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Problem Statement

6(a) Show $ℱ [\int_{- \infty}^{t} s (λ) d λ] = \frac{S (f)}{j 2 π f} if S (f_{0}) = 0$ .

6(b) If $S (f_{0}) \neq 0$ can you find $ℱ [\int_{- \infty}^{t} s (λ) d λ]$ in terms of $S (0)$ ?

Answer

a)Remember that dummy variable $λ$ was used in substitution such that $λ = t - t_{0}$

Then $s (λ) = s (t - t_{0}) = ℱ [S (f) - S (f_{0})]$

and $\int_{- \infty}^{t} s (λ) d λ = \int_{- \infty}^{t} ℱ [S (f) - S (f_{0})] d λ$

The problem statement says to make $S (f_{0}) = 0$ that makes the above equation simplify to

$\int_{- \infty}^{t} s (λ) d λ = \int_{- \infty}^{t} ℱ [S (f)] d t$

Taking the inverse Fourier Transform and changing the order of intgration

$\int_{- \infty}^{t} s (λ) d λ = \int_{- \infty}^{t} e^{j 2 π f t} d t \int_{- \infty}^{\infty} S (f) d f = \frac{e^{j 2 π f t}}{j 2 π f} \int_{- \infty}^{\infty} S (f) d f =$

Then

$\int_{- \infty}^{t} s (λ) d λ = \int_{\infty}^{\infty} S (f) \frac{e^{j 2 π f t}}{j 2 π f} d f = ℱ^{- 1} [\frac{S (f)}{j 2 π f}]$

Therefore it is demonstrated that $ℱ [\int_{- \infty}^{t} s (λ) d λ] = \frac{S (f)}{j 2 π f}$

b)If $S (f_{0}) \neq 0$

Then $\int_{- \infty}^{t} s (λ) d λ = \int_{- \infty}^{t} ℱ^{- 1} [S (f) - S (f_{0})] d λ = \int_{- \infty}^{t} \int_{- \infty}^{\infty} e^{j 2 π f t} [S (f) - S (f_{0})] d λ$

@@ Line 4: / Line 4: @@
 '''Problem Statement'''
-(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>.
+(a) Show <math> \mathcal{F}\left[ \int_{- \infty}^{t} s(\lambda ) \,d\lambda \right] = \frac{S(f)}{j2 \pi f} \mbox{ if } S(f_0) = 0 </math>.
-(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>?
+(b) If <math> S(f_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'''
@@ Line 15: / Line 15: @@
 and <math> \int_{- \infty}^{t} s(\lambda) \,d\lambda = \int_{- \infty}^{t}\mathcal{F}\left[ S (f)- S(f_0) \right] \,d\lambda \! </math>
 The problem statement says to make <math>S(f_0)=0 \!</math> that makes the above equation simplify to
 <math> \int_{- \infty}^{t} s(\lambda) \,d\lambda = \int_{- \infty}^{t}\mathcal{F}\left[ S (f) \right] \,dt \! </math>
+Taking the inverse Fourier Transform and changing the order of intgration
 <math> \int_{- \infty}^{t} s(\lambda ) \,d\lambda = \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>
+Then
 <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>
+Therefore it is demonstrated that <math> \mathcal{F}\left[ \int_{- \infty}^{t} s(\lambda ) \,d\lambda \right] = \frac{S(f)}{j2 \pi f} \!</math>
-In my notations <math>S(f_0)=S(f)|_{f=0 \!</math>
-The problem statement says let <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>
+b)If <math>S(f_0)\neq 0</math>
-<math>\mbox{ if } S(0) = 0\,\,\, \int_{-\infty}^{\infty} s(t) dt =0 \!</math>
+Then <math> \int_{- \infty}^{t} s(\lambda) \,d\lambda = \int_{- \infty}^{t}\mathcal{F}^{-1}\left[ S (f)- S(f_0) \right] \,d\lambda = \int_{- \infty}^{t}\int_{- \infty}^{\infty} e^{j2 \pi f t} [S (f)- S(f_0)] \,d\lambda \! </math>

Link title: Difference between revisions

Latest revision as of 21:58, 18 December 2009

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