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# Frequency rc 1 1 w gjw 1 1 1 w jw 1 1 1 w jw 1 1 1 1

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frequency Rc 1 = 1 w G(jw) = 1 1 1 + w jw = ) 1 ( 1 1 + w jw * ) 1 ( ) 1 ( 1 1 w jw w jw = 1 ) ( 1 2 1 2 1 w w j w jw = 1 ) )( 1 ( 1 2 1 1 w w w jw = + 1 ) ( ) 1 ( ) 1 )( 1 ( 2 1 1 w w w jw G(jw) = R(w) + j X(w) = + 1 ) ( 1 2 1 w w + j + 1 ) ( ) ( 2 1 1 w w w w w = 0 R(w) = 1 , X(w) = 0 w = R(w) = 1 = 0 , X(w) = - 2 = - 1 = 0 w = 1 w R(w) = 2 1 , X(w) = - 2 1 ) ( w G = [ ] [ ] 2 2 ) ( ) ( w X w R + w = 0 ) ( w G = 2 2 0 1 + = 1 w = ) ( w G = 2 2 0 0 + = 0

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w = 1 w ) ( w G = 2 2 ) 2 1 ( ) 2 1 ( + = ) 4 1 ( ) 4 1 ( + = 4 2 = 2 1 G(jw) = 1 1 1 + w w j ) ( w φ = 1 1 1 + w w j = 1 1 1 + w w j = 1 1 tan 0 w w = 0 - 1 tan 1 1 + w w = 1 1 tan w w ) ( w φ = 1 1 tan w w w = 0 ) ( w φ = 1 tan 0 = 0 w = ) ( w φ = 1 tan = -90 ˙ w = 1 w ) ( w φ = 1 tan 1 = -45 ˙ R(w) X(w) Real 1 w w 1 Im
13. Draw the polar plot for the transfer function G(s) = 2 ) 1 ( + s k , when k=4 (16 Marks) Calculate the phase and magnitude at w = 0.5, 1, 2. Solution : G(s) = 2 ) 1 ( + s K G(jw) = 2 ) 1 ( + jw K = 1 2 2 2 + + jw w j K = 1 2 2 + + jw w K = jw w K 2 ) 1 ( 2 + = jw w K 2 ) 1 ( 2 + * jw w jw w 2 ) 1 ( 2 ) 1 ( 2 2 = 2 2 2 2 2 2 ) 1 ( 2 ) ( w j w Kw j Kw K = 2 2 2 2 4 ) 1 ( 2 ) ( w w Kw j Kw K + G(jw) = R(w) + jX(w) w= (0,0) -45 2 1 -90 ˙ 2 1 w=0 (1,0) ) 2 1 , 2 1 (

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G(jw) = 2 2 2 2 4 ) 1 ( 2 ) ( w w Kw j Kw K + + j 2 2 2 4 ) 1 ( ) 2 ( w w Kw + w = 0 , R(w) = 4 X(w) = 1 0 = 0 w = 0.5 , R(w) = 1.92 X(w) = -2.56 w = 1 , R(w) = 0 X(w) = -2 w = 2 , R(w) = -0.48 X(w) = -0.64 w = , R(w) = 0 X(w) = 0 w = 0 , ) ( jw G = [ ] [ ] 2 2 ) ( ) ( w X w R + = 2 2 0 4 + = 4 w = 0.5 , ) ( jw G = 2 2 ) 56 . 2 ( ) 92 . 1 ( + = 3.2 w = 1 , ) ( jw G = 2 2 ) 2 ( 0 + = 2 w = 2 , ) ( jw G = 2 2 ) 64 . 0 ( ) 48 . 0 ( + = 0.8 w = , ) ( jw G = 2 2 0 0 + = 0 G(jw) = 2 ) 1 ( 4 + jw ) ( w φ = ) 1 ( ) 1 ( 4 + + jw jw G(jw) = w w 1 1 tan tan 0 = w w 1 1 tan tan 0 = -2 w 1 tan w = 0 , ) ( w φ = -2 0 tan 1 = 0 w = 0.5, ) ( w φ = -2 5 . 0 tan 1 = -53.13 w = 1 , ) ( w φ = -2 1 tan 1 = -90 ˙ w = 2 , ) ( w φ = -2 2 tan 1 = -127 ˙ w = , ) ( w φ = -2 1 tan = -2 * 90 ˙ = -180 ˙ 14. Draw the Bode diagram for the following transfer function. (16 Marks) G(jw) = 2 5(1 0.1 ) (1 0.5 )(1 0.6 50 ( ) 50 j w jw j w j jw + + + + -127 ˙ X(w) -180 ˙ w= (0,0) (-0.48,-0.64) w=2 (0,2) w=2 127 ˙ 53.13 ˙ -90 ˙ (1.92,-2.56) w=0.5 w=0 (4,0) R(w)
Solution: Article 8.3 (page no. 426) 15. Draw the Bode diagram for the following transfer function. (16 Marks) G(s) = 15434 386 2572 2 + + s s = ) 341 )( 3 . 45 ( 2572 + + s s Show that the magnitude of a (jw) is -15.6dB at w=10 and -30dB at w=200. Also show that the phase is -150 at w =700. Solution : G(s) = ) 341 )( 3 . 45 ( 2572 + + s s G(jw) = ) 341 )( 3 . 45 ( 2572 + + jw jw = ) 1 341 )( 1 3 . 45 ( 341 * 3 . 45 2572 + + jw jw = ) 1 341 )( 1 3 . 45 ( 17 . 0 + + jw jw Corner frequency c w = 45.3, 341 w < 45 = dec dB 0 w < w < 340 = dec dB 20 w > 340 = dec dB 40 w = 1 ) ( jw GH = 2 2 2 2 ) 341 1 ( 1 * ) 3 . 45 1 ( 1 17 . 0 + + = 0.17 = 20 log 0.17 = -15.4dB Corner frequency c w = 45.3, 341 4.5 34 45 341 450 3400 dec 5 4 & dec 5 4 &

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