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ELEC 361/W: Midterm exam Solution: Fall 2005

Professor: A. AmerTA: M. Ghazal

Q1: 1. True: According to the “Shifting property” of the FT2. False: Causality is not determined based on the input signal x(t)3. True: Using shifting and linearity properties of the FS4. True: If (x(t) is bounded and since |cos(1/t)| is bounded by 15. False: The fundamental period of this signal is 24 which is the

least common multiple of 3 (the fundamental period of the first term) and 8 (the fundamental period of the second term)

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Q2: Fourier Transform

We have

Taking the inverse transform, we get

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Q2: Solution

We have

Taking the inverse transform, we get

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Q2: Solution

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Q3: Convolution

only. itiesdiscontinu threehas )( if ? (b)

?)( (a)

10)(

10:0:1

)(

tty

ty

txth

elset

tx

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Q3 Solution: Step 1

(a) Write down the x(t) and h(t) functionally and graphically Note that h(t) is a scaled version of x(t)

dthxthtxty

elset

else

ttxth

elset

tx

)()()()()(

0:0:110

:0:1

)(

10:0:1

)(

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Q3 Solution: Step 2

Sketch h(-τ) and h(t-τ)

h(-τ) Rreflection around y-axis Chage t to τ

h(t-τ) = h(-τ+t) Add t to all axis points Move the graph away to the left

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Q3 Solution: Step 3

Slide h(t-τ) to the right and collect the overlap As you go, find

Limits for y(t) Limits for integration

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Q3 Solution: Step 3

tdd

tORtandtforty

tt

0

0-

][1)-)h(tx(

graph) from (findn integratiofor Limits000

)(for Limits

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Q3 Solution: Step 3

t

t

t

dd

tORtandtforty

][1)-)h(tx(

graph) from (findn integratiofor Limits110

)(for Limits

-t-

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Q3 Solution: Step 3

tdd

tORtandtforty

t 1][1)-)h(tx(

graph) from (findn integratiofor Limits1111

)(for Limits

11

-t

1

-t

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Q3 Solution: Step 4

Put the limits together to make y(t)

elsettt

t

t

ty

111

0

::::

01

)(

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Q3 Solution: Step 5 (b) find the first derivative of y with respect to t

both functionally and graphically

This function has 4 discontinuities, only when = 1, it has 3 discontinuities (two discontinuities become one)

Note that we know 0<1

elsettt

tty

111

0

::::

01

01

)(

Q4 Fourier Series

Q4 Solution (1) Graph both x[n] and x[n-1] to get g[n]

From the graph, we can write g[n] as Note that g[n] is periodic with N = 10 (2) From the expression for g[n], the FS coefficients are

k

knknng ]108[]10[][

kj

k eb8

102

1101

Q4 Solution (3) Since g[n] = x[n] – x[n-1], the FS coefficients ak and bk

are related by

Therefore,

k

kj

kk aeab 102

kj

kj

kj

kk

e

e

e

ba102

8102

102

1

1101

1