BME130 BIOMEDICAL SIGNALS AND SYSTEMS (Required for BME and BMEP) Catalog Data: BME130 Biomedical Signals and Systems (Credit Units: 4) Analysis of analog and digital biomedical signals; Fourier Series expansions; difference and differential equations; co
BME 130 Final Exam
Dec. 14, 2007
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(Last)
(First)
This is a closed books, closed notes exam. To get a full credit you need to show all of
your work. If you are unable to solve a problem, but show a good reasoning, you might be
eligible for a p
Lecture 02
Emotional Face Processing
6 October 2014
Jie Zheng
What makes us different
from other animals?
6 October 2014
2
Welcome to ECoG World
What makes us different from other animals?
6 October 2014
3
Welcome to ECoG World
What makes us different fro
10
Laplace Transforms
24 October 2014
Midterm Exam 1
Date: October 29, 2014, Wednesday, 4:00pm to 4:50pm
Closed book, closed note, no calculators, cell phone silenced and
put away
Bring pencils and erasers
Covers Homework Assignments 1 and 2, including MA
11
Laplace Transforms
31 October 2014
Properties of Laplace Transform
Linearity
L cfw_c 1 f 1 (t ) + c 2 f 2 (t ) = c 1F1 (s ) + c 2 F2 ( s ) for constants c 1 & c 2
Time shift
L cfw_ f (t T ) = e sT F( s )
Complex translation
L cfw_e at f (t ) = F( s + a
12
Laplace Transforms
3 November 2014
Poles & Stability
From the previous example,
4
7
1
Y (s) =
+
( s + 2) ( s + 1) ( s + 3)
y(t ) = 7 e t 4 e 2 t e 3t
y(t) exponentially decays to zero, so the system is stable
i.e., Re cfw_ak + s > 0 , where a1 = 2, a2
BME 130 Final Exam
Dec. 12, 2008
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work. If you are unable to solve a problem, but show good reasoning, you might be eligible
for a par
BME 130 Mid-term Exam
Oct. 20, 2008
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(Last)
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(First)
This is a closed books, closed notes exam. To get a full credit you need to show all of your work.
If you are unable to solve a problem, but show a good reasoning, yo
Problem 1
8
9
Problem 2
10
Problem 3
Problem 4
Problem 5
Problem 6
3
4
clear all
close all
%hw8_p3.m
f1 = 10;
f2 = 100;
Npt = 1001;
t = linspace(0,1,Npt);
x = sin(2*pi*f1*t) + 0.2*sin(2*pi*f2*t);
figure(1)
plot(t,x)
hold on
num = 1;
den = [1/(2*pi*20) 1];
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2.
(a) B pts.] Using basic properties of the convolution operator (linearity distributivity,
associativity and commutativity) simplify the system in Fig. 1, i.e. orprem them as
a single box with single input and single output. Assume that 91, !12, aud 93
BME 130 Final Exam
Dec. L4,2407
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(First)
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[Warm-up, 20 pts.] Answer the follovring true/false qumtions.
explanation.
(a) A system that has a single input
a^nd
If
applicable, give a oneline
a single output, but multiple
1. [Warm-up, 30 pts.] Answer the following true/false questions. If applicable, give a one-line
explanation.
(a) Most physiological systems are linear.
F
(b) To solve a second order ODE we need 2 initial conditions.
T
(c) The natural frequency of a spring
06
ImpulseResponse,Convolution
15October2014
Outline
Responsestoimpulses
Impulseresponse
Constructingoutputsfromimpulseresponses
Convolution
15October2014
2
Constructingx(t) with (t)
Area = 1
1
x( t )
x( k )pi (t k )
k 0
pi (t k )
lim
0
k
x( k ) pi ( t k
08
Discrete-Time LTI,
Impulse Response
20 October 2014
Outline
Discrete-Time LTI systems
Linearity example
Time-invariant example
Total Responses in Difference Equations
Impulse Responses
Discrete-time unit step function
Discrete-time unit impulse functio
Biomedical Signals and
Systems (BME130)
Lecture Notes
Zoran Nenadic, D.Sc.
Associate Professor
Biomedical Engineering
Electrical Engineering and Computer Science
University of California, Irvine, CA 92697
Last modified:
November 28, 2016
Turn off cell pho
Homework #8
9.1
part a)
syms k r tau PD(t) t
tau = 60;
k = 5000;
V = [0.1,0.2,0.3,0.4];
P0= 12.5;
ri = 0.05;
t1 = 1:0.1:180;
for i = 1:4
r = ri*(heaviside(t-15)-(heaviside(t-(15+(V(i)/ri);
P(t) = P0+dsolve(tau*(diff(PD,t) + PD(t) = k*r, PD(0)=0);
figure(i
BME130, Homework Book
Zoran Nenadic, D.Sc.
Department of Biomedical Engineering
University of California, Irvine, CA 92697
September 30, 2016
This document changes frequently. Please print only the part that you need.
The goal of these exercises is to tes
4. [30 pts.] Consider a canonical first order system given by
T 3205) + 2105) = MU) (1)
Where T and k are positive constants.
(3.) Write the system (1) in a statespace form, i.e. nd scalars a, b, c and :1 such that
:1':(t) = aa:(t) + bu(t)
W) = 658(1) + d
Problem 3
The unit step response: () = (1 )(). Note: zero initial conditions.
(a) Calculate the response to () = () + ( 2). First, we write:
]
[; [0,] ] = [; [0,] + [0,]
Where () ( 2) is the delayed version of the Heaviside function. It may be helpful to
Problem 5 [Use common sense] On Monday morning a student starts preparing for a BME130 exam. The
student has not been attending lectures, discussion sessions, office hours, etc., therefore the
student has no initial knowledge of the course material. Fortu
Solution
(a) 1 cos(n t) 1 0 1 cos(n t) 2 0 Fk [1 cos(n t)] 2 Fk
F
= 0 F n sin(n t) = 0 n t = 0, , 2, . It
max (t) = 2 . Alternatively: (t)
k
k
can be shown that n t = 0, 2, correspond to minima of (t), whereas n t = , 3,
correspond to maxima of (t). Th
Solution
(a) y(t) + by(t)
+ y(t) = u(t)
? . We know that if we have a spring-mass-damper sysp
tem described by: m
y (t)+by(t)+ky(t)
= u(t), then n := k/m, therefore by comparing
p
b
this equation to ? , we get n := 1/1 = 1 . Similarly, := 2bkm =
.
2
(b)