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A system has the characteristics such that the input | A system has the characteristics such that the input | ||
| − | x(t) = exp(2jt) yields the output y(t) = exp(-2jt) | + | x(t) = exp(2jt) yields the output y(t) = (t)exp(-2jt) |
and the input | and the input | ||
| − | x(t) = exp(-2jt) yields the output y(t) = exp(2jt). | + | x(t) = exp(-2jt) yields the output y(t) = (t)exp(2jt). |
What, then, is the system's output for the input signal x(t) = cos(2t)? | What, then, is the system's output for the input signal x(t) = cos(2t)? | ||
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Hence, the system's output y(t) to the input x(t) = cos(2t) will be equal to the input, yielding | Hence, the system's output y(t) to the input x(t) = cos(2t) will be equal to the input, yielding | ||
| − | y(t) = (1/2)[cos(2t) + jsin(2t) + cos(2t) - jsin(2t)] | + | y(t) = (1/2)(t)[cos(2t) + jsin(2t) + cos(2t) - jsin(2t)] |
| − | y(t) = (1/2)(2)[cos(2t)] | + | y(t) = (1/2)(2)(t)[cos(2t)] |
| − | y(t) = cos(2t) | + | y(t) = (t)cos(2t) |
Latest revision as of 11:07, 19 September 2008
A system has the characteristics such that the input
x(t) = exp(2jt) yields the output y(t) = (t)exp(-2jt)
and the input
x(t) = exp(-2jt) yields the output y(t) = (t)exp(2jt).
What, then, is the system's output for the input signal x(t) = cos(2t)?
Well, it appears that the system produces the inverse of the input to provide the output signal, effectively dividing one by whatever the input signal is. Also, we know that exp(2jt) = cos(2t) + jsin(2t) and exp(-2jt) = cos(2t) - jsin(2t) according to Euler's Formula. Considering these factors, it seems that the input x(t) = cos(2t) will result in the output signal y(t) = 1/(cos(2t)).
However, one notes that cos(2t) = (1/2)[cos(2t) + jsin(2t) + cos(2t) - jsin(2t)]
and since the system is linear the output of the system where cos(2t) is the input will be half the sum of the two outputs when the inputs were the two components of cos(2t), as seen in the expanded equality of cos(2t) above.
Hence, the system's output y(t) to the input x(t) = cos(2t) will be equal to the input, yielding
y(t) = (1/2)(t)[cos(2t) + jsin(2t) + cos(2t) - jsin(2t)]
y(t) = (1/2)(2)(t)[cos(2t)]
y(t) = (t)cos(2t)
