Sample Exam 2: Solutions
#1) We use the time invariance principle and first find the solution to
0,0)0(),sin()(
)( ≥==+ tytty
dt
tdy
At the end we will shift the solution obtained by 3 time units. The homogeneous solution is obtained from
t
hh
hCetyty
dt
tdy −
=⇒=+ )(0)(
)(
The particular solution satisfies
)cos()sin()()sin()(
)( tttytty
dt
tdy
pp
p
βα
+=⇒=+
Plugging this solution into the differential equation implies 50.
=
α
and 50.
−
=
β
so that the particular solution is
given by )cos(5.0)sin(5.0)( tttyp−=
The system response is given by
)cos(5.0)sin(5.0)()()( ttCetytyty t
ph −+=+= −
Its initial condition produces
t
hph etyCCyyy −
=⇒=⇒−=+== 5.0)(5.05.0)0()0(0)0(
Hence, the system response due to )sin(t is
)cos(5.0)sin(5.05.0)()()( ttetytyty t
ph −+=+= −
The system response due to )3sin( −t , by the time invariance, is given by
)3())3cos(5.0)3sin(5.05.0()()()( )3( −−−−+=+= −− tuttetytyty t
ph
The system zero-state response satisfies
0,0)0(),3sin()(
)( ≥=−=+ tytty
dt
tdy
zszs
zs
Using the previously obtained result, the zero-state response is given by
)3())3cos(5.0)3sin(5.05.0()( )3( −−−−+= −− tuttety t
zs
The zero-input response satisfies
0,0)0(,0)(
)( ≥==+ tyty
dt
tdy
zizi
zi
The zero-input response is given by
00)( ≥== −tCety t
zi
The system response in terms of its zero-state and zero-input components, is given by
)3())3cos(5.0)3sin(5.05.0()()()( )3( −−−−+=+= −− tuttetytyty t
zizs
The steady state response is practically obtained for large values of time, that is )3cos(5.0)3sin(5.0)(),()( −
−
−
≈
⇒= tttylargetfortyty ssss
According to the textbook definition of the transient response, we have
smalltfortyt
tr
y),()( =
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