Class Wiki - User contributions [en]

Convolution Theorem

2006-10-13T08:55:51Z

Sherna:

Convolution Theorem is as follows
*<math>\mathcal{F}^{-1}\left[ X(f)H(f)\right] = x(t)\times h(t) = \int_{-\infty}^{\infty}x(\lambda)h(t-\lambda)\,d\lambda</math>
*<math>\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}x(\lambda)h(t-\lambda)\,d\lambda\right)e^{-j2\pi ft}\,dt</math>
*<math>\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}X(f'')e^{j2\pi f \lambda}\,df\int_{-\infty}^{\infty}H(f')e^{j2 \pi f'(t-\lambda)}\,df'\right) e^{-j2 \pi f t}\,dt\,d\lambda </math>
*<math>\int_{-\infty}^{\infty}X(f'')\int_{-\infty}^{\infty}H(f')\int_{-\infty}^{\infty} e^{j2\pi (f'-f)t} \, dt \int_{-\infty}^{\infty} e^{j2\pi (f''-f')}\,d \lambda \, df' \, {df''}</math>
*<math>\int_{-\infty}^{\infty}X(f'')\int_{-\infty}^{\infty}H(f')\delta(f-f') \delta(f''-f') \, df' \, {df''}</math>
*<math>\int_{-\infty}^{\infty}X(f')H(f')\delta(f-f')\, {df'}</math>
*<math>X(f)H(f)\,</math>

*<math> x(t) = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}x(\lambda)e^{-j2\pi f \lambda}\,d\lambda \,e^{j2\pi f t}\,df</math>
Switching the order of integration
*<math> x(t) = \int_{-\infty}^{\infty}x(\lambda) \left(\int_{-\infty}^{\infty}e^{j2\pi f(t-\lambda)}\, df\right)\,d\lambda</math>
Taking note of the fact that the inner integral simplifies to <math>\int_{-\infty}^{\infty}e^{j2\pi f(t-\lambda)t}\,df = \delta(t-\lambda) = \delta(\lambda - t)</math>
*<math> \bar{x} = \sum_i (\bar{x}-\hat{a_i})\hat{a_i} = \sum_i \left(\sum_j x_ja_{ij}\right)\hat{a_i}</math>

Work in progress
 
..................

Convolution Theorem

2006-10-13T06:56:59Z

Sherna:

Convolution Theorem is as follows

*<math> x(t) = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}x(\lambda)e^{-j2\pi f \lambda}\,d\lambda \,e^{j2\pi f t}\,df</math>
Switching the order of integration
*<math> x(t) = \int_{-\infty}^{\infty}x(\lambda) \left(\int_{-\infty}^{\infty}e^{j2\pi f(t-\lambda)}\, df\right)\,d\lambda</math>
Taking note of the fact that the inner integral simplifies to <math>\int_{-\infty}^{\infty}e^{j2\pi f(t-\lambda)t}\,df = \delta(t-\lambda) = \delta(\lambda - t)</math>
*<math> \bar{x} = \sum_i (\bar{x}-\hat{a_i})\hat{a_i} = \sum_i \left(\sum_j x_ja_{ij}\right)\hat{a_i}</math>

Work in progress
 
..................

User:Sherna

2006-10-13T06:33:10Z

Sherna:

Currently a student taking Signals & Systems
Working on
*[[The Game]]
*[[Rayleigh's Theorem]]
*[[Convolution Theorem]]

Energy in a signal

2006-10-11T09:38:15Z

Sherna:

*[[Signals and systems|Signals and Systems]]
===Definition of Energy===
Energy is the ability or potential for something to create change. Scientifically energy is defined as total work done by a force. Work can be mathematically calculated as the line integral of force per infinatesimal unit distance,
: <math> W = \int \mathbf{F} \cdot \mathrm{d}\mathbf{s}</math>

Power represents a change in energy.
: <math> P(t) = \frac{dW}{dt} </math>

This means we can also write energy as
: <math> W = \int_{-\infty}^{\infty} P(t)\,dt</math>

===Energy of a Signal===
From circuit analysis we know that the power generated by a voltage source is,
: <math>P(t) = {v^2(t) \over R}</math>
Assuming that R is 1 then the total energy is just,
: <math>W = \int_{-\infty}^\infty |v|^2(t) \, dt</math>
This can be written using [http://en.wikipedia.org/wiki/Bra-ket_notation bra-ket] notation as
: <math> <v(t) | v(t)> \!</math> or <math> <v|v> \!</math>
By [[Rayleigh's Theorem]],
: <math> <v|v> = \int_{-\infty}^{\infty} |V(f)|^2\,df </math>
This implies that the energy of a signal can be found by the fourier transform of the signal,
: <math> W = \int_{-\infty}^{\infty} |V(f)|^2\,df </math>

Energy in a signal

2006-10-11T09:37:19Z

Sherna:

*[[Signals and systems|Signals and Systems]]
===Definition of Energy===
Energy is the ability or potential for something to create change. Scientifically energy is defined as total work done by a force. Work can be mathematically calculated as the line integral of force per infinatesimal unit distance,
: <math> W = \int \mathbf{F} \cdot \mathrm{d}\mathbf{s}</math>

Power represents a change in energy.
: <math> P(t) = \frac{dW}{dt} </math>

This means we can also write energy as
: <math> W = \int_{-\infty}^{\infty} P(t)\,dt</math>

===Energy of a Signal===
From circuit analysis we know that the power generated by a voltage source is,
: <math>P(t) = {v^2(t) \over R}</math>
Assuming that R is 1 then the total energy is just,
: <math>W = \int_{-\infty}^\infty |v|^2(t) \, dt</math>
This can be written using [http://en.wikipedia.org/wiki/Bra-ket_notation bra-ket] notation as
: <math> <v(t) | v(t)> \!</math> or <math> <v|v> \!</math>
By [[Rayleigh's Theroem]],
: <math> <v|v> = \int_{-\infty}^{\infty} |V(f)|^2\,df </math>
This implies that the energy of a signal can be found by the fourier transform of the signal,
: <math> W = \int_{-\infty}^{\infty} |V(f)|^2\,df </math>

Rayleigh's Theorem

2006-10-11T09:36:17Z

Sherna:

Rayleigh's Theorem is derived from the equation for Energy
*<math> W = \int_{-\infty}^{\infty}p(t)\,dt </math>
If we assume that the circuit is a Voltage applied over a load then <math> p(t)=\frac{x^2(t)}{R_L}</math>
for matters of simplicity we can assume <math>R_L = 1\, \Omega</math>
 
This leaves us with
*<math> W = \int_{-\infty}^{\infty}|x|^2(t)\,dt</math>
This is the same as the dot product so to satisfy the condition for complex numbers it becomes
*<math> W = \int_{-\infty}^{\infty}x(t)\,x^*(t)\,dt</math>
If we substitute <math> x(t) = \int_{-\infty}^{\infty}X(f)\,e^{j2\pi ft}\,df </math> and <math>x^*(t)= \int_{-\infty}^{\infty}X(f')\,e^{-j2\pi f't}\,df'</math>
 
 Substituting this back into the original equation makes it
*<math>W = \int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}X(f)\,e^{j2\pi ft}\,df\right) \,\left(\int_{-\infty}^{\infty}X^*(f')\,e^{-j2\pi f't}\,df'\right)\,dt</math>
*<math>W = \int_{-\infty}^{\infty}X(f)\,\int_{-\infty}^{\infty}X^*(f')\left(\int_{-\infty}^{\infty}e^{j2\pi (f-f')t}\,dt\right)\,df'\,df</math>
The time integral becomes <math> \delta (f-f') \,which \ is\ 0\ except\ for\ when\ f' = f</math>
This simplifies the above equation such that
*<math>W = \int_{-\infty}^{\infty}X(f)\,\int_{-\infty}^{\infty}X^*(f')\left(\delta (f-f') \right)\,df'\,df</math>
*<math>W = \int_{-\infty}^{\infty}X(f)\,X^*(f)\,df</math>
Proving that the energy in the time domain is the same as that in the frequency domain
*<math> W = \int_{-\infty}^{\infty}X(f)\,X^*(f)\,df = \int_{-\infty}^{\infty}x(t)\,x^*(t)\,dt</math>

User:Sherna

2006-10-11T09:36:08Z

Sherna:

Currently a student taking Signals & Systems
Working on
*[[The Game]]
*[[Rayleigh's Theorem]]

Rayliegh's Theorem

2006-10-11T09:35:02Z

Sherna:

Rayliegh's Theorem

2006-10-11T09:34:15Z

Sherna:

Rayliegh's Theorem is derived from the equation for Energy
*<math> W = \int_{-\infty}^{\infty}p(t)\,dt </math>
If we assume that the circuit is a Voltage applied over a load then <math> p(t)=\frac{x^2(t)}{R_L}</math>
for matters of simplicity we can assume <math>R_L = 1\, \Omega</math>
 
This leaves us with
*<math> W = \int_{-\infty}^{\infty}|x|^2(t)\,dt</math>
This is the same as the dot product so to satisfy the condition for complex numbers it becomes
*<math> W = \int_{-\infty}^{\infty}x(t)\,x^*(t)\,dt</math>
If we substitute <math> x(t) = \int_{-\infty}^{\infty}X(f)\,e^{j2\pi ft}\,df </math> and <math>x^*(t)= \int_{-\infty}^{\infty}X(f')\,e^{-j2\pi f't}\,df'</math>
 
 Substituting this back into the original equation makes it
*<math>W = \int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}X(f)\,e^{j2\pi ft}\,df\right) \,\left(\int_{-\infty}^{\infty}X^*(f')\,e^{-j2\pi f't}\,df'\right)\,dt</math>
*<math>W = \int_{-\infty}^{\infty}X(f)\,\int_{-\infty}^{\infty}X^*(f')\left(\int_{-\infty}^{\infty}e^{j2\pi (f-f')t}\,dt\right)\,df'\,df</math>
The time integral becomes <math> \delta (f-f') \,which \ is\ 0\ except\ for\ when\ f' = f</math>
This simplifies the above equation such that
*<math>W = \int_{-\infty}^{\infty}X(f)\,\int_{-\infty}^{\infty}X^*(f')\left(\delta (f-f') \right)\,df'\,df</math>
*<math>W = \int_{-\infty}^{\infty}X(f)\,X^*(f)\,df</math>
Proving that the energy in the time domain is the same as that in the frequency domain
*<math> W = \int_{-\infty}^{\infty}X(f)\,X^*(f)\,df = \int_{-\infty}^{\infty}x(t)\,x^*(t)\,dt</math>

Rayliegh's Theorem

2006-10-11T05:07:02Z

Sherna:

Rayliegh's Theorem is derived from the equation for Energy
*<math> W = \int_{-\infty}^{\infty}p(t)\,dt </math>
If we assume that the circuit is a Thevin equavalent then <math> p(t)=\frac{x^2(t)}{R_L}</math>
for matters of simplicity we can assume <math>R_L = 1 \Omega</math>

User:Sherna

2006-10-11T01:31:15Z

Sherna:

Currently a student taking Signals & Systems
Working on
*[[The Game]]
*[[Rayliegh's Theorem]]

Linear Time Invarient System

2006-10-10T03:53:50Z

Sherna:

*[[Signals and systems|Signals and Systems]]
==LTI systems==
LTI System theory is a powerful and widely used concept in electrical engineering. It has applictions in circuit anlysis, control theory , and our main topic of interest signal processing.

===LTI system properties===
A system is considered to be a Linear Time Invariant when it satisfies the two basic criteria implied in its name, one it must be linear and two it must be time invariant. A Linear system is characterized by two properties superposition (additivity) and scaling (homogeneity). The superposition principal says that for any linear system a linear combination of solutions to the system is also a solution to the same linear system. The principal of scaling implies that if you adjust your scale an input by N amount, your output will also be adjusted by N amount. An example of a linear system then would be,

::<math>x_1(t)</math>
::<math>x_2(t)</math>
::<math>y_1(t) = H(x_1(t))</math>
::<math>y_2(t) = H(x_2(t))</math>
::<math>Ay_1(t) + By_2(t) = H(Ax_2(t) + Bx_1(t))</math>
for any scalar values of A and B.

Time invariance of a system means that for any input <math>x(t)</math> by some amount of time T the out put will also be adjusted by that amount of time. This implies that for,

::<math>x(t - T)</math>
::<math>y(t - T) = H(x(t - T))</math>

 
 
Related Links
*[[The Game]]

The Game

2006-10-08T18:10:37Z

Sherna:

The Game as mentioned in class starts with a Linear time invariant system and the impulse response of that system
u(t)------->h(t)

[[user:sherna|Nathan Sherman]]

The Game

2006-10-08T18:09:59Z

Sherna:

The Game as mentioned in class starts with a Linear time invariant system and the impulse response of that system
u(t)------->h(t)

[[user:sherna]Nathan Sherman]

The Game

2006-10-08T18:09:22Z

Sherna:

The Game as mentioned in class starts with a Linear time invariant system and the impulse response of that system
u(t)------->h(t)

User:Sherna

2006-10-08T18:07:59Z

Sherna:

Currently a student taking Signals & Systems
Working on
*[[The Game]]

User:Sherna

2006-09-29T22:30:10Z

Sherna:

Currently a student taking Signals & Systems