1.)
a.)
1. By checking the final slop (– 40 db/dec), we know the difference of orders
(Denominator – Numerator) is 2. Besides, no rising for phase plot but all
the way through to – 180°, therefore, we conclude “no zeros” and with
damping ratio less than 0.707 (due to the peak on the magnitude plot).
2. The Transfer Function should have the form as:
K
 S 2   2ζ S  
 + 1
ω n  2  + 
 ω n   ω n  
2
3. Substitute S = iω into T.F., we have our F.R.F.
4. From 90° crossing, we know ω n ≅ 3 (rad/sec).
5. For ω << ω n , 20 log( FRF ) ≅ 20 log(
6. For ω = ω n , 20 log( FRF ) ≅ 20 log(
≅ 4.5 (db), ⇒ ξ ≅ 0.19
K
) ≅ −4 (db) . ⇒ K ≅ 5.7
32
K
) − 20 log(2ξ ) ≅ −4 − 20 log(2ξ )
32
.
7. Therefore, our T.F. is approximately:
5 .7
S + 1.14 S + 9
2
[Note: The Bode plot above is actually generated by T .F . =
b.) When ω = 4 , and from plot we know that 20 log(
5
]
S + S +9
2
Output
) ≅ − 4.13 (db) ,
Input
4.13
−
Output
= 10 20 ≅ 0.6216 ; hence, Output = 5 * 0.6216 ≅ 3.1 . From the phase
Input
plot, we know the φ ≅ −150 $ . Therefore, Output ≡ 3.1 sin( 4 t − 150 $ ) .
[Note: the phase angle is in “rad” for “4t” part.]
c.) Output = Input ⇒ find 0 (db). From Magnitude plot, we have
ω ≅ 2.1 & 3.5 (rad / sec) .
d.) From Magnitude plot, the “peak” happens at ω ≅ 2.9 (rad / sec) .
e.) One-tenth of the peak ⇒ − 20 (db) from the peak. From Magnitude plot, it
happens at ω ≅ 6.2 (rad / sec) .
f.) Exactly out of phase ⇒ Phase angle has 180 $ difference. From Phase plot, it
happens at ω ≅ ≥10 (rad / sec) .
2a.)
a.) T.F. is given.
b.) When ω = 4 , and from plot we know that 20 log(
Output
) ≅ −10.5 (db) ,
Input
10.5
−
Output
= 10 20 ≅ 0.3 ; hence, Output = 5 * 0.3 ≅ 1.5 . From the phase plot, we
Input
know the φ ≅ 9.9 $ . Therefore, Output ≡ 1.5 sin( 4 t + 9.9 $ ) .
[Note: the phase
angle is in “rad” for “4t” part.]
c.) Output = Input ⇒ find 0 (db). From Magnitude plot, we have NONE.
d.) From Magnitude plot, the “peak” happens at ω ≅ 4.91(rad / sec) .
e.) One-tenth of the peak ⇒ − 20 (db) from the peak. From Magnitude plot, it
happens at ω ≅ 28.7 (rad / sec) .
f.) Exactly out of phase ⇒ Phase angle has 180 $ difference. From Phase plot, we
have NONE.
2b.)
a.) T.F. is given.
b.) When ω = 4 , and from plot we know that 20 log(
−
Output
= 10
Input
42.4
20
Output
) ≅ − 42.4 (db) ,
Input
≅ 0.0076 ; hence, Output = 5 * 0.0076 ≅ 0.038 . From the phase
plot, we know the φ ≅ 8.5$ . Therefore, Output ≡ 0.038 sin(4 t + 8.5$ ) .
[Note:
the phase angle is in “rad” for “4t” part.]
c.) Output = Input ⇒ find 0 (db). From Magnitude plot, we have
ω ≅ 59 & ω ≥1000 (rad / sec) .
d.) From Magnitude plot, the “peak” happens at ω ≅ 82.7 (rad / sec) .
e.) One-tenth of the peak ⇒ − 20 (db) from the peak. From Magnitude plot, it
happens at ω ≅ 37.8 (rad / sec) .
f.) Exactly out of phase ⇒ Phase angle has 180 $ difference. From Phase plot, we
have NONE.
2c.)
a.) T.F. is given.
b.) When ω = 4 , and from plot we know that 20 log(
Output
) ≅ − 72.4 (db) ,
Input
72.4
−
Output
= 10 20 ≅ 0.00024 ; hence, Output = 5 * 0.00024 ≅ 0.012 . From the
Input
phase plot, we know the φ ≅ 143$ . Therefore, Output ≡ 0.012 sin( 4 t + 143$ ) .
[Note: the phase angle is in “rad” for “4t” part.]
c.) Output = Input ⇒ find 0 (db). From Magnitude plot, we have NONE.
d.) From Magnitude plot, the “peak” happens at ω ≅ 84 (rad / sec) .
e.) One-tenth of the peak ⇒ − 20 (db) from the peak. From Magnitude plot, it
happens at ω ≅ 21.1 & ω ≅ 332 (rad / sec) .
f.) Exactly out of phase ⇒ Phase angle has 180 $ difference. From Phase plot, we
have NONE.