BME 4380: Exam 2

complex numbers general form

z=a+jb

complex numbers polar form

z=Me^(jθ)

impedance

V=IZ

resistor impedance

Z(R)=R

phase shift for Z(R)

no phase shift

capacitor impedance

Z(C)=(-j/ωC)

phase shift for Z(C)

current leads voltage by 90deg

impedance for higher frequency on Z(C)

lower impedance (passes higher frequency easier)

inductor impedance

Z(L)=jωL

phase shift for Z(L)

voltage leads current by 90deg

impedance for higher frequency of Z(L)

higher impedance (passes low frequency easier)

impedance in series

Z(eq)=Z1+Z2+…+Zn

parallel impedance

(1/Z(eq))=(1/Z1)+(1/Z2)+…(1/Zn)

transfer function

H(ω)=(Vout/Vin)

what does transfer function tell you

how much signal amplified or reduced; how much phase shift

output signal

y(t)=|H|Acos(ωt+<H)

RC low pass transfer function

H(ω)=1/(1+jωRC)

RC low pass gain

|H|=1/(sqrt(1+(ωRC)²))

RC low pass phase

<H=-tan^-1(ωRC)

RC cutoff frequency

f_c=1/(2πRC)

gain at the cutoff frequency

-3dB (70% dropoff)

phase at cutoff frequency

-45deg

RC high pass transfer function

H(ω)=(jωRC)/(1+jωRC)

RC high pass gain

|H|=(ωRC)/(sqrt(1+(ωRC)²))

phase cutoff for RC high pass filter

45deg

y-axis for gain plot of Bode

20log₁₀(|H|) [dB]

y-axis for phase plot of Bode

phase [deg]

shared x-axis of Bode

log of frequency

RL low pass transfer function

H=1/(1+jω(L/R))

RL high pass transfer function

H=(jω(L/R)/(1+(jω(L/R))

RL high pass cutoff frequency

f_c=R/(2πL)

voltage across C in RC filter

low pass

voltage across R in RC filter

high pass

voltage across L in RL filter

high pass

voltage across R in RL filter

low pass

order of RLC filter

second order (both L and C)

RLC filter natural frequency

ω_n=1/(sqrt(LC))

RLC filter damping

ζ=(R/2)(sqrt(L/C))

underdamped (oscillation)

ζ<1

critically damped

ζ=1

overdamped

ζ>1

what does blocking high frequencies do?

lessens noise

gain

G=Vout/Vin

gain (in dB)

G=20log₁₀(Vout/Vin)

ideal amplifier characteristics

input impedance=infinity (doesn’t mess up signal); output impedance=0 (delivers signal perfectly)

differential amplifier

Vout=G(V₊-V_-)

op-amp golden rules

1) inputs are equal (V₊=V_-); 2) no current enters input (i=0)

unity gain buffer

Vout=Vin

inverting amplifier

(Vout/Vin)=-(R_f/Rin)

what does gain depend on for inverting amplifier

resistor ratio

non-inverting amplifier

(Vout/Vin)=1+(R2/R1)

gain for non-inverting amplifier

gain>=1

active filter

op-amp and filter

active filter equation

(Vout/Vin)=(-R2/R1)/(1+jωR2C)

summing amplifier

V₀=-(V1+V2+…+Vn) ← adds signals together

integrator amplifier

V₀=-(1/RC)int(Vin)dt ← turns signal into accumulated area

differentiator amplifier

V₀=-RC(dVin/dt) ← responds to how fast signal changes

band pass filter

only middle range of frequencies goes through

band reject filter

removes middle range of frequencies

active low pass filter components

R, C, and op-amp

what happens at low frequencies in active low pass filters

C does nothing → gain is constant

what happens at high frequencies in active low pass filters

C kills signal → output drops

cutoff frequency for active low pass filter

ω_c=1/(R2C) or f_c=1/(2πR2C)

first order active filter

one capacitor

first order active filter slope

-20dB/decade

second order active filter (Sallen-Key)

two capacitors

second order active filter slope

-40dB/decade

third order active filter

-60dB/decade

cascading

can stack filters to build higher order filters

total gain for cascading filters

total gain=multiply gains

total phase for cascading filters

total phase=add phases

filter design trade offs

noise vs speed & sharpness vs stability

if you want a steep cutoff, you need…

higher order

if you want less noise, you need…

lower cutoff

if you want a fast response, you need…

higher cutoff

Butterworth filter

flat response, no ripple, “default choice”

Bessel filter

no overshoot, smooth time response, slower filtering

Chebyshev filter

ripple in signal, very sharp cutoff, more distortion

cutoff frequency vs time response

t_rise=0.3/f_c (lower f_c means less noise but slower signal response)

filter design process

1) choose order 2) choose type 3) choose cutoff frequency 4) pick R (usually 10kΩ) 5) solve for C

“good” filter characteristics

sharp cutoff, flat passband, good time response

sampling (f_s)

turning continuous signal into numbers a computer can use

sampling period

T=1/f_s

Nyquist theorem

if signal has a maximum f (B), then f_s>=2B for the minimum sampling rate

what is the reality of the sampling frequency

usually use 5-10x higher than B

aliasing

when sampling is too low, high frequency signals looks like lower frequencies

alias frequency

f_alias=|f₀-kf_s| where k=round(f₀/f_s)

why should you apply a LPF before sampling

LPF removes high frequencies and prevents aliasing (anti-aliasing filter)

how many levels do digital signals have

two levels (low=0, high=1)

why are digital signals better

less noise problems, easier to store, faster and more reliable

binary numbers

base 2

AND gate

output is 1 only if both inputs are 1

OR gate

output is 1 if any input is 1

NOT gate

flips signal (1→0, 0→1)

XOR gate

output is 1 if inputs are different

AND boolean operator

A*B

OR boolean operator

A+B

NOT boolean operator

~A or A(bar)

De Morgan’s laws

1) (A+B(bar))=~(A*~B) 2) (A*B(bar))=(~A+~B)

truth table

shows outputs for all inputs; can find Boolean functions from this