ASN6 a,b- fixing: Difference between revisions

From Class Wiki
Jump to navigation Jump to search
No edit summary
No edit summary
 
(9 intermediate revisions by the same user not shown)
Line 4: Line 4:
'''Problem Statement'''
'''Problem Statement'''


6(a) Show <math> \mathcal{F}\left[ \int_{- \infty}^{t} s(\lambda ) \,d\lambda \right] = \frac{S(f)}{j2 \pi f} \mbox{ if } S(0) = 0 </math>. HINT: <math> S(0) = S(f) \vert _{_{f=0}} = \int_{- \infty}^{\infty} s(t)e^{-j2 \pi (f \rightarrow 0)t} \,dt = \int_{- \infty}^{\infty} s(t) \,dt </math>
6(a) Show <math> \mathcal{F}\left[ \int_{- \infty}^{t} s(\lambda ) \,d\lambda \right] = \frac{S(f)}{j2 \pi f} \mbox{ if } S(0) = 0 </math>. Hint: <math> S(0) = S(f) \vert _{_{f=0}} = \int_{- \infty}^{\infty} s(t)e^{-j2 \pi (f \rightarrow 0)t} \,dt = \int_{- \infty}^{\infty} s(t) \,dt </math>


6(b) If <math> S(0) \neq 0 </math> can you find <math> \mathcal{F}\left[ \int_{- \infty}^{t} s(\lambda ) \,d\lambda \right] </math> in terms of <math> \displaystyle S(0) </math>?
6(b) If <math> S(0) \neq 0 </math> can you find <math> \mathcal{F}\left[ \int_{- \infty}^{t} s(\lambda ) \,d\lambda \right] </math> in terms of <math> \displaystyle S(0) </math>?


'''Answer'''
'''Answer'''
a)<math>S(0)= S(f)|_{f=0} = \int_{-\infty}^{\infty} s(t)e^{- j 2 \pi f t} dt = \int_{-\infty}^{\infty} s(t) dt</math>


a)
Remember dummy variable <math> \lambda= t-t_0 </math> Then <math> s(\lambda)= s(t-t_0)</math>

Remember dummy variable <math> \lambda= t-t_0 \! </math> Then <math> s(\lambda)= s(t-t_0)= \mathcal{F}\left[ S (f)- S(f_0) \right] \! </math> and <math> \int_{- \infty}^{t} s(\lambda) \,d\lambda = \int_{- \infty}^{t}\mathcal{F}\left[ S (f)- S(f_0) \right] \,d\lambda \! </math>

<math>f_0=0 \!</math> where <math>S(0)= S(f)|_{f=0} = \int_{-\infty}^{\infty} s(t)e^{- j 2 \pi f t} dt = \int_{-\infty}^{\infty} s(t) dt \! </math>

<math>\mbox{ if } S(0) = 0\,\,\, \int_{-\infty}^{\infty} s(t) dt =0 \!</math>


<math> \int_{- \infty}^{t} s(\lambda) \,d\lambda = \int_{- \infty}^{t}\mathcal{F}\left[ S (f) \right] \,dt \! </math>

<math> \mathcal{F}^{-1}\left[ S (f)- S(f_0) \right] = \int_{- \infty}^{t} e^{j2 \pi f t} \,dt \int_{- \infty}^{\infty} S(f)\,df = \frac{ e^{j2 \pi f t}} {j2 \pi f }\int_{- \infty}^{\infty} S(f) \,df =\! </math>

<math>\int_{- \infty}^{t} s(\lambda ) \,d\lambda = \int_{\infty}^{\infty} S(f)\frac{ e^{j2 \pi f t}} {j2 \pi f }\,df = \mathcal{F }^{-1}\left[ \frac{S(f)}{j2 \pi f} \right] \! </math>

Therefore <math> \mathcal{F}\left[ \int_{- \infty}^{t} s(\lambda ) \,d\lambda \right] = \frac{S(f)}{j2 \pi f} \!</math>

Latest revision as of 19:55, 18 December 2009

Back to my home page


Problem Statement

6(a) Show . Hint:

6(b) If can you find in terms of ?

Answer

a)

Remember dummy variable Then and

where


Therefore