ASN3 - Class Notes October 5: Difference between revisions

Latest revision as of 22:34, 17 December 2009

When T is very large (approaching infinity) the quanity on the left transforms to be approximately the quanity on the right.

$\lim_{T \to \infty}$

$1 / T ⟶ d f$

$n / T ⟶ f$

$\sum_{n = - \infty}^{\infty} \frac{1}{T} ⟶ \int_{- \infty}^{\infty} () d f$

Using the relations above, can we make an unperiodic signal such as the one given below and make it periodic by taking the limit?

$x (t) = \lim_{T \to \infty} \frac{1}{T} (\int_{- \frac{T}{2}}^{\frac{T}{2}} x (t^{'}) e^{\frac{- j 2 π n t^{'}}{T}} d t^{'}) e^{\frac{j 2 π n t}{T}}$

note that $X (F) = ℱ [x (t)] \int_{- \frac{T}{2}}^{\frac{T}{2}} x (t^{'}) e^{\frac{- j 2 π n t^{'}}{T}} d t^{'}$

Using the Fourier Transform property along with $\lim_{T \to \infty} n / t = f$ then

$x (t) = \int_{- \infty}^{\infty} [\int_{- \infty}^{\infty} x (t^{'}) e^{- j 2 π f t^{'}} d t^{'}] e^{j 2 π f t} d f$

Reordering order of integration

$x (t) = \int_{- \infty}^{\infty} x (t^{'}) [\int_{- \infty}^{\infty} e^{j 2 π f (t^{'} - t)} d f] d t^{'}$

note that the defination of the delta function is $\int_{- \infty}^{\infty} e^{j 2 π f (t^{'} - t)} d f$

$x (t) = \int_{- \infty}^{\infty} x (t^{'}) δ_{(} t^{'} - t) d t^{'}$

Using such techniquies as we did above (refered to as The Game in Signals and Systems), similar equations can be manipulated to find its output of Linear Invarient System.

                     THE GAME
            LTI (Linear Time Invariant System) 
Input     LTI                             Output                                  Reason

$x (t) ⟶ \int_{- \infty}^{\infty} x (t^{'}) δ_{(} t^{'} - t) d t^{'}$ Superposition

$X (f) ⟶ \int_{- \infty}^{\infty} X (f^{'}) δ_{(} f^{'} - f) d f^{'}$ Superposition

@@ Line 1: / Line 1: @@
 [[Jodi Hodge|Back to my Home Page]]
+When T is very large (approaching infinity) the quanity on the left transforms to be approximately the quanity on the right.
+<math>           \lim_{T\to \infty}  </math>
@@ Line 7: / Line 11: @@
 <math> n/T        \longrightarrow f </math>
-<math> \sum_{n=-\infty}^\infty  \frac{1}{T}\longrightarrow  \int_{-\infty] ^ \infty }(   )df \! </math>
+<math> \sum_{n=-\infty}^\infty  \frac{1}{T}\longrightarrow  \int_{-\infty} ^ {\infty }(   )df \! </math>
+Using the relations above, can we make an unperiodic signal such as the one given below and make it periodic by taking the limit?
+<math> x(t)= \lim_{T\to \infty} \frac {1}{T} (\int_{-\frac{T}{2}}^{\frac{T}{2}} x(t')e^{\frac{ -j2 \pi nt'}{T}} dt' )e^{\frac{ j2 \pi nt}{T}} \!</math>
+note that
+<math> X(F)= \mathcal{F}[x(t)] \int_{-\frac{T}{2}}^{\frac{T}{2}} x(t')e^{\frac{ -j2 \pi nt'}{T}} dt' \!</math>
-Can we make an unperiodic signal and make it periodic by taking the limit?
+Using the Fourier Transform property along with  <math> \lim_{T\to \infty} n/t = f </math> then
-<math> x(t)= \lim_{T\to \infty \frac {1}{T} (\int_{-\frac{T}{2}}^{\frac{T}{2}} x(t')e^{\frac{ j2 \pi nt'}{T}} dt' )e^{\frac{ j2 \pi nt}{T}} \!</math>
+<math> x(t)=  \int_{-\infty} ^ {\infty} [\int_{-\infty} ^ {\infty} x(t')e^{ -j2 \pi ft'} dt'] e^{ j2 \pi ft}df \!</math>
-note that  f replaced with n/t and that
+Reordering order of integration
-<math> X(F)= \mathcal{F}[x(t)] \int_{-\frac{T}{2}}^{\frac{T}{2}} x(t')e^{\frac{ j2 \pi nt'}{T}} dt' \!</math>
-<math> X(F)=\mathcal{F}[x(t)]\!<math>
+<math> x(t)=  \int_{-\infty} ^ {\infty} x(t')[\int_{-\infty} ^ {\infty} e^{ j2 \pi f(t'-t)} df ]dt'  \!</math>
-'''THE GAME'''
+note that the defination of the delta function is <math>\int_{-\infty} ^ {\infty} e^{ j2 \pi f(t'-t)} df \!</math>
-LTI (Linear Time Invariant System)
+<math> x(t)=  \int_{-\infty} ^ {\infty} x(t')\delta_(t'-t) dt'  \!</math>
-Input     LTI                       Output         Reason
+Using such techniquies as we did above (refered to as The Game in Signals and Systems), similar equations can be manipulated to find its output of Linear Invarient System.
+                      THE GAME
+             LTI (Linear Time Invariant System)
+ Input     LTI                             Output                                  Reason
+<math> x(t)\longrightarrow  \int_{-\infty} ^ {\infty} x(t')\delta_(t'-t) dt'  \!</math>      Superposition
-<math> x2(t) \!</math> '''.''' <math> e^{\frac{ -j2 \pi mt}{T}} = \int_{-\frac{T}{2}}^{\frac{T}{2}}\sum_{n=0}^\infty a_n e^{\frac{ j2 \pi nt}{T}}e^{\frac{ -j2 \pi mt}{T}} dt =\sum_{n=0}^\infty a_n \int_{-\frac{T}{2}}^{\frac{T}{2}} e^{\frac{ j2 \pi (n-m)t}{T}} dt =\sum_{n=0}^\infty a_n T \delta_{mn} \!</math>
+<math> X(f)\longrightarrow  \int_{-\infty} ^ {\infty} X(f')\delta_(f'-f) df'  \!</math>      Superposition

ASN3 - Class Notes October 5: Difference between revisions

Latest revision as of 22:34, 17 December 2009

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