EEE 303A TEST1 - Full Solution

SOLUTIONS: EEE 303A TEST1
Solution 1
Part-1
1.1. H1 ( s ) = ( s+1)
(s+2)( s2 + s+16)
Putting s = jw, we get
H1 ( jw )= ( jw+ 1)
( jw+ 2)(−w2 + jw +16)
So, magnitude = |H1 ( jw )∨¿ √ (w2+ 1)
√ (w2+ 4) √ ((16−w2)+ w2 )
Phase = ∠ H1 ( jw ) =tan−1 w−tan−1 w
2 −¿ tan−1 w
(16−w2 )−πu(w2−16)¿
Asymptotic plot:
Asymptotes at corner frequencies of 1 rad/s(20dB/dec, 90deg), 2 rad/s
(-20dB/dec, -90deg) and √16 rad/s(-40dB/dec, -180deg)
The transfer function has quadratic terms in the denominator. On comparing the
quadratic terms with the standard form of s2 +2 ξ wn s+ wn
2, we get wn=¿ √16 and
ξ= 1
8 =0.125< 1. Hence, the system is underdamped. The asymptotic plot also
shows significant resonance.

Bode Plot: is generated using following MATLAB commands.
>>syms s
>>H=tf([1,1],[1,3,18,32])
>>bode(H)

1.2. H2 ( s )= s
(s+10)( s+1)
Putting s = jw, we get
H2 ( jw )= jw
( jw+ 10)( jw +1)
So, magnitude = | H2 ( jw ) ∨¿ w
√ (w2+100)√ (w2+1)
Phase = ∠ H2 ( jw ) = π
2 −tan−1 w
10 −¿ tan−1 w ¿
Asymptotic Plot:
Asymptotes at corner frequencies of 1rad/s (-20dB/dec, -90deg), 10rad/s (-
20dB/dec, -90deg) and 0 rad/s(20dB/dec, 90deg)
Bode Plot: is generated using following MATLAB commands.
>>syms s
>>H=tf([1,0],[1,11,10])

>>bode(H)
1.3. H3 ( s )= 1
(s−1)( s+ 1)
Putting s = jw, we get
H3 ( jw )= 1
( jw−1)( jw +1)
So, magnitude = |H3 ( jw )∨¿ 1
(1+ w2)
Phase = ∠ H3 ( jw ) =¿-π+tan−1 w
1 −¿ tan−1 w
1 ¿ = -π
Asymptotic Plot:
Asymptote at corner frequency of 1rad/s(-40dB/dec, -180deg)