Solutions
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description
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. - PowerPoint PPT Presentation
Transcript of Solutions
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Solutions
Q1: a) 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.
)()()()(
)()(
)()( if
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jYjX
tjtj
tjtj
Violates the assumption that x(t) and y(t) are different
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Q1b: SolutionDiscrete time Fourier series Continuous time Fourier Transform
For discrete time signals For continuous time signals
For periodic signals For aperiodic signals
Discrete in frequency, i.e., ak are discrete vales
Continuous in frequency, i.e.,X(jw) is a continuous function
Periodic in frequency Aperidoic in frequency
Nk
nNjkk enxN
a )/2(][1
dtetxjX tj )()(
b)
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:
5
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T
djXdttx
22
22
)(1
)(21)(
Q2: Solution
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8N with periodic is x[n]8 is LCM
82
432
)4
3cos(21for
82
42
)4
cos(21for
)4
3cos(21)
4cos(
21][
)42
cos(21)
42cos(
21)
4cos()
2cos(][
)cos(21)cos(
21)cos()cos(
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nx
nnnx
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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
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Q3: Solution (Linearity)
System is linear
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)()()(
)()2()()()2()()()2()2()()(
)()2()()()()()(
213
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2121
333
213
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tutbxtaxtbxtaxtutxtxty
tbxtaxtx
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
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Q3: Solution (Stability)
Note that because the system is not LTI, we could not use the test
System is BIBO stable
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Stable BIBO is System 2
)2()(
)2()()(
)( ifCheck
)( Assume
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Q4: Solution
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4
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2
)4
sin(21
221
221
c)
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,22
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21,1
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24
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Q5: Solution
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We have
Taking the inverse transform, we get
