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PowerPoint Presentation Fundamentals of Electric Circuits AC Circuits Chapter 15 Laplace transform 15 1 Introduction 15 2 Definition of the Laplace transform 15 3 Properties of the Laplace transform 1[.]

Fundamentals of Electric Circuits AC Circuits Chapter 15 Laplace transform 15.1 Introduction 15.2 Definition of the Laplace transform 15.3 Properties of the Laplace transform 15.4 The inverse Laplace transform 15.5 Application to circuits 15.6 Transfer functions 15.7 The convolution integral FUNDAMENTALS OF ELECTRIC CIRCUITS – AC Circuits 15.1 Introduction + Frequency-domain analysis:  limited to circuits with sinusoidal input + This chapter introduces the Laplace transform  Powerful tool for analyzing circuits with sinusoidal or non-sinusoidal inputs + The Laplace transform:  follows the same process of phasor transform o Use Laplace transform to transform the circuit from the time domain to the frequency domain o Use the inverse Laplace transform to transform results back to the time domain FUNDAMENTALS OF ELECTRIC CIRCUITS – AC Circuits 15.1 Introduction + The Laplace transform is significant for a number of reasons: o can be applied to a wider variety of inputs than phasor analysis o provide an easy way to solve circuit problems involving initial condition (allow to work with algebraic equation instead of differential equation) o capable of providing the total response of the circuit comprising both the natural and forced responses in one single operation FUNDAMENTALS OF ELECTRIC CIRCUITS – AC Circuits 15.2 Definition of the Laplace transform + Laplace transform  an integral transformation of a function f(t) from the time domain into the complex frequency domain, giving F(s) + One-side transform: L f t   F s     f t e  st dt s    j 0   j + Inverse Laplace transform: 1 1 st   L  f t   f t   F s e ds  2j   j + To find function f(t), we use:  Look-up table  Heaviside theorem FUNDAMENTALS OF ELECTRIC CIRCUITS – AC Circuits 15.3 Properties of the Laplace transform + Linearity: If F1(s) and F2(s) are the Laplace transform of f1(t) and f2(t), then La1 f1 t   a2 f t   a1 F1 s   a2 F2 s  + Scaling: If F(s) is the Laplace transform of f(t), then L f at   s F  a a + Time shift: If F(s) is the Laplace transform of f(t), then L f t  a u t  a   e  as F s  FUNDAMENTALS OF ELECTRIC CIRCUITS – AC Circuits 15.3 Properties of the Laplace transform + Frequency shift: If F(s) is the Laplace transform of f(t), then   L e  at f t   F s  a  + Time differentiation: If F(s) is the Laplace transform of f(t), the Laplace transform of its derivative is   L f ' t   sF s   f 0   + Time integration:   L f '' t   s F s   sf 0    f ' 0   If F(s) is the Laplace transform of f(t), the Laplace transform of its integral is t  L   f t dt   F s  0  s FUNDAMENTALS OF ELECTRIC CIRCUITS – AC Circuits 15.3 Properties of the Laplace transform + Frequency differentiation: If F(s) is the Laplace transform of f(t), then Ltf t    dF s  ds + Initial and final values: The initial value and final value properties  allow to find the initial value f(0) and the final value f(∞) of f(t) directly from its Laplace transform F(s) o The initial value theorem: o The final value theorem: + Convolution: lim f t   f 0    lim sF s  t 0  s  lim f t   f    lim sF s  t  L f1 t * f t   F1 s F2 s  s 0 FUNDAMENTALS OF ELECTRIC CIRCUITS – AC Circuits 15.3 Properties of the Laplace transform + Some Laplace transform pairs: f (t)  F(s)  (t)  1 u(t)  s s a t s eat  teat  sin(t)  s a2  s2  2 s cos(t)  s  2 FUNDAMENTALS OF ELECTRIC CIRCUITS – AC Circuits 15.3 Properties of the Laplace transform + Example1: Find the Laplace transform of the given function f(t) as below Solution f(t) f(t) 20V f(t) can be seen as the sum of function: 20.u(t) 20V f1(t) = 20.u(t) (V) f2(t) = - 20.u(t-50) (V) 50 t 50 t - 20.u(t-50) - 20V f(t) =f1(t) +f2(t) =20.u(t) - 20.u(t-50) (V) So the Laplace transform of f(t) is: L f t    20 20 50 s 20  e   e 50 s s s s  FUNDAMENTALS OF ELECTRIC CIRCUITS – AC Circuits 15.4 The inverse Laplace transform m N(s) b0  b1s b2s   bms  n D(s) a0  a1s a2s   ans + Suppose F(s) has the general form: Assume that m < n + Simple poles: s1 , s2 , …, sn f t   A1e s1t  A2 e s2t   An e snt where Ak  lim + Repeated poles: s1 = s2 = … = sn = s*  f t   A1  A2t   Ant n 1 e s t d 0  where An  lim s  s ds s  sk  N s  d n 1  n  Ds  s  s  A1  slim  s  ds    + Complex poles: s1,2 = -α ± jβ f t   Ak e t cost   k  where N s  D ' s  Ak  lim s   j  N s   Ak  k ' D s   N s   n  Ds  s  s      FUNDAMENTALS OF ELECTRIC CIRCUITS – AC Circuits 15.5 Application to circuits + Applying the Laplace transform to analyze circuits:  Transform the circuit from the time domain to the s domain  Solve the circuit using any circuit analysis technique  Take the inverse transform of the solution and thus obtain the solution in the time domain + Advantage:  The Laplace transform can be used readily to solve 1st & 2nd order circuit  Complete response (transient and steady state) of a network is obtained by using the Laplace transform FUNDAMENTALS OF ELECTRIC CIRCUITS – AC Circuits 15.5 Application to circuits +Transform the circuit from the time domain to the s domain: For an inductor or di(t) v(t)  L dt V(s)  sLI (s)  Li(0 ) For a capacitor: i(t)  C or dv(t) dt  v(0 ) V(s)  I (s)  sC s For a resistor: v(t)  Ri(t) i(0 ) I (s)  V(s)  sL s V(s)  RI(s) I (s)  sCV(s)  Cv(0 ) FUNDAMENTALS OF ELECTRIC CIRCUITS – AC Circuits 15.5 Application to circuits + Example 6: Find v0(t) in the circuit if v0(0) = 5V 10et u(t)V 2 (t)A Solution  Transform the circuit to the s domain  Apply the nodal analysis: 10 10 s1  V0   25 s  35 N s V V s 1 V0     0.5   10 s  1s  2 Ds  10 10 s  s1  1 Ds      v0 t   A1e t  A2 e  2t u t  s2  2 N s  10 N s  15 D ' s   s  A1  lim '   10 A2  lim '   15 s  1 D s  s   D s       v0 t   10e  t  15e 2t u t V 2A FUNDAMENTALS OF ELECTRIC CIRCUITS – AC Circuits 15.5 Application to circuits + Example 7: In the circuit, the switch moves from position a to position b at t = Find i(t) for t > I0 Solution The initial current through the inductor is: i(0) = I0 For t > 0, transform the circuit to the s domain Using mesh analysis, we have:  I s   I s R  sL   LI  V0 0 s LI V0 I0 V0 / L    R  sL sR  sL  s  R / L ss  R / L  Applying the inverse Laplace transform: Transient response  V0  t / V0  i t    I  e  u t  R R  Natural response  Forced response R L FUNDAMENTALS OF ELECTRIC CIRCUITS – AC Circuits 15.5 Application to circuits R1  10 + Example 8: In the circuit, the switch moves from position to position at t = Find i(t) for t > if e1(t) = 100sin(103t) V, e2(t) = 100e-20t V Solution R2  100 K XC  10 e1 (t) Find the initial condition of the capacitor: U c    e2 (t) E1  jX c   100   450V R1  jX c 100 uc t   sin 103 t  450 V  uc 0    50V □ For t > 0, transform the circuit to the s domain: u (0) uc 0   s E2 s     s  10 N s s I s      i t   A1e  20t  A2 e 100t 1t A s  20s  100 Ds  R2  sC N s  N s  A1  lim '  0.25 A2  lim '  1.75  i t    0.25e 20t  1.75e 100t 1t  s  100 D s  s  20 D s  C     sC R2 E2 (s) FUNDAMENTALS OF ELECTRIC CIRCUITS – AC Circuits 15.5 Application to circuits L1  2H Solution Find the initial condition of two inductors: iL1 0    E i 0    1A  iL 0    L1  0.5 A R2 R3 R1  R2  R3 R2  10 R1  5 E s   L1iL1 0    MiL 0    L2iL 0    MiL1 0   R1  R2  sL1  sL2  sM 0.75s  I s    i t   0.67  0.083e 7.5t 1t A ss  7.5   M  1H E  10V K For t > 0, transform the circuit to the s domain: I s   R3  10 L2  2H + Example 9: In the circuit, the switch is opened at t = Find i(t) for t > sL1 L1i L1 (0) Mi L2 (0) R1 10 s sM Mi L1(0) Li L2 (0) sL2 R2