Fdt Bode Esercizi Crop
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Esercizi sulla forma di BODE della f.d.t. Scrivere nella forma adatta per il tracciamento dei diagrammi di BODE ( forma di BODE) le seguenti fdt:
Ø a) 24)(s10)10(sG(s)
++
=
SOLUZIONE Mettendo in evidenza 10 al numeratore e scomponendo il denominatore e mettendo in evidenza al denominatore due volte il 4 si ottiene:
)44s)4(4
4s4(
1)10s100(
4)4)(s(s
)1010s10(*10
G(s)
44
10++
+=
++
+=
0,25s)0,25s)(1(10,1s)6,25(1
0,25s)0,25s)(116(10,1s)100(1G(s)
+++
=++
+=
Per tracciare il diagramma di BODE occorre sostituire la variabile s con jw, perciò:
20,25jw)(10,1jw)6,25(1G(s)
++
=
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Ø b) 5)4)(ss(s2)5(sG(s)
+++
=
SOLUZIONE Mettendo in evidenza 2 al numeratore e al denominatore il 4 e il 5 si ottiene:
0,2s)0,25s)(120s(10,5s)10(1
)55
5s)5(
44
4ss4(
)22
2s2(*5
G(s)++
+=
++
+=
0,2s)0,25s)(1s(10,5s)(1 0,5G(s)
+++
=
Per tracciare il diagramma di BODE occorre sostituire la variabile s con jw, perciò:
0,2jw)0,25jw)(1jw(10,5jw)(1 0,5G(s)
+++
=
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4
Ø c) 0,5s)3s)(5(1s1)5(sG(s) 2 ++
+=
SOLUZIONE Mettendo in evidenza al denominatore il 5 si ottiene:
0,1s)3s)(1(1s1)(s
s)5
0,5553s)5((1s
1)5(sG(s) 22 ++
+=
++
+=
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ES.1 Tracciare il grafico del modulo e della fase della fdt seguente:
10s)s)(1(10,4s)20(1G(s)++
+=
SOLUZIONE Si sostituisce la s con jw:
10jw)jw)(1(10,4jw)20(1G(jw)
+++
=
Il guadagno statico Go = 20 in dB risulta: Go[dB]= 20* Log 20= 26 dB
Le pulsazioni d’angolo valgono h w1= 1/0.4=2.5 rad/s i w2= 1 rad/s i w3= 1/10=0,1 rad/s A queste pulsazioni associamo una freccia rivolta verso l’alto o verso il basso, a seconda che provengano da zeri o da poli ed in corrispondenza la spezzata sale ( h ) di 20 dB/dec (pendenza p= +1) o scende ( i ) di 20 dB/dec (p=-1). Nel caso di zeri o poli doppi la pendenza vale rispettivamente: Ø + 40dB/dec (p=+2); Ø - 40 dB/dec (p= -2)
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Il diagramma del modulo si costruisce nel seguente modo:
• si traccia una semiretta // all’asse delle w di ordinata 26 dB ( corrispondente al guadagno statico Go=20 ) da w=0 fino a w3= 0,1 rad/sec dove è posizionata la prima freccia (i).
• Da w3 il diagramma prosegue con pendenza -2 ( -20 dB/dec) fino a w2 (i)
• Da w2 (i) cambia la pendenza da -2 a - 4 (-40 dB/dec) fino a w1 (h).
• Da w1 (h) prosegue, essendo la freccia rivolta verso l’alto, con pendenza +2 ( 20 dB/dec) che sommata alla precedente - 4 fa -2 (-20 dB/dec)
Il diagramma degli sfasamenti si costruisce dividendo per 10 e moltiplicando per 10 ogni pulsazione d’angolo, si ha:
• 0,1 w1= 0,25 rad/s a ϕ = 0° • 10 w1= 25 rad/s a ϕ = +90° • 0,1 w2= 0,1 rad/s a ϕ= 0° • 10 w2= 10 rad/s a ϕ= -90° • 0,1 w3= 0,01 rad/s a ϕ= 0° • 10 w3= 1 rad/s a ϕ= -90°
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Es.2 Tracciare il grafico del modulo e della fase della fdt seguente:
4s)0,5s)(1s(12s)10(1G(s)
+++
=
SOLUZIONE
4jw)0,5jw)(1(12jw)10(1
jw4jw)0,5jw)(1jw(12jw)10(1G(jw)
+++
⋅=++
+=
1
Per facilitare il tracciamento del modulo di Bode consideriamo G(jw) = G1(jw)*G2(jw) come prodotto di 2 funzioni che disegniamo separatamente. Per ottenere il modulo complessivo si sommano le ordinate dei singoli grafici. G1(jw) fdt di un polo nell’origine G2(jw) G0=10 G0[dB]= 20*Log10= 20 dB Le pulsazioni d’angolo di G valgono h w1= 1/2= 0,5 rad/s i origine w*2= 1 rad/s i w3= 1/0,5= 2 rad/s i w4= ¼ =0,25 rad/s
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Es.3 Tracciare il grafico del modulo e della fase della fdt seguente:
10s)0,1s)(1(1s)20s(1G(s)++
+=
SOLUZIONE
=++
+=
10jw)0,1jw)(1(1jw) jw(120G(jw)
Per facilitare il tracciamento del modulo di Bode consideriamo G(jw) = G1(jw)*G2(jw) come prodotto di 2 funzioni che disegniamo separatamente. Per ottenere il modulo complessivo si sommano le ordinate dei singoli grafici. G1(jw) fdt di uno zero nell’origine G2(jw) G20=20 G20[dB]= 20*Log20= 26 dB Le pulsazioni d’angolo di G valgono h origine w*1= 1 rad/s h w2= 1/1= 1 rad/s i w3= 1/0,1= 10 rad/s i w4= 1/10 =0,1 rad/s
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Es.4 Tracciare il grafico del modulo e della fase della fdt seguente:
10s)2s)(1(1s)0,1(1G(s)
−+−
=
SOLUZIONE
10jw)2jw)(1(1jw)0,1(1G(jw)−+
−=
G0=0,1 G0[dB]= 20*Log 0,1= -20 dB Le pulsazioni d’angolo di G valgono h (parte immaginaria negativa) w*1= 1 rad/s i w2= 1/2= 0.5 rad/s i (parte immaginaria neg) w*3= 1/10 =0,1 rad/s La costante di tempo a numeratore , essendo negativa per w >10 w1, produce uno sfasamento negativo di -90°; per lo stesso motivo la costante di tempo a denominatore produce uno sfasamento di +90°
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Es.5 Tracciare il grafico del modulo e della fase della fdt seguente:
22s)5s)(10.4s)(1(15s)50(1G(s)
+++−
=
SOLUZIONE
22jw)5jw)(10.4jw)(1(15jw)50(1G(jw)
+++−
=
G0=50 G0[dB]= 20*Log 50= 34 dB Le pulsazioni d’angolo di G valgono h (parte immaginaria neg) w*1= 1/5= 0.2 rad/s i w2= 1/0.4= 2.5 rad/s i w3= 1/5 =0,2 rad/s ii w4= 1/2 =0,5 rad/s La costante di tempo a numeratore , essendo negativa per w >10 w1, produce uno sfasamento negativo di -90°;
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Es.6 Tracciare il grafico del modulo e della fase della fdt seguente:
22
2
s)0.1s)(1(1s10s)10(1G(s)
+++
=
SOLUZIONE
22
2
jw)0.1jw)(1(1jw10jw)10(1G(jw)
+++
=
G0=10 G0[dB]= 20*Log 10= 20 dB Le pulsazioni d’angolo di G valgono hh w1= 1/10= 0.1 rad/s ii origine w*2= 1 rad/s i w3= 1/0.1= 10 rad/s ii w4= 1 rad/s
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Es.7 Tracciare il grafico del modulo e della fase della fdt seguente:
2
2
50s)5s)(10.5s)(1(110sG(s)
+−+=
SOLUZIONE
2
2
50jw)5jw)(10.5jw)(1(110jwG(jw)
+−+=
G0=10 G0[dB]= 20*Log 10= 20 dB Le pulsazioni d’angolo di G valgono hh origine w*1= 1 rad/s i (parte immaginaria neg) w*2= 1/0.5 = 2 rad/s i w3= 1/5 = 0.2 rad/s ii w4= 1/50= 0.02 rad/s
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Es.8 Tracciare il grafico del modulo e della fase della fdt seguente:
4s))(1s(110s)-20(1-G(s) 2 +−
=
SOLUZIONE
4jw)jw)(1jw)(1(110jw)-20(1-G(jw)
++−=
G0=20 G0[dB]= 20*Log20= 26 dB Le pulsazioni d’angolo di G valgono h (parte immaginaria neg) w*1= 1/10= 0.1 rad/s i (parte immaginaria neg) w*2= 1 rad/s i w3= 1 rad/s i w4= ¼= 0.25 rad/s
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Es.9 Tracciare il grafico del modulo e della fase della fdt caratterizzata da poli e zeri complessi coniugati seguente:
13)4s2)(s(s78G(s) 2 +++
=
SOLUZIONE
La G(s) ha un polo reale s+2=0 p1= -2
e due poli complessi coniugati che si ottengono risolvendo l’equazione (1) s2+ 4s +13=0
p2=-2+j3 p2=-2-j3
scriviamo il trinomio nella forma canonica: (2) s2 + 2ζωn s + ωn
2 Uguagliando i coefficienti della (1) e della (2) si ha:
2ζωn= 4 ωn
2= 13 ωn = rad.q.(13) = 3.61 rad/s ζ = 4/(2ωn) = 2/3.61= 0.554
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Mettiamo la G(s) nella forma di Bode:
)0.076s0.3s0.5s)(1(13
)13s
134s
1313)13(
2s
222(
7813)4s2)(s(s
78G(s)
2
22
+++=
=+++
=+++
=
Il guadagno statico vale G0=3 G0[dB]= 20*Log3= 9.54 dB Le pulsazioni d’angolo di G valgono i w1 = 1/0.5= 2 rad/s ( per il polo reale) i w*n= 3.61 rad/s ( pulsazione naturale per i due poli complessi coniugati) Il grafico sarà tracciato tenendo presente che per valori di Ø w < 2 rad/s il grafico è costante è pari a 9,54
dB Ø w=2 rad/s si avrà una pendenza di -20 dB/dec
dovuta al polo reale Ø w=3,61 rad/s si avrà un’ulteriore incremento di
pendenza negativa di - 40 dB/dec a causa dei poli complessi coniugati Ø wn =0 3,61 rad/s dovuta ai poli complessi
coniugati , non si ha un effetto di risonanza in quanto lo scostamento massimo del diagramma
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asintotico da quello effettivo è di soli 0,89 dB. Infatti essendo ζ = 0,554 si ha Ø -20 Log10 2ζ= - 0,89 dB
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Es.10 Tracciare il grafico del modulo e della fase della fdt caratterizzata da poli e zeri complessi coniugati seguente:
2
2
10s)2s)(1(1)s10(1G(s)
+++
=
SOLUZIONE
La G(s) ha 3 poli reali 1+2s = 0 p1= -1/2 1+10s = 0 p2=p3= -1/10 E due zeri immaginari coniugati (1) 1+s2=0
z1=-rad.q(-1)=-j z2=rad.q.(-1)= +j scriviamo il trinomio nella forma canonica: (2) s2 + 2ζωn s + ωn
2 Uguagliando i coefficienti della (1) e della (2) si ha:
2ζωn= 0 ωn
2= 1 ωn = 1 rad/s ζ = 0 Il guadagno statico vale G0=10 G0[dB]= 20*Log10= 20dB Le pulsazioni d’angolo di G valgono i w1 = 1/2= 0.5 rad/s ( per i poli reali)
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ii w2 = 1/10= 0.1 rad/s ( per i poli reali) hh w*n= 1 rad/s ( pulsazione naturale per i due zeri immaginari coniugati)
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Es.11 Tracciare la risposta in frequenza della seguente fdt:
3)6)(s(s5)2)(s(sG(s)
++++
=
SOLUZIONE
Mettiamo la G(s) nella forma di BODE:
)3s)(1
6s(1
)5s)(1
2s0.55(1
)3s
33)3(
6s
666(
)5s
55)5(
2s
222(
G(s)++
++=
++
++=
Il guadagno statico vale G0=0.55 G0[dB]= 20*Log 0.556= -5.1 dB Le pulsazioni d’angolo di G valgono h w1 = 2 rad/s h w2 = 5 rad/s i w3= 6 rad/s i w4= 3 rad/s
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Es.12 Tracciare la risposta in frequenza della seguente fdt:
0,05)6)(ss(s3)20(sG(s)
+++
=
SOLUZIONE
Mettiamo la G(s) nella forma di BODE:
=++
+=
+++
=s)
0,051(1*0.05*s)
616(1*s
s)313(1*20
0,05)6)(ss(s3)20(sG(s)
s)0,05
1s)(161s(1
s)31200(1
G(s)++
+=
Il guadagno statico vale G0=200 G0[dB]= 20*Log 200= 46 dB Le pulsazioni d’angolo di G valgono h w1 = 3 rad/s i origine w*2 = 1 rad/s i w3= 6 rad/s i w4= 0.05 rad/s
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Es.13 Tracciare la risposta in frequenza della seguente fdt:
3)7)(s(ss5)30(sG(s) 2 ++
−=
SOLUZIONE
Mettiamo la G(s) nella forma di BODE:
s)31s)(1
71(1s
s)51-7.14(1-
s)31s)(1
713(1*7*s
s)51-5(1*30-
G(s)22 ++
=++
=
Il guadagno statico vale G0=7.14 G0[dB]= 20*Log 7.14= 17 dB Le pulsazioni d’angolo di G valgono h (parte immaginaria neg) w*1 = 5 rad/s ii origine w*2 = 1 rad/s i w3= 7 rad/s i w4= 3 rad/s
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Modulo
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Es.14 Tracciare la risposta in frequenza della seguente fdt:
3)2)(s1)(s(s3)s(sG(s)
+++−
=
SOLUZIONE
Mettiamo la G(s) nella forma di BODE:
1/3s)1/2s)(1s)(1(1*3*2*11/3s)-s(1*3-G(s)
+++=
1/3s)1/2s)(1s)(1(11/3s)-0.5s(1-G(s)
+++=
Il guadagno statico vale G0=0.5 G0[dB]= 20*Log 0.5= -6 dB Le pulsazioni d’angolo di G valgono h (origine) w*1 = 1 rad/s h (parte immaginaria neg) w*2 = 3 rad/s i w3 = 1 rad/s i w4= 2 rad/s i w5= 3 rad/s
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Modulo
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Es.15 Determinare il modulo e l’argomento della fdt seguente per Ø w=0 rad/s Ø w=1 rad/s Ø w=10 rad/s Ø w= ∞ rad/s
4)3)(s(s20G(s)
++=
SOLUZIONE
Mettiamo la G(s) nella forma di BODE:
0.25s)0.33s)(1(11.67
1/4s)1/3s)(14(1*320G(s)
++=
++=
0.25jw)0.33jw)(1(11.67G(jw)
++=
Il modulo
22 (0.25w)1 (0.33w)11.67
0.25jw)(10.33jw)(11.67G(jw)
++=
=++
=
L’argomento w)arctg(0.25w)arctg(0.33G(jw) −−=∠
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Ø w=0 rad/s
1.670)*(0.251 0)*(0.331
1.67G(j0)22
=++
=
0arctg(0)arctg(0)G(j0) =−−=∠ Ø w=1 rad/s
1.541.03*1.05
1.67(0.25)1 (0.33)1
1.67G(j1)22
==++
=
°=−−=∠ -32.3)arctg(0.25)arctg(0.33G(j1) Ø w=10 rad/s
0.182.7*3.45
1.6710)*(0.251 10)*(0.331
1.67G(j10)22
==
=++
=
°=−−=∠ -141.310)*arctg(0.2510)*arctg(0.33G(j10) Ø w= ∞ rad/s
01.67)*(0.251 )*(0.331
1.67)G(j22
=∞
==∞+∞+
=∞
°=∞−∞−=∞∠ -180)*arctg(0.25)*arctg(0.33)G(j
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