1/. (a) the response is the sum of the responses to 10
u
(
t
) and 20cos(50
t
)
DC gain is 40dB = 0.01
response to 10
u
(
t
) = (0.01)(10) = 0.1
at
ω
= 50 the gain is 20dB = 0.1 and the phase shift is about +85”
response to 20cos(50
t
) = (0.1) 20cos(50
t
+ 85”)
total response = 0.1 + 2cos(50
t
+ 85”)
(b) The usual rules apply. Whenever the input is a sinusoidal the steady state output will
be the input multiplied by some gain and shifted by some amount.
at
ω
= 10000 the gain is 30dB = 0.0316 and the phase shift is about 89”
the response is (0.0316)500sin(10000
t
+ (34  89)”) = 15.811sin(10000
t
55”)
(c) From the slopes of the asymptotes and their intersections, it can be concluded that
there is a zero with
ω
C
= 5 and a pair of complex poles with
ω
N
= 400. The height of the
peak (above the intersection of the asymptotes) is about 20dB. The value of
ζ
can be
calculated from this height.
05
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 Winter '08
 John Bryant
 Phase Shift

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