Expert answer:RLC filters

Answer & Explanation:I have this project that is due in 24 hours, I have done the first part
but parts two and three are not complete. There is .xyz = .306
I need this to be typed in a Microsoft word. only the second and third
part. These are the answers for part one and the project tasks. project tasks.docx EE321ProjectPrelimSolutionsFall2015.pdf
project_tasks.docx

ee321projectprelimsolutionsfall2015.pdf

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EE 321
Project Solutions – Tasks 1-3
Fall 2015
+
v i (t )
1 ( )
–
2 ( )
3 ( )
+
A circuit with input voltage vi (t ) , and output voltage v0 (t ) , is shown below. The complex impedances
in the circuit represent either a resistor ( ( ) = ), an inductor ( ( ) = ), or a capacitor ( ( ) =
1/ ).
4 ( )
v 0 (t )
–

Figure 1. Generic circuit with input voltage ( ), and output voltage 0 ( ).
Part 1.
1. Use s-domain techniques and determine the transfer function
0 ( )
( ) =
( )
in terms of the impedances 1 , 2 , 3 , and 4 .
Solution: Use KVL around the two loops, with 1 ( ) and 2 ( ) the loop currents.
( ) = 1 ( 1 + 2 ) − 2 ( 2 )
0 = − 1 2 + 2 ( 2 + 3 + 4 )
Writing as a matrix equation and using Cramer’s rule yields
0 ( )
2 4
( ) =
=
.
( ) 1 ( 2 + 3 + 4 ) + 2 ( 3 + 4 )
2. Let 1 ( ) = 1 , 2 ( ) = 1/ 2 , 3 ( ) = 3 , and 4 ( ) = 1/ 4 , corresponding to the circuit
below. Using Kirchhoff’s voltage and currents laws, derive the 3rd order differential equation for
v0 (t ) . Assume that there is no energy stored in the capacitors or inductor at time t = 0 − . From
the differential equation, find a state variable representation, and specify the state variable
matrices A, B, C, D, to generate the voltages v0 (t ) and v L1 (t ) . From the differential equation,
( )
determine the transfer function, ( ) = 0( ). Verify that it is equivalent to the result in 1) for the
choice of impedances in 2).

+
–
+
+
1 ( )
–
–
Solution: Define 1 and 2 as the two loop currents, and use KVL for each loop, and
KCL at the top node in the circuit.
1
+ 2
= 1

0 = − 2 + 2 3 + 4
2
4
+ 4
.
1 = 2 + 4 = 2

Use the second equation in the third, and then take the derivative of the (resulting) third equation in the
first equation, yielding
3 4
1 2 + 1 4 2 4
1 4
1
1
+
�
�
+
+
4 =
.
3
2

1 2 4 3

1 2
1 2 4 3
1 2 4 3
Taking the Laplace transform (with all-zero initial conditions) yields the same transfer function as
above, with the appropriate values of , namely
1
0
0 ( )
1 2 4 3
=
= 3
.
( ) =
2
1
+ 2 + 1 + 0
( ) 3 + � 1 2 + 1 4 � 2 + 1 +
1 2 4 3
1 2
1 2 4 3
The state variables are then 1 = 0 = 4 , 2 =
 x   0
 1 
 x 2  =  0
 x   − a
 3  0
0

, and 3 =
2 0
2
, so that
0  x1   0 
   
0
1  x2  +  0 vi .
− a1 − a2  x3   a0 
One output voltage is simply found as v0 = x1 . Using KVL around the outer loop of the circuit yields
1

= 1 + 3 4 0 + 0 ⇒ 1 = − 3 4 2 − 1 .
The state variables output equation is then
 x1 
0
0    0 
 v0   y1   1
  =   = 
 x2  +  vi .
 vL 
 1   y2   − 1 R3C4 0  x   1 
 3
3. Let the circuit elements have parameter values L1 = 0.1 H, 2 = 1.4 − × 0.4 , 3 =
1300 + × 100 ohms, and 4 = 1.0 + × 0.2 , where = 0. , with xyz the final
three digits of your student ID number. The value of 0.xyz satisfies 0 ≤ 0.xyz ≤ 1 , so the 2
capacitance value lies in the range [1, 1.4] μF, the 4 capacitance value lies in the range [1, 1.2]
μF, and the resistor value lies in the range [1300, 1400] ohms. Use Matlab to determine the
response of the circuit ( v0 (t ) ) over a suitable time interval (roughly [0, 0.010] sec) to input
vi (t ) = u(t ) V, where u(t) is the unit-step signal. Accurately plot the response, v0 (t ) , and
determine (from the numerical response) the 100% rise time, percent overshoot, and 2% settling
time. (Note that the default rise time value that some Matlab functions provide is the 10% to 90%
rise time, so make sure your numerical result matches your plot and corresponds to the 100% rise
time.) Also, accurately plot v L1 (t ) . Verify that v0 (t ) and v L1 (t ) approach the correct steady-state
values as t → ∞ .
Solution: A Matlab function to generate the circuit unit step response is listed below, as well as the step
responses for .xyz = 0 and .xyz = 1. For the latter case, the 100% rise time is actually larger than the 2%
settling time.
function projectFall2015(L1,C2,R3,C4,alpha)
% Fall 2015 EE 321 Project
%
part 1: Generate unit step response
R3=R3+100*alpha;C2=C2 – 0.4e-6*alpha;C4=C4+.2e-6*alpha;
A=[0 1 0;0 0 1;-1/(L1*C2*R3*C4) -1/(L1*C2) -(1/(R3*C2) + 1/(R3*C4))];
B=[0;0;1/(L1*C2*R3*C4)];C=[1 0 0;-1 -R3*C4 0];D=[0;1];
x0=[0;0;0];
t=[0:1e-5:12e-3];
vi=ones(1,length(t));
[y,x]=lsim(A,B,C,D,vi,t,x0);
figure(1)
plot(t,y(:,1),t,y(:,2))
xlabel(‘Time, t, sec’)
ylabel(‘v_0(t) and v_{L_1}(t)’)
title(‘Unit-Step Response, .xyz = 1’)
…
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