NAS-Ch4-Application of Laplace Transform
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Chapter 4 discusses the application of the Laplace transform in solving differential equations, emphasizing its superiority in simplifying complex calculations by transforming the time domain into the s-domain. Key concepts include the definitions and properties of singularity functions such as the unit step and impulse functions, as well as their Laplace transforms. The chapter also covers circuit analysis techniques using Laplace transforms and the convolution theorem for two functions.
NAS-Ch4-Application of Laplace Transform
- 1. Network Analysis and Synthesis Chapter 4: Application of Laplace Transform By Dr. K Hussain Associate Professor and Head Dept. of EE, SITCOE
- 2. Contents • Solution of differential equation using Laplace transform, Unit step, Impulse & ramp functions • Laplace transform of singular & shifted function, Convolution integral • Concept of complex frequency • Transform impedance & transform admittance, Series & parallel combination of these transform networks.
- 3. Laplace Transform An integral transform mapping functions from the time domain to the Laplace domain or s‐domain f Time‐domain functions are functions of time, t Laplace‐domain functions are functions of s s is a complex variable
- 4. Laplace Transforms – Motivation We’ll use Laplace transforms to solve differential equations Differential equations in the time domain difficult to solve Apply the Laplace transform Transform to the s‐domain Differential equations become algebraic equations easy to solve Transform the s‐domain solution back to the time domain Transforming back and forth requires extra effort, but the solution is greatly simplified
- 5. The Laplace Transform (LT) method is much superior to the classical method due to the following reasons. 1. Laplace transformation transforms exponential and trigonometric functions into algebraic functions. 2. Laplace transformation transforms differentiation and integration into multiplication and division respectively. 3. It transforms integro-differential equations into algebraic equations which are much simpler to handle. 4. The arbitrary constants need not be determined separately. Complete solution will be obtained directly.
- 6. Laplace Transform The Laplace transform provides a useful method of solving certain types of differential equations when certain initial conditions are given, especially when the initial values are zero. It is also very useful in the area of circuit analysis. It is often easier to analyze the circuit in its Laplace form, than to form differential equations. The techniques of Laplace transform are not only used in circuit analysis, but also in •Proportional-Integral-Derivative (PID) controllers •DC motor speed control systems •DC motor position control systems •Second order systems of differential equations (underdamped, overdamped and critically damped)
- 7. Definition of Laplace Transform The Laplace transform ℒ, of a function f(t) for t > 0 is defined by the following integral over 0 to ∞: The resulting expression is a function of s, which we write as F(s). In words we say "The Laplace Transform of f(t) equals function F of s". ℒ {f(t)}=F(s) ℒ {g(t)}=G(s) Similarly, the Laplace transform of a function g(t) would be written: and write: ℒ {f(t)}=∫0 ∞ e−stf(t)dt Unilateral or one‐sided transform Lower limit of integration is Assumed that the time domain function is zero for all negative time, i.e., g(t)<0
- 8. • The two singularity functions of interest are 1. unit step function, u(t) and its construct: the gate function 2. delta or unit impulse function, (t) and its construct: the sampling function Laplace Transforms of Singularity Functions • Singularity functions are either not finite or don't have finite derivatives everywhere
- 9. Step Function A useful and common way of characterizing a linear system is with its step response The system’s response (output) to a unit step input The unit step function or Heaviside step function: 01 00 )( t t tu
- 10. Extensions of Unit Step Function • A more general unit step function is u(t-a) at at atu 1 0 )( 1 t0 a 1 t0 +T • The gate function can be constructed from u(t) – a rectangular pulse that starts at t= and ends at t=+T : u(t-) – u(t--T) – like an on/off switch
- 11. Unit Step Function – Laplace Transform Using the definition of the Laplace transform The Laplace transform of the unit step is Note that the unilateral Laplace transform assumes that the signal being transformed is zero for t<0 Equivalent to multiplying any signal by a unit step £{f(t)}=1/s
- 12. Impulse Function Another common way to describe a dynamic system is with its impulse response System output in response to an impulse function input Impulse function: If a unit step function u(t) is differentiated with respect to t, the derivative is zero for time t greater than zero, and is infinite for time t equal to zero. An infinitely tall, infinitely narrow pulse and where the symbol δ(t) (delta) is used to represent the unit impulse. Mathematically, the function is defined as
- 13. Delta or Unit Impulse Function, (t) • The delta or unit impulse function, (t) – Mathematical definition – Graphical illustration 0 0 0 1 0 )( tt tt tt 1 t0 (t) t0
- 14. Extensions of the Delta Function • An important property of the unit impulse function is its sampling property – Mathematical definition 00 0 0 )( 0 )()( tttf tt tttf f(t) t0 t0 f(t) (t-t0)
- 15. The Laplace transform of the unit impulse function is
- 16. Unit Ramp Function – Laplace Transform Could easily evaluate the transform integral Requires integration by parts Alternatively, recognize the relationship between the unit ramp and the unit step Unit ramp is the integral of the unit step The unit ramp function is a useful input signal for evaluating how well a system tracks a constantly‐ increasing input 0 00 )( tt t tu The unit ramp function:
- 17. Apply the integration by parts, Laplace transform of unit Ramp function is 1 ! )()()( n n s n sFtuttf In general 66 5 2 1 120!5 )()()(,5 !1 )()()(,1 1!0 )()()(,0 ss sFtuttfn s sFttutfn ss sFtutfn
- 18. If L{f(t)}=F(s) when s>a then, In words, the substitution (s−a) for s in the transform corresponds to the multiplication of the original function by eat. Proof: L{eatf(t)}=F(s−a) Laplace Transform of Shifted Function First (frequency) Shifting Property: F(s−a)=L{eatf(t)} F(s)=∫0 ∞e−st f(t)dt F(s-a)=∫0 ∞e−(s-a)t f(t)dt F(s-a)=∫0 ∞e−st+at f(t)dt F(s-a)=∫0 ∞e−st eat f(t)dt F(s-a)= eat ∫0 ∞e−st f(t)dt Therefore
- 19. Second (Time) Shifting Property If L{f(t)}=F(s) and g(t)= f(t−a) t>a 0 t<a then, L{g(t)}=e−asF(s) Proof: g(t)= f(t−a) t>a 0 t<a L{g(t)}=∫0 ∞e−st g(t)dt L{g(t)}=∫0 ae−st (0)dt+ ∫a ∞e−st f(t-a)dt L{g(t)}=∫a ∞e−st f(t-a)dt
- 20. Let z=t−a t=z+a dt=dz L{g(t)}=e−asF(s)Therefore L{g(t)}=∫a ∞e−st f(t-a)dt L{g(t)}=∫0 ∞e−sz−sa f(z)dz L{g(t)}=∫0 ∞e−sze−sa f(z)dzL{g(t)}=∫0 ∞e−s(z+a)f(z)dz when t=a, z=0 when t=∞, z=∞ L{g(t)}=e−sa∫0 ∞e−szf(z)dz L{g(t)}=e−asL{f(t−a)}L{g(t)}=e−asL{f(z)}
- 24. Convolution Theorem states that if we have two functions, taking their convolution and then Laplace is the same as taking the Laplace first (of the two functions separately) and then multiplying the two Laplace Transforms. Theorem: For any two functions f(t) and g(t) with Laplace transforms F(s) and G(s) we have L(f ∗ g) = F(s) · G(s). Proof: We start by writing L(f ∗ g) as the convolution integral followed by the Laplace integral Convolution Integral
- 25. Next, we change the order of integration Finally, change variables in the inner integral: substitute v = t−u, dv = dt, (u =a constant)
- 26. Q. Find y(t) from the given functions using convolution integral. Solution: To evaluate the convolution integral we will use the convolution property of the Laplace Transform: We need the Laplace Transforms of f(t) and h(t), but we can look them up in the tables: So, (t) is step function
- 27. We can now write y(t) (which is implicitly zero for t<0) We can look up both of these terms in the tables.
- 28. Solution of Differential Equation using Laplace Transform
- 29. s-Domain Circuit Analysis Time domain (t domain) Complex frequency domain (s domain) Linear Circuit Differential equation Classical techniques Response waveform Laplace Transform Inverse Transform Algebraic equation Algebraic techniques Response transform L L-1 Laplace Transform L Transformed Circuit
- 30. Kirchhoff’s Laws in s-Domain s domain i2 (t) t domain Kirchhoff’s current law (KCL) i1(t) i2 (t) i3 (t) i4 (t) 0 Kirchhoff’s voltage law (KVL) v1(t) v2 (t) v3 (t) 0 V1(s) V2 (s) V3(s) 0 i1(t) i3(t) i4 (t) I1(s) I2 (s) I3 (s) I4 (s) 0 v2(t) v4 (t) 5v1(t) v3 (t) v (t)
- 31. Signal Sources in s Domain L L t domain s domain + v(t) _ vS (t) + i(t) Voltage Source: v(t) vS (t) i(t) depends on circuit _ + V(s) + I(s) Voltage Source: VS (s) V (s) VS (s) I(s) depends on circuit _ v(t) + iS (t) i(t) Current Source: i(t) iS(t) v(t) depends on circuit IS (s)V(s) + _ I(s) Current Source: I(s) IS (s) V (s) depends on circuit --
- 32. Time and s-Domain Element Models L L L + vR (t) _ R vR (t) RiR(t) Resistor: + VR (s) _ R Impedance and Voltage Source for Initial Conditions Time Domain s-Domain VR (s) RIR (s) Resistor: + v (t)L _ L iL(t) dt L L di (t) v (t) L Inductor: _ + + V (s)L _ Ls LLi (0) IL (s) LLi (0) VL (s) LsIL (s) Inductor: _ + v (t)C C iC (t) 0 C t CC C v (0) v (t) 1 i ( )d Capacitor: _ + + V (s)C _ 1 Cs C s v (0) IC (s) Cs C s C C v ( 0 ) V (s) 1 I (s) Capacitor: IR (s)iR (t)
- 33. Impedance and Voltage Source for Initial Conditions R R R R I (s) V (s) Z (s) L L L L Ls with i (0) 0 I (s) V (s) Z (s) with v (0) 0 I (s) Cs V (s) 1 Z (s) C C C C • Impedance Z(s) Z (s) voltage transform current transform with all initial conditions set to zero • Impedance of the three passive elements
- 34. Time and s-Domain Element Models L L L + vR (t) _ R R R Ri (t) 1 v (t) Resistor: + VR (s) _ R Admittance and Current Source for Initial Conditions Time Domain s-Domain R R RI (s) 1 V (s) Resistor: + v (t)L _ L iL(t) Inductor: 0 t LL L iL (0) i (t) 1 v ( )d _ + vC (t) C ic(t) dt C C dv (t) i (t) C Capacitor: Inductor: Ls L L iL(0) I (s) 1 V (s) + V (s)L _ Ls s iL(0) IL (s) C CvC (0) I (s) CsVC (s) s Capacitor: _ + VC (s) 1 Cs CCv (0) I (s)C iR (t) IR (s)
- 35. Admittance and Current Source for Initial Conditions R R R I (s) 1 V (s) R Y (s) L L L L with i (0) 0 V (s) Ls I (s) 1 Y (s) Cs with v (0) 0 V (s) I (s) Y (s) C C C C • Admittance Y(s) Y (s) current transform 1 voltage transform Z (s) with all initial conditions set to zero • Admittance of the three passive elements
- 36. L + VA(t) i(t) L + V (s)L _LLi (0) V (s) R +VA s Example: Solve for Current Waveform i(t). I(s) Ls _ + s R L VA By KVL: V (s) V (s) 0 Resistor: VR (s) RI(s) Inductor: VL (s) LsI(s) LiL (0) L RI(s) LsI(s) Li (0) 0 s VA VA L VA R VA R iL (0) s s R L s R L iL(0) s(s R L) s R L I (s) V LL V iL (0)e u(t A e R R Rt Rt Inverse Transform: i(t) A forced response natural response -- R R
- 37. Consider the RL circuit shown in Fig.(a). Assume that the switch is closed at time t = 0 and assume that the current i at the time of switching is zero. The transform circuit in s domain is shown in Fig.(b). From this, Taking inverse LT Thus, inductor current rises exponentially with time constant L/R. Analysis of RL Circuit using Laplace Transform Method
- 38. Inductor voltage decreases exponentially with time constant L/R. The current and voltage transients are shown in Fig. Voltage across the inductor is given by Taking inverse LT
- 39. Example: Initially relaxed series RL circuit with R = 100 Ω and L = 20 H has dc voltage of 200 V applied at time t = 0. Find (a) the equation for current and voltages across different elements (b) the current at time t = 0.5 s and 1.0 s and (c) the time at which the voltages across the resistor and inductor are equal. Solution: Transform circuit for time t > 0 is shown. Therefore, (a) current = i(t) = 2 (1-e-5t) A Voltage=VR = R i(t) = 200 (1-e-5t) V Voltage VL (c) Let t1 be the time at which VR(t) = VL(t). Then This gives t1 = 0.1386 s 200 (1- e-5t1)= 200 e-5t1 i.e., e-5t1 = 0.5 (b) i(0.5) = 2 (1- e-2.5) = 1.8358 A and i(1.0) = 2 (1-e-5) = 1.9865 A
- 40. Use Partial fraction method and then Apply inverse Laplace Transform Example: Solution: This is series R-L circuit.
- 41. Voltage across the capacitor Taking inverse LT, we get the capacitor voltage as The circuit current and the voltage across the capacitor vary as shown in Fig. The transform circuit for time t > 0 is shown in Fig.(b). From this Taking inverse LT Consider the RC circuit shown in Fig. (a). Assume that the switch is closed at time t = 0 and assume that the voltage across the capacitor at the time of switching is zero. Analysis of RC Circuit using Laplace Transform Method
- 42. Example: In the RC circuit shown in fig (a)below, the capacitor has an initial charge q0 = 2500 μC. At time t = 0, the switch is closed. Find the circuit current for time t > 0. Solution: Transform circuit for time t > 0 is shown in Fig. (b) Taking inverse LT, current i(t) = 15 e-2000t A Fig (a) Fig (b) From Fig (b)
- 43. Consider the RLC series circuit shown in Fig. (a). Assume that there is no initial charge on the capacitor and there is no initial current through the inductor. The switch is closed at time t = 0. Transform circuit for time t > 0 is shown in Fig. (b). Using the transform circuit, expression for the current is obtained as The roots of the denominator polynomial are Where and Analysis of RLC Circuit using Laplace Transform Method
- 44. Depending on whether , , or , the discriminant value will be positive, zero or negative and three different cases of solutions are possible. The value of R, for which the discriminant is zero, is called the critical resistance, RC. Then If the circuit resistance R > RC, then Thus If the circuit resistance R < RC, then
- 46. The plot of this current transient is shown in Fig. In this case, the current is said to be critically damped.
- 47. As seen in Equation, α will be a negative number. Thus, for this under damped case, the current is oscillatory and at the same time it decays. For this case, the roots are complex conjugate, s1=α+jβ and s2=α-jβ Taking inverse LT, we get Case (iii): Waveform shown is a exponentially decaying sinusoidal wave.
- 48. Example: For the RLC circuit shown, find the expression for the transient current when the switch is closed at time t = 0. Assume initially relaxed circuit conditions. Solution: The transform circuit is shown in Fig. (b) The roots of the denominator polynomial are
- 49. This is an example for over-damped. Taking inverse LT, we get current
- 50. Example: Taking the initial conditions as zero, find the transient current in the circuit shown in Fig. (a) when the switch is closed at time t = 0. Solution The transform circuit is shown in Fig (b). The roots of the denominator polynomial are Fig (b) It can be seen that Fig (a)
- 51. This is an example for under-damped. Taking inverse LT, we get
- 52. Network Functions Input Signal Transform Network function Zero-State ResponseTransform I(s) Y(s) V(s) 1 Z(s) V (s) • Transfer function relates an input and response at different ports in the circuit V (s) VT (s) Voltage Transfer Function 2 Circuit in the zero-state V(s) I(s) Circuit in the zero-state V1or I1 V2 or I2 Input Output _ + V2 TV (s)V1 In Out I1(s) 1 I (s) TI (s) Current Transfer Function 2 2 V (s) I (s) YT (s) Transfer Admittance I1(s) 1 V2 (s) TZ (s) Transfer Impedance _ + 1V 2I TY (s) In Out 1I V2 TZ (s) In Out I1 TI (s) In Out I2 Driving-point function relates the voltage and current at a given pair of terminals called a port
- 53. Calculating Network Functions Z1 Z2 V2(s)V1(s)_ + Y1 Y2I1(s) 2I (s) 1 1 2 2 Z2 (s)V (s) V V (s) Z (s) Z (s) T (s) • Driving-point impedance ZEQ (s) Z1(s) Z2 (s) • Voltage transfer function: 2 V (s) Z2 (s) 1 2 1Z (s) Z (s) V (s) 1 1 2 2 Y2(s)I (s) I I (s) Y (s) Y (s) T (s) • Driving-point admittance YEQ (s) Y1(s) Y2 (s) • Current transfer function: Y (s) Y1(s) Y2 (s) I2 (s) 2 I1(s)
- 54. Circuit Analysis in s-Domain The equivalent impedance Zeq(s) of two impedances Z1(s) and Z2(s) in series is V(s) Z2 Z1 + - Same current flows V(s) Z1(s)I s Z2 (s)I (s) Zeq (s)I s The equivalent admittance Yeq(s) of two admittances Y1(s) and Y2(s) in parallel is + s) Y1 Y2 - V( Same voltage I(s) Y1(s)V(s) Y2(s)V(s) Yeq(s)V(s) I(s) Basic Rules: I1(s) Yeq(s) Y1(s) Y2(s) Zeq (s) Z1(s) Z2 (s)
- 55. Limitation of Laplace Transform • Only be used to solve differential equations with known constants. • An equation without the known constants, then this method is useless. Laplace Transform is powerful tool using in differential areas of mathematics, physics, and engineering. But
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