1. 11/5/2015
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Test
Solution
15th April 2015
Question 1 (a)
• x1(t) and x2(t) are periodic signals of period
T1=2π/12 and T2=2 π /16 sec repectively.
• The ratio of period, T1/T2 = 4/3 is a rational
number.
• Therefore, x(t) is a periodic signal of new
period T= π/2 sec.
Question 1 (b)
• i.
• ii.
841.0)1sin()12sin()2()1sin(
5
5
==−=−−∫−
dttt δ
2
1
2
1
2
1
)2(
2
1
)42( 0)22()2()2(
===−=− −−
∞
∞−
∞
∞−
−−−−
∫ ∫ eedttedtte tt
δδ
Question 1 (c)
• Let t = 1: y(1) = 7x(0) + 6
• The system is with memory because present
output depends on past input.
• The system is causal because present output does
not depend on future input.
• The system is time-invariant because output
does not explicitely depends on when the input
is applied to the system.
• The system is non-linear because the
superposition principle is not applicable to the
system
Question 2 (a)i Question 2 (a)ii
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Question 2 (a)iii Question 2 (b)i
)5(
2
1
)3(
2
1
)2()1()(1 −+−−−−−= trtrtrtrtx
Question 2 (b)ii
elsewhere
t
t
t
t
t
tx
53
32
21
;
;
;
;
0
2
5
2
1
1
)(1
<<
<<
<<
+−
−
=
Question 3 (a)
• Therefore x2(t) is Energy signal because 0<E<∞ and P = 0.
<<
<<−+
=
Elsewhere
t
tt
tx
;0
10;1
01;1
)(2
dtdttI ∫∫ ++=
−
1
0
2
0
1
2
|1||1|
3
4
111
3
1
=
++−=
I
I
3
4
|)(|lim 2
2 == ∫−∞→
dttxE
T
TT
joules
0|)(|
2
1
lim 2
2 == ∫−∞→
dttx
T
P
T
TT
watt
Question 3 (b)i
Write the Kirchoff’s voltage law around the loop: 0)()()( =−− tvtvtv RL
Write the voltage across the inductor:
dt
tdi
LtvL
)(
)( =
Write the voltage across the resistor: )()( tRitvR =
Substitute into first equation: )()(
)(
tvtRi
dt
tdi
L =+
Divide all by L : )(
1
)(
)(
tv
L
ti
L
R
dt
tdi
=+
Question 3 (b)ii
)5)(100()(1000000)( =+′ titi
The homogeneous solution is
t
h Aeti 1000000
)( −
=
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Question 3 (b)ii
The particular solution form is
Btip =)( and 0)( =′ tip
Substitute )(tip and )(tip
′ into the above differential equation
0005.0
1000000
500
50010000000
==
=+
B
B
Therefore, the particular solution is
0005.0)( =tip
And the total solution is
0005.0)( 1000000
+= − t
Aeti
Question 3 (b)ii
Substitute the given initial condition, 1.0)0( =−
i
0995.00005.01000.0
1.00005.0
=−=
=+
A
A
Hence, the total solution is
0005.00995.0)( 1000000
+= − t
eti Ampere
Question 4 (a)i
Write the differential equation:
t
tetytyty −
=+′+′′ )(5)(2)(
Write the characteristic equation:
0522
=++ αα
Write the charateristic roots:
211 j+−=α , 212 j−−=α
Write the homogeneous solution:
teCteCty tt
h 2sin2cos)( 21
−−
+=
And its derivative is:
teCteCteCteCty tttt
h 2cos22sin2sin22cos)( 2211
−−−−
+−−−=′
Question 4 (a)i
Setting 0=t and substituting 1)0( =−
y and 0)0( =′ −
y in these equation yields:
11 C=
21 20 CC +−=
The solution of these two simultaneous equations are:
2
1
1
2
1
=
=
C
C
The zero-Input Response is therefore:
tetety tt
zi 2sin
2
1
2cos)( −−
+=
Question 4 (a)ii
Assume the particular solution base on input form:
t
p ePtPty −
+= )()( 21
tt
p ePtPePty −−
+−=′ )()( 211
ttt
p ePePtPePty −−−
−++−=′′ 121
3
1 )()(
Sustitute )(typ
′′ , )(typ
′ and )(typ in the diffential equation:
ttttttttt
teePtePePtePePePePtPeP −−−−−−−−−
=+++−+−++− 21211121
3
1 55222)(
Equating coefficients of similar terms on both sides of this expression yields:
152: 111 =+−−
PPPte t
05222: 22121 =++++−−
PPPPPe t
Question 4 (a)ii
Solving the unknown coefficient from the above equation yield:
0
4
1
2
1
=
=
P
P
Therefore, we can write the particular solution:
t
p tety −
=
4
1
)(
And the zero-state response is given by:
ttt
zs teteCteCty −−−
++=
4
1
2sin2cos)( 21
where its derivative is :
tttttt
zs teeteCteCteCteCty −−−−−−
−++−−−=′
4
1
4
1
2cos22sin2sin22cos)( 2211
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Question 4 (a)ii
Setting 0=t and substituting all initial conditions set to zero, 0)0( =−
y and 0)0( =′ −
y in these
equation yields:
10 C=
4
1
20 21 ++−= CC
Solving these two simultaneous equations:
4
1
0
2 21
1
−=
=
+− CC
C
Question 4 (a)ii
The solution of these two simultaneous equations are:
8
1
0
2
1
−=
=
C
C
Therefore, the Zero State Response is :
tt
zs tetety −−
+−=
4
1
2sin
8
1
)( for 0>t
Question 4 (a)iii
The Total Response of the system is given by:
)()()( tytyty zszi +=
ttt
tetetety −−−
++=
4
1
2sin
8
3
2cos)( for 0>t
Question 5 (a)
i. tt
h eCeCty 697.0
2
302.4
1)( −−
+=
ii. tCtCtyh 3sin3cos)( 21 +=
Question 5 (b)
i.
t
p etPtPty 3
21 )4sin4cos()( −
+=
ii. tPtPePPtPtPty t
p 5sin5cos)( 54
3
301
2
2 +++++= −
Question 5 (c)
tnbtnaatf n
n
n 00
1
0 sincos)( ωω ++= ∑
∞
=
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Question 5 (d)
• i. The Fundamental period is T0 = 2 sec and the
angular frequency is ω0 = π rad/s
• ii Determine DC value of g(t)
∫ ==
2
0
0
2
1
)(
2
1
dttga
Determine an fourier coefficient
∫=
2
0
cos)(
2
2
tdtntgan π
0sin
1
cos)1(
2
1
2
1
=== ∫ tn
n
tdtnan π
π
π
Question 5 (d)
Determine bn fourier coefficient
∫=
2
0
sin)(
2
2
tdtntgbn π
( )ππ
π
π
π
π nn
n
tn
n
tdtnbn cos2cos
1
cos
1
sin)1(
2
1
2
1
−−=−== ∫
=−
=
=
oddn
n
evenn
bn
;
2
;0
π
The Trigonometric Fourier series is
tn
n
tg
oddn
π
π
sin
1
2
2
1
)( ∑
∞
=
−=
