Assignment- 2
Mayank Sharma (2019EEP2310)
Q1)
Non Inverting Buck -Boost Transfer Function
Q1)
Non-inverting buck-boost converter is operated at
D=0.7,
Vin=50 V,
L=150 μH,
C= 300 μF.
R= 12 Ω
Code for bode plot
clc;
clear all;
% Information given in question
Vin = 50;
R = 12;
L = 150e-6;
C = 300e-6;
D = 0.7;
s = tf('s');
De = 1-D;
wo = (De)/(sqrt(L*C));
Q = (De*R)/(sqrt(L/C));
wz=((De^2)*R)/(D*L) ;
Gvdo = (Vin)/(De^2) ;
Gvgo = (D)/(De);
% Control to Output Transfer Function TF1
TF1= (Gvdo*(1-(s/wz)) )/( ( (s^2)/(wo^2)) + (s/(Q*wo))+ 1 );
% Line to Output Transfer Function
TF2= Gvgo /( ( (s^2)/(wo^2)) + (s/(Q*wo))+ 1 );
grid on;
hold on;
options = bodeoptions;
options.FreqUnits = 'kHz';
bode(TF1,TF2,options);
Since we have to comment on open loop transfer function stability , that depends upon
position of poles for that transfer function (not dealing with any closed loop system ) ,
and from bode plot we can comment on stability for CLTF
Gvd(s) Simple Transfer Function → Poles are in LHP hence OLTF for Gvd(s) is stable .
Gvg(s) Simple Transfer Function → Poles are in LHP hence OLTF for Gvg(s) is
stable.
But, If we consider a feedback system having its plant function as Gvg(s) or Gvd(s)
We can conclude from bode plot
Close Loop stability considering plant as Gvd(s) – As phase margin is negative (-84.4o)
system would be unstable. (wgc=17.3 kHz)
Close Loop stability considering plant as Gvg(s) – As phase margin is Positive
(8.8o)system would be stable . (wgc=0.409 kHz)
Q 2 a)
Duty is varied from 0.4 to 0.8 in steps of 0.1 with Ro=12 Ω and Vin=50 V.
Code for bode plot
clc;
clear all;
% Information given in question
Vin = 50;
R = 12;
L = 150e-6;
C = 300e-6;
D = 0.4; % Starting value of D
s = tf('s');
for D = 0.4:0.1:0.8
De = 1-D;
wo = (De)/(sqrt(L*C));
Q = (De*R)/(sqrt(L/C));
wz=((De^2)*R)/(D*L) ;
% Control to Output Transfer Function TF1
Gvdo = (Vin)/(De^2) ;
TF1= (Gvdo*(1-(s/wz)) )/( ( (s^2)/(wo^2)) + (s/(Q*wo))+ 1 );
grid on;
hold on;
options = bodeoptions;
options.FreqUnits = 'kHz';
bode(TF1,options);
end
Bode Plot of Gvd(s)
CLTF=Gvd(s)/(1+Gvd(s)) Pole Zero Mapping for cross verification
From Bode Plot as D increases Gain crossover frequency (wgc) increases
From 1st page a zero in RHP is located at
𝐷2 ∗ 𝑅
𝑤𝑧 =
𝐷∗𝐿
so as D increases wz moves towards complex conjugate poles ,or wgc and wz come
closer , hence effect of RHP zero ,in phase of system (phase would become more
negative near wgc ), increases which will make system more unstable .
Also in Bode plot phase margin is most negative for highest D value so from there as
well we conclude as D increases system become more unstable .
This can be verified from pole zero plot of CLTF=Gvd(s)/(1+Gvd(s)) as D increases close
loop system pole move away from origin hence will create unstability.
Q 2 b)
Load resistance is varied from 12 Ω to 48 Ω in steps of 12 Ω with D=0.5 and Vin=50 V.
Code for bode plot
clc;
clear all;
% Information given in question
Vin = 50;
L = 150e-6;
C = 300e-6;
D = 0.5;
R = 12; % Starting Value of R
s = tf('s');
De = 1-D;
for R =12:12:48
wo = (De)/(sqrt(L*C));
Q = (De*R)/(sqrt(L/C));
wz=((De^2)*R)/(D*L) ;
% Control to Output Transfer Function TF1
Gvdo = (Vin)/(De^2) ;
TF1= (Gvdo*(1-(s/wz)) )/( ( (s^2)/(wo^2)) + (s/(Q*wo))+ 1 );
grid on;
hold on;
options = bodeoptions;
options.FreqUnits = 'kHz';
bode(TF1,options);
end
Bode Plot of Gvd(s)
CLTF=Gvd(s)/(1+Gvd(s)) Pole Zero Mapping for cross verification
From Bode Plot as R increases Gain crossover frequency (wgc) is decreasing.
A zero in RHP is located at
𝐷2 ∗ 𝑅
𝑤𝑧 =
𝐷∗𝐿
so as R increases wz moves away from complex conjugate poles ,or wgc and wz goes
away from each other , hence effect of RHP zero, in phase of system (phase would
become less negative near wgc ),decreases which will make system more stable .
Also in Bode plot phase margin is most negative for Low R value so from there as well
we conclude as R increses system become more stable .
This can be verified from pole zero plot of CLTF=Gvd(s)/(1+Gvd(s)) as R increses close loop
system pole move towards origin hence will create more stablity.
Q 2 c)
Input voltage is varied from 10 V to 100 V in steps of 10 V with Vo=70 V and Ro=10 Ω.
Code
clc;
clear all;
% Information given in question
L = 150e-6;
C = 300e-6;
Vo = 70;
R = 10;
s = tf('s');
for Vin =10:10:100
D=Vo/(Vo + Vin );
De = 1-D;
wo = (De)/(sqrt(L*C));
Q = (De*R)/(sqrt(L/C));
wz=((De^2)*R)/(D*L) ;
% Control to Output Transfer Function TF1
Gvdo = (Vin)/(De^2) ;
TF1= (Gvdo*(1-(s/wz)) )/( ( (s^2)/(wo^2)) + (s/(Q*wo))+ 1 );
grid on;
hold on;
options = bodeoptions;
options.FreqUnits = 'kHz';
bode(TF1,options);
end
Bode Plot of Gvd(s)
CLTF=Gvd(s)/(1+Gvd(s)) Pole Zero Mapping for cross verification
From Bode Plot as Vin increases Gain crossover frequency (wgc) is decreasing.
A zero in RHP is located at
𝐷2 ∗ 𝑅
𝑤𝑧 =
𝐷∗𝐿
so as Vin increases, D=Vo/(Vo+Vin) decreases hence wz increases ,moves away from
complex conjugate poles ,or wgc and wz goes away from each other , hence effect of
RHP zero , in phase of system (phase would become less negative near wgc ),decreases
which will make system more stable .
Also in Bode plot phase margin is most negative for Low Vin value so from there as well
we conclude as Vin increases system become more stable .
This can be verified from pole zero plot of CLTF=Gvd(s)/(1+Gvd(s)) as Vin increses close
loop system pole move towards origin hence will create more stablity.