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1
Homework 5
1.
For a first order system with
τ
= 10 ms, up to what frequency will the system attenuate
the signal by less than 2.0%?
M(
ω
) = 1/[1 + (
ωτ
)
2
]
½
= 1/[1 + (0.01
ω
)
2
]
½
The easiest way to do the problem is like this.
1  M(
ω
)
≤
0.020
1  1/[1 + (0.01
ω
)
2
]
½
≤
0.020
1  0.020
≤
1/[1 + (0.01
ω
)
2
]
½
0.980
≤
1/[1 + (0.01
ω
)
2
]
½
1 + (0.01
ω
)
2
]
½
≤
1/0.980 =
1.020
1 + (0.01
ω
)
2
≤
1.041
(0.01
ω
)
2
= 10
4
ω
2
≤
0.041
ω
2
≤
10,400
ω
≤
102 rad/s
f
≤
16.24 Hz
Alternatively, we can approach the problem like this.
We know that at
ωτ
= 0.1, M(
ω
)
≈
1
M(
ω
) = 1/[1 + (0.1)
2
]
½
= 1/[1 + 0.01]
½
= 1/[1.01]
½
= 1/1.005 = 0.995
The output is attenuated 0.5% (99.5% of the signal amplitude is passed).
Since
τ
= 0.01 sec, this corresponds to
ω
= 0.1/0.01 = 10,
which corresponds to f =
ω
/2
π
= 1.6 Hz.
We know that at
ωτ
= 1, M(
ω
)
≈
0.7
M(
ω
) = 1/[1 + (1)
2
]
½
= 1/[1 +1]
½
= 1/[2]
½
= 1/1.414 = 0.707
The output is attenuated almost 30%.
Since
τ
= 0.01 sec, this corresponds to
ω
= 1/0.01 = 100,
which corresponds to f =
ω
/2
π
= 16 Hz.
The answer lies between these.
Make a plot in this range.
M(
ω
) > 0.98 below 3.2 Hz.
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0.95
0.96
0.97
0.98
0.99
1.00
02468
1
0
1
2
1
4
1
6
f (1/sec)
M(
ω
)
Alternatively, one can guess a few frequencies, approaching the answer.
The fastest is to guess
in halfway there steps.
For example, guess
ωτ
= 0.3 (halfway between 0.1 and 1 on the log
scale), at which M(
ω
) = 0.96.
Close, but need a smaller number.
Next guess
ωτ
= 0.2, and find
that M(
ω
) = 0.98.
This corresponds to
ω
= 0.2/0.01 = 20, which corresponds to f =
ω
/2
π
= 3.2
Hz.
2.
Plot the following function vs. t, y(t) = A cos(
ω
t + 45°) for f = 0.16 = 1/2
π
, A = ln5.
For f = 1/2
π
,
ω
= 2
π
f = 1, T = 2
π = 6.28
A = ln5 = 1.6
The phase is positive, so the curve is shifted
left
.
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This note was uploaded on 11/28/2010 for the course ENES enes100 taught by Professor Staff during the Spring '10 term at Maryland.
 Spring '10
 staff

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