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Solutions

Solutions. Q1: False. The Fourier transform is unique. If two signals have the same Fourier transform, then there is a confusion when trying to find the inverse transform. . Violates the assumption that x(t) and y(t) are different. Q1b: Solution. b). Q1c: Solution.

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Solutions

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  1. Solutions Q1: False. The Fourier transform is unique. If two signals have the same Fourier transform, then there is a confusion when trying to find the inverse transform. Violates the assumption that x(t) and y(t) are different

  2. Q1b: Solution b)

  3. Q1c: Solution • c) The signal strength before and after the Fourier transform and Fourier series remains the same. This is evident from Parseval’s Relations:

  4. Q2: Solution

  5. Q3: Solution (Memory & Time Invariance) • System is not memoryless because y(t) depends on x(t-2) for some t • System is not time-invariant, because y1(t) ≠ y2(t) below

  6. Q3: Solution (Linearity) • System is linear

  7. Q3: Solution (Causality) • System is causal. Because the system is not LTI, we cannot use the h(t) = 0 for t < 0 test. However, we can use the following test: A linear non-TI system is causal if x(t) = 0 for t < t0 then y(t) = 0 for t < t0 For our system, the above is true for t0=0 as in the definition. So system is causal • We can also see that the output y(t) depends on current x(t) and past x(t-2) values of the input

  8. Q3: Solution (Stability) • Note that because the system is not LTI, we could not use the test • System is BIBO stable

  9. Q4: Solution

  10. Q5: Solution • We have • Taking the inverse transform, we get

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