Baixe Chapter 14 - The Laplace Transform e outras Provas em PDF para Automação, somente na Docsity! Problems Section 14-2: Laplace Transform P14.2-1 ( ) ( ) ( ) ( ) ( ) ( ) 1 1 2 2 1 1 2 2cos A f t A F s AsF ss sf t t F s s ωω ω ⎫=⎡ ⎤⎣ ⎦ ⎪ ⇒ =⎬ += ⇒ = ⎪+ ⎭ L P14.2-2 ( ) 11 1 2 1! 1Fn n nt s t s s s+ + ⎡ ⎤ ⎡ ⎤= = =⎣ ⎦ ⎣ ⎦L L 1 ! = P14.2-3 ( ) ( ) ( ) ( ) ( ) ( ) ( ) [ ] ( ) ( ) 1 1 2 2 1 1 2 2 1 2 3 1 1 2 22 2 Linearity: a a Here 1 1 3 1 1 1so 3 t f t a f t F s a F s a a f t e F s s f t t F s s F s s s − + = +⎡ ⎤⎣ ⎦ = = ⎡ ⎤= = =⎡ ⎤⎣ ⎦ ⎣ ⎦ + = = =⎡ ⎤⎣ ⎦ = + + L L L L L P14.2-4 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 2 2 3 1 2 3 1 ( ) 1 1 1 1 , 1 1 bt bt bt f t A e u t A f t f t e u t u t e u t f t f t F s F s s s b AbF s AF s A F s F s A s s b s s b − − − = − = = − = − = + − = = + ⎡ ⎤ 3 ∴ = = + = − =⎡ ⎤⎣ ⎦ ⎢ ⎥+ +⎣ ⎦ 1 Section 14-3: Impulse Function and Time Shift Property P14.3-1 ( ) ( ) ( )f t A u t u t T= − −⎡ ⎤⎣ ⎦ ( ) ( ) ( ) ( )1 sTsT eA AeF s A u t A u t T A s s s −− − = − − = − =⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦L L P14.3-2 ( ) ( ) ( ) ( ) ( ) ( )at atf t u t u t T e F s e u t u t T⎡ ⎤= − − ⇒ = − −⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦⎣ ⎦L ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 sT s a T at eu t u t T es F s s ae g t G s a − − ⎫− − − =⎡ ⎤ −⎪⎣ ⎦ ⇒ =⎬ −⎪⎡ ⎤ = −⎣ ⎦ ⎭ L L P14.3-3 (a) ( ) ( )3 2 3 F s s = + (b) ( ) ( ) ( ) ( )sT sTf t t T F s e t eδ δ− −= − ⇒ = =⎡ ⎤⎣ ⎦L (c) ( ) ( ) ( ) ( )2 2 22 5 55 8 48 16 254 5 F s s ss ss = = = 1+ ++ + ++ + P14.3-4 ( ) ( ) ( ) ( )( (0.5 0.5)) 0.5 ( 0.5)0.5 0.5 0.5t t tg t e u t e u t e e u t− − + − − − −= − = − = − ( ) ( ) ( ) 0.5 0.50.5 0.5 0.5 ( 0.5) 0.5 ( 0.5) 0.5 0.50.5 0.5 1 1 ss t t s t ee ee e u t e e u t e e e u t s s −− − − − − − − − − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤− = − = = =⎣ ⎦ ⎣ ⎦ ⎣ ⎦ + + L L L P14.3-5 ( ) ( ) ( ) 2 sT s sT e et T tu t T e u t t u t T T T T − − −−⎡ ⎤ ⎡ ⎤− − = − = − = −⎡ ⎤⎣ ⎦⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ L L L T s 1 P14.4-2 ( ) ( ) ( )( ) 2 2 * 3 2 2 1 2 1 3 4 2 1 1 1 1 1 s s s s a a bF s s s s s s j s j s j s j s − + − + = = = + + + + + + − + + + − + + +1 + where ( ) ( ) ( ) 2 2 2 * 2 1 4 1 1 1 2 1 3 4 3 2 1 1 2 21 3 2 2 s sb s s s s ja j s s j s j a j − + = = + + =− − + − = = + + + −=− + = − − =− + Then ( ) 3 32 2 42 2 1 1 j j F s s j s j s − + − − = + + 1+ − + + + Next ( ) ( )22 15 23 2 2 and tan 126.932 2 m θ − ⎛ ⎞ ⎜ ⎟ = − + = = =⎜ ⎟ ⎜ ⎟− ⎝ ⎠ ° From Equation 14.5-8 ( ) ( ) ( )5 cos 127 4t tf t e t e u− −⎡ ⎤= + ° +⎣ ⎦ t P14.4-3 ( ) ( ) ( )2 2 5 1( ) 1 21 2 1 s A B CF s s ss s s − = = + + + −+ − + where ( )21 2 5 1 5 12 and 1 2 1s s s sB C s s=− = − − = = = − + = Then ( ) ( ) ( ) 2 21 1 91 1 2s s dA s F s ds s=− =− −⎡ ⎤= + = =⎣ ⎦ − − Finally ( ) ( ) ( )22 1 2 1( ) 2 + 1 21 t t tF s f t e t e e u t s ss − −− ⎡ ⎤= + + ⇒ = − +⎣ ⎦+ −+ 2 P14.4-4 ( ) ( ) ( ) ( ) ( ) ( )222 1 1 11 2 2 1 11 1 1 A Bs CY s ss s s ss s + = = = + +⎡ ⎤+ + + + ++ + +⎣ ⎦ where 2 1 1 1 2 2 s A s s =− = = + + Next ( ) ( ) ( ) ( ) ( ) 2 22 2 1 1 1 2 2 ( 1) 1 2 21 2 2 1 1 2 B s C s s Bs C s s s ss s s B s B C s C 2 + = + ⇒ = + + + + + + + ++ + + ⇒ = + + + + + + Equating coefficients: 2s : 0 1 1 : 0 2 1 B B s B C C = + ⇒ = − = + + ⇒ =− Finally ( ) ( ) ( ) ( )2 1 cos 1 1 1 t tsY s y t e e t u t s s − −1 + ⎡ ⎤= − ⇒ = −⎣ ⎦+ + + P14.4-5 ( ) ( ) ( ) ( ) ( ) ( ) ( )2 22 2 3 11 2 11 2 5 1 4 1 4 s s F s ss s s s s + − + = = + + ++ + + + + + + ( ) ( ) ( ) ( )cos 2 sin 2t t tf t e e t e t u− − −⎡ ⎤= − +⎣ ⎦ t P14.4-6 ( ) ( )( ) ( ) 2 s+3 s s+1 2 1 2 A B CF s s s s s = = + + + + + where ( ) ( )( ) ( ) ( ) ( ) ( ) ( )0 1 10 2 3 2 3 3, 1 4 1 2 2s s ss s s A sF s B s F s s s s s= =− =−= + + = = = = + = = + + + − and ( ) ( ) ( )( )2 2 2 3 2 1 1s s s s F s C s s=− =− + + = = + = Finally ( ) ( ) ( ) ( )23 4 1 3 41 2 t tF s f t e e u t s s s − −−= + + ⇒ = − + + + 3 P14.4-7 ( ) ( ) ( ) ( ) [ ] ( ) ( ) ( ) ( ) ( 2 2 2 2 1 2 2 1 2 1 , tan 2 cos sin cos cos j j at at at cs ca d c jd c jdF s s a j s a js a me me dwhere m c d cs a j s a j )f t e c t d t u t e c d t u t m e t u t θ θ ω ω ωω θ ω ω ω ω ω θ ω θ − − − − − + − ⎡ ⎤+ − = = +⎢ ⎥+ − + ++ + ⎣ ⎦ ⎡ ⎤ = + = + =⎢ ⎥+ − + +⎣ ⎦ ⎡ ⎤∴ = − = + + = +⎣ ⎦ P14.4-8 (a) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) 22 2 21 2 2 8 38 3 1 4 13 2 2 9 3 8 2 2, 8, 3 & 3 7.33 3 6.33 tan 38.4 , 8 6.33 10.85 8 10.85 cos 3 42.5t ssF s s s s a c ca d d m f t e t u t ω ω θ − ° − −− = = × + + + + − − ∴ = = = − =− ⇒ = = − ⎛ ⎞∴ = = = + =⎜ ⎟ ⎝ ⎠ ⇒ = + (b) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 22 2 1 1 1 1 2 33 3 1Given , first consider . 2 17 2 17 2 1 16 3 4Identify 1, 0, 4 3 3 4. Then | | 3 4, tan 90 0 So ( ) (3 4) sin 4 . Next, 1 . Fin s t s eF s F s s s s s s a c and d d m d f t e t u t F s e F s f t f t ω ω θ − − − − = = = × + + + + + + −⎛ ⎞= = = − = ⇒ =− = = = =−⎜ ⎟ ⎝ ⎠ = = ⇒ = − ally ° ( ) ( ) ( )( 1)(3 4) sin 4 1 1tf t e t u t− −∴ = −⎡ ⎤⎣ ⎦ − P14.4-9 (a) ( ) ( ) ( ) 2 2 2 5 11 1 s A B CF s s ss s s − = = + + ++ + where ( ) ( ) ( )20 1 5 1| 5 and 1 | 4 1 1s s A sF s C s F s= = 5 − − − = = = − = + = − = 6 Multiply both sides by ( )21s s + ( ) ( )22 5 5 1 1 4s s Bs s s B− = − + + + + ⇒ = 4 Section 14-6: Solution of Differential Equations Describing a Circuit P14.6-1 KVL: 42 1050 0.001 2 tdii v dt − ×+ + = e for t ≥ 0 The capacitor current and voltage are related by ( )62.5 10 dvi dt −= × 42 10 1 = 2e V tv − × , (0) 1 A, (0) 8 Vi v= = Taking the Laplace transforms of these equations yields [ ] ( ) ( ) 4 6 250 ( ) 0.001 ( ) (0) ( ) 2 10 ( ) 2.5 10 ( ) 0 I s s I s i V s s I s sV s v− + − + = + × = × −⎡ ⎤⎣ ⎦ Solving for I(s) yields ( ) ( ) ( ) ( ) 2 4 8 4 44 4 4 1.4 10 1.6 10 10 2 10 4 1010 2 10 4 10 s s A BI s s s ss s s + × − × = = + + + × + ×+ + × + × 4 C + where ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 4 8 8 4 4 84 4= 10 4 = 10 2 4 8 8 4 4 84 4= 2 10 4 = 2 10 2 4 4 = 4 10 1.4 10 1.6 10 2 10 2 10 3 10 32 10 4 10 s 1.4 10 s 1.6 10 .4 10 1 2 10 2 10 5s 10 s 4 10 s 1.4 4 10 s s s s s s sA s I s s s B s I s C s I s − − − × − × − × + × − × − × − = + = = = ×+ × + × + × − × × = + × = = = ×+ + × + × = + × = ( ) ( ) 4 8 8 84 4 4 = 4 10 10 s 1.6 10 8.8 10 22 6 10 15s 10 s 2 10 s − × − × × = = ×+ + × Then ( ) ( ) ( )4 4 410 t 2x10 4x104 4 4 12 3 1 5 22 15 10e 3e 22e A 10 2 10 4 10 15 t tI s i t u t s s s − − −⎡ ⎤= − + + ⇒ = − + +⎣ ⎦+ + × + × 14-1 P14.6-2 We are given ( ) 160cos 400v t t= . The capacitor is initially uncharged, so ( )C 0 0 Vv = . Then ( ) ( )160cos 400 0 00 160 A 1 i × − = = KCL yields C C3 10 100 d v v i dt − + = Apply Ohm’s law to the 1 Ω resistor to get C C 1 v v i v v i − = ⇒ = − Solving yields ( )41010 1600 cos 400 6.4 10 sin 400d i i tdt + = − × t Taking the Laplace transform yields ( ) ( ) ( ) ( ) ( ) 2 2 22 2 6.4 10 4001600s( ) (0) 1010 ( ) s 400 400 s I s i I s s × − + = − + + so ( ) 7 2 2 160 1600s 2.5 10( ) 1010 1010 (400) I s s s s − × = + + ⎡ ⎤+ +⎣ ⎦ Next ( ) 7 * 2 2 1600 2.5 10 1010 400 4001010 (400) s A B s s j s js s − × = + + + + −⎡ ⎤+ +⎣ ⎦ B where ( ) 7 22 = 1010 1600 2.5 10 23.1 400 s sA s − − ×= = + − , ( ) ( ) 7 7 * 5 = 400 1600 2.5 10 2.56 10 1.4 11.5 27.2 and 11.5 27.2 1010 400 8.69 10 68.4s j s xB j s s j ° ° − − × ∠ = = = − + − × ∠ B j= + Then ( ) 136.9 11.5 27.2 11.5 27.2 1010 400 400 j jI s s s j s j − + = + + + + − Finally ( ) ( ) ( )1010 1010 136.9 2 11.5 cos 400 2 27.2 sin 400 for 0 136.9 23.0 cos 400 54.4sin 400 for 0 t t i t e t t t e t t t − − = + − = + − > > 14-2 P14.6-3 C (0) 0v = 3 c c c3 c 15 10 10cos 2 2 20cos 21 10 30 v i t d v v td v dti dt − ⎫+ × = ⎪ ⇒ + =⎬⎛ ⎞= ×⎜ ⎟ ⎪⎝ ⎠ ⎭ Taking the Laplace Transform yields: ( ) ( ) ( ) ( ) ( )( ) * C C C C2 2 200 2 20 4 22 4 s s AsV s v V s V s s ss s − + = ⇒ = = + + 2 2 B B s j s j+ + + −+ + where ( )( )2 = 2 = 2 20 40 20 5 5 5 5 5*5, and 4 8 2 2 1 2 2 2s s j s sA B j s s s j j− − − = = = − = = = + = − + + − − 2 B j Then ( ) ( ) ( )2C C 5 5 5 5 5 2 2 2 2 5 5 cos 2 sin 2 2 2 2 t j j V s v t e t t s s j s j − + −−= + + ⇒ = − + + + + − V P14.6-4 L c c L L12 2 8 and d i d vv i i C dt dt − + + = − = − Taking the Laplace transform yields ( ) ( ) ( ) ( )L Lc L 812 2 0V s I s sI s i s − + + − = −⎡ ⎤⎣ ⎦ ( ) ( ) ( )c cL 0I s C sV s v= − −⎡ ⎤⎣ ⎦ c L(0) 0, (0) 0v i= = 14-3 ( ) ( )040 0.328 A 16 4 9 v i −⎛ ⎞− = =⎜ ⎟+⎝ ⎠ The capacitor voltage and inductor current are continuous so ( ) ( )0 0v v+ = − and ( ) ( )0 0i i+ = − . After the switch opens ( ) ( ) ( ) ( ) ( ) ( ) ( )2 0 9 0 9 0.32814.7540 29.508 0.4 0.4 0.4 0.4 v id dv t L i t R i t i dt dt + + = + ⇒ + = + = + = Substituting these initial conditions into the Laplace transformed differential equation gives ( ) ( ) ( ) ( )2 12529.508 0.328 25 0.328 156.25s I s s s I s I s s ⎡ ⎤− + + − + =⎡ ⎤⎣ ⎦⎣ ⎦ ( ) ( ) ( )2 12525 156.25 29.508 0.328 25 0.328s s I s ss+ + = + + + ( ) so ( ) ( )( )( ) ( )( ) ( ) ( ) 2 2 2 2 2 0.328 29.508 25 0.328 125 25 156.25 0.328 29.508 25 0.328 125 0.471 23.6 0.8 12.512.5 12.5 s I s s s s s s ss s s + + + = + + + + + − = = ++ + + + Taking the inverse Laplace transform ( ) ( )12.50.8 23.6 0.471 A for 0ti t e t t−= + − ≥ so ( ) ( )12.5 0.328 A for 0 0.8 23.6 0.471 A for 0t t i t e t t− ≤⎧ = ⎨ + − ≥⎩ (checked using LNAP 10/11/04) P14.6-7 KCL: 1 67 5 tv i e−+ = KVL: 1 14 3 0 4 di dii v v i dt dt 3+ − = ⇒ = + Then 6 6 4 3 357 2 5 4 t t di i didt i e i e dt − − + + = ⇒ + = Taking the Laplace transform of the differential equation: 35 1 35 1( ) (0) 2 ( ) ( ) 4 6 4 ( 2)( 6 s I s i I s I s s s − + = ⇒ = )s+ + + Where we have used . Next, we perform partial fraction expansion. (0) 0i = 2 1 1 1 where and ( 2) ( 6) 2 6 6 4 2 4s s A B A B s s s s s s=− = − = + = = = = − + + + + + + 6 1 1 Then 2 635 3535 1 35 1( ) ( ) A, 0 16 2 16 6 16 16 t tI s i t e e s s − −= − ⇒ = − + + t ≥ P14.6-8 Apply KCL at node a to get 1 2 1 1 1 2 1 2 2 48 24 d v v v d v v v dt dt − = ⇒ + = Apply KCL at node b to get 2 2 1 2 2 2 1 2 50cos 2 1 0 3 60 20 24 30 24 v t v v v d v d v v v dt dt − − + + + = ⇒ − + + = cos 2 t Take the Laplace transforms of these equations, using 1 2(0) 10 V and (0) 25 Vv v= = , to get ( ) ( ) 2 1 2 1 2 2 25 60 1002 ( ) 2 ( ) 10 and ( ) 3 ( ) 4 s ss V s V s V s s V s s + + + − = − + + = + Solve these equations using Cramer’s rule to get ( ) ( ) ( ) ( )( ) ( ( )( )( ) ) ( )( )( ) 2 2 22 2 2 3 2 2 25 60 1002 10 2 25 60 100 10 44 2 (3 ) 2 4 1 4 25 120 220 240 4 1 4 s ss s s s ss V s s s s s s s s s s s s ⎛ ⎞+ + + +⎜ ⎟ + + + + ++⎝ ⎠= = + + − + + + + + + = + + + Next, partial fraction expansion gives ( ) * 2V 2 2 1 4 A A B Cs s j s j s s = + + + + − + + where ( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 2 2 * 3 2 2 1 3 2 2 4 25 120 220 240 240 240 6 6 1 4 2 40 6 6 25 120 220 240 115 23 15 34 4 25 120 220 240 320 16 60 34 1 s j s s s s s jA j s s s j A j s s sB s s s s sC s s =− =− =− + + + − −= = + + − − = − + + + = = = + + + + + −= = = −+ + = + Then ( )2 6 6 6 6 23 3 16 3 2 2 1 j jV s s j s j s s 4 + − = + + + + − + + Finally 4 2 23 16( ) 12cos 2 12sin 2 V 0 3 3 t tv t t t e e t− −= + + + ≥ P14.7-5 Node equations: ( ) ( ) ( ) ( ) ( )a C a a C 1 6 6 6 V s V s V s V s V s s s s − + = ⇒ = + 6 6s+ + ( ) ( ) ( ) ( ) C CC C 6 66 1 36 63 0 4 4 V s V sV s ss ss V s s s ⎛ ⎞− +− ⎜ ⎟+ +⎝ ⎠+ + + + 2 − = ) After quite a bite of algebra: ( ) ( )( )( 2 C 6 56 132 2 3 s sV s s s s + + = 5+ + + Partial fraction expansion: ( )( )( ) 2 44 1 6 56 132 93 3( )c 3 2 5 2 3 s sV s s s s s s s + + = = − 5 + + + + + + + Inverse Laplace transform: 2 3 544 1( ) 9 V, 0c 3 3 t t tv t e e e t− − −= − + ≥ 14-3 P14.7–6 Write a node equation in the frequency domain: ( ) ( ) ( ) 2 2 1 1o o o 1 2 22 1010 5 10 5 10 5 01 11 1 R R s R C RV s V ss C V s R R s ss sCs R CR C R ⎡ ⎤ + −⎢ ⎥ ⎢ ⎥+ − + = ⇒ = − = − +⎢ ⎥⎛ ⎞ +⎢ ⎥+⎜ ⎟⎜ ⎟ ⎢ ⎥⎣ ⎦⎝ ⎠ Inverse Laplace transform: ( ) 22 2 1000o 1 1 10 5 10 10 5 V for 0t R C t R R v t e e t R R − − ⎡ ⎤⎛ ⎞ ⎡ ⎤= − + − = − − >⎢ ⎥⎜ ⎟ ⎣ ⎦⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦ 14-4 P14.7-7 Here are the equations describing the coupled coils: ( ) ( ) ( ) ( ) 1 2 1 1 1 1 2 1 2 2 1 2 2 2 1 2 1 2 ( ) ( ) 3 ( ) 2 ( ) 3 3 ( ) ( ) 9 ( ) ( ) ( ) 2 2 ( ) 3 ( ) 2 ( ) 8 di div t L M V s s I s sI s s I s sI s dt dt di div t L M V s s I s sI s sI s sI s dt dt = + ⇒ = − + − = + = + ⇒ = − + − = + − − Writing mesh equations: ( ) ( ) ( ) ( ) ( ) 1 2 1 1 2 1 2 1 2 1 2 2 1 2 1 2 2 1 2 55 2 ( ) ( ) 2 ( ) ( ) 3 ( ) ( ) 9 3 2 2 9 ( ) ( ) 1 ( ) 3 ( ) ( ) 9 ( ) 2 ( ) 8 ( ) 2 1 1 I s I s V I s I s s I s sI s s I s I s s V s V s I s s I s sI s sI s sI s I s s I s I == + + + + + − ⇒ + + + = + = + ⇒ + − = + − + ⇒ − + = Solving the mesh equations for I2(s): ( ) ( )( )2 2 15 8 3 1.6 0.64 2.36 = + 0.26 1.54 + 0.26 + 1.545 9 2 s sI s s s s ss s + + = = + ++ + Taking the inverse Laplace transform: 0.26 1.54 2( ) 0.64 2.36 A for 0 t ti t e e t− −= + > P14.7-8 t<0 time domain frequency domain Mesh equations in the frequency domain: ( ) ( ) ( )( ) ( ) ( ) ( )1 1 2 1 1 212 2 26 6 6 0 3 3I s I s I s I s I s I ss s+ − + + = ⇒ = − 14-5 P14.7-11 We will determine , the Laplace transform of the output, twice, once from the given equation and once from the circuit. From the given equation for the output, we have ( )oV s ( )o 10 5 100 V s s s = + + Next, we determine from the circuit. For , we represent the circuit in the frequency domain using the Laplace transform. To do so we need to determine the initial condition for the capacitor. ( )oV s 0t ≥ When and the circuit is at steady state, the capacitor acts like an open circuit. Apply KCL at the noninverting input of the op amp to get 0t < ( ) ( ) 1 3 0 0 0 v v R − − = ⇒ − = 3 V The initial condition is ( ) ( )0 0 3v v+ = − = V Now we can represent the circuit in the frequency domain, using Laplace transforms. Apply KCL at the noninverting input of the op am to get ( ) ( ) 6 1 2 3 10 V s V s s s R s − − = Solving gives ( ) 6 1 6 6 1 1 103 2 2 1 10 10 s R V s s s s s R R + = = + ⎛ ⎞ ⎛ ⎞ + +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ Apply KCL at the inverting input of the op amp to get ( ) ( ) ( ) ( ) ( )o 2o 62 1 2 11 1 1000 1000 1000 10 V s V s R RV s V s V s R s s R ⎛ ⎞ ⎜ ⎟ − ⎛ ⎞ ⎛ ⎞⎜ ⎟ = ⇒ = + = + +⎜ ⎟ ⎜ ⎟⎜ ⎟⎛ ⎞⎝ ⎠ ⎝ ⎠⎜ ⎟+⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ 2 The expressions for Vo(s) must be equal, so 2 6 1 10 5 2 11 100 1000 10 R s s s s R ⎛ ⎞ ⎜ ⎟ ⎛ ⎞⎜ ⎟ + = + +⎜ ⎟⎜ ⎟+ ⎛ ⎞⎝ ⎠⎜ ⎟+⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ Equating coefficients gives 2 21 5 41000 R R+ = ⇒ = kΩ and 6 1 1 10 100 10 kR R = ⇒ = Ω (checked using LNAPTR 7/31/04) P14.7-12 For t < 0, The input is constant. At steady state, the capacitor acts like an open circuit and the inductor acts like a short circuit. The circuit is at steady state at time 0t = − so ( )C 0v − = 0 and ( )L 0i B− = The capacitor voltage and inductor current are continuous so ( ) ( )C C0 0v v+ = − ) and . ( ) (L L0 0i i+ = − For t < 0, represent the circuit in the frequency domain using the Laplace transform as shown. is the node voltage at the top node of the circuit. Writing a node equation gives ( )CV s ( ) ( ) ( )C C C V s V sA B B C sV s s R s L s + = + + + so ( ) 2 C A L s R R LC s V s s R L s + + = Then ( )C 2 2 1 1 A A R L CV s R LC s L s R s s RC LC = = + + + + and ( ) ( )CL 2 1 1 A V s B BLCI s L s s s s s s RC LC = + = + ⎛ ⎞ + +⎜ ⎟ ⎝ ⎠ a.) When 12 , 4.5 H, F, 5 mA and 2 mA 9 R L C A B= Ω = = = = − , then ( ) ( )L 2 5 40 10 2 3 7 7 144.5 2 2 I s s s ss s s s − = + = + ++ + − + Taking the inverse Laplace transform gives ( ) 4 0.5L 5 53 mA for 0 7 7 t ti t e e t− −= + − ≥ b.) When , then 1 , 0.4 H, 0.1 F, 1 mA and 2 mAR L C A B= Ω = = = = − ( ) ( ) ( ) ( )L 2 22 25 2 25 2 1 5 1 510 25 5 5 I s s s ss s s s s s ⎛ ⎞− − = + = + = − +⎜ ⎟ ⎜ ⎟++ + + +⎝ ⎠s + Taking the inverse Laplace transform gives ( ) ( )5 5L 1 5 mA for 0t ti t t e e t− −= − + − ≥ c.) When , then 1 , 0.08 H, 0.1 F, 0.2 mA and 2 mAR L C A B= Ω = = = = − ( ) ( ) ( ) ( ) ( )L 2 22 2 2 25 2 1.8 0.2 2 1.8 5 100.2 0.1 10 125 5 10 5 10 5 10 s sI s s s ss s s s s s − − − − − + = + = + = − − + + + + + + + +2 2 Taking the inverse Laplace transform gives ( ) ( ) ( )( )5L 1.8 0.2cos 10 0.1sin 10 mA for 0ti t e t t t−= − − + ≥ P14.7-14 For t < 0, The input is 12 V. At steady state, the capacitor acts like an open circuit. Notice that v(t) is a node voltage. Express the controlling voltage of the dependent source as a function of the node voltage: va = −v(t) Writing a node equation: ( ) ( ) ( )12 3 0 8 4 4 v t v t v t −⎛ ⎞ ⎛ ⎞− + + −⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎝ ⎠ = ( ) ( ) ( ) ( )12 2 6 0 4 Vv t v t v t v t− + + − = ⇒ = − ( ) ( )0 0 4v v+ = − = − V For t < 0, represent the circuit in the frequency domain using the Laplace transform as shown. ( )V s is a node voltage. Express the controlling voltage of the dependent source in terms of the node voltages ( ) ( )aV s V s= − Writing a node equation gives ( ) ( ) ( ) ( ) 6 3 4 0.75 8 4 40 V s V s ss V s V s s − ⎛ ⎞+ + + =⎜ ⎟ ⎝ ⎠ Solving gives ( ) ( ) ( ) ( ) 10 10 4 2 2 4 1 15 4 2 5 5 5 5 s V s V s s s s s s s s s − ⎛ ⎞− = − ⇒ = − = + − = − +⎜ ⎟5s− − − − ⎝ ⎠− Taking the inverse Laplace transform gives ( ) ( )52 1 V for 0tv t e t= − + ≥ This voltage becomes very large as time goes on. P14.7-15 For t < 0, the voltage source voltage is 2 V and the circuit is at steady state. At steady state, the capacitor acts like an open circuit. ( ) 3 3 2 00 0.0 10 10 40 10 i −− = = × + × 4 mA and ( ) ( )( )3 3C 0 40 10 0.04 10 1.6 Vv −− = × × = The capacitor voltage is continuous so ( ) ( )C C0 0v v+ = − o . For t > 0, the voltage source voltage is 12 V. Represent the circuit in the frequency domain using the Laplace transform as shown. ( ) ( )C and V s V s are node voltages. Writing a node equation gives ( ) ( ) ( ) ( ) ( ) ( ) C C C C C63 3 12 1.6 12 1.60 4 0.08 0 0.5 1010 10 40 10 V s V s V ss s V s s V s V s s s s − − ⎛ ⎞ ⎛ ⎞+ + = ⇒ − + − +⎜ ⎟ ⎜ ⎟×× × ⎝ ⎠ ⎝ ⎠ C = ( )( ) ( ) ( ) ( )C C 48 80 48 1.6 600 9.6 80.08 5 0.128 0.08 5 62.5 62.5 s sV s s V s s s s s s s s + + − + = + ⇒ = = = + + + + Taking the inverse Laplace transform gives ( ) 62.5C 9.6 8 V for 0tv t e t−= − ≥ The 40 kΩ resistor, 50 kΩ resistor and op amp comprise an inverting amplifier so ( ) ( ) ( )62.5 62.5o C50 50 9.6 8 12 10 V for 040 40 t tv t v t e e t− −= − = − − = − + ≥ so ( )o 62.5 2 V for 0 12 10 V for 0t t v t e t− − ≤⎧ = ⎨− + ≥⎩ (checked using LNAP 10/11/04) P14.7-16 For t < 0, the voltage source voltage is 5 V and the circuit is at steady state. At steady state, the capacitor acts like an open circuit. Using voltage division twice ( ) 32 300 5 5 0 32 96 120 30 v − = − = + + .25 V V and ( ) ( )0 0 0.25v v+ = − = For t > 0, the voltage source voltage is 20 V. Represent the circuit in the frequency domain using the Laplace transform as shown. We could write mesh or node equations, but finding a Thevenin equivalent of the part of the circuit to the left of terminals a-b seems promising. Using voltage division twice ( )oc 32 20 30 20 5 4 1 V 32 96 120 30 V s s s s −⎛ ⎞ ⎛ ⎞= − =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠ s = Section 14-8: Transfer Function and Impedance P14.8-1 1 2 11 1 2 1 2 1 1 2 2 1 1 ( ) where and1 1 1 R Z RC sH s Z Z 2R Z Z R C sR C s = = = R C s = + + ++ ( ) ( )( ) ( ) ( ) ( )( ) ( ) 2 1 1 1 1 2 2 2 1 2 1 2 2 2 2 2 1 2 1 2 1 Let and then 1 1 1 When constant, as required. 1 R s R C R C H s R s s R R s RH s R R s R R τ τ τ τ τ τ τ τ τ τ + = = = + + + + = = ⇒ = = = + + + 1 1 2 2we require R C R C∴ = P14.8-2 1 2 1Let and then the input impedance isZ R Z R Ls Cs = + = + ( ) ( ) 2 1 2 2 1 2 1 1 1 2 1 LR R Ls LCs RC s Z Z Cs RZ s R Z Z LCs RCsR R Ls Cs ⎛ ⎞⎛ ⎞ ⎛ ⎞+ + + +⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎜ ⎟= = = + +⎜ ⎟+ + + ⎜ ⎟ ⎝ ⎠ + + 2Now require : 2 then LRC RC L R C Z R + = ⇒ = = R P14.8-3 The transfer function is ( ) 2 2 1 2 1 2 2 1 2 1 1 1 1 R R C s R C H s R R R R s R C s R R C + = = + + + + Using 1 22 , 8 and 5 FR R C= Ω = Ω = gives ( ) 0.1 0.125 H s s = + The impulse response is ( ) ( ) ( )0.1250.1 Vth t H s e u t-1L −⎡ ⎤= =⎣ ⎦ . The step response is 1 ( ) ( ) ( ) ( ) 0.1250.1 0.8 0.8 0.8 1 V 0.125 0.125 tH s e u t s s s s s -1 -1 -1L = L = L − ⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥− = −⎢ ⎥⎢ ⎥ ⎢ ⎥+ +⎣ ⎦⎢ ⎥⎣ ⎦ ⎣ ⎦ (Checked using LNAP, 12/29/02) P14.8-4 The transfer function is: ( ) ( ) ( ) 4 2 2 12 1212 8 164 tH s t e u t s ss −⎡ ⎤= = =⎣ ⎦ + ++ L The Laplace transform of the step response is: ( ) ( ) ( )2 2 3 12 34 44 4 H s k s ss s s −= = + + ++ + s The constant k is evaluated by multiplying both sides of the last equation by . ( )24s s+ ( ) ( ) ( )2 23 312 4 3 4 3 4 12 4 4 s s ks s k s k s k ⎛ ⎞⎟⎜= + − + + = + + + + ⇒ =−⎟⎜ ⎟⎜⎝ ⎠ 3 4 The step response is ( ) ( )1 43 33 V 4 4 tH s e t u t s L − − ⎡ ⎤ ⎛ ⎞⎛ ⎞⎟⎜ ⎟⎢ ⎥ ⎜= − + ⎟⎟⎜ ⎜ ⎟⎟⎜⎢ ⎥ ⎜ ⎟⎝ ⎠⎝ ⎠⎣ ⎦ P14.8-5 The transfer function can also be calculated form the circuit itself. The circuit can be represented in the frequency domain as We can save ourselves some work be noticing that the 10000 ohm resistor, the resistor labeled R and the op amp comprise a non-inverting amplifier. Thus ( ) ( )a c1 10000 RV s V s ⎛ ⎞⎟⎜= + ⎟⎜ ⎟⎜⎝ ⎠ Now, writing node equations, 2 ( ) ( ) ( ) ( ) ( ) ( )c i o a oc 0 and 01000 5000 V s V s V s V s V s CsV s Ls − − + = + = Solving these node equations gives ( ) 1 50001 1000 10000 1 5000 1000 R C LH s s s C L ⎛ ⎞⎟⎜ + ⎟⎜ ⎟⎜⎝ ⎠= ⎛ ⎞⎛⎟ ⎟⎜ ⎜+ +⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎝ ⎠⎝ ⎞ ⎠ Comparing these two equations for the transfer function gives ( ) ( )1 12000 or 5000 1000 1000 s s s s C C ⎛ ⎞ ⎛ ⎞⎟ ⎟⎜ ⎜+ = + + = +⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎝ ⎠ ⎝ ⎠ ( ) ( )5000 50002000 or 5000s s s s L L ⎛ ⎞ ⎛ ⎞⎟ ⎟⎜ ⎜+ = + + = +⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎝ ⎠ ⎝ ⎠ 1 1000 1 10000 5000 15 106 C R L + ⎛ ⎝⎜ ⎞ ⎠⎟ = × The solution isn’t unique, but there are only two possibilities. One of these possibilities is ( )1 2000 0.5 F 1000 s s C C μ ⎛ ⎞⎟⎜ + = + ⇒ =⎟⎜ ⎟⎜⎝ ⎠ ( )s L s L+⎛⎝⎜ ⎞ ⎠⎟ = + ⇒ = 5000 5000 1 H ( ) 6 6 1 50001 15 10 10000 11000 0.5 10 R R ⎛ ⎞⎟⎜ + = × ⇒ = Ω⎟⎜ ⎟⎜⎝ ⎠× 5 k (Checked using LNAP, 12/29/02) 3 ( ) ( )( ) o i 1212 RsV s R L s LH s RV s R L s s L ++ = = = ++ + + From the given step response: ( ) ( ) ( ) ( ) ( ) 4 0.5 0.5 2 20.5 1 4 4 tH s s se u t H s s s s s s − 4s + +⎡ ⎤= + = + = ⇒ =⎣ ⎦ + + + L Comparing these two forms of the transfer function gives: 2 12 2 4 6 H, 12 12 4 R LL L R R L L ⎫= ⎪ +⎪ ⇒ = ⇒ = =⎬+ ⎪= ⎪⎭ Ω (Checked using LNAP, 12/29/02) P14.8-10 Mesh equations: ( ) ( ) ( ) ( ) ( ) 1 1 2 1 1 1 1 1 10 V s R I s I s Cs Cs Cs R R I s I s Cs Cs ⎛ ⎞= + + −⎜ ⎟ ⎝ ⎠ ⎛ ⎞= + + −⎜ ⎟ ⎝ ⎠ 2 Solving for I2(s): ( ) ( ) 2 1 2 1 2 12 ( ) V s CsI s R R Cs Cs Cs ⎛ ⎞ ⎜ ⎟ ⎝ ⎠= ⎛ ⎞ ⎛ ⎞+ + −⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 1 Then gives ( ) ( )o 2V s R I s= ( ) ( )( ) [ ] [ ] ( ) 0 1 2 1 1 22 2 1 1 2 2 1 1 4 12 2 2 V s RCs sH s V s R Cs RCs RC R CR C s s RR C RR C = = = + + − ⎡ ⎤+⎢ ⎥+ + ⎢ ⎥ ⎣ ⎦ 6 P14.8-11 Let 2 1 1 1 1 x x R RCsZ RCsR Cs Z R L s ⎛ ⎞ ⎜ ⎟ ⎝ ⎠= = ++ = + Then ( ) ( ) 2 2 2 1 1 2 x x x x x x 2 x x x21 x x x 1 1 1 R V Z RRCs RV Z Z L RCs L R RC s R RR L s RCs V L C L R RCV R Rs s L RC L RC += = = + + ++ + + = + + + + + + P14.8-12 Node equations: ( ) ( )1 out1 in 1 11 1 1 out1 1 2 2 out 2 2 0 1 0 1 V VV V sC R C s V R C sV V R VV V R C sV R sC − − + = ⇒ + = + − − = ⇒ = − 1 1 in out Solving gives: ( ) 1 1 2 2out 2 2in 1 2 1 2 2 2 1 1 1 2 1 2 1 1 1 1 s R C s R CVH s V R R C C s R C s s s R C R R C C − − = = = + + + + 7 P14.8-13 Node equations in the frequency domain: 1 11 1 2 3 0iV V V VV R R R 0− −+ + = 0 1 1 2 3 3 1 1 1 iV VV 1R R R R R ⎛ ⎞ ⇒ + + − =⎜ ⎟ ⎝ ⎠ 1 2 0 1 2 2 0 2 0V sC V V sC R V R − − = ⇒ = − After a little algebra: ( ) 0 3 2 2 3 2 1 3 2 1 2 1i V RH s V sC R R sC R R sC R R R − = = + + + P14.8-14 ( ) ( ) 2 1 1 ( ) 1 1 o i V s Cs LCH s RV s Ls R s s Cs L LC = = = + + + + L, H C, F R, Ω H(s) 2 0.025 18 ( )( )2 20 20 9 20 4 5s s s s = + + + + 2 0.025 8 ( )22 2 20 20 4 20 2 4s s s = + + + + 1 0.391 4 ( )( )2 2.56 2.56 4 2.56 0.8 3.2s s s s = + + + + 2 0.125 8 ( )22 4 4 4 4 2s s s = + + + 8 vo(t) = 7.07 cos(100t +165°) V. c. Here’s the circuit represented in the frequency domain, using The Laplace transform (assuming zero initial conditions). Writing a node equation at the inverting input node of the op amp gives ( ) ( )o o 3 3 6 1 014 10 10 1010 V s V ss s + + × ×× = ( ) ( ) 3 o 10 100 0 4 s V s s + + = ( ) ( )o 250 2.5 2.5 100 100 V s s s s s − = = + + + Finally, ( ) ( ) ( )1002.5 1 Vtov t e u t−= − P14.8-17 Represent the circuit in the frequency domain using the Laplace transform as shown. (Set the initial conditions to zero to calculate the step response.) First, ( ) ( ) ( ) 2 2 2 2 2 2 1 1 || 1 1 R L s R L sC sR L s C s C L s C R sR L s C s × + + + = = + ++ + Next, using voltage division, 11 ( ) ( )( ) ( ) 2 2 2 2o 2 2i 2 1 2 12 2 2 1 1 2 1 2 1 22 1 1 1 1 1 2 4 4 29 R L s C L s C R s R L sV s H s R L sV s R L s R C L s C R sR C L s C R s Rs R C R LC s L R R C R R s ss s R LC R LC + + + + = = = + + + + ++ + + + + = = + + + ++ + Using ( )i 1V s s = gives ( ) ( ) ( ) ( ) ( ) ( ) o 22 2 2 2 22 2 2 4 0.1379 0.1379 1.4483 4 294 29 0.1379 0.1379 1.4483 2 5 0.1379 2 50.1379 0.3449 2 5 2 5 H s s sV s s s s ss s s s s s s s s s + − + = = = + + ++ + − + = + + + + = − + + + + + Taking the inverse Laplace transform ( ) ( ) ( )( ) ( ) 2 o 2 0.1379 0.1379cos 5 0.3448sin 5 0.1379 0.3713 cos 5 111.8 V t t v t e t t e t − − = + − + = + − ° (checked using LNAP 10/15/04) P14.8-18 First, we determine the transfer function corresponding to the step response. Taking the Laplace transform of the given step response ( ) ( ) ( )( ) ( ) (( )( ) ) ( )( ) o 50 20 0.667 20 1.667 501 0.667 1.667 50 20 50 20 1000 50 20 H s s s s s s s V s s s s s s s s s s s + + + + − + = = + − = + + + + = + + Consequently, ( ) ( )( ) ( )( ) o i 1000 50 20 V s H s V s s s = = + + 12 Next, we determine the transfer function of the circuit. Represent the circuit in the frequency domain using the Laplace transform as shown. (Set the initial conditions to zero to calculate the transfer function.) Apply KVL to the left mesh to get ( ) ( ) ( ) ( ) ( )ii 1 a a a 1 V s V s L s I s K I s I s K L s = + ⇒ = + Next, using voltage division, ( ) ( ) ( ) ( )( ) ( )o a o2 2 1 R RV s K I s V s V s L s R L s R K L s = ⇒ = + + + i K Then, the transfer function of the circuit is ( ) ( )( ) ( )( ) 1 2o i 2 1 2 1 R K L LV s R KH s V s L s R L s K R Ks s L L = = = ⎛ ⎞⎛+ + + +⎜ ⎟⎜⎜ ⎟⎜ ⎝ ⎠⎝ ⎞ ⎟⎟ ⎠ Comparing the two transfer functions gives ( )( ) ( ) 1 2 2 1 1000 50 20 R K L L H s s s R Ks s L L = = + + ⎛ ⎞⎛ + +⎜ ⎟⎜⎜ ⎟⎜ ⎝ ⎠⎝ ⎞ ⎟⎟ ⎠ We require 1 2 1000 R K L L = and either 2 50 R L = and 1 20 K L = or 2 20 R L = and 1 50 K L = . These equations do not have a unique solution. One solution is L1 = 0.1 H, L2 = 0.1 H, R = 5 Ω and K = 2 V/A (checked using LNAP 10/15/04) 13 ( ) ( )( ) 2o i 3 1 1 2 K R C LI s H s V s 2R R Rs s L R R C = = ⎛ ⎞+⎛ ⎞ + +⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ Comparing the two transfer functions gives ( ) ( ) 22 3 1 1 2 80 5 K R C L H s 2R R Rs s s L R R = = ⎛ ⎞++ ⎛ ⎞ + +⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠C We require ( ) 1 2 1 2 40 105 2 40 10 R R C R R C C + + = = ⇒ = × 5 mF , 3 205 4 H R L L L = = ⇒ = and ( )2 80 80 V/V 10 0.025 4 K K K R C L = = ⇒ = . (checked using LNAP 10/15/04) P14.8-21 First, ( ) ( ) ( ) ( ) ( ) ( ) ( )6.5cos 2 22.6 6.5 cos 22.6 cos 2 6.5 sin 22.6 sin 2 6cos 2 2.5sin 2t t t+ ° = ° − ° = −t t Consequently, the impulse response can be written as ( ) ( ) ( )( ) ( )2o 6cos 2 2.5sin 2 Vtv t e t t u t−= − The transfer function is ( ) ( ) ( ) ( )2 2 2 22 2 2 3 2 6 13 66 2.5 6 133 2 3 2 3 2 s sH s s ss s s + + = − = = 13s + + ++ + + + + + The Laplace transform of the step response is ( ) ( ) ( ) ( ) ( )2 222 2 2 6 13 1 1 1 3 3 2 6 13 26 13 3 2 3 2 3 2 H s s s s s s s s s s ss s s s s s + + = = − = − = − + × + ++ + 2 2+ + + + + + 16 Taking the inverse Laplace transform gives the step response: ( ) ( ) ( )( )( ) ( ) ( )( )2 2o 1 1.5sin 2 cos 2 1 1.803 cos 2 123.7 Vt tv t e t t u t e t− −= + − = + − ° P14.8-22 Taking the Laplace transform of the step response, ( ) ( ) ( ) ( )2 2 1 3 1 1 6 9 33 3 H s s s s s ss s ⎡ ⎤ + = − + = − =⎢ ⎥ ++ +⎢ ⎥⎣ ⎦ 23s s + The transfer function is ( ) ( )2 9 3 H s s = + Taking the inverse Laplace transform gives the impulse response: ( ) ( )3o 9 Vtv t t e u t−= (checked using LNAP 10/15/04) P14.8-23 Represent the circuit in the frequency domain using the Laplace transform as shown. (Set the initial conditions to zero to calculate the transfer function.) First, ( ) ( )a 1 iV sI s L s R = + The equivalent impedance of the parallel capacitor and inductor is 2 2 2 2 2 1 1|| 1 1 R RC sR C s R C sR C s × = = ++ Next, using voltage division, ( ) ( ) ( ) ( )( )( ) ( ) 3 2 33 3 2 3 o a a 2 2 3 2 3 1 2 3 2 3 3 21 K R R R C sR R R R C s V s K I s K I s V sR R R R R C s L s R R R R R C sR R C s ++ = = = + + + + ++ + i 17 Then, the transfer function of the circuit is ( ) ( )( ) ( ) ( )( ) 2o i 1 2 3 2 3 1 5 0.5 5 2. K s L R CV s s H s V s s sR R R s s L R R C ⎛ ⎞ +⎜ ⎟⎜ ⎟ +⎝ ⎠= = = + +⎛ ⎞+⎛ ⎞ + +⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ 5 Using ( )i 1V s s = gives ( ) ( ) ( )( )( )o 5 0.5 0.2 1.8 1.6 5 2.5 5 2.5 H s s V s s s s s s s s + − = = = + + + + + + Taking the inverse Laplace transform ( ) ( ) ( )5 2.5o 0.2 1.8 1.6 Vt tv t e e u t− −= − + (checked using LNAP 10/15/04) 18 Section 14-10: Stability P14.10-1 a. From the given step response: ( ) ( ) ( ) ( ) 1003 71 4 100 tH s e u t s s −⎡ ⎤= − =⎢ ⎥ +⎣ ⎦ L 5 s From the circuit: ( ) ( ) 55 R H sR LH s RR Ls s s s L = ⇒ = ++ + ⎛ ⎞+⎜ ⎟ ⎝ ⎠ Comparing gives 75 15 5 0.2 H100 R RL R L L ⎫= ⎪ = Ω⎪ ⇒⎬+ =⎪= ⎪⎭ b. The impulse response is ( ) ( )1 10075 75 100 th t e u t s − −⎡ ⎤= =⎢ ⎥+⎣ ⎦ L c. ( ) 100 75 3 45 100 100 4 2jω ω = = = ∠− + H ° ( ) ( )o 3 1545 5 0 45 V 4 2 4 2 ω ⎛ ⎞= ∠ ° ∠ ° = ∠− °⎜ ⎟ ⎝ ⎠ V ( ) ( )2.652 cos 100 45 Vov t t= − ° (Checked using LNAP, 12/29/02) P14.10-2 The transfer function of this circuit is given by ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 5 5 10 20 205 5 1 2 2 2 2 tH s e t u t H s s s s s s s − − −⎡ ⎤= − + = + + = ⇒ =⎣ ⎦ + + + + L 22 This transfer function is stable so we can determine the network function as ( ) ( ) ( ) ( )2 2 20 20 2 2s j s j H s s jω ω ω ω= = = = = + + H The phasor of the output is 14-1 ( ) ( ) ( ) ( ) ( )o 2 2 20 205 45 5 45 12.5 45 V 2 2 2 2 45j ω = ∠ ° = ∠ ° = ∠− + ∠ ° V ° The steady-state response is ( ) ( )o 12.5cos 2 45 Vv t t= − ° (Checked using LNAP, 12/29/02) P 14.11-3 The transfer function of the circuit is ( ) ( ) 1 5 2 3030 ( ) 5 tH s t e u t s − −⎡ ⎤= =⎣ ⎦ + L . The circuit is stable so we can determine the network function as ( ) ( ) ( ) ( )2 2 30 30 5 5s j s j H s s jω ω ω ω= = = = = + + H The phasor of the output is ( ) ( ) ( ) ( ) ( )o 2 2 30 3010 0 10 0 8.82 62 V 5 3 5.83 31j ω = ∠ ° = ∠ ° = ∠− + ∠ ° V ° The steady-state response is ( ) ( )o 8.82cos 3 62 Vv t t= − ° P14.10-4 ( ) ( ) ( ) ( )( ) 8 320 40 1.03 41 10240040 1.03 41 8 320 8 320 t tH s e e u t s s s s − −⎡ ⎤= + − = + − =⎣ ⎦ + + + + L s s s ) so ( ) ( ) ( 102400 8 32 H s s s = + + 0 The poles of the transfer function are 1 8 rad/ss = − and 2 320 rad/ss = − , so circuit is stable. Consequently, ( ) ( ) ( ) ( ) 102400 40 8 320 1 1 8 320 s j H s j j j j ω ω ω ωω ω= = = = + + ⎛ ⎞ ⎛+ +⎜ ⎟ ⎜ ⎝ ⎠ ⎝ H ⎞ ⎟ ⎠ 14-2 The network function has poles at 8 and 320 rad/s and has a low frequency gain equal to 32 dB = 40. Consequently, the asymptotic magnitude Bode plot is P14.10-5 ( ) ( ) ( ) ( )( ) 2 6 60 60 24060 2 6 2 t tH s e e u t s s s − −⎡ ⎤= − = − =⎣ ⎦ 6s s+ + + + L so ( ) ( )( ) 240 6 2 sH s s s = + + The poles of the transfer function are 1 2 rad/ss = − and 2 6 rad/ss = − , so circuit is stable. Consequently, ( ) ( ) ( ) ( ) 240 20 2 6 1 1 2 6 s j j jH s j j j j ω ω ωω ω ωω ω= = = = + + ⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ H The network function has poles at 2 and 6 rad/s. The asymptotic magnitude Bode plot has a gain equal to 40 = 32 dB between 2 and 6 rad/s. Consequently, the asymptotic magnitude Bode plot is 14-3 ( ) ( ) ( ) ( )o 2 2 0.4 10 4812 0 9.6 37 V 1 2101 5 j j jj ω = ∠ = = ∠− +⎛ ⎞+⎜ ⎟ ⎝ ⎠ V ° Back in the time domain, the steady state response is ( ) ( )o 9.6 cos 10 37 Vv t t= − ° (checked using LNAP 10/12/04) 14-6 Section 14.12 How Can We Check…? P14.12-1 ( ) ( ) 2.1 15.93 6 2t tL L dv t i t e e dt − −= = − − ( ) ( ) 2.1 15.91 0.092 0.575 75 t t C C di t v t e e dt − −= = − − ( ) ( ) 2.1 15.91 12 12 6 2t tR Lv t v t e e− −= − = + + ( ) ( ) ( )( ) 2.1 15.9 2 12 1 0.456 0.123 6 L C t t R v t v t i t e e− − − + = = + − ( ) ( ) 2.1 15.93 1 0.548 0.4526 C t t R v t i t e e− −= = + + Thus, ( ) ( ) ( ) ( ) ( )1 212 0 and +L R R C Rv t v t i t i t i t− + + = = 3 as required. The analysis is correct. P14.12-2 ( ) ( )1 2 18 20 and 3 3 4 4 I s I s s s = = − − 1 KVL for left mesh: 12 1 18 18 206 03 3 32 4 4 4 s s s s s ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ + + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟− − − ⎝ ⎠ ⎝ ⎠ = (ok) KVL for right mesh: 18 20 20 186 3 43 3 3 3 4 4 4 4 s s s s ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ − − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟− − − − ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 0= (ok) The analysis is correct. P14.12-3 Initial value of IL (s): 2 lim 2 1 5 ss s s s + = →∞ + + (ok) Final value of IL (s): 2 lim 2 0 0 5 ss s s s + = → + + (ok) Initial value of VC (s): ( ) ( )2 lim 20 2 0 5 s s s s s s − + = →∞ + + (not ok) Final value of VC (s): ( ) ( )2 lim 20 2 8 0 5 s s s s s s − + = − → + + (not ok) Apparently the error occurred as VC (s) was calculated from IL (s). Indeed, it appears that VC (s) was calculated as ( )20 LI ss− instead of ( ) 20 8 LI ss s − + . After correcting this error ( ) 2 20 2 8 5C sV s s s s s +⎛ ⎞= − +⎜ ⎟+ +⎝ ⎠ . Initial value of VC (s): ( ) ( )2 lim 20 2 8 8 5 s s s ss s s ⎛ ⎞− + ⎜ ⎟+ = ⎜ ⎟→ ∞ + +⎝ ⎠ (ok) Final value of VC (s): ( ) ( )2 lim 20 2 8 0 0 5 s s s ss s s ⎛ ⎞− + ⎜ ⎟+ = ⎜ ⎟→ + +⎝ ⎠ (ok) 2