Fourier Transform Properties: Difference between revisions

From Class Wiki
Jump to navigation Jump to search
No edit summary
No edit summary
Line 1: Line 1:
[[Max Woesner|<b><u>Max Woesner</u></b>]]<br><br>
[[Max Woesner|<b><u>Max Woesner</u></b>]]<br><br>
Find <math>\mathcal{F}[cos(w_0t)g(t)]\!</math><br>
'''Find <math>\mathcal{F}[cos(w_0t)g(t)]\!</math><br>'''
Recall <math> w_0 = 2\pi f_0\!</math>, so <math>\mathcal{F}[cos(w_0t)g(t)] = \mathcal{F}[cos(2\pi f_0t)g(t)] = \int_{-\infty}^{\infty}cos(2\pi f_0t)g(t)e^{-j2\pi ft}dt\!</math><br>
Recall <math> w_0 = 2\pi f_0\!</math>, so <math>\mathcal{F}[cos(w_0t)g(t)] = \mathcal{F}[cos(2\pi f_0t)g(t)] = \int_{-\infty}^{\infty}cos(2\pi f_0t)g(t)e^{-j2\pi ft}dt\!</math><br>
Also recall <math> cos(\theta) = \frac{1}{2}(e^{j\theta} + e^{-j\theta})\!</math>,so <math>\int_{-\infty}^{\infty}cos(2\pi f_0t)g(t)e^{-j2\pi ft}dt = \int_{-\infty}^{\infty} \frac{1}{2}[e^{j2\pi f_0t}+e^{-j2\pi f_0t}]g(t)e^{-j2\pi ft}dt\!</math><br>
Also recall <math> cos(\theta) = \frac{1}{2}(e^{j\theta} + e^{-j\theta})\!</math>,so <math>\int_{-\infty}^{\infty}cos(2\pi f_0t)g(t)e^{-j2\pi ft}dt = \int_{-\infty}^{\infty} \frac{1}{2}[e^{j2\pi f_0t}+e^{-j2\pi f_0t}]g(t)e^{-j2\pi ft}dt\!</math><br>

Revision as of 21:29, 17 October 2009

Max Woesner

Find [cos(w0t)g(t)]
Recall w0=2πf0, so [cos(w0t)g(t)]=[cos(2πf0t)g(t)]=cos(2πf0t)g(t)ej2πftdt
Also recall cos(θ)=12(ejθ+ejθ),so cos(2πf0t)g(t)ej2πftdt=12[ej2πf0t+ej2πf0t]g(t)ej2πftdt
Now 12[ej2πf0t+ej2πf0t]g(t)ej2πftdt=12ej2π(ff0)tg(t)dt+12ej2π(f+f0)tg(t)dt=12G(ff0)+12G(f+f0)
So [cos(w0t)g(t)]=12[G(ff0)+G(f+f0)]


Nick Christman

Find [10tg(t)ej2πft0]

To begin, we know that

[10tg(t)ej2πft0]=10tg(t)ej2πft0ej2πftdt=10tg(t)ej2πf(t0t)dt

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


Because of this definition, our problem has now been simplified significantly:

[10tg(t)ej2πft0]=10tg(t)δ(tt0)dt=10t0g(t0)

Therefore,

[10tg(t)ej2πft0]=10t0g(t0)