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The problem with the code provided was that the sampling rate was much too large for the function in question. With the sampling rate this low, it is impossible to get an accurate representation of what the function actually looks like. To fix this problem, just decrease the time between samples (Ts). As shown below, I chose a Ts value of 0.001. It provided a resolution fine enough to easily observe what the function is supposed to look like. | The problem with the code provided was that the sampling rate was much too large for the function in question. With the sampling rate this low, it is impossible to get an accurate representation of what the function actually looks like. To fix this problem, just decrease the time between samples (Ts). As shown below, I chose a Ts value of 0.001. It provided a resolution fine enough to easily observe what the function is supposed to look like. | ||
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Latest revision as of 08:31, 11 September 2008
Fixing the Bug
The problem with the code provided was that the sampling rate was much too large for the function in question. With the sampling rate this low, it is impossible to get an accurate representation of what the function actually looks like. To fix this problem, just decrease the time between samples (Ts). As shown below, I chose a Ts value of 0.001. It provided a resolution fine enough to easily observe what the function is supposed to look like.
%Ben Moeller %Fixing the bug F0 =13; T0 =1/F0 Ts = 0.001; t = 0:Ts:13*T0 x = real(exp(j*(2*pi*F0*t-pi/2))); plot(t,x)
13 Cycles of the Signal
