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Transcript of Revision Bode 1
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Filter Design/1
Frequency Response of Filters
An important group of systems are filters. The
characteristics of a filter are shown by its response to
sinusoids of various frequencies.
Example, the spectral content of speech over telephone
channel 300 3100 Hz. A filter that limits the spectrum
of the signal is used.
A filter has a passband and a stop-band frequencies and
is classified into lowpass, highpass or bandpass types.
The transmission of a signal must be distortionless,
except only for a possible scaling of amplitude and aconstant time delay.
Suppose we have a filter with response H(s). Then the
output due to an exponential ste isstesH )( .
Setting js = , we gettjtj ejHe )( and tjtj ejHe )( .
From these,
])(Re[2
)()()cos(2
tj
tjtj
ejH
ejHejHt
+
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We can express )( jH in polar form as
)(|)(|)( jHejHjH = .
The response due to the input ttf cos)( = is then
)](cos[|)(|)( jHtjHty += .
Similarly, the output due to the input )cos()( += ttf
can be shown to be given by
)](cos[|)(|)( jHtjHty ++=
Note that the result is valid only for stable systems as we
have set js = . For marginally stable and unstable
systems, the convergence region does not include the
imaginary axis.
We see that a sinusoid of frequency when passed
through a filter will have its amplitude increase by
|)(| jH and its phase shifted by )( jH .
|)(| jH is called the amplitude response of the filter, and
)( jH is called the phase response. Both of them
together form the frequency response of the filter.
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Example: (a) Find the frequency response of a system
with5
1.0)(
++
=s
ssH .
(b) Find the output if input is )5010cos( t .
a) Amplitude response25
01.0|)(|
2
2
+
+=
jH
Phase response =
5tan
1.0tan)( 11
jH
b) For input )5010cos( +t , 894.025100
01.0100|)10(| =
++
=jH
=
= 26
5
10tan
1.0
10tan)( 11jH
so )265010cos(894.0)( += tty
= 0.894cos(10t- 24).
Causal Sinusoidal Inputs
Now we consider the case of the causal sinusoid. First
we look at the causal exponential input )()( tuetf tj= .
Then -
jssF
=
1)( .
The output is given by
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))()...()((
)()(
)(
)()()()(
21 jssss
sPsF
sQ
sPsHsFsY
n ===
where n are the natural roots of the system
characteristic eqn.
Suppose the partial fraction expansion of the right hand
term yields coefficients nkkk ,...,, 21 for
)),..((),( 21 nsss . The coefficient corresponding to
the last term )( js is -
)()(
)(
jHsQ
sP
js
==
.
Hence
= +
=
n
i i
i
js
jH
s
ksY
1
)()(
and =
+=n
i
tjti ejHtuekty
i
1
)()()(
.
For a stable system, all the characteristic modes will
decay to 0, and thus the steady state output from a
causal sinusoid is also )](cos[|)(|)( jHtjHty += .
Bode Plots
The amplitude and phase response plot of a system can
be sketched easily if we use a logarithmic scale for.
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Such plots are called Bode plots, and the approach uses
the asymptotic behavior of the amplitude and the phase
responses.
Consider the transfer function
))((
))(()(
322
1
21
bsbsbss
asasKsH
+++++
=
where 322 bsbs ++ has complex conjugate roots.
Rewriting it,
++
+
+
+
=11
11
)(
3
2
3
2
1
21
31
21
sb
b
b
s
b
ss
a
s
a
s
bb
aKasH .
Thus, in terms of frequency,
++
+
+
+
=
3
2
3
2
1
21
31
21
)(11
11
)(
b
j
b
bj
b
jj
a
j
a
j
bb
aKajH
.
The amplitude response is the quotient of the product of
the absolute values of the complex terms. The phase
response is just the sum of phases of each term in
numerator minus the sum of the phases of each term in
the denominator.
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