solution

omg = 0.1:0.1:1000;
num = [10,0];den = [1,20,100];
sys = tf(num,den);
figure(1),bode(omg,sys);grid on; % using "bode" to get the Bode diagram
Magnitude (dB)
Phase (deg)
0
-20
-40
-60
90
45
0
-45
-90
10 -1
Asymptote:
10 0
j10ω
H( jω)
=
( jω
+ 10)
2
10 1
10 2
10 jω
= ⋅
jω+ 10 jω+
10
From the expression, we see the frequency response have: zero at jω = 0 , pole at
jω =− 10 and pole at jω =− 10 .
(1). Magnitude asymptote
First consider the part
10
jω + 10
10
20log10 0
10 = . For high frequencies ( ω ω p
10 3
. For low frequencies ( ω ω p ), the dB-scale magnitude is
20log
), magnitude should be 10
10
, which
jω
is a straight line with the slope of -20dB and go through (10,0). The magnitude
asymptote from
10
jω + 10
is a pair of straight lines with corner frequency at ω p = 10 ,