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)
