ECEN3714 Network Analysis Lecture #21 2 March 2015 Dr. George Scheets n Read 14.7 n Problems: 14.5,...

ECEN3714 Network AnalysisLecture #21 2 March 2015Dr. George Scheetswww.okstate.edu/elec-eng/scheets/ecen3714


Read 14.7 Problems: 14.5, 7, & 55 Quiz #6 this Friday Quiz #5 Results

Hi = 10, Low = 4.0, Average = 7.89Standard Deviation = 2.02



Problems: 14.11, 22,37 No Quizzes or Labs week prior to Spring Break Exam #2 on 3 April Note: Lab Practicum dropped

10 Labs & 1 Design Problem Still worth 220 points total in the end ≈ 1/3 of grade



Problems: 14.51, 52, 57 Next Quiz is week after Spring Break Exam #2 on 3 April



Problems: 14.58, 59, 72 Next Quiz is week after Spring Break Exam #2 on 3 April

TransformsTransforms

X(s) = x(t) e-st dt

0-

∞Laplace

s = σ +jω

∞

X(f) = x(t) e-j2πft dt

-∞Fourier

Got a Laplace Transform?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]y(t) = x(t) + y(t-1): H(s) = 1/[1 – e–s]

σ = 0

Frequency Responseσ = 0 axis

Re(s) = σ

Im(s) = jω

|V(s)|

ω = 0

Generating a Square Wave...Generating a Square Wave...

0

1.5

-1.50 1.0

5 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


5 cycle per second square wave generated using 100 sinusoids.

Max frequency at (N*10 – 5) Hz = 995 Hz

0

1.5

-1.50 1.0

5 Hz square wave after Single Pole Low Pass Filtering159.2 Hz half power frequency

5 Hz square wave after Single Pole Low Pass Filtering159.2 Hz half power frequency

0

1.5

-1.50 1.0

Not much visible change.Blue = Error = Output(t) – Input(t)


5 cycle per second square wave generated using 100 sinusoids.

Max frequency at (N*10 – 5) Hz = 995 Hz

0

1.5

-1.50 1.0

5 Hz square wave after Single Pole High Pass Filtering159.2 Hz half power frequency

5 Hz square wave after Single Pole High Pass Filtering159.2 Hz half power frequency

0

1.5

-1.50 1.0

FiltersFilters Low Pass, High Pass, & Band Pass

All are LTI if made from R's, C's, and/or L's LTI Systems have an impulse response h(t)

δ(t) in? h(t) is output. 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)

If not LTI, h(t) does not exist We'll hit ☺ in more detail later

OpAmpsOpAmps

High Gain Devices Typically, Voltage Gain > 10,00 Vout(t) = [Vp(t) – Vn(t)]*Voltage Gain

High Input Impedance Zin typically > 1 MΩ

Integrator: H(jω) = -1/jωRCIntegrator: H(jω) = -1/jωRC

ω

|H(ω)|

10

10000 100

5 Hz Square Wave In...5 Hz Square Wave In...

0

1.5

-1.50 1.0

This one made up of 100 sinusoids.Fundamental frequency of 5 Hz &

next 99 harmonics (15, 25, ..., 995 Hz).

5 Hz Triangle Out...5 Hz Triangle Out...

0

1.5

-1.50 1.0

-50

50

Differentiator: H(jω) = -jωRCDifferentiator: H(jω) = -jωRC

ω

|H(ω)|

1

10000 100

5 Hz Square Wave In...5 Hz Square Wave In...

0

1.5

-1.50 1.0

This curve made up of 100 sinusoids.Fundamental frequency of 5 Hz &

next 99 harmonics (15, 25, ..., 995 Hz).

Spikes Out...Spikes Out...

0

1.5

-1.50 1.0

ECEN3714 Network Analysis Lecture #21 2 March 2015 Dr. George Scheets n Read 14.7 n Problems: 14.5,...

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