(New page: == Definition of a CT LTI system == if <math>\,y(t)=Cx(t)</math> where C is any constant, then <math>\,h(t)=k\delta (t)</math> == Example of Laplace transform== Use the definition of ...)
 
 
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<math> H(s)=\mathcal{L}\left \{ x(t) \right \} \ = \int_{-\infty}^{\infty} h(t) e^{-st}\, dt  </math>
 
<math> H(s)=\mathcal{L}\left \{ x(t) \right \} \ = \int_{-\infty}^{\infty} h(t) e^{-st}\, dt  </math>
  
if our system is <math>h(t)=k\delta (t)\,</math>
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if our system is <math>h(t)=k\delta (t)\,</math>, then
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<math>H(s)=\int_{-\infty}^{\infty} k \delta (t) e^{-st}\, dt  </math>
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<math>H(s)=\int_{-\infty}^{\infty} k \delta (t) e^{-st}\, dt  </math>
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<math>H(s)=k\int_{-\infty}^{\infty}\delta (t) e^{-st}\, dt  </math>
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<math>H(s)=k*2e^0 = 2\,</math>
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== Compute the new response ==
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Using the formula :<math>y(t)=H(j\omega)x(t)\,</math> given that
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<math>x(t)=\frac{1+j}{2}e^{j\pi t}+\frac{1+j}{2}e^{-j\pi t}+\frac{1+j}{2j}e^{j2\pi t}+\frac{-1-j}{2j}e^{-j2\pi t}</math>
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so simply multiply everything by my <math>H(jw)\,</math>
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<math>y(t)=(1+j)e^{j\pi t}+(1+j)e^{-j\pi t}+\frac{1+j}{j}e^{j2\pi t}+\frac{-1-j}{j}e^{-j2\pi t}</math>

Latest revision as of 13:56, 25 September 2008

Definition of a CT LTI system

if $ \,y(t)=Cx(t) $ where C is any constant, then

$ \,h(t)=k\delta (t) $


Example of Laplace transform

Use the definition of laplace transforms to convert an h(t) function to an H(s) function:

$ H(s)=\mathcal{L}\left \{ x(t) \right \} \ = \int_{-\infty}^{\infty} h(t) e^{-st}\, dt $

if our system is $ h(t)=k\delta (t)\, $, then

$ H(s)=\int_{-\infty}^{\infty} k \delta (t) e^{-st}\, dt $

$ H(s)=\int_{-\infty}^{\infty} k \delta (t) e^{-st}\, dt $

$ H(s)=k\int_{-\infty}^{\infty}\delta (t) e^{-st}\, dt $

$ H(s)=k*2e^0 = 2\, $


Compute the new response

Using the formula :$ y(t)=H(j\omega)x(t)\, $ given that

$ x(t)=\frac{1+j}{2}e^{j\pi t}+\frac{1+j}{2}e^{-j\pi t}+\frac{1+j}{2j}e^{j2\pi t}+\frac{-1-j}{2j}e^{-j2\pi t} $

so simply multiply everything by my $ H(jw)\, $

$ y(t)=(1+j)e^{j\pi t}+(1+j)e^{-j\pi t}+\frac{1+j}{j}e^{j2\pi t}+\frac{-1-j}{j}e^{-j2\pi t} $

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang