HW3 solutions – ECE351

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Problem 2.32 From

given that ][][][][][][ 321231 −+−−−+++= nnnnnnh δδδδδ (a) ][][][ 123 −−= nnnx δδ ][][][ nhnxny ∗=

{ }

][][][][][][][][][][][][][][][

][][][][][

423527713423224162332316913

123123

−−−+−−++=−−−+−−−−−−+−−−+++=

−−=∗−−=

nnnnnnnnnnnnnnn

nhnhnhnn

δδδδδδδδδδδδδδδ

δδ

(b) ][][][][][][][ 21131 −+−+++=−−+= nnnnnununx δδδδ

][][][ nhnxny ∗=

{ }

][][][][][][][][][][][

][][][][][

53225156142211211

−+−+−+−+++++=−+−+++=

∗−+−+++=

nnnnnnnnhnhnhnh

nhnnnn

δδδδδδδ

δδδδ

HW3 solutions – ECE351

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The plot of the y(t) for part (a), the time-shifting version of original h(t) with some multiplication in the amplitude (if any), is shown as follows:

Note that: For the plots of parts (b) and (c), we do the same process.

Problem 2.40

(a) τττ dtxytxtytytxtm ∫∞

∞−−=∗=∗= )()()()()()()(

Now we consider the value of )(tm over different range of time t by moving x ( t - τ ). Case 1: When t +1 < 0, this implies that

1−<t Therefore, 0=)(tm Case 2: When t +1 < 2, this implies that

11 <≤− t

Therefore,

HW3 solutions – ECE351

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Case 3: When t +1 < 4, this implies that

31 <≤ t

Therefore,

Case 4: When t -1 < 4, this implies that

53 <≤ t

Therefore,

Case 5: When t -1 > 4, this implies that

5≥t

Therefore, 0=)(tm Finally, we can say that

HW3 solutions – ECE351

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(g) τττ dtygtytgtgtytm ∫∞

∞−−=∗=∗= )()()()()()()(

Now we consider the value of )(tm over different range of time t by moving y ( t - τ ). Case 1: When t < -1, 0=)(tm Case 2: When t < 1, this implies that

11 <≤− t Therefore,

Case 3: When t-2 < 1, this implies that

31 <≤ t Therefore,

502502 21

2

2

1.)(.)( −−=+= ∫∫ −

−

−tddtm

t

tττττ

HW3 solutions – ECE351

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] Case 4: When t-4 < 1, this implies that

53 <≤ t Therefore,

Case 5: When t-4 ≥ 1, this implies that

5≥t Therefore, 0=)(tm Finally, we can say that

HW3 solutions – ECE351

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(h) τττ dtcytctytm ∫∞

∞−−=∗= )()()()()(

Now we consider the value of )(tm over different range of time t by moving c ( t - τ ). Case 1: When t +2 < 0, this implies that

2−<t Therefore, 0=)(tm

Case 2: When t +2 < 2, this implies that

02 <≤− t Therefore, 1=)(tm ( from )(δ at 2+= tτ ) Case 3: When t < 1, this implies that

10 <≤ t Therefore,

HW3 solutions – ECE351

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Case 4: When t < 2, this implies that

21 <≤ t Therefore,

Case 5: When t-1 < 2, this implies that

32 <≤ t Therefore,

Case 6: When t < 4, this implies that

43 <≤ t Therefore, Case 7: When t-1 < 4, this implies that

54 <≤ t Therefore,

Case 8: When t-2 < 4, this implies that

65 <≤ t Therefore,

HW3 solutions – ECE351

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Case 9: When t-2 ≥ 4, this implies that

6≥t Therefore, 0=)(tm Finally, we can say that Problem 2.41 [Refer to textbook, page 67, for the impulse response h(t) of an RC circuit.]

HW3 solutions – ECE351

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Thus,

τττττ τ deRC

pdthpthtptpthty RCtp

)/()()()()()()()()()( −−∞

∞−

∞

∞− ∫∫ =−=∗=∗=1

Now we consider the value of )(ty p over different range of time t. Case 1: When t < 0, 0=)(ty p Case 2: When t < T, this implies that

Tt <≤0 Therefore,

Case 3: When t ≥ T, we have

Finally, we can say that

HW3 solutions – ECE351

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HW3 solutions – ECE351

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HW3 solutions – ECE351

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