EE 370
Chapter V: Angle Modulation
Two additional Examples for Sketching FM and PM Signals
Example 1: Sketch the FM signal that results when modulating the message signal m(t)
shown below with kf = 2(2) and c = 2 (10) rad/s.
m(t)
2
1
0.3
t
–4
–7
–10
–2
2
4
8 14
14 6
5
6
–1
–2
fi(t) in Hz
is equal to
10 10
11
12 12
12
12 14
10
6 6
6 10 10 10
10
t
Phase Discontinuity here because of delta in m(t)
Phase Discont. = 0.3*kf = 1.2 rad
Example 2: Sketch the PM signal that results when modulating the message signal m(t)
shown below with kp = 2 and c = 2 (14) rad/s.
To sketch the PM signal, we can compute dm(t)/dt and sketch the frequency modulated
signal when dm(t)/dt is input to an FM block similar to Example 1.
EE 370
Chapter V: Angle Modulation
m(t)
8
5.2
4
1.5
t
–10
–7
–4
–2
2
–2
3
4
6
7
8
–4
–6.5
–8
dm (t )
 m (t )
dt
8
4
3.7
1.5
–10
–8
–6
–4
–2
2
4
5
6
–1.2
–4
–8
fi(t) in Hz
is equal to
18
18 14
14
6
6 14
14 16
16 14 14 14 14 12
12 16 16
t
Phase discont. because of discont. in
m(t) [delta function in dm(t)/dt]
Phase Discont. = 1.5*kp = 3 rad
Phase discont. because of discont. in
m(t) [delta function in dm(t)/dt]
Phase Discont. = 3.7*kp = 7.4 rad
Phase discont. because of discont. in
m(t) [delta function in dm(t)/dt]
Phase Discont. = – 1.2*kp = – 2.4 rad