OCTOBER 19-20, 2012 - YMCA University of Science & Technology

jωT
/ 2
e [ 2 j sin( ωT
/ 2) ]
[ cos( ωT)
+ j sin( ωT)
] − 0.07 [(1
− cos( ωT)
+ j sin( ωT)
]
Proceedings of the National Conference on
Trends and Advances in Mechanical Engineering,
YMCA University of Science & Technology, Faridabad, Haryana, Oct 19-20, 2012
H(
ωT)
= (6)
Substituting
∏ / 2
j = e
j in equation (6)
j(
ωT
/ 2+∏
/ 2)
e [2sin( ωT
/ 2)]
H(
ωT)
=
cos( ωT)
− 0.07(1 − cos( ωT)
+ j
[ sin( ωT)
− 0.07sin( ωT)
]
(7)
H ( ω T )
=
(1 .07
j ( ω T / 2 + ∏ / 2 )
e [2 sin( ω T / 2)]
cos( ω T ) − 0 .07 ) + j0 .93 sin( ω T )
(8)
Thus the magnitude response
H(
ωT)
= [2sin( ωT
/ 2)]
(1.07cos( ωT)
− 0.07) + j 0.93sin( ωT)
(9)
Magnitude in dB = H ( ωT
)
20 log 10
And the Phase Response is
Θ
( ω ) =
ω T
2
+
∏
2
−
tan
− 1
0 .93 sin( ω T )
1 .07 cos( ω T ) − 0 .07
(10)
Possible solution to this problem can be obtained in the following two steps.
First, an IIR filter is designed; satisfying the magnitude specifications then it is connected in cascade to increase
the gain of the filter and reducing the side lobes in the pass band of the filter. This filter satisfies the amplitude
specifications and it has linear phase in the pass band [6].
7. Results
The desired magnitude response of band pass filter is obtained in the specified range of GSM. The band pass
filter is designed which is having a 3 dB bandwidth of 200 kHz of each channel. The phase response of the filter
is piecewise linear in the desired range. Thus the phase linearity is preserved in the pass band of the filter.
Fig. 2. Magnitude response of band pass filter
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