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Here is the original code given: | Here is the original code given: | ||
<pre> | <pre> | ||
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</pre> | </pre> | ||
| − | [[Image:hw2_B1_bhorst_ECE301Fall2008mboutin.png|frame|The original code output]] | + | [[Image:hw2_B1_bhorst_ECE301Fall2008mboutin.png|frame|center|The original code output]] |
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Note how even though the output from the original code (the blue stars) resembled a sinusoidal function, it was not nearly the correct information to display from the signal. | Note how even though the output from the original code (the blue stars) resembled a sinusoidal function, it was not nearly the correct information to display from the signal. | ||
| − | [[Image:hw2_B2_bhorst_ECE301Fall2008mboutin.png|frame|The adjusted code output (red line is the higher resolution signal)]] | + | [[Image:hw2_B2_bhorst_ECE301Fall2008mboutin.png|frame|center|The adjusted code output (red line is the higher resolution signal)]] |
Revision as of 10:36, 9 September 2008
Here is the original code given:
F0 =13; T0 =1/F0; Ts = 0.07; t = 0:Ts:13*T0; x = real(exp(j*(2*pi*F0*t-pi/2))); plot(t,x)
The original code output
The problem is that the cycle of signal is 1/13 which is approximately .077 which means that a timestep of .07 for sampling is not going to yield accurate results. If we decrease the timestep, the resolution of the result will improve dramatically.
The adjusted code is shown here:
F0 =13; T0 =1/F0; Ts = 0.07; t = 0:Ts:13*T0; x = real(exp(j*(2*pi*F0*t-pi/2))); %------------Added section------------ Ts2 = Ts/1000; t2 = 0:Ts2:13*T0; x2 = real(exp(j*(2*pi*F0*t2-pi/2))); %-----------/Added section------------ plot(t2,x2,'r',t,x,'b*') %tweaked to show both
Note how even though the output from the original code (the blue stars) resembled a sinusoidal function, it was not nearly the correct information to display from the signal.
The adjusted code output (red line is the higher resolution signal)
