solution
solution 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 ,
as shown (dashed line). Similarly, for part jω jω + 10 jω jω 20log ≈ 20log jω + 10 10 , 10 10 at low frequencies ( ω ω p ), which is a straight line with the slope of 20dB and go through (10,0). At high frequencies ( ω ω p asymptote from (dotted line). jω jω 20log ≈ 20log = 0 . So, the magnitude jω+ 10 jω ), 10 10 jω jω + 10 is also a pair of straight lines intersect at ω = ω p , as shown At last, applying compose the dashed line and dotted line to get the final magnitude asymptote, as shown (solid line). gain factor gain factor 0 -10 -20 -30 -40 10 -1 0 -10 -20 -30 -40 10 -1 (2). Phase asymptote The phase asymptote from ( ω ω 10 ) and p 10 0 10 0 10 jω + 10 10 1 ω 10 1 ω 10 2 10 2 10 3 10 3 contains three segments: 0 for low frequencies π − for high frequencies ( ω 10ω p ), and at frequencies around 2
- Page 1 and 2: Problem 1 (Chapter 8, P. 22) T 0 =
- Page 3 and 4: Problem 2 (Chapter 8, P. 23) From (
- Page 5 and 6: for every value of k . (2) From the
- Page 7: 5 5 5 ⎡ ⎛ × kπ × ⎞ ⎤ ⎢
- Page 11 and 12: asym_1 = 20*log10(ym); figure(2),su
- Page 13 and 14: The dB-scale magnitude asymptote fr
- Page 15: omg2 = omg_p+1:1:10000; ym = zeros(
as shown (dashed line).<br />
Similarly, for part<br />
jω<br />
jω<br />
+ 10<br />
jω jω<br />
20log ≈ 20log<br />
jω<br />
+ 10 10<br />
, 10 10<br />
at low frequencies<br />
( ω ω p ), which is a straight line with the slope of 20dB and go through (10,0). At high<br />
frequencies ( ω ω p<br />
asymptote from<br />
(dotted line).<br />
jω jω<br />
20log ≈ 20log = 0 . So, the magnitude<br />
jω+ 10 jω<br />
), 10 10<br />
jω<br />
jω<br />
+ 10<br />
is also a pair of straight lines intersect at ω = ω p , as shown<br />
At last, applying compose the dashed line and dotted line to get the final magnitude<br />
asymptote, as shown (solid line).<br />
gain factor<br />
gain factor<br />
0<br />
-10<br />
-20<br />
-30<br />
-40<br />
10 -1<br />
0<br />
-10<br />
-20<br />
-30<br />
-40<br />
10 -1<br />
(2). Phase asymptote<br />
The phase asymptote from<br />
( ω ω 10 ) and<br />
p<br />
10 0<br />
10 0<br />
10<br />
jω + 10<br />
10 1<br />
ω<br />
10 1<br />
ω<br />
10 2<br />
10 2<br />
10 3<br />
10 3<br />
contains three segments: 0 for low frequencies<br />
π<br />
− for high frequencies ( ω 10ω p ), and at frequencies around<br />
2
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