Midterm 1 Flashcards
Causal Signal
Signal starts on or after time 0
Non Causal Signal
Signal starts before time 0
Anti Causal Signal
Signal stops before time 0
Analog
Continuous range of values for amplitude
Digital
Finite number of values for amplitude (ex. 0 or 1)
Time shift of signal
Right: T>0
Left: T< 0
x(t-T)
Time scale of signal
Compression by a, a>1
Expansion by 1/a, 0<a<1
x(at)
Flip signal
signal flips across vertical axis
x(-t)
Periodic Signal
Exists repeating for all of t, continuous forever signal, is non causal
Additivity
if x1-> y1
x2->y2
then x1+x2 -> y1+y2
Homogeneity
if x->y then kx->ky
Superposition/Linearity
Order shouldn’t matter so
k1x1 + k2x2 -> k1y1 + k2y2
Additivity and homogeneity both work for system
Time-invariant
if x(t) -> y(t)
then x(t-T) should give y(t-T)
doesn’t matter if x is processed by system first or time delay
Time-varying
Order of system process or time delay matter, will not yield same answer
u(t) plotting
Practice graphing u(t-2), u(3t), sin(t)u(t), etc..
Relationship between Step Function (u(t)) & Step Impulse (delta(t))
du(t)/dt = delta(t)
Five steps to model Biomedical System
- Schematic of sys.
- Identify input & output
- Establish sys. equations
- Specify initial conditions
- Simplify equations to describe input-output relationship
Conservation of Electric Charge
Sum current @ node = 0. KCL, resistor, capacitor, inductor
Conservation of Electric Energy
Sum voltage in loop = 0. KVL, resistor, capacitor, inductor
Conservation of Momentum
Sum of Forces = 0, mechanical sys, springs, dampers, mass, etc.
Conservation of Mass
Sum of Q = 0, Fluid mechanical, can convert to electrical ~= sum of current at node = 0. Q –> i, P –> v, R –> R
Laplace under Zero State initial conditions
D–>s
ex)(D^2 + 5D + 6)y(t) = (D+1)x(t)
–>
(s^2 + 5s +6)y(s) = (s+1)x(s)
Full Laplace question
- x(t) -> X(S)
- concert sys. to freq. domain
- solve sys. in freq. domain for Y(S)
- Y(S) -> y(t) (could include partial fraction expansion)
Transfer function
H(S) = Y(S)/X(S)
First Order Reverse Engineering
H(S) = k/(s-p)
Steady state: -k/p
t_1/2 = -ln(2)/p
solve for p, solve for k, get H(S)
Second Order
Undamped, Underdamped, Critically Damped, Overdamped
H(S) = k/(s-p1)(s-p2)
Undamped
p1 & p2 are imaginary & conjugate roots
p1,2 = +-jw
Oscillation: T = 2pi/w
Decay Constant = 0
Steady state = k/p1p2
Underdamped
p1 & p2 are complex & conjugate roots
p1,2 = sigma +- jw
Oscillation T = 2pi/w
Decay = sigma (from e^sigma*t)
Steady state = k/p1p2
Critically Damped
p1 & p2 are real & equal roots
steady state: k/p^2
Overdamped
p1 & p2 are real & distinct roots
steady state: k/p1p2
Positive Feedback System
- amplification
- enhanced sensitivity
- unstable
Hn(S) = G(S) / (1 - H(S)G(S)
Negative Feedback System
- increased accuracy/reliability
- reduced sensitivity to disturbances/noise
- used for automatic control
- stable
Hn(S) = G(S) / (1 + H(S)G(S)
Stable
Real parts of all poles are smaller than zero Re(p_n) < 0
Unstable
Real part of a pol is larger than 0, Re(p_n) > 0
Real part of a repeated pole is 0, Re(p1=p2) = 0
Marginally Stable
Real part of non-repeated pole is zero, Re(p_n) =0
BIBO stability
Real part for all poles in net transfer function are less than 0
If all subsystems are internally stable –> BIBO stable
but BIBO stable does not mean all subs are internally stable
BIBO Steps
- Derive net transfer function
- Determine Poles of Net Transfer function
- Investigate values of k which make system stable
Gain
G = lim (s->0) H(s)
Single frequency response
- H(s) -> H(jw), s = jw
- |H(jw)|
- <H(jw)
Bode Plot Practice
Graph practice
