Class lecture notes October 5 - HW3

Max Woesner

Homework #3 - Class lecture notes October 5

The following notes are my interpretation of the material covered in class on October 5, 2009

Nonperiodic Signals

In real life, most systems have a finite time period and can be fairly easily evaluated as periodic. However, we still want to be able to work with nonperiodic signals.
A nonperiodic signal can be thought of as periodic signal with an infinite period. To deal with such signals we can take the limit as the period $T\!$ goes to infinity, or
$\lim _{T\to \infty }\sum _{n=-\infty }^{\infty }({\frac {1}{T}}\int _{-{\frac {T}{2}}}^{\frac {T}{2}}x(t^{'})e^{\frac {-j2\pi nt^{'}}{T}}dt^{'})e^{\frac {j2\pi nt}{T}}\!$, where ${\frac {1}{T}}\int _{-{\frac {T}{2}}}^{\frac {T}{2}}x(t^{'})e^{\frac {-j2\pi nt^{'}}{T}}dt^{'}\!$ is the $\alpha _{n}\!$ term.
We want to remove the restriction $x(t)=x(t+T)\!$, which we can do as follows.

${\frac {1}{T}}\longrightarrow df\!$

${\frac {n}{T}}\longrightarrow f\!$

$\sum _{n=-\infty }^{\infty }{\frac {1}{T}}\longrightarrow \int _{-\infty }^{\infty }(\qquad )df\!$

$\alpha _{n}\longrightarrow X(f)\!$

So $x(t)=\lim _{T\to \infty }\sum _{n=-\infty }^{\infty }{\frac {1}{T}}(\int _{-{\frac {T}{2}}}^{\frac {T}{2}}x(t^{'})e^{\frac {-j2\pi nt^{'}}{T}}dt^{'})e^{\frac {j2\pi nt}{T}}\!$
Using ${\frac {n}{T}}=f\!$ and the information above, we can rewrite the equation.
$x(t)=\int _{-\infty }^{\infty }(\int _{-\infty }^{\infty }x(t^{'})e^{-j2\pi ft^{'}}dt^{'})e^{j2\pi ft}df\!$, where $\int _{-\infty }^{\infty }x(t^{'})e^{-j2\pi ft^{'}}dt^{'}=X(f)\!$
So $x(t)=\int _{-\infty }^{\infty }X(f)e^{+j2\pi ft}df=\ <X(f)|e^{j2\pi ft}>\!$ This is the inverse Fourier transform.
Also, $X(f)=\int _{-\infty }^{\infty }x(t)e^{-j2\pi ft}dt=\ <x(t)|e^{j2\pi ft}>\!$ This is the Fourier transform.
So $x(t)={\mathcal {F}}^{-1}[X(f)]\!$ and $X(f)={\mathcal {F}}[x(t)]\!$
From above, $x(t)=\int _{-\infty _{f}}^{\infty }(\int _{-\infty _{t^{'}}}^{\infty }x(t^{'})e^{-j2\pi ft^{'}}dt^{'})e^{j2\pi ft}df\!$, so
$x(t)=\int _{-\infty _{t^{'}}}^{\infty }x(t^{'})(\int _{-\infty _{\delta (t-t^{'})}}^{\infty }e^{j2\pi f(t-t^{'})}df)dt^{'}\!$, where $\int _{-\infty }^{\infty }e^{j2\pi f(t-t^{'})}df=\ <e^{j2\pi ft}|e^{j2\pi ft^{'}}>\!$ with respect to f.
Similarly, $X(f)=\int _{-\infty _{t}}^{\infty }(\int _{-\infty _{f^{'}}}^{\infty }X(f^{'})e^{+j2\pi ft^{'}}df^{'})e^{-j2\pi ft}dt\!$, where $\int _{-\infty _{f^{'}}}^{\infty }X(f^{'})e^{+j2\pi ft^{'}}df^{'}=x(t)\!$,so
$X(f)=\int _{-\infty _{f^{'}}}^{\infty }X(f^{'})(\int _{-\infty _{t}}^{\infty }e^{j2\pi t(f^{'}-f)}dt)df^{'}\!$
Now $\delta (f^{'}-f)=\delta (f-f^{'})\!$ which is the projection $<e^{j2\pi ft}|e^{j2\pi f^{'}t}>\!$ with respect to t.

The Linear Time Invariant System Game

Recall the Linear Time Invariant System Game that can be used to help us understand the impulse response of a linear time invariant system.

Input$-----\longrightarrow \,\!$	Linear Time Invariant System	$\longrightarrow \,\!$Output	Reason
$\delta (t)\!$	$-------\longrightarrow \,\!$	$h(t)\!$	Given
$\delta (t-t_{0})\!$	$-------\longrightarrow \,\!$	$h(t-t_{0})\!$	Time Invariance
$x(t_{0})\delta (t-t_{0})\!$	$-------\longrightarrow \,\!$	$x(t_{0})h(t-t_{0})\!$	Proportionality
$\int _{-\infty }^{\infty }x(t_{0})\delta \ (t-t_{0})dt_{0}\!$	$-------\longrightarrow \,\!$	$\int _{-\infty }^{\infty }x(t_{0})h(t-t_{0})dt_{0}\!$	Superposition

where $\int _{-\infty }^{\infty }x(t_{0})\delta \ (t-t_{0})dt_{0}=x(t)\!$ for any $x(t)\!$ and $\int _{-\infty }^{\infty }x(t_{0})h(t-t_{0})dt_{0}\!$ is the convolution integral.

We can expand the game further.

Input$-----\longrightarrow \,\!$	Linear Time Invariant System	$\longrightarrow \,\!$Output	Reason
$\delta (t)\!$	$-------\longrightarrow \,\!$	$h(t)\!$	Given
$\delta (t-t_{0})\!$	$-------\longrightarrow \,\!$	$h(t-t_{0})\!$	Time Invariance
$x(t_{0})\delta \ (t-t_{0})\!$	$-------\longrightarrow \,\!$	$x(t_{0})h(t-t_{0})\!$	Proportionality
$x(t)\!$	$-------\longrightarrow \,\!$	$\int _{-\infty }^{\infty }x(t_{0})h(t-t_{0})dt_{0}\!$	Superposition
$e^{j2\pi ft}\!$	$-------\longrightarrow \,\!$	$\int _{-\infty }^{\infty }e^{j2\pi ft_{0}}h(t-t_{0})dt_{0}\!$	Superposition

Let $\lambda \ =t-t_{0}$, so $t_{0}=t-\lambda \$ and $dt_{0}=-d\lambda \$
Therefore $\int _{-\infty }^{\infty }e^{j2\pi ft_{0}}h(t-t_{0})dt_{0}=\int _{+\infty }^{-\infty }h(\lambda )e^{j2\pi f(t-\lambda )}(-d\lambda )=e^{j2\pi ft}\int _{-\infty }^{\infty }h(\lambda )e^{-j2\pi f\lambda }d\lambda \!$
This tells us that $e^{j2\pi ft}\!$ is the eigenfunction and $\int _{-\infty }^{\infty }h(\lambda )e^{-j2\pi f\lambda }d\lambda \!$ is the eigenvalue of all linear time invariant systems.
Also, the eigenvalue $\int _{-\infty }^{\infty }h(\lambda )e^{-j2\pi f\lambda }d\lambda \!$ is $H(f)\!$, which equals the Fourier transform of $h(t)\!$, or $H(f)={\mathcal {F}}[h(t)]\!$
We can expand the game even further.

Input$-----\longrightarrow \,\!$	Linear Time Invariant System	$\longrightarrow \,\!$Output	Reason
$\delta (t)\!$	$-------\longrightarrow \,\!$	$h(t)\!$	Given
$\delta (t-t_{0})\!$	$-------\longrightarrow \,\!$	$h(t-t_{0})\!$	Time Invariance
$x(t_{0})\delta \ (t-t_{0})\!$	$-------\longrightarrow \,\!$	$x(t_{0})h(t-t_{0})\!$	Proportionality
$x(t)\!$	$-------\longrightarrow \,\!$	$\int _{-\infty }^{\infty }x(t_{0})h(t-t_{0})dt_{0}\!$	Superposition
$e^{j2\pi ft}\!$	$-------\longrightarrow \,\!$	$H(f)e^{j2\pi ft}\!$	Superposition
$X(f)e^{j2\pi ft}\!$	$-------\longrightarrow \,\!$	$X(f)H(f)e^{j2\pi ft}\!$	Proportionality
$\int _{-\infty }^{\infty }X(f)e^{j2\pi ft}df\!$	$-------\longrightarrow \,\!$	$\int _{-\infty }^{\infty }X(f)H(f)e^{j2\pi ft}df\!$	Superposition

where $\int _{-\infty }^{\infty }X(f)e^{j2\pi ft}df=x(t)\!$ and $\int _{-\infty }^{\infty }X(f)H(f)e^{j2\pi ft}df={\mathcal {F}}^{-1}[H(f)X(f)]\!$
This is helpful because in frequency space, when I go through a linear time invariant system, it multiplies by the transfer function, compared to time space, which convolves the impulse response, and we would all prefer to do multiplication rather than convolution.