(→Solution) |
(→Solution) |
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4. We now have something like this | 4. We now have something like this | ||
<math>aje^{j{\pi \over 4}t} + bje^{j{\pi \over 4}t}</math> | <math>aje^{j{\pi \over 4}t} + bje^{j{\pi \over 4}t}</math> | ||
| − | Also, the function is based on sin, because of Euler's formula: | + | Also, the function is based on sin, because of Euler's formula: <math> \sin(t) = {e^{jt} - e^{-jt} \over 2j}</math> Notice the j in there. |
5. This suggests that the signal is <math>{1 \over 2} \sin(\omega_0t)</math> | 5. This suggests that the signal is <math>{1 \over 2} \sin(\omega_0t)</math> | ||
With the information above, we can derive the answer: <math>{1 \over 2}\sin({\pi \over 4}t)</math> | With the information above, we can derive the answer: <math>{1 \over 2}\sin({\pi \over 4}t)</math> | ||
Revision as of 17:05, 25 September 2008
Homework 4 Ben Horst: 4.1 :: 4.3 :: 4.5
Problem
1. The signal has only two non-zero Fourier coefficients (-1 and 1).
2. The fundamental period of the signal is 8.
3. The function is sinusoidal.
4. Both Fourier coefficients are non-real.
5. The signal has a maximum value of 0.5 and a minimum of -0.5.
Solution
Problem shown worked by steps from above:
1. We know that the answer must be of the following form
$ ae^{j\omega_0t} + be^{j\omega_0t} $
2. From this we can find omega naught
$ \omega_0 = {2\pi \over 8} $
Therefore we have $ ae^{j{\pi \over 4}t} + be^{j{\pi \over 4}t} $
3. The final answer must have the form of c*sin(\omega_0t) or c*cos(\omega_0t)
4. We now have something like this
$ aje^{j{\pi \over 4}t} + bje^{j{\pi \over 4}t} $
Also, the function is based on sin, because of Euler's formula: $ \sin(t) = {e^{jt} - e^{-jt} \over 2j} $ Notice the j in there.
5. This suggests that the signal is $ {1 \over 2} \sin(\omega_0t) $
With the information above, we can derive the answer: $ {1 \over 2}\sin({\pi \over 4}t) $
