Homework Four: Difference between revisions

From Class Wiki
Jump to navigation Jump to search
No edit summary
No edit summary
Line 14: Line 14:
<br/>
<br/>


After some factoring and combinting of like terms we get:  
After some factoring and combinting <math> \Longleftarrow </math> "combining-Kevin" of like terms we get:  


<br/>
<br/>
Line 38: Line 38:


<math> \mathcal{F} \left[ \int_{- \infty}^{\infty}10^{t}g(t)e^{j2 \pi ft_0} \,dt \right] = 10^{t_0}g(t_0) </math>
<math> \mathcal{F} \left[ \int_{- \infty}^{\infty}10^{t}g(t)e^{j2 \pi ft_0} \,dt \right] = 10^{t_0}g(t_0) </math>
<br>
Other than the one typo looks good - Kevin

Revision as of 21:41, 2 November 2009

Fourier Transform Properties


Nick Christman

1. Find [10tg(t)ej2πft0dt]

To begin, we know that

[10tg(t)ej2πft0dt]=(10tg(t)ej2πft0dt)ej2πftdt

After some factoring and combinting "combining-Kevin" of like terms we get:


[10tg(t)ej2πft0dt]=(10tg(t)dt)ej2πf(t0t)dt

But recall that ej2πf(t0t)dtδ(t0t) or δ(tt0)


Because of this definition (and some "math magic") our problem has been simplified significantly:

[10tg(t)ej2πft0dt]=10tg(t)δ(tt0)dt=10t0g(t0)

Therefore,

[10tg(t)ej2πft0dt]=10t0g(t0)
Other than the one typo looks good - Kevin