(→unit impulse response) |
(→Part) |
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:<math>H(s)=10 e^{0 s} + e^{-1 s} \, </math> | :<math>H(s)=10 e^{0 s} + e^{-1 s} \, </math> | ||
:<math>H(s)=10 + e^{- s} \, </math>, where s =jw | :<math>H(s)=10 + e^{- s} \, </math>, where s =jw | ||
| − | ==Part == | + | ==Part B== |
Compute the response of your system to the signal you defined in Question 1 using H(s) and the Fourier series coefficients of your signal. | Compute the response of your system to the signal you defined in Question 1 using H(s) and the Fourier series coefficients of your signal. | ||
:<math>x(t)=e^{.5 j t \pi}-e^{-.5 j t \pi}+3 \, </math> | :<math>x(t)=e^{.5 j t \pi}-e^{-.5 j t \pi}+3 \, </math> | ||
:<math>Response = H(s)*x(t) \,</math> | :<math>Response = H(s)*x(t) \,</math> | ||
:<math>x(t)=(e^{.5 j t \pi}-e^{-.5 j t \pi}+3)(10+e^{-jw}) \, </math> | :<math>x(t)=(e^{.5 j t \pi}-e^{-.5 j t \pi}+3)(10+e^{-jw}) \, </math> | ||
| − | :<math>x(t)=10e^{.5 j t \pi}-10e^{-.5 j t \pi}+30 +e^{.5 j t \pi}e^{-jw}-e^{-.5 j t \pi}e^{-jw}+ | + | :<math>x(t)=10e^{.5 j t \pi}-10e^{-.5 j t \pi}+30 +e^{.5 j t \pi}e^{-jw}-e^{-.5 j t \pi}e^{-jw}+3e^{-jw} \, </math> |
| − | :<math>x(t)=10e^{.5 j t \pi}-10e^{-.5 j t \pi}+30 +e^{.5 j t \pi | + | :<math>x(t)=10e^{.5 j t \pi}-10e^{-.5 j t \pi}+30 +e^{.5 j (t-1) \pi}-e^{-.5 j (t-1) \pi}+3e^{-jw} \, </math> |
Revision as of 06:02, 25 September 2008
CT LTI system
The system is:
- $ y(t)=10x(t)+x(t-1) $
unit impulse response
Obtain the unit impulse response h(t) and the system function H(s) of your system. :
- $ d (t) => System =>10 d (t) + d(t-1)\, $
- $ h(t)=10d(t) +d(t-1)\, $
- $ H(s)=\int_{-\infty}^{\infty} h(t)e^{-s t}dt $
- $ H(s)=\int_{-\infty}^{\infty} (10d(t) +d(t-1))e^{-s t}dt $
Using the shifting property,
- $ H(s)=10 e^{0 s} + e^{-1 s} \, $
- $ H(s)=10 + e^{- s} \, $, where s =jw
Part B
Compute the response of your system to the signal you defined in Question 1 using H(s) and the Fourier series coefficients of your signal.
- $ x(t)=e^{.5 j t \pi}-e^{-.5 j t \pi}+3 \, $
- $ Response = H(s)*x(t) \, $
- $ x(t)=(e^{.5 j t \pi}-e^{-.5 j t \pi}+3)(10+e^{-jw}) \, $
- $ x(t)=10e^{.5 j t \pi}-10e^{-.5 j t \pi}+30 +e^{.5 j t \pi}e^{-jw}-e^{-.5 j t \pi}e^{-jw}+3e^{-jw} \, $
- $ x(t)=10e^{.5 j t \pi}-10e^{-.5 j t \pi}+30 +e^{.5 j (t-1) \pi}-e^{-.5 j (t-1) \pi}+3e^{-jw} \, $
