = (−α,ω0) jω + α ( jω + α ) + ω02 Im{s} ω 2 −α α=2, ω0=100 rad/sec Re{s} (−α,−ω0) |H(jω)| ω ∠H(jω) ∞/(∞∞)=0 ∞ π/2-(π/2+π/2)=-π/2 √(α2+ω02)/[α√(α2+4ω02)]=0.26 ω0 atn(ω0/α)-[0+atn(2ω0/α)]=-0.01 α/(α2+ω02)=2e-4 0 0-(atn(ω0/α)-atn(ω0/α))=0 √(α2+ω02)/[α√(α2+4ω02)]=0.26 -ω0 -atn(ω0/α)-[-atn(2ω0/α)+0]=0.01 ∞/(∞∞)=0 -∞ −π/2-(-π/2-π/2)=π/2 ECSE303 Chap. 9 MR - 3/12/2010 76 H ( jω ) = α=2, ω0=100 rad/sec ( jω + α ) + ω02 (−α,ω0) α +ω α α 2 + 4ω 1 2 jω + α 2 0 2 0 −α 2 Im{s} ω Re{s} (−α,−ω0) −ω0 ECSE303 Chap. 9 MR - 3/12/2010 ω0 77 Example A third-order system with transfer function H (s) = ( s + 1)( s − 2) ( s + 3)( s + 2 − j 2)( s + 2 + j 2) , Re{s} > − 2 has frequency response ( jω + 1)( jω − 2) H ( jω ) = ( jω + 3)( jω + 2 − j 2 )( jω + 2 + j 2 ) ECSE303 Chap. 9 MR - 3/12/2010 78 H ( jω ) = ECSE303 Chap. 9 MR - 3/12/2010 ( jω + 1)( jω − 2) ( jω + 3)( jω + 2 − j 2 )( jω + 2 + j 2 ) 79 ( jω + 1)( jω − 2) H ( jω ) = ( jω + 3)( jω + 2 − j 2 )( jω + 2 + j 2 ) ECSE303 Chap. 9 MR - 3/12/2010 80 Suggested exercises 9.7, 9.10, 9.11, 9.23 ECSE303 Chap. 9 MR - 3/12/2010 81 ECSE-303A Signals and Systems I Friday, March 12, 2010 LECTURE 24 Section 9.5: Properties of the Laplace transform Chapter 9: The Laplace transform 9.1 The Laplace transform 9.2 The region of convergence of the Laplace transform 9.3 The inverse Laplace transform 9.4 Geometric evaluation of the Fourier transform from the pole-zero plot 9.5 Properties of the Laplace transform 9.7 Analysis and characterization of LTI systems using the Laplace transform 9.8 System function algebra and block diagram representations 9.9 The unilateral Laplace transform NIB Application of Laplace Transforms to circuit analysis NIB Transform circuit diagrams 9.5 Properties of the Laplace Transform Linearity The Laplace transform is linear If LT x1 (t ) ↔ X 1 ( s ), s ∈ ROC1 LT x2 (t ) ↔ X 2 ( s ), s ∈ ROC 2 then LT ax1 (t ) + bx2 (t ) ↔ aX 1 ( s ) + bX 2 ( s ), s ∈ ROC = ROC1 ∩ ROC2 ECSE303 Chap. 9 MR - 3/12/2010 84 Time-Shifting LT If x(t ) ↔ X ( s ), s ∈ ROC LT Then x(t − t0 ) ↔ e − st0 X ( s ), s ∈ ROC Example (linearity and time-shifting) Unit pulse of duration T h (t ) = u (t ) − u (t − T ) ECSE303 Chap. 9 MR - 3/12/2010 1 1 − sT H (s) = − e ss 1 − e − sT = , ∀s ∈ C s 85 COMMENT The unit pulse of duration T is made h (t ) = u (t ) − u (t − T ) of two signal components, each 1 1 − sT having ROC = Re{s}>0 H (s) = The new ROC results from a polezero cancellation. Using a Taylor expansion e − sT −e ss 1 − e − sT , ∀s ∈ C = s = 1 − sT + ( sT ) 2 ! − ( sT ) 3 ! 2 3 1 − e − sT H (s) = s sT − ( sT ) 2 ! + ( sT ) 3 ! 2 = ECSE303 Chap. 9 MR - 3/12/2010 3 s Having no poles, the new ROC is the s plane 86 Shifting in the s-Domain LT If x(t ) ↔ X ( s ), s ∈ ROC then e x(t ) ↔ X ( s − s0 ), s ∈ ROC+Re{s 0 } s0t LT the new ROC equals the original one shifted by Re{s0}, to the right if this number is positive, to the left otherwise. ECSE303 Chap. 9 MR - 3/12/2010 87 LT e x (t ) ↔ X ( s − s0 ), s ∈ ROC+Re{s 0 } s0t Example s From cos(bt ) ↔ 2 s + b2 LT Re {s} > 0 if we are looking for the Laplace transform of e − at cos(bt ) then e − at LT cos(bt ) ↔ ECSE303 Chap. 9 MR - 3/12/2010 s+a (s + a) 2 +b 2 Re {s + a} > 0, or Re {s} > −a 88 Time-Scaling LT If x(t ) ↔ X ( s ), s ∈ ROC then 1 s x(at ) ↔ X ( a ), s ∈ aROC a LT where the new ROC is the original one, expanded or contracted by |a|, and flipped around the imaginary axis if a<0. ECSE303 Chap. 9 MR - 3/12/2010 89 Example LT x(−2t ) ↔ 0.5 X (−0.5s ), s ∈ −2ROC a=-2 1 e u (t ) ↔ , Re {s} > 1 s −1 LT −1 ⎛1⎞ 1 −2 t = e u ( −2t ) ↔ ⎜ ⎟ , Re {s} < −2 ⎝ 2 ⎠ −s 2 − 1 s + 2 t Example a=-2 LT Im{s} ROC X ( s ) ROC -2 ECSE303 Chap. 9 MR - 3/12/2010 Im{s} 1 Re{s} -2ROC -2 ROC X (−0.5s) 4 Re{s} 90 Conjugation LT If then x(t ) ↔ X ( s ), s ∈ ROC LT ∗ x (t ) ↔ X ∗ ( s∗ ), s ∈ ROC Thus for x(t) real, X(s) = X*(s*) If x(t) is real and thus has a pole (or zero) at s=s0, then X(s) must also have a pole (or zero) at the complex conjugate point s=s0*. The complex poles and zeros of the Laplace transform of a real signal always come in conjugate pairs. e e ECSE303 Chap. 9 MR - 3/12/2010 jω0t 1 , Re {s} > 0 u (t ) ↔ s − jω0 − jω0t LT LT u (t ) ↔ 1 , Re {s} > 0 s + jω0 91 Example The damped/amplified sine ω0 e sin(ω0t ) ↔ Re{s} > −α 2 2 ( s + α ) + ω0 LT ω0 −α t e sin(ω0t ) ↔ Re{s} > −α ( s + α + jω )( s + α − jω ) −α t LT has poles s = -α ± jω0 ECSE303 Chap. 9 MR - 3/12/2010 92 Convolution LT If x1 (t ) ↔ X 1 ( s ), s ∈ ROC1 LT x2 (t ) ↔ X 2 ( s), s ∈ ROC2 then LT x1 (t ) ∗ x2 (t ) ↔ X 1 ( s ) X 2 ( s ), s ∈ ROC ⊇ ROC1 ∩ ROC2 Note that the resulting ROC includes the intersection of the two original ROC's, but in some special cases, the ROC may become larger as a result of a polezero cancellation. ECSE303 Chap. 9 MR - 3/12/2010 93 Example The response of the LTI system with impulse response h ( t ) = ⎡ e −2t + e − t ⎤ u ( t ) ⎣ ⎦ to the input x ( t ) = − e −2 t u ( t ) + δ ( t ) is given by the product of their Laplace transform Y(s): LT 2s + 3 h(t ) ↔ H ( s ) = , Re{s} > −1 ( s + 2)( s + 1) −1 x(t ) ↔ X ( s ) = + 1, Re{s} > −2 s+2 s +1 , Re{s} > −2 = s+2 LT ECSE303 Chap. 9 MR - 3/12/2010 94 Y (s) = H (s) X (s) (2 s + 3) ( s + 1) = ( s + 2)( s + 1) ( s + 2) s : {Re{s} > −2} ∩ {Re{s} > −1} = {Re{s} > −1} Actually, this is a special case: although the intersection of the two input ROCs is Re{s}>-1, the output ROC becomes instead Re{s}>-2 due a polezero cancellation (2 s + 3) Y ( s) = , Re{s} > −2 2 ( s + 2) By partial fractions and inverse transformation y (t ) = 2e −2 t − te −2 t u( t ) ECSE303 Chap. 9 MR - 3/12/2010 95 To get a better understanding of the meaning of a pole-zero cancellation, let’s solve the same problem by convolution. Then h ( t ) = ⎡ e −2t + e − t ⎤ u ( t ) ⎣ ⎦ x ( t ) = −e u ( t ) + δ ( t ) −2 t = x1 ( t ) + x2 ( t ) ⇒ y1 ( t ) = ⎡ e −2t − te −2t − e − t ⎤ u ( t ) ⎣ ⎦ y 2 ( t ) = h ( t ) = ⎡ e −2 t + e − t ⎤ u ( t ) ⎣ ⎦ The pole-zero cancellation has its origin from a cancellation of two exponential signal components y (t ) = 2e −2 t − te −2 t u( t ) ECSE303 Chap. 9 MR - 3/12/2010 96 Differentiation in the Time Domain LT x(t ) ↔ X ( s ), s ∈ ROC dx(t ) LT ↔ sX ( s ), s ∈ ROC1 ⊇ ROC dt Again, ROC1 can be larger than ROC when there is a pole-zero cancellation at s=0 . Example LT The ramp function x ( t ) = tu ( t ) ↔ X ( s ) = 1 , Re {s} > 0 s2 LT dx ( t ) 1 = u ( t ) ↔ X ( s ) = , Re {s} > 0 dt s LT d 2 x (t ) ECSE303 Chap. 9 = δ ( t ) ↔ X ( s ) = 1, ∀s 2 MR - 3/12/2010 dt 97 X (s) Differentiation in the S Domain +∞ ∫ x ( t ) e − st dt −∞ LT x(t ) ↔ X ( s ), s ∈ ROC LT dX ( s ) tx(t ) ↔− , s ∈ ROC ds Example 1 e u (t ) ↔ Re {s} > − a s+a LT 1 − at te u ( t ) ↔ Re {s} > − a 2 (s + a) − at . LT LT t e u (t ) ↔ 2 − at ECSE303 Chap. 9 MR - 3/12/2010 2 (s + a) 3 Re {s} > − a 98 From this, we find a new identity LT LT 1 1 − at − at e u (t ) ↔ e u (t ) ↔ s+a s+a LT LT 1 1 − at − at te u ( t ) ↔ te u ( t ) ↔ 2 2 s + a) ( (s + a) LT t 2 e − at u ( t ) ↔ LT t 3e − at u ( t ) ↔ . ECSE303 Chap. 9 MR - 3/12/2010 2 (s + a) 3 6 (s + a) 4 ⇒ LT t 2 − at 1 e u (t ) ↔ 3 2 (s + a) LT t 3 − at 1 e u (t ) ↔ 4 6 (s + a) LT t n −1 1 − at e u (t ) ↔ n ( n − 1) ! (s + a) 99 Integration in the Time Domain LT x(t ) ↔ X ( s ), s ∈ ROC LT 1 t ∫−∞ x(τ )dτ ↔ s X (s), s ∈ ROC1 ⊇ ROC ∩ {Re{s}>0} Example the unit step response of an LTI system being the running integral of its impulse response: s ( t ) = ∫ h (τ ) dτ t −∞ 1 S (s) = H (s) s ECSE303 Chap. 9 MR - 3/12/2010 with the appropriate ROC 100 The Initial and Final Value Theorems Assuming x(t<0)=0, these theorems apply only to signals with finite power at t=0 (no impulse function or higher order discontinuities) and finite energy. . The initial-value theorem states that () x 0+ = lim sX ( s ) s →+∞ and the final-value theorem states that lim x ( t ) = lim sX ( s ) t →+∞ s →0 Let’s prove them ECSE303 Chap. 9 MR - 3/12/2010 101 The Final Value Theorem We consider the Laplace transform of the derivative Integration by parts of x(t) ∞ dx ( t ) − st ⎡ dx ( t ) ⎤ =∫ LT ⎢ e dt = sX ( s ) − x ( 0− ) ⎥ dt ⎦ 0− dt ⎣ We take the limit of this equality as s approaches zero ⎡∞ ⎤ dx ( t ) − st Lim ⎢ ∫ e dt ⎥ = Lim ⎡ sX ( s ) − x ( 0− ) ⎤ ⎣ ⎦ s →0 ⎢ 0− dt ⎥ s →0 ⎣ ⎦ dx ( t ) dt = Lim ⎡ sX ( s ) − x ( 0− ) ⎤ ∫− dt ⎦ s →0 ⎣ 0 ∞ Lim ⎡ x ( t ) ⎤ − x ( 0− ) = Lim ⎡ sX ( s ) − x ( 0− ) ⎤ ⎣ ⎦ ⎣ ⎦ t →∞ ECSE303 Chap. 9 MR - 3/12/2010 s →0 Lim ⎡ x ( t ) ⎤ = Lim ⎡ sX ( s ) ⎤ ⎣ ⎦ ⎣ ⎦ t →∞ s →0 102 The Initial Value Theorem We consider the Laplace transform of the derivative of x(t) ⎡ dx ( t ) ⎤ ∞ dx ( t ) − st =∫ LT ⎢ e dt = sX ( s ) − x ( 0− ) ⎥ dt ⎦ 0− dt ⎣ Consider two cases: x(t) continuous or discontinuous at t=0 Case 1: x(t) continuous at t=0 x(0-)=x(0+) We take the limit of this equality as s approaches ⎡ ∞ dx ( t ) − st ⎤ infinity Lim ⎢ ∫ e dt ⎥ = Lim ⎡ sX ( s ) − x 0− ⎤ ⎦ s →∞ s →∞ ⎣ dt − ⎢0 ⎥ ⎣ ⎦ () 0 = Lim ⎡ sX ( s ) − x ( 0− ) ⎤ ⎣ ⎦ s →∞ ECSE303 Chap. 9 MR - 3/12/2010 Lim ⎡ sX ( s ) ⎤ = x ( 0− ) = x ( 0+ ) ⎣ ⎦ s →∞ 103 The Initial Value Theorem Case 2: x(t) has a finite discontinuity (step) at t=0 dx(t)/dt=[x(0+)-x(0-)]δ(t) ⎡ dx ( t ) ⎤ ∞ dx ( t ) − st =∫ LT ⎢ e dt = sX ( s ) − x ( 0− ) ⎥ dt ⎦ 0− dt ⎣ We take the limit of this equality as s approaches infinity ⎡ ∞ dx ( t ) − st ⎤ Lim ⎢ ∫ e dt ⎥ = Lim ⎡ sX ( s ) − x ( 0− ) ⎤ ⎣ ⎦ s →∞ ⎢ 0− dt ⎥ s →∞ ⎣ ⎦ + ∞ ⎡0 dx ( t ) − st ⎤ + − − st Lim ⎢ ∫ ⎡ x ( 0 ) − x ( 0 ) ⎤ δ ( t ) e dt + ∫ e dt ⎥ = Lim ⎡ sX ( s ) − x ( 0− ) ⎤ ⎣ ⎦ ⎦ s →∞ s →∞ ⎣ dt − + ⎢0 ⎥ 0 ⎣ ⎦ Lim ⎡ x ( 0+ ) − x ( 0− ) ⎤ = Lim ⎡ sX ( s ) − x ( 0− ) ⎤ ⎦ s →∞ ⎣ ⎦ s →∞ ⎣ ECSE303 Chap. 9 MR - 3/12/2010 x ( 0+ ) = Lim ⎡ sX ( s ) ⎤ ⎣ ⎦ s →∞ 104 () x 0+ = lim sX ( s ) s →+∞ lim x ( t ) = lim sX ( s ) Example t →+∞ s →0 Find the initial value of the signals below with finite energy and Laplace transform . X (s) = () x0 + ( 10 1 − e − st0 s = lim sX ( s ) s →+∞ 10 = lim s s →+∞ s = 10 ECSE303 Chap. 9 MR - 3/12/2010 ) 10 X (s) = , Re {s} > −3 s+3 () x 0+ = lim sX ( s ) s →+∞ 10 = lim s s →+∞ s + 3 = 10 105 () x 0+ = lim sX ( s ) s →+∞ lim x ( t ) = lim sX ( s ) Example t →+∞ s →0 Find the final value of the step response of the system below u (t ) s +1 s+4 y (t ) y ( +∞ ) = lim ⎡ sY ( s ) ⎤ ⎣ ⎦ s →0 ⎡ 1 ( s + 1) ⎤ = lim ⎢ s ⎥ s →0 ⎢ s ( s + 4) ⎥ ⎣ ⎦ 1 = 4 ECSE303 Chap. 9 MR - 3/12/2010 106 Example The initial and final value theorem can also be used to make a quick check on the correctness of a Laplace transform, for example e u (t ) + e −2 t −2 t 2 s 2 + 5s + 12 cos ( 3t ) u ( t ) ↔ 2 s + 2 s + 10 ( s + 2 ) LT ( ) Knowing that x(0+)=2, we can apply the initial value theorem ⎡ ⎤ ⎡ 2s3 ⎤ 2 s 2 + 5s + 12 ⎥ = l im ⎢ 3 ⎥ = 2 l im ⎢ s 2 s →∞ ⎢ s + 2 s + 10 ( s + 2 ) ⎥ s →∞ ⎣ s ⎦ ⎣ ⎦ ( ) Knowing that x(∞)=0, we apply the final value theorem to find ECSE303 Chap. 9 MR - 3/12/2010 ⎡ ⎤ 2s 2 + 5s + 12 ⎡12s ⎤ ⎥ = l im ⎢ l im ⎢ s 2 ⎥=0 s →0 s →0 ⎣ 20 ⎦ ⎢ s + 2 s + 10 ( s + 2 ) ⎥ ⎣ ⎦ ( ) 107 ECSE303 Chap. 9 MR - 3/12/2010 108 ECSE303 Chap. 9 MR - 3/12/2010 109 ECSE303 Chap. 9 MR - 3/12/2010 110 ECSE303 Chap. 9 MR - 3/12/2010 111 Suggested exercises 9.8, 9.9, 9.12-9.14 ECSE303 Chap. 9 MR - 3/12/2010 112 ...
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