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

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Proceedings <strong>of</strong> the National Conference on<br />

Trends and Advances in Mechanical Engineering,<br />

<strong>YMCA</strong> <strong>University</strong> <strong>of</strong> <strong>Science</strong> & <strong>Technology</strong>, Faridabad, Haryana, Oct <strong>19</strong>-<strong>20</strong>, <strong>20</strong>12<br />

4. Problem Description<br />

When a mobile user moves from one region to another region, a different channel is allotted to the user<br />

depending upon the availability <strong>of</strong> the channel. A Band Pass IIR filter is designed which is used to give the<br />

response <strong>of</strong> a particular channel. Two band pass filters are used to give the responses <strong>of</strong> two available channels.<br />

5. System Modeling<br />

The filter consists <strong>of</strong> two stages connected in cascading. The single stage is designed in the direct form two. In<br />

the first stage x (n) is input and y 11 (n) is the output and Z -1 represents the delay element. The output <strong>of</strong> delay<br />

element is multiplied by the coefficients <strong>of</strong> numerator and denominator respectively and then fed the adder. The<br />

output <strong>of</strong> first stage y 11 (n) is fed to the input <strong>of</strong> second stage. The output <strong>of</strong> second stage is represented by y 1 (n).<br />

Thus the system is designed using cascaded structure while the individual stage is designed using direct form<br />

two structure.<br />

Fig. 1. Structure <strong>of</strong> band pass filter<br />

6. Mathematical Analysis<br />

The filter is having two stages in cascade form while the individual stage is designed in the direct form two<br />

structure.<br />

The band pass filter transfer function is<br />

− 1<br />

1 − z<br />

H ( z ) =<br />

(1)<br />

−1<br />

− 2<br />

1 − 0 .07 z + 0 .07 z<br />

Substituting<br />

j T<br />

e ω for z in equation (1)<br />

H<br />

1 − e<br />

= (2)<br />

1 − 0.07 e + 0.07 e<br />

− jω<br />

T<br />

( ω T )<br />

− jω<br />

T<br />

− 2 jω<br />

T<br />

Rearranging the terms to produce positive and negative exponents <strong>of</strong> equal magnitude in equation (2)<br />

H ( ωT<br />

) =<br />

H ( ωT<br />

) =<br />

e<br />

e<br />

− jωT<br />

jωT<br />

/ 2 jωT<br />

/ 2 − jωT<br />

/ 2<br />

[ e − e ]<br />

jωT<br />

− jωT<br />

[ e − 0.07 + 0.07e<br />

]<br />

−<br />

jωT<br />

/ 2 jωT<br />

/ 2 − jωT<br />

/ 2<br />

e [ e − e ]<br />

jωT<br />

− jω<br />

[ e − 0.07( 1 − e )]<br />

T<br />

(3)<br />

(4)<br />

Substituting inside the parentheses, the relation e<br />

j ω T = cos( ωT<br />

) + j sin( ωT<br />

)<br />

jωT<br />

/ 2<br />

e [ cos( ωT<br />

/ 2) + jsin(<br />

ωT<br />

/ 2) − cos( ωT<br />

/ 2) + j sin( ωT<br />

/ 2) ]<br />

H(<br />

ωT)<br />

=<br />

[ cos( ωT)<br />

+ jsin(<br />

ωT)<br />

] − 0.07 [(1<br />

− cos( ωT)<br />

+ j sin( ωT)<br />

]<br />

(5)<br />

386

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