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==1. Creating two DT signals (one periodic and one non-periodic) from a periodic CT signal== | ==1. Creating two DT signals (one periodic and one non-periodic) from a periodic CT signal== | ||
| − | Let x(t) = sin (t), which is a periodic CT signal | + | Let <math>x(t) = sin (2\pi t),</math> which is a periodic CT signal |
| − | + | <math>x(t) = sin (2\pi t)</math> | |
[[Image:Sin1_ECE301Fall2008mboutin.jpg]] | [[Image:Sin1_ECE301Fall2008mboutin.jpg]] | ||
| Line 10: | Line 10: | ||
[[Image:Samp0_ECE301Fall2008mboutin.jpg]] | [[Image:Samp0_ECE301Fall2008mboutin.jpg]] | ||
| + | |||
| + | ==Periodic Signal== | ||
Sampling every <math>t = \pi</math> | Sampling every <math>t = \pi</math> | ||
[[Image:Samp pi_ECE301Fall2008mboutin.jpg]] | [[Image:Samp pi_ECE301Fall2008mboutin.jpg]] | ||
| − | Sampling every t = 1 | + | This discrete time signal was produced from a CT sine wave by sampling at a frequency of <math>\frac{1}{\pi}</math>. |
| − | [[Image: | + | |
| + | As can be seen from the graph, the values of x[n] are periodic because they repeat after every period of <math>t = 2\pi</math>. | ||
| + | |||
| + | Therefore, <math>x[n + 2\pi] = x[n]</math> | ||
| + | |||
| + | However, this still does not fulfill the requirement as <math>N = 2\pi</math> is not an integer. For the signal to become periodic, the CT waveform has to be modified to <math>x(t) = sin(0.5\pi t)</math> and sampled at a frequency of 1 Hz. Upon modification, <math>x[n + 4] = x[n]</math> | ||
| + | |||
| + | <math>x(t) = sin(0.5\pi t)</math> sampled at 1 Hz[[Image:Samp4_ECE301Fall2008mboutin.jpg]] | ||
| + | |||
| + | ==Non Periodic Signal== | ||
| + | Sampling every t = 2 | ||
| + | [[Image:Samp2_ECE301Fall2008mboutin.jpg]] | ||
| + | |||
| + | For this discrete time signal which was produced by sampling the same sine wave at a frequency of 0.5, the values of x[n] are non-periodic because the discrete time signal is scattered all over the place with no indication of a pattern. | ||
| + | Therefore, <math>x[n + k] \neq x[n]</math> | ||
| + | |||
| + | ==2. Create a periodic signal by summing shifted copies of a non-periodic signal== | ||
| + | |||
| + | Using Cheng Chen's non periodic CT signal from HW1, y(t) = t. | ||
| + | |||
| + | |||
| + | MATLAB CODING | ||
| + | |||
| + | >> t1 = 0:0.1:5; | ||
| + | |||
| + | >> t2 = 5.001:0.1:10; | ||
| + | |||
| + | >> t3 = 10.001:0.1:15; | ||
| + | |||
| + | >> t = 0:0.1:15; | ||
| + | |||
| + | >> y1 = (t1); | ||
| + | |||
| + | >> y2 = (t2 - 5); %Referred to Wei Jean's code because I was getting an error and I was not sure | ||
| + | |||
| + | >> y3 = (t3 - 10); %why I still needed to shift when the t1,t2, and t3 values were defined like they are. | ||
| + | |||
| + | >> y_sum = [y1 y2 y3]; | ||
| + | |||
| + | >> plot(t,y_sum) | ||
| + | |||
| + | >> | ||
| + | |||
| + | |||
| + | [[Image:Awsaw_ECE301Fall2008mboutin.jpg]] | ||
| + | |||
| + | As can be seen from the graph the signal is periodic. It repeats every 5 seconds. Therefore x(t + 5) = x(t). | ||
| + | |||
| + | Therefore when y(t) = t is copied and shifted periodically by an infinite number of times a periodic signal can be created. | ||
Latest revision as of 09:24, 11 September 2008
Contents
1. Creating two DT signals (one periodic and one non-periodic) from a periodic CT signal
Let $ x(t) = sin (2\pi t), $ which is a periodic CT signal
$ x(t) = sin (2\pi t) $
Sampling every t = 0.01
Periodic Signal
Sampling every $ t = \pi $
This discrete time signal was produced from a CT sine wave by sampling at a frequency of $ \frac{1}{\pi} $.
As can be seen from the graph, the values of x[n] are periodic because they repeat after every period of $ t = 2\pi $.
Therefore, $ x[n + 2\pi] = x[n] $
However, this still does not fulfill the requirement as $ N = 2\pi $ is not an integer. For the signal to become periodic, the CT waveform has to be modified to $ x(t) = sin(0.5\pi t) $ and sampled at a frequency of 1 Hz. Upon modification, $ x[n + 4] = x[n] $
$ x(t) = sin(0.5\pi t) $ sampled at 1 Hz
Non Periodic Signal
Sampling every t = 2
For this discrete time signal which was produced by sampling the same sine wave at a frequency of 0.5, the values of x[n] are non-periodic because the discrete time signal is scattered all over the place with no indication of a pattern. Therefore, $ x[n + k] \neq x[n] $
2. Create a periodic signal by summing shifted copies of a non-periodic signal
Using Cheng Chen's non periodic CT signal from HW1, y(t) = t.
MATLAB CODING
>> t1 = 0:0.1:5;
>> t2 = 5.001:0.1:10;
>> t3 = 10.001:0.1:15;
>> t = 0:0.1:15;
>> y1 = (t1);
>> y2 = (t2 - 5); %Referred to Wei Jean's code because I was getting an error and I was not sure
>> y3 = (t3 - 10); %why I still needed to shift when the t1,t2, and t3 values were defined like they are.
>> y_sum = [y1 y2 y3];
>> plot(t,y_sum)
>>
As can be seen from the graph the signal is periodic. It repeats every 5 seconds. Therefore x(t + 5) = x(t).
Therefore when y(t) = t is copied and shifted periodically by an infinite number of times a periodic signal can be created.
