1

EECS 215: Introduction to Electronic Circuits FA15

Homework Set 1 Solutions

© Fred Terry 9/17/15

Problem 1 Solution

I will use one m-file for all parts of this problem. There are some comments in the code on specific

issues. Keys to the problem:

I set up time t in seconds, frequency in cycle/second (Hz) and scale the plots for the desired

units of time.

We have to use ‘.*’ not ‘*’ alone to multiply arrays on a point but point basis.

%problem 1 clear all; close all; f0=1e4; t=-2:1e-3:1; % in ms spaced by 1e-3 ms t=t*1e-3; % convert t into seconds f1=1e3; f2=500; % part a y0=5*cos(2*pi*f0*t); plot(t*1e6,y0,'linewidth',2); % multiplying t by 1e6 for plot converts to

microseconds ylabel('Voltage'); xlabel('t (\mus)'); title('Problem 1(a)'); grid; % part b m=0.25; y1=y0.*(1+m*cos(2*pi*f1*t)); % note the .* figure; plot(t*1e6,y1,'linewidth',2); ylabel('Voltage'); xlabel('t (\mus)'); title('Problem 1(b)'); grid; % part c y2=y0.*(1+m*cos(2*pi*f1*t)).*(1+m*cos(2*pi*f2*t)); figure; plot(t*1e6,y2,'linewidth',2); ylabel('Voltage'); xlabel('t (\mus)'); title('Problem 1(c)'); grid;

2

-2000 -1500 -1000 -500 0 500 1000-5

-4

-3

-2

-1

0

1

2

3

4

5

Voltage

t (s)

Problem 1(a)

3

-2000 -1500 -1000 -500 0 500 1000-8

-6

-4

-2

0

2

4

6

8

Voltage

t (s)

Problem 1(b)

4

-2000 -1500 -1000 -500 0 500 1000-8

-6

-4

-2

0

2

4

6

8

Voltage

t (s)

Problem 1(c)

5

Problem 2 Solution

First I’ll read and plot the data using m-file:

%problem 2 clear all; close all; data=csvread('prob2_data.csv'); plot(data(:,1),data(:,2)); % column 1 is x, column 2 is y grid;

Resulting in the plot

Note that this looks parabolic with a minimum that appears to be at about (0.2, 1.2) integrating by eye

through the noise. Thus we would guess a fitted curve of the form 2

0.2 1.2y C x .

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 11

1.5

2

2.5

3

3.5

4

4.5

5Raw Data Only

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Also note that we seem to have y=4.5 at x=-1 (again fitting by eye for the noise). Thus we estimate:

2

2

2

2 2

0.2 1.2

4.5 1 0.2 1.2

4.5 1.2 1.2 2.292

2.292 0.2 1.2 2.292 2.692 1.292

y C x

C

C

y x x x

Using the following m-file

%problem 2 clear all; close all; data=csvread('prob2_data.csv'); x=data(:,1); y=data(:,2); plot(x,y); % column 1 is x, column 2 is y grid; C=(4.5-1.2)/(-1.2)^2 yfit=C*(x-0.2).^2+1.2; figure; plot(x,y); hold on; plot(x,yfit,'r','linewidth',2); grid; legend('data','fit');

We get

7

Note that the fit appears to be good. This is all l expected for this problem; however, you could use

numerical procedures to get a least-squares fit. One of several solutions is shown below and yields a

least-squares result of (to an excessive number of digits):

22.301193248761393 0.919310931047964 + 1.291204901139047y x x

Note that these coefficients are pretty close to those we found from eye-ball procedures (Chi-by-eye to

numerical folks).

%problem 2 clear all; close all; data=csvread('prob2_data.csv'); x=data(:,1); y=data(:,2); plot(x,y); % column 1 is x, column 2 is y grid; title('Raw Data Only'); C=(4.5-1.2)/(-1.2)^2 yfit=C*(x-0.2).^2+1.2; figure;

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 11

1.5

2

2.5

3

3.5

4

4.5

5 by Eye Approach

data

fit

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plot(x,y); hold on; plot(x,yfit,'r','linewidth',2); grid; legend('data','fit'); title('\chi by Eye Approach'); % least squares pfit=polyfit(x,y,2) % fit quadratic to data yfit2=pfit(1)*x.^2+pfit(2)*x+pfit(3); figure; plot(x,y); hold on; plot(x,yfit2,'r','linewidth',2); grid; legend('data','fit2'); title('Non-linear Least Squares Using Polyfit');

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 11

1.5

2

2.5

3

3.5

4

4.5

5Non-linear Least Squares Using Polyfit

data

fit2

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Problem 3 Solution

I’m going to do the general set-up to solve both (a) and (b), then I will look at these individual solutions.

First I will reorder the equations for my convenience:

1 3 1

4 5 4 3 2 3

0 3 3 1 3 2 3 4

2 4

5

0 1 2

10 5 5 0

10 2 8 2 0

4 1 5 2 8 0

1

L

V K i i K i

V K i i K i i K i i

K I K i K i i K i i K i i

i i mA

i I

I i i

I will then group terms and put knowns on the right hand side of the equations.

1 3

2 3 4 5

1 2 3 4 0

1 2 0

2 4

5

10 5 10

2 10 10 2 10

5 2 16 8 4 0

0

1

L

K i K i V

K i K i K i K i V

K i K i K i K i K I

i i I

i i mA

i I

Now include all the “0” terms in order:

10

1 2 3 4 5 0

1 2 3 4 5 0

1 2 3 4 5 0

1 2 3 4 5 0

1 2 3 4 5 0

1 2 3

10 (0) 5 (0) (0) (0) 10

0 2 10 10 2 (0) 10

5 2 16 8 (0) 4 0

1 1 0 (0) (0) 1 0

0 1 0 ( 1) (0) 0 1

0 0 0 (

K i i K i i i I V

i K i K i K i K i I V

K i K i K i K i i K I

i i i i i I

i i i i i I mA

i i i

4 5 00) (1) 0 Li i I I Now map this row by row to a 6x6 matrix equation

1

2

3

4

3

5

0

1010000 0 5000 0 0 0

100 2000 10000 10000 2000 0

05000 2000 16000 8000 0 4000

01 1 0 0 0 1

100 1 0 1 0 0

0 0 0 0 1 0 L

i V

i V

i

i

i A

I I

a) For this part, 0LI , so we can solve the above problem purely numerically in Matlab. This

file:

%problem3 clear all; a=[1e4 0 -5000 0 0 0; 0 2000 -1e4 1e4 -2000 0; -5000 -2000 16e3 -8000 0 4000; 1 -1 0 0 0 -1; 0 1 0 -1 0 0; 0 0 0 0 1 0]; vvec=transpose([10 -10 0 0 1e-3 0]); % transpose make vvec a column vector format long; % to see high precision result ivec=a\vvec % doing ivec=inv(a)*vvec would also work

Yields:

ivec =

0.000086956521739

-0.001521739130435

-0.001826086956522

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-0.002521739130435

0

0.001608695652174

Which is in the form:

1

2

3

4

5

0

i

i

i

i

i

I

b) I can do this part in one of two ways. First, I can use Matlab to get the matrix inverse and

manually do row/column evaluation. This is total fine. The inverse matrix values are messy here,

but the result will be the same as in my more elegant solution below.

If I use the symbolic tool box of Matlab, I can have it do the work for me and produce and exact

answer. The new m-file below:

%problem3 clear all; a=[1e4 0 -5000 0 0 0; 0 2000 -1e4 1e4 -2000 0; -5000 -2000 16e3 -8000 0 4000; 1 -1 0 0 0 -1; 0 1 0 -1 0 0; 0 0 0 0 1 0]; vvec=transpose([10 -10 0 0 1e-3 0]); % transpose make vvec a column vector format long; % to see high precision result ivec=a\vvec % doing ivec=inv(a)*vvec would also work %% get inv(a) to do part b numerically inv(a) %% part b done using symbolics IL=sym('IL','real'); % make IL a symbolic term vvec2=transpose([10 -10 0 0 1e-3 -IL]); % transpose make vvec a column vector ivec2=a\vvec2; eval(ivec2) % eval just cleans up the fractions

Produces:

1/11500 - (7*IL)/23

- (31*IL)/46 - 7/4600

- (14*IL)/23 - 21/11500

- (31*IL)/46 - 29/11500

12

-IL

(17*IL)/46 + 37/23000

Again in the format:

1

2

3

4

5

0

1/11500 7* / 23

   31* / 46 7 / 4600

  14* / 23 21/11500

  31* / 46 29 /11500

                    

  17* / 46 37 / 23000

L

L

L

L

L

L

i

i

I

I

I

I

Ii

I

i

i

I