ECEN3714 Network Analysis Lecture #9 2 February 2015 Dr. George Scheets n Read 13.8 n Problems: 13.16a, 19a, & 31c n Exam #1, Lecture 11, Friday 6 February Open Book & Notes, Closed Instructor u Chapter 13, Forward & Inverse Laplace Transforms u Simple R, L, & C circuits n Quiz 2 Results Hi = 8.0, Low = 5.6, Average = 5.60 Standard Deviation = 1.31

V(s) = 10s/D(s) D(s) = s s + 25 = (s + 5) 2 Re(s) D(s) -5 n Denominator has two real roots at s = -5 n Using Partial Fraction Expansion, can write as V(s) = -50/(s+5) /(s+5) u These two terms have transform pairs in the tables

V(s) = -50/(s+5) /(s+5) ↕ v(t) = -50te -5t + 10e -5t t v(t) 10

V(s) has 2 nd order pole at -5 Zero at 0 Re(s) V(s) = 10s/(s+5) 2 -5 Re(s) = σ Im(s) = jω x x -5 |V(s)| The figure above is a plot along the real axis.

3D Plot of |V(s)| = |10s/(s+5) | ω σ ω = 0 axis = real axis

v(t) = -50te -5t + 10e -5t t v(t) 10 Re(s) = σ Im(s) = jω x x -5 |V(s)| Frequency Content (magnitude) along σ = 0 axis

Fourier Transform n Is built into the Laplace Transform u Provided nothing much happens before t = 0 u Set σ = 0 and examine jω axis n Got a waveform x(t)? u F.T. indicates sinusoids needed to generate x(t) n Got a filter transfer function? u F.T. indicates frequencies system will pass or block.

V(s) = 10s/D(s) D(s) = s 2 + 2s + 25 Re(s) D(s) n Roots exist even if equation doesn't hit zero u They'll be complex

V(s) = 10s/(s 2 + 2s + 25) ↕ v(t) = 10.2e -t cos(4.899t π) t v(t) 10.2

V(s) has poles at -1 + j4.899 Zero at 0 s V(s) = __10s___ (s 2 +2s+25) Im(s) = jω x x Re(s) = σ The figure above is a plot along the real axis.

3D Plot of |V(s)| = 10s (s 2 +2s+25) ω σ ω = 0 axis = real axis

v(t) = 10.2e -t cos(4.899t π) t v(t) 10.2 Im(s) = jω x x Re(s) = σ Frequency Content (magnitude) along σ = 0 axis

V(s) = 10s/(s+5) 2 Re Im x x -5 t v(t) 10 v(t) = -50te -5t + 10e -5t Stability Issues: Location of poles on Real axis sets decay rate.

V(s) = 10s/(s2+2s+25) Im x x Re t 10.2 v(t) = 10.2e -t cos(4.899t π) Stability Issues: Location of poles on Real axis sets decay rate.

V(s) = 10s/(s2+2s+25) Im x x Re t 10.2 v(t) = 10.2e -t cos(4.899t π) Stability Issues: Complex conjugate poles indicate oscillation. Location of poles on Imaginary axis sets oscillation rate.

S Domain Voltage & Currents n Resistor (v = iR) V(s) = I(s) R I(s) = V(s)/R n Capacitor (i = C dv/dt) I(s) = sCV(s) – Cv(0 - ) V(s) = I(s)/(sC) – v(0 - )/s n Inductor (v = L di/dt) V(s) = sLI(s) – Li(0 - ) I(s) = V(s)/(sL) – i(0 - )/s