ASN6 a,b- Prove given Fourier Transform property: Difference between revisions

Latest revision as of 13:21, 19 December 2009

Problem Statement

6(a) Show ${\mathcal {F}}\left[\int _{-\infty }^{t}s(\lambda )\,d\lambda \right]={\frac {S(f)}{j2\pi f}}{\mbox{ if }}S(f_{0})=0$ .

6(b) If $S(f_{0})\neq 0$ can you find ${\mathcal {F}}\left[\int _{-\infty }^{t}s(\lambda )\,d\lambda \right]$ in terms of $\displaystyle S(f_{0})$ ?

Answer

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

Then $s(\lambda )=s(t-t_{0})={\mathcal {F}}\left[S(f)-S(f_{0})\right]\!$

and $\int _{-\infty }^{t}s(\lambda )\,d\lambda =\int _{-\infty }^{t}{\mathcal {F}}\left[S(f)-S(f_{0})\right]\,d\lambda \!$

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

$\int _{-\infty }^{t}s(\lambda )\,d\lambda =\int _{-\infty }^{t}{\mathcal {F}}\left[S(f)\right]\,dt\!$

Taking the inverse Fourier Transform and changing the order of intgration

$\int _{-\infty }^{t}s(\lambda )\,d\lambda =\int _{-\infty }^{t}e^{j2\pi ft}\,dt\int _{-\infty }^{\infty }S(f)\,df={\frac {e^{j2\pi ft}}{j2\pi f}}\int _{-\infty }^{\infty }S(f)\,df\!$

Then

$\int _{-\infty }^{t}s(\lambda )\,d\lambda =\int _{\infty }^{\infty }S(f){\frac {e^{j2\pi ft}}{j2\pi f}}\,df={\mathcal {F}}^{-1}\left[{\frac {S(f)}{j2\pi f}}\right]\!$

Therefore it is demonstrated that ${\mathcal {F}}\left[\int _{-\infty }^{t}s(\lambda )\,d\lambda \right]={\frac {S(f)}{j2\pi f}}\!$

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

Then

$\int _{-\infty }^{t}s(\lambda )\,d\lambda =\int _{-\infty }^{t}{\mathcal {F}}^{-1}\left[S(f)-S(f_{0})\right]\,d\lambda =\int _{-\infty }^{t}\int _{-\infty }^{\infty }e^{j2\pi ft}[S(f)-S(f_{0})]\,d\lambda \,d\lambda \!$

$\int _{-\infty }^{t}s(\lambda )\,d\lambda =\int _{-\infty }^{\infty }e^{j2\pi ft}S(f)\,d\lambda \,d\lambda -\int _{-\infty }^{t}\int _{-\infty }^{\infty }e^{j2\pi ft}S(f_{0})\,d\lambda \,d\lambda \!$

$\int _{-\infty }^{t}s(\lambda )\,d\lambda =\int _{-\infty }^{\infty }{\frac {e^{j2\pi ft}}{j2\pi f}}S(f)\,d\lambda -\int _{-\infty }^{\infty }{\frac {e^{j2\pi ft}}{j2\pi f}}S(f_{0})\,d\lambda \!$

Using $dt=d\lambda \!$ and taking the Fourier transform of the equation

answer is ${\mathcal {F}}\left[\int _{-\infty }^{t}s(\lambda )\,d\lambda \right]={\frac {e^{j2\pi ft}}{j2\pi f}}S(f)-{\frac {e^{j2\pi ft}}{j2\pi f}}S(f_{0})\!$

@@ Line 2: / Line 2: @@
-Problem Statement
+'''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(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(f_0) </math>?
+'''Answer'''
-Solution
+a)Remember that dummy variable <math> \lambda \!</math>  was used in substitution such that <math> \lambda= t-t_0 \! </math>
-<math>
-\mathcal{F} \left[ g(t-t_{0})e^{j2 \pi f_{0}t} \right] = \int_{- \infty}^{\infty} \left[ g(t-t_{0})e^{j2 \pi f_{0}t} \right] e^{-j2 \pi ft} \,dt
-</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>
-And
+The problem statement says to make <math>S(f_0)=0 \!</math> that makes the above equation simplify to
-<math>
-\int_{- \infty}^{\infty} g(t-t_{0})e^{-j2 \pi (f-f_{0})t} \,dt = \int_{- \infty}^{\infty} g(\lambda )e^{-j2 \pi (f-f_{0})(\lambda + t_{0})} \,dt
-</math>
+<math> \int_{- \infty}^{t} s(\lambda) \,d\lambda = \int_{- \infty}^{t}\mathcal{F}\left[ S (f) \right] \,dt \! </math>
-And
+Taking the inverse Fourier Transform and changing the order of intgration
-<math>
-\int_{- \infty}^{\infty} g(\lambda )e^{-j2 \pi (f-f_{0})(\lambda + t_{0})} \,dt = \int_{- \infty}^{\infty} g(\lambda )e^{-j2 \pi (f-f_{0})\lambda } e^{-j2 \pi (f-f_{0})t_{0}} \,dt =  </math>
+<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>
-And
-<math>
-\int_{- \infty}^{\infty} g(\lambda )e^{-j2 \pi (f-f_{0})\lambda } e^{-j2 \pi (f-f_{0})t_{0}} \,dt = e^{-j2 \pi (f-f_{0})t_{0}} \left[ \int_{- \infty}^{\infty} g(\lambda )e^{-j2 \pi (f-f_{0})\lambda }  \,dt \right]
-</math>
+Then
-Result
+<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>
-<math>
-\mathcal{F} \left[ g(t)e^{j2 \pi f_{0}t} \right] = G(f-f_{0})e^{-j2 \pi (f-f_{0})t_{0}}
+Therefore it is demonstrated that <math> \mathcal{F}\left[ \int_{- \infty}^{t} s(\lambda ) \,d\lambda \right] = \frac{S(f)}{j2 \pi f} \!</math>
-</math>
+b) If <math>S(f_0)\neq 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\,d\lambda  \! </math>
+<math> \int_{- \infty}^{t} s(\lambda) \,d\lambda = \int_{- \infty}^{\infty} e^{j2 \pi f t}S (f)\,d\lambda\,d\lambda - \int_{- \infty}^{t}\int_{- \infty}^{\infty} e^{j2 \pi f t} S(f_0) \,d\lambda\,d\lambda \! </math>
+<math> \int_{- \infty}^{t} s(\lambda) \,d\lambda = \int_{- \infty}^{\infty}\frac{ e^{j2 \pi f t}} {j2 \pi f } S (f)\,d\lambda - \int_{- \infty}^{\infty}\frac{ e^{j2 \pi f t}} {j2 \pi f } S (f_0)\,d\lambda \! </math>
+Using <math> dt=d\lambda \!</math> and taking the Fourier transform of the equation
+answer is <math>\mathcal{F}\left[ \int_{- \infty}^{t} s(\lambda ) \,d\lambda \right] =  \frac{ e^{j2 \pi f t}} {j2 \pi f }S(f) - \frac{ e^{j2 \pi f t}} {j2 \pi f }S(f_0) \! </math>

ASN6 a,b- Prove given Fourier Transform property: Difference between revisions

Latest revision as of 13:21, 19 December 2009

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