ASN6 a,b- fixing: Difference between revisions

Latest revision as of 20:55, 18 December 2009

Problem Statement

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

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

Answer

a)

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

$f_{0} = 0$ where $S (0) = S (f) |_{f = 0} = \int_{- \infty}^{\infty} s (t) e^{- j 2 π f t} d t = \int_{- \infty}^{\infty} s (t) d t$

$if S (0) = 0 \int_{- \infty}^{\infty} s (t) d t = 0$

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

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

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

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

@@ 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>. 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>
+(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>
 (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'''
-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>
-Remember dummy variable <math> \lambda= t-t_0 </math> Then <math> s(\lambda)= s(t-t_0)</math>
+a)
+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>

ASN6 a,b- fixing: Difference between revisions

Latest revision as of 20:55, 18 December 2009

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