ENSC 327
Midterm
Oct 31, 2011
School of Engineering Science
Simon Fraser University
Burnaby, BC
Canada
INSTRUCTIONS
1. Open book.
2. Answer all FOUR questions. They are of equal value.
3. Time allowed: 110 minutes.
4. Most definitions/notations in this paper are consistent with those in the lecture notes and are not
repeated here.
5. There is no need to re-derive results that are already in the notes, text, or in the solutions to the
tutorials and assignments. Just quote them in your solution.
6. You may find the following definitions and formulae useful:
=1 KHz 10=
3
=
Hz; 1 MHz 10 6
Hz; 1 GHz 109 Hz
Analytical expression – a mathematical equation describing a physical phenomenon in terms of the
relevant parameters.
e jθ =
cos θ + j sin θ ; j2 =
−1
(e − eβa ) ,
b
d βx 1
e= β e β x , ∫e
βx
=
dx βb
e b e a +b ,
e a ⋅=
dx a β
=
cos( a + b) cos( a ) cos(b) − sin( a ) sin(b) , 2cos( a )cos(b=
) cos( a + b ) + cos( a − b )
=
sin( a + b) sin( a ) cos(b) + cos( a ) sin(b) , 2sin( a )cos(b=
) sin( a + b ) + sin( a − b )
b
1
∫ cos(2π fx )dx
= ( sin(2π fb) − sin(2π fa ) ) ,
a 2π f
b
1
∫ sin(2π fx )dx
= ( cos(2π fa ) − cos(2π fb) )
a 2π f
, Question 1
Consider the signal m(t ) = sin 2 (2πf m t ) , where f m > 0 . Determine
(a) the Hilbert transform of m(t ) ;
(b) the positive pre-envelope m p (t ) of m(t )
Solution
First we rewrite m(t ) as
1 − cos(4π f m t )
= =
m(t ) sin 2
(2π f m t )
2
The corresponding Fourier Transform is
1 δ ( f − 2 fm ) + δ ( f + 2 fm )
δ ( f ) −
1
=
M( f )
2 2 2
(5 marks)
(a) The Hilbert Transform of M ( f ) is
Mˆ ( f ) = − j sgn( f ) M ( f )
1 δ ( f − 2 fm ) + δ ( f + 2 fm )
− j sgn( f )δ ( f ) + j sgn( f ) ⋅
=
2 2
1 δ ( f − 2 fm ) + δ ( f + 2 fm )
= j 2 sgn( f ) ⋅
2 2j
1 δ ( f − 2 fm ) − δ ( f + 2 fm )
= −
2 2j
Taking inverse Fourier Transform of Mˆ ( f ) yields
1
mˆ (t ) = − sin ( 2π 2 f mt )
2
(10 marks)
(a) The positive pre-envelope is
2 M ( f ), f > 0;
=
M p ( f ) =M ( f ), f 0;
0,
otherwise
1 1
= δ ( f ) + δ ( f − 2 fm )
2 2
Midterm
Oct 31, 2011
School of Engineering Science
Simon Fraser University
Burnaby, BC
Canada
INSTRUCTIONS
1. Open book.
2. Answer all FOUR questions. They are of equal value.
3. Time allowed: 110 minutes.
4. Most definitions/notations in this paper are consistent with those in the lecture notes and are not
repeated here.
5. There is no need to re-derive results that are already in the notes, text, or in the solutions to the
tutorials and assignments. Just quote them in your solution.
6. You may find the following definitions and formulae useful:
=1 KHz 10=
3
=
Hz; 1 MHz 10 6
Hz; 1 GHz 109 Hz
Analytical expression – a mathematical equation describing a physical phenomenon in terms of the
relevant parameters.
e jθ =
cos θ + j sin θ ; j2 =
−1
(e − eβa ) ,
b
d βx 1
e= β e β x , ∫e
βx
=
dx βb
e b e a +b ,
e a ⋅=
dx a β
=
cos( a + b) cos( a ) cos(b) − sin( a ) sin(b) , 2cos( a )cos(b=
) cos( a + b ) + cos( a − b )
=
sin( a + b) sin( a ) cos(b) + cos( a ) sin(b) , 2sin( a )cos(b=
) sin( a + b ) + sin( a − b )
b
1
∫ cos(2π fx )dx
= ( sin(2π fb) − sin(2π fa ) ) ,
a 2π f
b
1
∫ sin(2π fx )dx
= ( cos(2π fa ) − cos(2π fb) )
a 2π f
, Question 1
Consider the signal m(t ) = sin 2 (2πf m t ) , where f m > 0 . Determine
(a) the Hilbert transform of m(t ) ;
(b) the positive pre-envelope m p (t ) of m(t )
Solution
First we rewrite m(t ) as
1 − cos(4π f m t )
= =
m(t ) sin 2
(2π f m t )
2
The corresponding Fourier Transform is
1 δ ( f − 2 fm ) + δ ( f + 2 fm )
δ ( f ) −
1
=
M( f )
2 2 2
(5 marks)
(a) The Hilbert Transform of M ( f ) is
Mˆ ( f ) = − j sgn( f ) M ( f )
1 δ ( f − 2 fm ) + δ ( f + 2 fm )
− j sgn( f )δ ( f ) + j sgn( f ) ⋅
=
2 2
1 δ ( f − 2 fm ) + δ ( f + 2 fm )
= j 2 sgn( f ) ⋅
2 2j
1 δ ( f − 2 fm ) − δ ( f + 2 fm )
= −
2 2j
Taking inverse Fourier Transform of Mˆ ( f ) yields
1
mˆ (t ) = − sin ( 2π 2 f mt )
2
(10 marks)
(a) The positive pre-envelope is
2 M ( f ), f > 0;
=
M p ( f ) =M ( f ), f 0;
0,
otherwise
1 1
= δ ( f ) + δ ( f − 2 fm )
2 2
