1 / 8

System Realization

System Realization. Laplace transforms with zero-valued initial conditions Capacitor. Inductor Resistor. +. v ( t ). –. +. v ( t ). +. –. v ( t ). –. Passive Circuit Elements. Transfer Function. First-Order RC Lowpass Filter. R. +. +. x ( t ). C. y ( t ). i ( t ).

gbrannon
Download Presentation

System Realization

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. System Realization

  2. Laplace transforms with zero-valued initial conditions Capacitor Inductor Resistor + v(t) – + v(t) + – v(t) – Passive Circuit Elements Transfer Function

  3. First-Order RC Lowpass Filter R + + x(t) C y(t) i(t) Time domain R + + X(s) Y(s) I(s) Laplace domain

  4. First-Order RC Highpass Filter C + + x(t) R y(t) i(t) Time domain + + X(s) R Y(s) I(s) Frequency response is also an example of a notch filter Laplace domain

  5. Laplace transforms with non-zeroinitial conditions Capacitor Inductor Passive Circuit Elements

  6. Operational Amplifier • Ideal case: model this nonlinear circuit as linear and time-invariant Input impedance is extremely high (considered infinite) vx(t) is very small (considered zero) _ + vx(t) + _ + y(t) _

  7. Operational Amplifier Circuit • Assuming that Vx(s) = 0, • How to realize a gain of –1? • How to realize a gain of 10? H(s) I(s) Zf(s) _ F(s) Z(s) + Vx(s) + _ + + _ Y(s) _

  8. Differentiator • A differentiator amplifies high frequencies, e.g. high-frequency components of noise: H(s) = s for all values of s (see next slide) Frequency response is H(f) = j 2 p f  | H( f ) |= 2 p | f | • Noise has equal amounts of low and high frequencies up to a physical limit • A differentiator may amplify noise to drown out a signal of interest • In analog circuit design, one would generally use integrators instead of differentiators

More Related