ECEN3714 Network Analysis Lecture #36 13 April 2015 Dr. George Scheets n Read 15.1 (thru example 15.4) Problems: Olde Quiz #10 Problems: Olde Quiz #10 n Quiz #9 this Friday (opamps) n Quiz #8 Results Hi = 9.0, Low = 4.9, Ave = 6.37 Standard Deviation = 1.23

ECEN3714 Network Analysis Lecture #37 15 April 2015 Dr. George Scheets n Read 15.1 (example 15.4 to end) n Problems: Quiz #11 (Fourier Series) n Quiz #9 this Friday (opamps)

ECEN3714 Point Wise (670 points max) n In the Classroom (450 points) u 280 points down, 170 to go (2 quizzes & 1 final) n In the Lab (130 x = 220 points) u 80 points down, 50 to go (lab 9, 10, design project) u down, 84.6 to go n Syllabus A > 90% (603), B > 80% (536), C > 70% (469), D > 60% (402) n Class will be curved. Previously… A > 85% (570), B > 73% (489), C > 61% (409), D > 49% (328) u This year's most likely different… … but probably in the same ball park n To estimate where you are now [2(Test & Quiz Ave.) + (Lab Ave.)]/3

Catastrophic Point Loss on Final n You have an OpAmp with negative feedback… u And V - not pinned to a specific value u And you fail to treat V + = V - n You have current entering/exiting an OpAmp input n You show a current entering a device… u And have the input side labeled "-" & output "+"

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 = axis Re(s) = σ Im(s) = jω |V(s)| ω = 0

Generating a Square Wave Hz + 15 Hz + 25 Hz + 35 Hz (1) cos2*pi*5t - (1/3)cos2*pi*15t + (1/5)cos2*pi*25t + (1/7)cos2*pi*35t) ( ) terms are a n 5 Hz is the Fundamental Frequency This waveform has 3rd, 5th, and 7th harmonics. Need to multiply by 4/π to get a 2 v peak-to-peak waveform.

8 vp Sine + 4 vp Cosine Magnitude = ( ) 0.5 = vp angle? = atan(8/4) = 63.43º angle? = atan(-8/4) = º What we want

Cosine with 2 Different Phase Shifts Magnitude = ( ) 0.5 = angle = atan(8/4) = 63.43º angle = atan(-8/4) = º What we want

Cosine with 2 Different Phase Shifts Magnitude = ( ) 0.5 = angle = atan(8/4) = 63.43º angle = atan(-8/4) = º

Integrals of Products n Comes back with a number (function of β) u Tells how "alike" x(α) and y(α) are u Tells how much of y(α) is in x(α) F & Vice-Versa z(β) = x(α) y(α,β) dα

Time Domain Convolution n z(t1) positive? x(α) & y(t1-α) "alike", tend to have same sign n z(t1) negative? x(α) & y(t1-α) "opposites", tend to have opposite sign n z(t1) near zero? x(α) & y(t1-α) not very similar z(t) = x(α) y(t-α) dα +∞ -∞

Time Domain Correlation n z(τ1) positive? x(t) & y(t+τ1) "alike", tend to have same sign, have some predictability n y(τ1) negative? x(t) & y(t+τ1) "opposites", tend to have opposite sign, have some predictability n y(τ1) near zero? x(t) & y(t+τ1) not very similar. Little or no predictability. z(τ) = x(t) y(t+τ) dt +∞ -∞

Laplace Transform F(s) = f(t) e -st dt 0-0- ∞ n F(s) tells how much of e -st is in f(t) n F(3) tells how alike f(t) and e -3t are u Over the time interval 0- to infinity

Fourier Transform F(ω) = f(t) e -jωt dt - ∞ ∞ n F(ω) tells how much of e -jωt is in f(t) n F(3) tells how alike f(t) and e -j3t are u Over the time interval - u Over the time interval - ∞ to +∞ u u Where e -j3t = cos(3t) + jsin(3t)

Fourier Series a n = 2/T f(t) cos(nω o t) dt 0 n a 3 tells how alike f(t) and n a 3 tells how alike f(t) and cos(3ω o t) are u Over one period, T u 1/T = average u 2 = scaling factor to get power correct T