00:01
A signal is described by the function d t equals to c e to the power t by toe.
00:08
We have to calculate the fourier transform for this function that is a omega and sketch and interpret the results and for b part how are d t and a omega are affected by the change in door.
00:37
First for transform, the formula we will be using is ak is 1 by 2 pi integration from minus infinity to plus infinity, f of x, e to the power minus ik x dx.
00:52
Here the function is a omega equals to 1 by 2 pi from minus infinity to plus infinity, c e to the power minus t by toe e to the power minus i omega t this is equals to c by two pi c taking outside of the integration from minus infinity to plus infinity e to the power minus t by toe e to the power minus omega -t can be written as cos of omega -t minus i -s -rine omega -t.
01:40
Opening the bracket and multiplying by the exponent, e by 2 -py, the integration from minus infinity, e to the power minus t by toe, cos omega -t minus of integration from minus infinity to plus infinity, e to the power minus t by t sine omega t.
02:07
This is the imaginary part since e to the power minus t by tau is an even function due to symmetric about the y -axis and sine is an odd function the product of them will be an odd function and when integrating between the equal and opposite limits the result will be zero so the remaining equation we have is the integration of minus infinity to plus infinity e to the power minus t by toe, cos of omega t.
02:49
Here cost is an even function and e to the power minus t by tau is also an even function.
02:55
Their product becomes even.
02:57
And when integrating between the equal and opposite units, the result will be equal to twice the integral and integrating between the zero and the upper limit, e to the power minus t by toe, cos omega, t, d t.
03:24
So, on integrating, we have, since that integral doesn't have a closed form solution, we can take it as a equals to 1 by 2 and b equals to omega...