ECEN3714 Network Analysis Lecture #21 2 March 2015 Dr. George Scheets n Read 14.7 n Problems: 14.5, 7, & 55 n Quiz #6 this Friday n Quiz #5 Results Hi = 10, Low = 4.0, Average = 7.89 Standard Deviation = 2.02

ECEN3714 Network Analysis Lecture #24 9 March 2015 Dr. George Scheets n Problems: 14.11, 22,37 n No Quizzes or Labs week prior to Spring Break n Exam #2 on 3 April n Note: Lab Practicum dropped u 10 Labs & 1 Design Problem u Still worth 220 points total in the end ≈ 1/3 of grade

ECEN3714 Network Analysis Lecture #25 11 March 2015 Dr. George Scheets n Problems: 14.51, 52, 57 n Next Quiz is week after Spring Break n Exam #2 on 3 April

ECEN3714 Network Analysis Lecture #26 13 March 2015 Dr. George Scheets n Problems: 14.58, 59, 72 n Next Quiz is week after Spring Break n Exam #2 on 3 April

Transforms X(s) = x(t) e -st dt 0-0- ∞ Laplace s = σ +jω ∞ X(f) = x(t) e -j2πft dt -∞-∞ Fourier

Got a Laplace Transform? Re(s) = σ Im(s) = jω |V(s)| The Fourier Transform of x(t) is on the jω axis.* *Provided x(t) = 0; t < 0

y(t) = x(t) + y(t-1): H(s) = 1/[1 – e –s ] σ = 0 Frequency Response σ = 0 axis Re(s) = σ Im(s) = jω |V(s)| ω = 0

Generating a Square Wave Hz + 15 Hz + 25 Hz + 35 Hz cos2*pi*5t - (1/3)cos2*pi*15t + (1/5)cos2*pi*25t - (1/7)cos2*pi*35t) 5 cycle per second square wave generated using 4 sinusoids

Generating a Square Wave... 5 cycle per second square wave generated using 100 sinusoids. Max frequency at (N*10 – 5) Hz = 995 Hz

5 Hz square wave after Single Pole Low Pass Filtering Hz half power frequency Not much visible change. Blue = Error = Output(t) – Input(t)

Generating a Square Wave... 5 cycle per second square wave generated using 100 sinusoids. Max frequency at (N*10 – 5) Hz = 995 Hz

5 Hz square wave after Single Pole High Pass Filtering Hz half power frequency

Filters n Low Pass, High Pass, & Band Pass u All are LTI if made from R's, C's, and/or L's n LTI Systems have an impulse response h(t) u δ(t) in? h(t) is output. n If LTI, for any input x(t), In Time Domain: x(t)☺h(t) = y(t) In Laplace Domain: X(s)H(s) = Y(s) In Frequency Domain: X(jω)H(jω) = Y(jω) n If not LTI, h(t) does not exist n We'll hit ☺ in more detail later

OpAmps n High Gain Devices u Typically, Voltage Gain > 10,00 u Vout(t) = [Vp(t) – Vn(t)]*Voltage Gain n High Input Impedance u Zin typically > 1 MΩ

Integrator: H(jω) = -1/jωRC ω |H(ω)|

5 Hz Square Wave In This one made up of 100 sinusoids. Fundamental frequency of 5 Hz & next 99 harmonics (15, 25,..., 995 Hz).

5 Hz Triangle Out

Differentiator: H(jω) = -jωRC ω |H(ω)|

5 Hz Square Wave In This curve made up of 100 sinusoids. Fundamental frequency of 5 Hz & next 99 harmonics (15, 25,..., 995 Hz).

Spikes Out