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4.1 Laplace Transforms, Inverse Transforms & Circuit Applications
PRC Registered Electrical Engineer REE
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4.1 Laplace Transforms, Inverse Transforms & Circuit Applications
Key Takeaways Explain This Topic AI
The direct Laplace transform L { f ( t ) } = F ( s ) = ∫ 0 ∞ f ( t ) e − s t d t \mathcal{L}\{f(t)\} = F(s) = \int_0^\infty f(t)e^{-st}\,dt L { f ( t )} = F ( s ) = ∫ 0 ∞ f ( t ) e − s t d t converts time-domain differential equations into algebraic s s s -domain equations.
Operational properties including first shifting L { e − a t f ( t ) } = F ( s + a ) \mathcal{L}\{e^{-at}f(t)\} = F(s+a) L { e − a t f ( t )} = F ( s + a ) , second shifting L { f ( t − a ) u ( t − a ) } = e − a s F ( s ) \mathcal{L}\{f(t-a)u(t-a)\} = e^{-as}F(s) L { f ( t − a ) u ( t − a )} = e − a s F ( s ) , and differentiation L { f ′ ( t ) } = s F ( s ) − f ( 0 − ) \mathcal{L}\{f'(t)\} = sF(s) - f(0^-) L { f ′ ( t )} = s F ( s ) − f ( 0 − ) enable transient system modeling.
The Initial Value Theorem ( lim s → ∞ s F ( s ) = f ( 0 + ) \lim_{s \to \infty} sF(s) = f(0^+) lim s → ∞ s F ( s ) = f ( 0 + ) ) and Final Value Theorem ( lim s → 0 s F ( s ) = f ( ∞ ) \lim_{s \to 0} sF(s) = f(\infty) lim s → 0 s F ( s ) = f ( ∞ ) ) determine boundary states without full inverse transformation.
Inverse Laplace transformation utilizes partial fraction expansion across real distinct, repeated, and complex conjugate pole pairs to restore time-domain signals.
In electrical circuit analysis, s s s -domain impedance models ( Z R = R Z_R = R Z R = R , Z L = s L Z_L = sL Z L = s L , Z C = 1 s C Z_C = \frac{1}{sC} Z C = s C 1 ) and initial condition generators facilitate transient voltage and current derivations.
Last updated: August 2026
4.1 Laplace Transforms, Inverse Transforms & Circuit Applications
The Laplace transform is one of the most powerful analytical tools in electrical engineering and applied mathematics. It converts complex continuous-time differential equations governing dynamic physical systems into simple linear algebraic equations in the complex frequency domain ( s s s -domain, where s = σ + j ω s = \sigma + j\omega s = σ + jω ). For the PRC Registered Electrical Engineer (REE) Licensure Examination, mastering Laplace transforms is essential for solving circuit transients, transfer functions, stability criteria, and control system responses.
- 1. Definition and Region of Convergence
- Direct Unilateral Laplace Transform
For a time-domain function f ( t ) f(t) f ( t ) defined for t ≥ 0 t \ge 0 t ≥ 0 , the unilateral direct Laplace transform is defined by integral: L { f ( t ) } = F ( s ) = ∫ 0 − ∞ f ( t ) e − s t d t \mathcal{L}\{f(t)\} = F(s) = \int_{0^-}^\infty f(t) e^{-st} \, dt L { f ( t )} = F ( s ) = ∫ 0 − ∞ f ( t ) e − s t d t
where s = σ + j ω s = \sigma + j\omega s = σ + jω is the complex frequency parameter in rad/s \text{rad/s} rad/s , and lower limit 0 − 0^- 0 − captures potential instantaneous discontinuities (such as impulse functions δ ( t ) \delta(t) δ ( t ) ) at t = 0 t = 0 t = 0 .
Region of Convergence (ROC)
The Region of Convergence (ROC) consists of all points in the complex s s s -plane for which the defining integral converges absolutely. For causal signals, the ROC is a half-plane Re { s } > σ 0 \text{Re}\{s\} > \sigma_0 Re { s } > σ 0 , where σ 0 \sigma_0 σ 0 is the axis of convergence.
2. Standard Laplace Transform Pairs
The table below summarizes elementary transform pairs routinely encountered in engineering mathematics and circuit transient problems:
Time-Domain Signal f ( t ) , t ≥ 0 f(t), t \ge 0 f ( t ) , t ≥ 0 s s s -Domain Transform F ( s ) = L { f ( t ) } F(s) = \mathcal{L}\{f(t)\} F ( s ) = L { f ( t )} Region of Convergence (ROC)
- Unit Impulse δ ( t ) \delta(t) δ ( t ) 1 1 1 All s s s
- Unit Step u ( t ) u(t) u ( t ) 1 s \frac{1}{s} s 1 Re { s } > 0 \text{Re}\{s\} > 0 Re { s } > 0
- Ramp t t t 1 s 2 \frac{1}{s^2} s 2 1 Re { s } > 0 \text{Re}\{s\} > 0 Re { s } > 0
Polynomial t n ( n = 1 , 2 , 3 , … ) t^n \quad (n = 1, 2, 3, \dots) t n ( n = 1 , 2 , 3 , … ) n ! s n + 1 \frac{n!}{s^{n+1}} s n + 1 n ! Re { s } > 0 \text{Re}\{s\} > 0 Re { s } > 0
Exponential e − a t e^{-at} e − a t 1 s + a \frac{1}{s + a} s + a 1 Re { s } > − a \text{Re}\{s\} > -a Re { s } > − a
Damped Ramp t e − a t t e^{-at} t e − a t 1 ( s + a ) 2 \frac{1}{(s + a)^2} ( s + a ) 2 1 Re { s } > − a \text{Re}\{s\} > -a Re { s } > − a
Sine Wave sin ( ω t ) \sin(\omega t) sin ( ω t ) ω s 2 + ω 2 \frac{\omega}{s^2 + \omega^2} s 2 + ω 2 ω Re { s } > 0 \text{Re}\{s\} > 0 Re { s } > 0
Cosine Wave cos ( ω t ) \cos(\omega t) cos ( ω t ) s s 2 + ω 2 \frac{s}{s^2 + \omega^2} s 2 + ω 2 s Re { s } > 0 \text{Re}\{s\} > 0 Re { s } > 0
Damped Sine e − a t sin ( ω t ) e^{-at} \sin(\omega t) e − a t sin ( ω t ) ω ( s + a ) 2 + ω 2 \frac{\omega}{(s + a)^2 + \omega^2} ( s + a ) 2 + ω 2 ω Re { s } > − a \text{Re}\{s\} > -a Re { s } > − a
Damped Cosine e − a t cos ( ω t ) e^{-at} \cos(\omega t) e − a t cos ( ω t ) s + a ( s + a ) 2 + ω 2 \frac{s + a}{(s + a)^2 + \omega^2} ( s + a ) 2 + ω 2 s + a Re { s } > − a \text{Re}\{s\} > -a Re { s } > − a
Hyperbolic Sine sinh ( a t ) \sinh(at) sinh ( a t ) a s 2 − a 2 \frac{a}{s^2 - a^2} s 2 − a 2 a Re { s } > ∣ a ∣ \text{Re}\{s\} > \vert a\vert Re { s } > ∣ a ∣
Hyperbolic Cosine cosh ( a t ) \cosh(at) cosh ( a t ) s s 2 − a 2 \frac{s}{s^2 - a^2} s 2 − a 2 s Re { s } > ∣ a ∣ \text{Re}\{s\} > \vert a\vert Re { s } > ∣ a ∣
3. Operational Theorems & Properties
Linearity L { a f 1 ( t ) + b f 2 ( t ) } = a F 1 ( s ) + b F 2 ( s ) \mathcal{L}\{a f_1(t) + b f_2(t)\} = a F_1(s) + b F_2(s) L { a f 1 ( t ) + b f 2 ( t )} = a F 1 ( s ) + b F 2 ( s )
First Shifting Theorem (Frequency Shift) L { e − a t f ( t ) } = F ( s + a ) \mathcal{L}\{e^{-at} f(t)\} = F(s + a) L { e − a t f ( t )} = F ( s + a )
Second Shifting Theorem (Time Shift) L { f ( t − a ) u ( t − a ) } = e − a s F ( s ) for a ≥ 0 \mathcal{L}\{f(t - a) u(t - a)\} = e^{-as} F(s) \quad \text{for } a \ge 0 L { f ( t − a ) u ( t − a )} = e − a s F ( s ) for a ≥ 0
Differentiation in Time Domain
First Derivative: L { f ′ ( t ) } = s F ( s ) − f ( 0 − ) \mathcal{L}\{f'(t)\} = s F(s) - f(0^-) L { f ′ ( t )} = s F ( s ) − f ( 0 − )
Second Derivative: L { f ′ ′ ( t ) } = s 2 F ( s ) − s f ( 0 − ) − f ′ ( 0 − ) \mathcal{L}\{f''(t)\} = s^2 F(s) - s f(0^-) - f'(0^-) L { f ′′ ( t )} = s 2 F ( s ) − s f ( 0 − ) − f ′ ( 0 − )
n n n -th Derivative: L { f ( n ) ( t ) } = s n F ( s ) − s n − 1 f ( 0 − ) − s n − 2 f ′ ( 0 − ) − ⋯ − f ( n − 1 ) ( 0 − ) \mathcal{L}\{f^{(n)}(t)\} = s^n F(s) - s^{n-1} f(0^-) - s^{n-2} f'(0^-) - \dots - f^{(n-1)}(0^-) L { f ( n ) ( t )} = s n F ( s ) − s n − 1 f ( 0 − ) − s n − 2 f ′ ( 0 − ) − ⋯ − f ( n − 1 ) ( 0 − )
Integration in Time Domain L { ∫ 0 t f ( τ ) d τ } = F ( s ) s \mathcal{L}\left\{ \int_0^t f(\tau) \, d\tau \right\} = \frac{F(s)}{s} L { ∫ 0 t f ( τ ) d τ } = s F ( s )
Differentiation in s s s -Domain (Frequency Differentiation) L { t ⋅ f ( t ) } = − d d s F ( s ) \mathcal{L}\{t \cdot f(t)\} = -\frac{d}{ds} F(s) L { t ⋅ f ( t )} = − d s d F ( s ) L { t n ⋅ f ( t ) } = ( − 1 ) n d n d s n F ( s ) \mathcal{L}\{t^n \cdot f(t)\} = (-1)^n \frac{d^n}{ds^n} F(s) L { t n ⋅ f ( t )} = ( − 1 ) n d s n d n F ( s )
Convolution Theorem
The Laplace transform of the time-domain convolution of two signals equals the algebraic product of their individual transforms: L { f 1 ( t ) ∗ f 2 ( t ) } = L { ∫ 0 t f 1 ( τ ) f 2 ( t − τ ) d τ } = F 1 ( s ) ⋅ F 2 ( s ) \mathcal{L}\{f_1(t) * f_2(t)\} = \mathcal{L}\left\{ \int_0^t f_1(\tau) f_2(t - \tau) \, d\tau \right\} = F_1(s) \cdot F_2(s) L { f 1 ( t ) ∗ f 2 ( t )} = L { ∫ 0 t f 1 ( τ ) f 2 ( t − τ ) d τ } = F 1 ( s ) ⋅ F 2 ( s )
Boundary Value Theorems
Initial Value Theorem (IVT): If f ( t ) f(t) f ( t ) and f ′ ( t ) f'(t) f ′ ( t ) are Laplace transformable, then: f ( 0 + ) = lim t → 0 + f ( t ) = lim s → ∞ s F ( s ) f(0^+) = \lim_{t \to 0^+} f(t) = \lim_{s \to \infty} s F(s) f ( 0 + ) = t → 0 + lim f ( t ) = s → ∞ lim s F ( s )
Final Value Theorem (FVT): If all poles of s F ( s ) s F(s) s F ( s ) lie strictly in the left-half of the s s s -plane (system is stable), then: f ( ∞ ) = lim t → ∞ f ( t ) = lim s → 0 s F ( s ) f(\infty) = \lim_{t \to \infty} f(t) = \lim_{s \to 0} s F(s) f ( ∞ ) = t → ∞ lim f ( t ) = s → 0 lim s F ( s )
4. Inverse Laplace Transform & Partial Fractions
To convert an algebraic s s s -domain expression F ( s ) = P ( s ) Q ( s ) F(s) = \frac{P(s)}{Q(s)} F ( s ) = Q ( s ) P ( s ) back into the time domain f ( t ) f(t) f ( t ) , Partial Fraction Expansion decomposes proper rational functions into sum terms matching standard table forms.
Case I: Distinct Real Poles
If Q ( s ) = ( s − p 1 ) ( s − p 2 ) … ( s − p n ) Q(s) = (s - p_1)(s - p_2) \dots (s - p_n) Q ( s ) = ( s − p 1 ) ( s − p 2 ) … ( s − p n ) with distinct real roots: F ( s ) = A 1 s − p 1 + A 2 s − p 2 + ⋯ + A n s − p n F(s) = \frac{A_1}{s - p_1} + \frac{A_2}{s - p_2} + \dots + \frac{A_n}{s - p_n} F ( s ) = s − p 1 A 1 + s − p 2 A 2 + ⋯ + s − p n A n
Residue formula for coefficient A k A_k A k : A k = lim s → p k ( s − p k ) F ( s ) A_k = \lim_{s \to p_k} (s - p_k) F(s) A k = s → p k lim ( s − p k ) F ( s )
Case II: Repeated Real Poles
If Q ( s ) Q(s) Q ( s ) contains a repeated factor ( s − p ) m (s - p)^m ( s − p ) m of multiplicity m m m : F ( s ) = A 1 s − p + A 2 ( s − p ) 2 + ⋯ + A m ( s − p ) m + other terms F(s) = \frac{A_1}{s - p} + \frac{A_2}{(s - p)^2} + \dots + \frac{A_m}{(s - p)^m} + \text{other terms} F ( s ) = s − p A 1 + ( s − p ) 2 A 2 + ⋯ + ( s − p ) m A m + other terms
Residue formula for A m − k A_{m-k} A m − k : A m − k = 1 k ! lim s → p d k d s k [ ( s − p ) m F ( s ) ] A_{m-k} = \frac{1}{k!} \lim_{s \to p} \frac{d^k}{ds^k} \left[ (s - p)^m F(s) \right] A m − k = k ! 1 s → p lim d s k d k [ ( s − p ) m F ( s ) ]
Case III: Complex Conjugate Poles
If Q ( s ) Q(s) Q ( s ) contains quadratic term ( s + a ) 2 + ω 2 (s + a)^2 + \omega^2 ( s + a ) 2 + ω 2 with roots s = − a ± j ω s = -a \pm j\omega s = − a ± jω : B s + C ( s + a ) 2 + ω 2 = B ( s + a ) + ( C − a B ) ( s + a ) 2 + ω 2 = B s + a ( s + a ) 2 + ω 2 + C − a B ω ω ( s + a ) 2 + ω 2 \frac{B s + C}{(s + a)^2 + \omega^2} = \frac{B(s + a) + (C - a B)}{(s + a)^2 + \omega^2} = B \frac{s + a}{(s + a)^2 + \omega^2} + \frac{C - a B}{\omega} \frac{\omega}{(s + a)^2 + \omega^2} ( s + a ) 2 + ω 2 B s + C = ( s + a ) 2 + ω 2 B ( s + a ) + ( C − a B ) = B ( s + a ) 2 + ω 2 s + a + ω C − a B ( s + a ) 2 + ω 2 ω
Taking the inverse transform yields: f ( t ) = B e − a t cos ( ω t ) + C − a B ω e − a t sin ( ω t ) f(t) = B e^{-at} \cos(\omega t) + \frac{C - a B}{\omega} e^{-at} \sin(\omega t) f ( t ) = B e − a t cos ( ω t ) + ω C − a B e − a t sin ( ω t )
5. s s s -Domain Electrical Circuit Models
By transforming circuit elements into s s s -domain impedances along with initial condition energy sources, Kirchhoff's Laws (KVL and KCL) can be applied directly using standard linear algebra.
Element Time-Domain Relationship s s s -Domain Impedance Z ( s ) Z(s) Z ( s ) Initial Condition Source
Resistor v ( t ) = R ⋅ i ( t ) v(t) = R \cdot i(t) v ( t ) = R ⋅ i ( t ) Z R ( s ) = R Z_R(s) = R Z R ( s ) = R None
Inductor v ( t ) = L d i d t v(t) = L \frac{di}{dt} v ( t ) = L d t d i Z L ( s ) = s L Z_L(s) = s L Z L ( s ) = s L Series voltage source L i ( 0 − ) L i(0^-) L i ( 0 − ) (positive at entry node)
Capacitor i ( t ) = C d v d t i(t) = C \frac{dv}{dt} i ( t ) = C d t d v Z C ( s ) = 1 s C Z_C(s) = \frac{1}{s C} Z C ( s ) = s C 1 Series voltage source v ( 0 − ) s \frac{v(0^-)}{s} s v ( 0 − ) (matching initial polarity)
- Step-by-Step Worked Numerical Examples
- Example 4.1.1: Inverse Laplace Transform via Partial Fraction Expansion
Problem: Determine the inverse Laplace transform f ( t ) = L − 1 { F ( s ) } f(t) = \mathcal{L}^{-1}\{F(s)\} f ( t ) = L − 1 { F ( s )} for: F ( s ) = 10 s + 40 s 2 + 6 s + 25 F(s) = \frac{10 s + 40}{s^2 + 6 s + 25} F ( s ) = s 2 + 6 s + 25 10 s + 40
Solution:
Analyze the denominator quadratic: Discriminant Δ = 6 2 − 4 ( 1 ) ( 25 ) = 36 − 100 = − 64 < 0 \Delta = 6^2 - 4(1)(25) = 36 - 100 = -64 < 0 Δ = 6 2 − 4 ( 1 ) ( 25 ) = 36 − 100 = − 64 < 0 . The poles are complex conjugate pairs.
Complete the square for the denominator: s 2 + 6 s + 25 = ( s 2 + 6 s + 9 ) + 16 = ( s + 3 ) 2 + 4 2 s^2 + 6 s + 25 = (s^2 + 6 s + 9) + 16 = (s + 3)^2 + 4^2 s 2 + 6 s + 25 = ( s 2 + 6 s + 9 ) + 16 = ( s + 3 ) 2 + 4 2
Rewrite the numerator in terms of ( s + 3 ) (s + 3) ( s + 3 ) : 10 s + 40 = 10 ( s + 3 ) + 10 10 s + 40 = 10(s + 3) + 10 10 s + 40 = 10 ( s + 3 ) + 10
Split into standard damped cosine and sine forms: F ( s ) = 10 ( s + 3 ) + 10 ( s + 3 ) 2 + 4 2 = 10 ⋅ s + 3 ( s + 3 ) 2 + 4 2 + 10 4 ⋅ 4 ( s + 3 ) 2 + 4 2 F(s) = \frac{10(s + 3) + 10}{(s + 3)^2 + 4^2} = 10 \cdot \frac{s + 3}{(s + 3)^2 + 4^2} + \frac{10}{4} \cdot \frac{4}{(s + 3)^2 + 4^2} F ( s ) = ( s + 3 ) 2 + 4 2 10 ( s + 3 ) + 10 = 10 ⋅ ( s + 3 ) 2 + 4 2 s + 3 + 4 10 ⋅ ( s + 3 ) 2 + 4 2 4 F ( s ) = 10 ⋅ s + 3 ( s + 3 ) 2 + 4 2 + 2.5 ⋅ 4 ( s + 3 ) 2 + 4 2 F(s) = 10 \cdot \frac{s + 3}{(s + 3)^2 + 4^2} + 2.5 \cdot \frac{4}{(s + 3)^2 + 4^2} F ( s ) = 10 ⋅ ( s + 3 ) 2 + 4 2 s + 3 + 2.5 ⋅ ( s + 3 ) 2 + 4 2 4
Apply inverse transform matching table pairs: L − 1 { s + 3 ( s + 3 ) 2 + 4 2 } = e − 3 t cos ( 4 t ) \mathcal{L}^{-1}\left\{ \frac{s + 3}{(s + 3)^2 + 4^2} \right\} = e^{-3t} \cos(4t) L − 1 { ( s + 3 ) 2 + 4 2 s + 3 } = e − 3 t cos ( 4 t ) L − 1 { 4 ( s + 3 ) 2 + 4 2 } = e − 3 t sin ( 4 t ) \mathcal{L}^{-1}\left\{ \frac{4}{(s + 3)^2 + 4^2} \right\} = e^{-3t} \sin(4t) L − 1 { ( s + 3 ) 2 + 4 2 4 } = e − 3 t sin ( 4 t ) f ( t ) = 10 e − 3 t cos ( 4 t ) + 2.5 e − 3 t sin ( 4 t ) for t ≥ 0 f(t) = 10 e^{-3t} \cos(4t) + 2.5 e^{-3t} \sin(4t) \quad \text{for } t \ge 0 f ( t ) = 10 e − 3 t cos ( 4 t ) + 2.5 e − 3 t sin ( 4 t ) for t ≥ 0
Example 4.1.2: s s s -Domain Transient Response of Series RLC Circuit
Problem: A series RLC circuit consists of R = 6 Ω R = 6\ \Omega R = 6 Ω , L = 1 H L = 1\text{ H} L = 1 H , and C = 0.04 F = 1 25 F C = 0.04\text{ F} = \frac{1}{25}\text{ F} C = 0.04 F = 25 1 F . A step voltage source v ( t ) = 100 u ( t ) V v(t) = 100 u(t)\text{ V} v ( t ) = 100 u ( t ) V is connected at t = 0 t = 0 t = 0 . Assuming zero initial energy ( i ( 0 − ) = 0 A , v C ( 0 − ) = 0 V i(0^-) = 0\text{ A}, v_C(0^-) = 0\text{ V} i ( 0 − ) = 0 A , v C ( 0 − ) = 0 V ), derive the current response i ( t ) i(t) i ( t ) .
Solution:
Formulate the s s s -domain KVL equation: V ( s ) = I ( s ) [ R + s L + 1 s C ] V(s) = I(s) \left[ R + s L + \frac{1}{s C} \right] V ( s ) = I ( s ) [ R + s L + s C 1 ] Substitute V ( s ) = 100 s V(s) = \frac{100}{s} V ( s ) = s 100 , R = 6 R = 6 R = 6 , L = 1 L = 1 L = 1 , and 1 C = 25 \frac{1}{C} = 25 C 1 = 25 : 100 s = I ( s ) [ 6 + s + 25 s ] = I ( s ) [ s 2 + 6 s + 25 s ] \frac{100}{s} = I(s) \left[ 6 + s + \frac{25}{s} \right] = I(s) \left[ \frac{s^2 + 6s + 25}{s} \right] s 100 = I ( s ) [ 6 + s + s 25 ] = I ( s ) [ s s 2 + 6 s + 25 ]
Solve algebraically for I ( s ) I(s) I ( s ) : I ( s ) = 100 / s ( s 2 + 6 s + 25 ) / s = 100 s 2 + 6 s + 25 I(s) = \frac{100/s}{(s^2 + 6s + 25)/s} = \frac{100}{s^2 + 6s + 25} I ( s ) = ( s 2 + 6 s + 25 ) / s 100/ s = s 2 + 6 s + 25 100
Complete the square in the denominator: s 2 + 6 s + 25 = ( s + 3 ) 2 + 4 2 s^2 + 6s + 25 = (s + 3)^2 + 4^2 s 2 + 6 s + 25 = ( s + 3 ) 2 + 4 2
Format I ( s ) I(s) I ( s ) for inverse transformation: I ( s ) = 100 ( s + 3 ) 2 + 4 2 = 100 4 ⋅ 4 ( s + 3 ) 2 + 4 2 = 25 ⋅ 4 ( s + 3 ) 2 + 4 2 I(s) = \frac{100}{(s + 3)^2 + 4^2} = \frac{100}{4} \cdot \frac{4}{(s + 3)^2 + 4^2} = 25 \cdot \frac{4}{(s + 3)^2 + 4^2} I ( s ) = ( s + 3 ) 2 + 4 2 100 = 4 100 ⋅ ( s + 3 ) 2 + 4 2 4 = 25 ⋅ ( s + 3 ) 2 + 4 2 4
Take inverse Laplace transform: i ( t ) = 25 e − 3 t sin ( 4 t ) A for t ≥ 0 i(t) = 25 e^{-3t} \sin(4t) \text{ A} \quad \text{for } t \ge 0 i ( t ) = 25 e − 3 t sin ( 4 t ) A for t ≥ 0
Example 4.1.3: Boundary Analysis via Initial and Final Value Theorems
Problem: A control system closed-loop output in the frequency domain is given by: Y ( s ) = 5 s + 20 s ( s 2 + 4 s + 13 ) Y(s) = \frac{5 s + 20}{s (s^2 + 4 s + 13)} Y ( s ) = s ( s 2 + 4 s + 13 ) 5 s + 20
Find the initial value y ( 0 + ) y(0^+) y ( 0 + ) and the steady-state final value y ( ∞ ) y(\infty) y ( ∞ ) .
Solution:
Apply Initial Value Theorem (IVT): y ( 0 + ) = lim s → ∞ s Y ( s ) = lim s → ∞ s [ 5 s + 20 s ( s 2 + 4 s + 13 ) ] = lim s → ∞ 5 s + 20 s 2 + 4 s + 13 y(0^+) = \lim_{s \to \infty} s Y(s) = \lim_{s \to \infty} s \left[ \frac{5 s + 20}{s (s^2 + 4 s + 13)} \right] = \lim_{s \to \infty} \frac{5 s + 20}{s^2 + 4 s + 13} y ( 0 + ) = s → ∞ lim s Y ( s ) = s → ∞ lim s [ s ( s 2 + 4 s + 13 ) 5 s + 20 ] = s → ∞ lim s 2 + 4 s + 13 5 s + 20 Divide numerator and denominator by s 2 s^2 s 2 : y ( 0 + ) = lim s → ∞ 5 / s + 20 / s 2 1 + 4 / s + 13 / s 2 = 0 + 0 1 + 0 + 0 = 0 y(0^+) = \lim_{s \to \infty} \frac{5/s + 20/s^2}{1 + 4/s + 13/s^2} = \frac{0 + 0}{1 + 0 + 0} = 0 y ( 0 + ) = s → ∞ lim 1 + 4/ s + 13/ s 2 5/ s + 20/ s 2 = 1 + 0 + 0 0 + 0 = 0
Apply Final Value Theorem (FVT): Check pole stability: Poles of s Y ( s ) s Y(s) s Y ( s ) are roots of s 2 + 4 s + 13 = 0 ⟹ s = − 2 ± j 3 s^2 + 4s + 13 = 0 \implies s = -2 \pm j3 s 2 + 4 s + 13 = 0 ⟹ s = − 2 ± j 3 , which both reside in the left-half s s s -plane (stable system). y ( ∞ ) = lim s → 0 s Y ( s ) = lim s → 0 s [ 5 s + 20 s ( s 2 + 4 s + 13 ) ] = lim s → 0 5 s + 20 s 2 + 4 s + 13 y(\infty) = \lim_{s \to 0} s Y(s) = \lim_{s \to 0} s \left[ \frac{5 s + 20}{s (s^2 + 4 s + 13)} \right] = \lim_{s \to 0} \frac{5 s + 20}{s^2 + 4 s + 13} y ( ∞ ) = s → 0 lim s Y ( s ) = s → 0 lim s [ s ( s 2 + 4 s + 13 ) 5 s + 20 ] = s → 0 lim s 2 + 4 s + 13 5 s + 20 Substitute s = 0 s = 0 s = 0 : y ( ∞ ) = 5 ( 0 ) + 20 0 2 + 4 ( 0 ) + 13 = 20 13 ≈ 1.5385 y(\infty) = \frac{5(0) + 20}{0^2 + 4(0) + 13} = \frac{20}{13} \approx 1.5385 y ( ∞ ) = 0 2 + 4 ( 0 ) + 13 5 ( 0 ) + 20 = 13 20 ≈ 1.5385
Conclusion: y ( 0 + ) = 0 y(0^+) = 0 y ( 0 + ) = 0 and y ( ∞ ) = 1.5385 y(\infty) = 1.5385 y ( ∞ ) = 1.5385 .