solution
solution solution
Code: omg = 0.01:0.01:10000; num = [20,0];den = [1,20,10000]; sys = tf(num,den); figure(1),bode(omg,sys);grid on Magnitude (dB) Phase (deg) 0 -10 -20 -30 -40 90 45 0 -45 -90 10 1 Asymptote: Bode Diagram 10 2 Frequency (rad/sec) j20ω H( jω) = ( jω+ 10 − j99.5)( jω+ 10 + j99.5) ω j j20ω = = 500 2 2 ⎡ ω ω ⎤ 10000 ⎢1− + j ⎛ ω ⎞ ω 10000 500 ⎥ 1− ⎜ ⎟ + j ⎣ ⎦ ⎝100 ⎠ 500 From the expression, we see the frequency response have: zero at jω = 0 , a pair of complex conjugate poles at jω =− 10 + j99.5 and jω = −10 − j99.5 . (1). Magnitude asymptote 10 3
The dB-scale magnitude asymptote from zero j 500 ω is a straight line with a slope of 20dB and go through (500,0), as shown (dashed line). The dB-scale magnitude asymptote from pole ⎛ ω ⎞ ω 1− ⎜ ⎟ + j ⎝100 ⎠ 500 2 is a pair of straight lines and ω p = 100 . For low frequency ( ω ω p ), it’s a straight line with a slope of 0dB. For high frequency ( ω ω p ), it’s a straight line with a slope of -40dB. These two asymptotes intersect at ω = ω p , as shown (dotted line). Then, applying composition rules to get the final magnitude asymptote: dashed line plus dotted line, as shown (solid line). gain factor gain factor 50 0 -50 -100 -10 -20 -30 -40 -50 -60 10 0 10 0 (2). Phase asymptote 10 1 10 1 10 2 ω 10 2 The phase asymptote from zero j 500 ω is a straight line with a slope of 0 at phase π equals , as shown (dashed line). 2 The phase asymptote from pole ω ⎛ ω ⎞ ω 1− ⎜ ⎟ + j ⎝100 ⎠ 500 2 10 3 10 3 10 4 10 4 contains three segments and
- 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 and 8: 5 5 5 ⎡ ⎛ × kπ × ⎞ ⎤ ⎢
- Page 9 and 10: as shown (dashed line). Similarly,
- Page 11: asym_1 = 20*log10(ym); figure(2),su
- Page 15: omg2 = omg_p+1:1:10000; ym = zeros(
The dB-scale magnitude asymptote from zero j<br />
500<br />
ω is a straight line with a slope of<br />
20dB and go through (500,0), as shown (dashed line).<br />
The dB-scale magnitude asymptote from pole<br />
⎛ ω ⎞ ω<br />
1−<br />
⎜ ⎟ + j<br />
⎝100 ⎠ 500<br />
2<br />
is a pair of straight<br />
lines and ω p = 100 . For low frequency ( ω ω p ), it’s a straight line with a slope of<br />
0dB. For high frequency ( ω ω p ), it’s a straight line with a slope of -40dB. These<br />
two asymptotes intersect at ω = ω p , as shown (dotted line).<br />
Then, applying composition rules to get the final magnitude asymptote: dashed line<br />
plus dotted line, as shown (solid line).<br />
gain factor<br />
gain factor<br />
50<br />
0<br />
-50<br />
-100<br />
-10<br />
-20<br />
-30<br />
-40<br />
-50<br />
-60<br />
10 0<br />
10 0<br />
(2). Phase asymptote<br />
10 1<br />
10 1<br />
10 2<br />
ω<br />
10 2<br />
The phase asymptote from zero j<br />
500<br />
ω is a straight line with a slope of 0 at phase<br />
π<br />
equals , as shown (dashed line).<br />
2<br />
The phase asymptote from pole<br />
ω<br />
⎛ ω ⎞ ω<br />
1−<br />
⎜ ⎟ + j<br />
⎝100 ⎠ 500<br />
2<br />
10 3<br />
10 3<br />
10 4<br />
10 4<br />
contains three segments and
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