DFT Exploration by harrde - Revision history

Harrde: /* Explanation */

2007-12-07T07:10:39Z

Explanation

← Older revision		Revision as of 00:10, 7 December 2007
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	===Explanation===		===Explanation===
	The DFT only gives us an approximation of the actual Fourier Transform, as was found in the last homework. With few sampling points the DFT hardly resembles the expected trasfrom. This is because the edge effects have a greater affect with fewer sample points. ~~But as~~ we ~~increase~~ the ~~number of~~ sample points the DFT ~~more closely resembles~~ the expected ~~transform~~.		The DFT only gives us an approximation of the actual Fourier Transform, as was found in the last homework. With few sampling points the DFT hardly resembles the expected trasfrom. This is because the edge effects have a greater affect with fewer sample points.
			From the example above we see that the more sampling points we take the closer the DFT resembles the actual transform. e.g. with 30 sample points the DFT has "spikes" with height of about 12 but with 500 sample points they have a height of 250, which is obviously closer the expected Dirac delta funtions with infinite height.

2007-12-07T06:54:17Z

← Older revision		Revision as of 23:54, 6 December 2007
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	[[Image:DH13_5.jpg]]		[[Image:DH13_5.jpg]]
	[[Image:DH13_6.jpg]]		[[Image:DH13_6.jpg]]

			===Explanation===
			The DFT only gives us an approximation of the actual Fourier Transform, as was found in the last homework. With few sampling points the DFT hardly resembles the expected trasfrom. This is because the edge effects have a greater affect with fewer sample points. But as we increase the number of sample points the DFT more closely resembles the expected transform.

2007-12-07T06:38:45Z

Solution

2007-12-07T06:35:23Z

← Older revision		Revision as of 23:35, 6 December 2007
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	==Solution==		==Solution==
			Here is the MATLAB code and resulting figures:
			<pre>
			f = 3; % Sampling freq.
			T = 1/f; % Sampling period
			t = 0:.01:5;
			N2 = 500; % Number of sampling points
			N3 = 30;
			t2 = 0:T:N2*T;
			t3 = 0:T:N3*T;

			x = sin(2pit); % Signal that is sampled
			x2 = sin(2pit2);
			x3 = sin(2pit3);

			X2 = fft(x2); % DFT of long signal
			X3 = fft(x3); % DFT of short signal

			figure(1) %Original signal
			plot(t(1:500),x(1:500))
			xlabel('Time (s)')
			ylabel('x(t)')
			title('Original Input Signal')

			figure(2)
			plot(t2(1:15),x2(1:15)) % Sampled signal
			xlabel('Time (s)')
			ylabel('x(t)')
			title('Sampled Input Signal')

			figure(3) %DFT of long signal
			plot(t2/(N2TT),abs(X2))
			xlabel('Frequency (s)')
			ylabel('X(F)')
			title('DFT of 500 Samples')

			figure(4) % DFT of short signal
			plot(t3/(N3TT),abs(X3))
			xlabel('Frequency (s)')
			ylabel('X(F)')
			title('DFT of 30 Samples')

			figure(5) % Shifted DFT of long signal
			XS2=fftshift(X2);
			f2=-1/(2T):1/(N2T):1/(2*T);
			plot(f2,abs(XS2))
			xlabel('Frequency (s)')
			ylabel('X(F)')
			title('Shifted DFT of 500 Samples')

			figure(6) % Shifted DFT of short signal
			XS3=fftshift(X3);
			f3=-1/(2T):1/(N3T):1/(2*T);
			plot(f3,abs(XS3))
			xlabel('Frequency (s)')
			ylabel('X(F)')
			title('Shifted DFT of 30 Samples')
			</pre>

			[[DH13_1.jpg]]
			[[DH13_2.jpg]]
			[[DH13_3.jpg]]
			[[DH13_4.jpg]]
			[[DH13_5.jpg]]
			[[DH13_6.jpg]]

2007-12-06T00:24:50Z

← Older revision		Revision as of 17:24, 5 December 2007
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	[[User:Harrde\|Back to my page]]		[[User:Harrde\|Back to my page]]

			==Problem Statement==
			Sample <math> sin( 2 \pi nt) </math> at 3Hz, take the DFT, and explain the results.

			==Solution==

2007-11-13T01:43:53Z

New page

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