Mark's Homework - Revision history

Pridma: Removing all content from page

2007-11-12T18:50:13Z

Removing all content from page

← Older revision		Revision as of 11:50, 12 November 2007
Line 1:		Line 1:
	~~==Problem==~~
	~~Sample <math>sin(2 \pi t)</math> at 3Hz and take the DFT. Explain the results.~~

	~~==MATLAB script==~~
	~~<pre>~~
	~~clear all;~~
	~~close all;~~
	~~f = 1; %sine wave frequency~~
	~~fs = 3; %sample frequency~~
	~~tmax = 600/fs; %go to tmax seconds -- 600 points~~
	~~T = 1/fs;~~
	~~t = 0:T:tmax;~~
	~~N=length(t);~~
	~~x = sin(2pif*t); % My signal to transform~~
	~~t2 = 0:.01:tmax;~~
	~~x2 = sin(2pif*t2);~~
	~~X=fft(x);~~
	~~Xs((N+1)/2:N) = X(1:(N+1)/2); %The fftshift didn't work for me~~
	~~Xs(1:(N+1)/2) = X((N+1)/2:N); %So this manually shifts the FFT~~
	~~f = -1/(2T):1/(NT):1/(2T)-1/(NT);~~
	~~figure(1)~~
	~~plot(t, x, '-o', t2, x2, 'r')~~
	~~xlabel('Time (s)')~~
	~~ylabel('x(t)')~~
	~~title('Input Signal')~~
	~~legend('Sampled signal', 'Original signal');~~
	~~figure(2)~~
	~~plot(f, abs(Xs))~~
	~~xlabel('Frequency')~~
	~~ylabel('X(f)')~~
	~~title('Graph of F[sin(2 \pi t)]')~~

	~~</pre>~~

	~~==Graphs==~~
	~~[[Image:DFT20071109Fig1.png\|thumb\|left\|694px\| Figure 1. The sampled signal with only 7 sample points.]]~~

	~~[[Image:DFT20071109Fig2.png\|thumb\|left\|694px\| Figure 2. The DFT of Figure 1.]]~~

	~~[[Image:DFT20071109Fig3.png\|thumb\|left\|894px\| Figure 3. The same as Figure 1 but with 600 sample points, and zoomed in.]]~~

	~~[[Image:DFT20071109Fig4.png\|thumb\|left\|694px\| Figure 4. The DFT of Figure 3.]]~~

	~~<br clear="all"/>~~

	~~==Explanation==~~

	In Figure 1. I took only 7 sample points. As you can see, the DFT (Figure 2) looks nothing like what it should look like (a spike at -1 and 1 Hz). This occurs because the edge effects account for 28% of the points. If you take a whole bunch more points, like that in Figure 3, the DFT looks much more what it should be as in Figure 4. With this many more points, the edge effects are much less noticeable, because the edge points are a small fraction of the total amount of sample points.

Pridma: /* MATLAB script */

2007-11-09T20:49:01Z

MATLAB script

← Older revision		Revision as of 13:49, 9 November 2007
Line 24:		Line 24:
	ylabel('x(t)')		ylabel('x(t)')
	title('Input Signal')		title('Input Signal')
	legend('~~Original~~ signal', '~~Sampled~~ signal');		legend('Sampled signal', 'Original signal');
	figure(2)		figure(2)
	plot(f, abs(Xs))		plot(f, abs(Xs))

Pridma: New page: ==Problem== Sample

sin(2 \pi t)

at 3Hz and take the DFT. Explain the results. ==MATLAB script==

 clear all; close all; f = 1; %sine wave frequency fs = 3; %sample frequ...

2007-11-09T20:48:04Z

New page: ==Problem== Sample <math>sin(2 \pi t)</math> at 3Hz and take the DFT. Explain the results. ==MATLAB script== <pre> clear all; close all; f = 1; %sine wave frequency fs = 3; %sample frequ...

New page

==Problem==
Sample <math>sin(2 \pi t)</math> at 3Hz and take the DFT. Explain the results.

==MATLAB script==
<pre>
clear all;
close all;
f = 1; %sine wave frequency
fs = 3; %sample frequency
tmax = 600/fs; %go to tmax seconds -- 600 points
T = 1/fs;
t = 0:T:tmax;
N=length(t);
x = sin(2*pi*f*t); % My signal to transform
t2 = 0:.01:tmax;
x2 = sin(2*pi*f*t2);
X=fft(x);
Xs((N+1)/2:N) = X(1:(N+1)/2); %The fftshift didn't work for me
Xs(1:(N+1)/2) = X((N+1)/2:N); %So this manually shifts the FFT
f = -1/(2*T):1/(N*T):1/(2*T)-1/(N*T);
figure(1)
plot(t, x, '-o', t2, x2, 'r')
xlabel('Time (s)')
ylabel('x(t)')
title('Input Signal')
legend('Original signal', 'Sampled signal');
figure(2)
plot(f, abs(Xs))
xlabel('Frequency')
ylabel('X(f)')
title('Graph of F[sin(2 \pi t)]')

</pre>

==Graphs==
[[Image:DFT20071109Fig1.png|thumb|left|694px| Figure 1. The sampled signal with only 7 sample points.]]

[[Image:DFT20071109Fig2.png|thumb|left|694px| Figure 2. The DFT of Figure 1.]]

[[Image:DFT20071109Fig3.png|thumb|left|894px| Figure 3. The same as Figure 1 but with 600 sample points, and zoomed in.]]

[[Image:DFT20071109Fig4.png|thumb|left|694px| Figure 4. The DFT of Figure 3.]]

<br clear="all"/>

==Explanation==

In Figure 1. I took only 7 sample points. As you can see, the DFT (Figure 2) looks nothing like what it should look like (a spike at -1 and 1 Hz). This occurs because the edge effects account for 28% of the points. If you take a whole bunch more points, like that in Figure 3, the DFT looks much more what it should be as in Figure 4. With this many more points, the edge effects are much less noticeable, because the edge points are a small fraction of the total amount of sample points.