(New page: Exam 1, Problem 6 - Summer 2008 a) By Inspection th linear constant-coefficient differential equation that describes the LTI system is: y(t)=1/7x(t)-(1/7)(dy(t)/dt) b) Show that the...)
 
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Exam 1, Problem 6 - Summer 2008
 
Exam 1, Problem 6 - Summer 2008
  
 
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(Formatting to follow after dinner...)
  
 
a) By Inspection th linear constant-coefficient differential equation that describes the LTI system is:
 
a) By Inspection th linear constant-coefficient differential equation that describes the LTI system is:

Revision as of 18:13, 30 June 2008

Exam 1, Problem 6 - Summer 2008

(Formatting to follow after dinner...)

a) By Inspection th linear constant-coefficient differential equation that describes the LTI system is:

y(t)=1/7x(t)-(1/7)(dy(t)/dt)


b) Show that the impulse response to this LTI system is given by h(t)=e^(-7t)u(t)

This means that x(t) = delta(t) and y(t) = e^(-7t)u(t)

e^(-7t)u(t) = (1/7)delta(t) - (1/7)d(e^(-7t)u(t)/dt

Differentiating (1/7)d(e^(-7t)u(t)/dt requires use of the chain rule.

This portion of the equation becomes:

(1/7)(-7)(e^(-7t))u(t) - (1/7)delta(t)(e^(-7t))

-(1/7)delta(t)(e^(-7t)) is -(1/7)delta(t)(e^(-7t)) evaluated at t=0 or -(1/7)delta(t)*1

Plugging that back in yields:

e^(-7t)u(t) = (1/7)delta(t) - (1/7)(-7)(e^(-7t))u(t) - (1/7)delta(t)

This equation simplifies to:

0=0 indicating that it is correct.


c) Find H(s) at s=jw for the LTI system with impuls response h(t)=e^(-7t)u(t)

H(s) = Integral(-infinity to infinity)h(t)*e^(-st)dt

H(s) = Integral(-infinity to infinity)e^(-7t)u(t)*e^(-st)dt

Limits change to 0 to infinity and u(t) drops out.

H(s) = Integral(0 to infinity)e^(-7t)*e^(-st)dt

H(s) = Integral(0 to infinity)e^(-7t-st)dt

H(s) = e^(-7t-st)/(-7-s) evaluated from 0 to inifinity

H(s) = e^(-7*infinity-s*infinity)/(-7-s) - e^(0)/(-7-s)

H(s) = e^(-7*infinity-s*infinity)/(-7-s) goes to 0

H(s) = 1/(7+s)

H(jw) = 1/(7+jw)



d)Find the output y(t) of the above LTI system when the input x(t) is e^(j7t)

y(t) = x(t) convolved with h(t)

x(t) = e^(j7t)

h(t) = e^(-7t)u(t)

Using the properties of convolution we will choose to convolve using x(t-tau) and h(tau) for simplicity.

y(t) = Integral(-infinity to infinity)e^(j7(t-tau))e^(-7tau)u(tau)dtau

Drop the u(t) and change the integration limits.

y(t) = Integral(0 to infinity)e^(j7(t-tau))e^(-7tau)dtau

Simplify the exponentials.

y(t) = Integral(0 to infinity)e^(-7tau + j7t - j7tau)dtau

y(t) = (e^(-7tau + j7tau - j7tau))/(-7-j7) evaluated from 0 to inifity

y(t) = e^(-infinity)/(-7-j7) + e^(j7t)/(7+7j)

y(t) = e^(j7t)/(7+7j)

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