Fourier Transform Properties: Difference between revisions

From Class Wiki
Jump to navigation Jump to search
No edit summary
No edit summary
Line 15: Line 15:
<math> \mathcal{F}[10^{t}g(t)e^{j2 \pi ft_0}] = \int_{-\infty}^{\infty}10^{t}g(t)e^{j2 \pi ft_0}e^{-j2 \pi ft}\,dt = \int_{-\infty}^{\infty}10^{t}g(t)e^{j2 \pi f(t_{0}-t)}\,dt</math>
<math> \mathcal{F}[10^{t}g(t)e^{j2 \pi ft_0}] = \int_{-\infty}^{\infty}10^{t}g(t)e^{j2 \pi ft_0}e^{-j2 \pi ft}\,dt = \int_{-\infty}^{\infty}10^{t}g(t)e^{j2 \pi f(t_{0}-t)}\,dt</math>


<br/>
But recall that, <math>e^{j2 \pi f(t_{0}-t)} \equiv \delta (t_{0}-t) \mbox{ or } \delta (t-t_{0})</math>
 
 
Because of this definition, our problem has now been simplified significantly: <br/>
 
<math> \mathcal{F}[10^{t}g(t)e^{j2 \pi ft_0}] = \int_{-\infty}^{\infty}10^{t}g(t) \delta (t-t_{0})\,dt = 10^{t_0}g(t_0)</math> <br/>
 
Therefore,
 
<math> \mathcal{F}[10^{t}g(t)e^{j2 \pi ft_0}] = 10^{t_0}g(t_0) </math>


But recall that, <math>e^{j2 \pi f(t_{0}-t} \equiv \delta (t_{0}-t) \mbox{ or } \delta (t-t_{0})</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>
Now <math>\int_{-\infty}^{\infty} \frac{1}{2}[e^{j2\pi f_0t}+e^{-j2\pi f_0t}]g(t)e^{-j2\pi ft}dt = \frac{1}{2}\int_{-\infty}^{\infty}e^{-j2\pi (f-f_0)t}g(t)dt+\frac{1}{2}\int_{-\infty}^{\infty}e^{-j2\pi (f+f_0)t}g(t)dt = \frac{1}{2}G(f-f_0)+ \frac{1}{2}G(f+f_0)\!</math><br>
So <math>\mathcal{F}[cos(w_0t)g(t)] = \frac{1}{2}[G(f-f_0)+ G(f+f_0)]\!</math>




----
----

Revision as of 15:12, 16 October 2009

Max Woesner

Find ℱ[cos(w0t)g(t)]
Recall w0=2πf0, so ℱ[cos(w0t)g(t)]=ℱ[cos(2πf0t)g(t)]=∫−∞∞cos(2πf0t)g(t)e−j2πftdt
Also recall cos(θ)=12(ejθ+e−jθ),so ∫−∞∞cos(2πf0t)g(t)e−j2πftdt=∫−∞∞12[ej2πf0t+e−j2πf0t]g(t)e−j2πftdt
Now ∫−∞∞12[ej2πf0t+e−j2πf0t]g(t)e−j2πftdt=12∫−∞∞e−j2π(f−f0)tg(t)dt+12∫−∞∞e−j2π(f+f0)tg(t)dt=12G(f−f0)+12G(f+f0)
So ℱ[cos(w0t)g(t)]=12[G(f−f0)+G(f+f0)]


Nick Christman

Find ℱ[10tg(t)ej2πft0]

To begin, we know that

ℱ[10tg(t)ej2πft0]=∫−∞∞10tg(t)ej2πft0e−j2πftdt=∫−∞∞10tg(t)ej2πf(t0−t)dt

But recall that, ej2πf(t0−t)≡δ(t0−t) or δ(t−t0)


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

ℱ[10tg(t)ej2πft0]=∫−∞∞10tg(t)δ(t−t0)dt=10t0g(t0)

Therefore,

ℱ[10tg(t)ej2πft0]=10t0g(t0)