Download Lab2 - ECE232

27.02.2012
ECE 232 Lab2
Second Order Circuits
Preliminary Work:
1) Consider the series RLC circuit of Fig.1.
R = 50kΩ potantiometer.
C = 4.7 nF
L = 0.1 H
Vin(t) =Vpeak. u(t) volts
Fig. 1
a. Show that the differential equation for Vc(t) is given as,
d2
Rd
1
1
V (t)+
VC (t)+
VC (t)=
Vin (t)
2 C
dt
L dt
LC
LC
The characteristic equation for this differential equation is,
s2 +
R
1
s+
=0
L
LC
or by substitution of variables it can be written as
The roots of the characteristic equation are denoted by s1 and s2 when they are distinct and by s0 when
the roots are repeated. The characteristic equation can also be written as
b. When the roots if the characteristic equation are repeated the response is called critically damped.
Determine the value of the resistor R=R0 for the critically damped response.
c. Determine the values of the currents and voltages at t  0  , t  0  and t   for all circuit elements.
d. For R>R0 (overdamped response):
i. Solve the differential equation and show that the variables Vc(t) and iL(t) are given below.
VC (t )  V peak 
i L (t ) 
V peak
s1  s 2
( s 2 e s1t  s1e s2t ) ,
1 V peak
(e s1t  e s2t )
L s1  s 2
where s1 and s2 are the real distinct roots of the characteristic equation.
ii. Determine VL(t), VR(t), ic(t), iR(t) using the above results.
iii. Determine and sketch VL(t), Vc(t), VR(t), ic(t), iL(t) and iR(t) for R=3R0.
e. For R=R0 (critically damped response):
i. Show that the variables Vc(t) and iL(t) are given as,
VC (t ) V peak V peak (s0 t 1) e s0t ,
i L (t ) 
V peak
L
t e s0 t
where s0 is the only (repeated) root of the characteristic equation.
ii. Repeat Part 1.d.ii for R=R0.
iii. Repeat Part 1.d.iii for R=R0.
f. For R<R0 (underdamped response):
i. Show that the variables Vc(t) and iL(t) are given as,
VC (t )  V peak  V peak
i L (t ) 
V peak
dL
 0 t
d
e sin ( d t  arctan
),
d

e t sin ( d t )
where s1     j d and s 2     j d are the complex conjugate roots of the characteristic
equation and  0 
1
.
LC
ii. Repeat Part 1.d.ii.
iii. Repeat Part 1.d.iii for R=R0/4.
2) Consider the parallel RLC circuit of Fig. 2.
R1 = 10 KΩ
C = 0.1 μF
L = 0.1 H
Fig. 2
a. Obtain the differential equation for Vc(t).
b. Determine the value of R=R0 for the critically damped response.
c. Determine the value of Vc(t) at t  0  , t  0  and t   .
EXPERIMENTAL WORK
1) Set up the circuit of Fig. 1. Adjust the square wave output of the signal generator so that Vin(t) is a 2
Vp-p square wave with 150Hz frequency.
a. Determine experimentally R=R0 for the critically damped response.
b. Observe and plot (Ac components only) Vin(t), VL(t), VR(t) and VC(t), also note the DC levels for the
following cases:
i. R=3R0 ,
ii. R=R0 ,
iii. R=R0/4
2) Set up the circuit of Fig.2. and repeat Part 1.
3) Set up the following circuit for f = 10kHz. Observe and plot V0(t) for the following cases:
a. S1 and S2 open. b. S1 closed and S2 open. c. S1 and S2 closed.
R1 = 10 KΩ
R2 = 10 KΩ
C = 0.1 μF
L = 0.1 H
1a.
C = 4.7 nF
L = 0.1 H
Vin(t) =Vpeak. u(t) volts
KVL around the loop gives;

,
,
,


or we can write this equation in the operator format

Call
which has the characteristic equation
,
,
,
1b. If the roots are repeated;


R0 = 2
L
C
For L=0.1H, C=4.7nF 
1c. For
there is no initial condition stated here for any circuit element, therefore
, also since step fuction u(t) is used in this
experiment, it is zero until t reaches zero. Furthermore since u(t) is a bounded function, there is no sudden
change in inductor current and capacitor voltage when t = 0+ . Both inductor current and capacitor voltage
are unchanged. Both must be continuous.
VC (0- ) = 0 Volt
Since the current over the resistor is zero the voltage over the resistor will also be zero
VR (0+ ) = 0 Volt
However due to Kirchoff’s Voltage Law,
Vs (0+ ) = Vpeak ´ u(0+ ) = VC (0+ )+VR (0+ )+VL (0+ ) = Vpeak
hence,
VL (0+ ) = Vpeak
1d. I.
(overdamped response)
The differential equation for the circuit is
The characteristics equation of the homogenous eqn. is
1
L
, R0 = 2
LC
C
, w0 =
,

, so we have 2 real roots.
-2a ± 4(a 2 - w02 )
R
R2
1
 s1,2 =
= -a ± a 2 - w02 = - ±
2
2
2L
4L LC
soln. of the homogenous equation
a particular soln.
and Vx is a constant DC term put into eqn., to get;
d2
Rd
1
1
1
V+
Vx +
Vx =
Vin (t) =
Vpeak
2 x
dt
L dt
LC
LC
LC
As Vx is a constant term,
1
1
Vx =
Vpeak
LC
LC
,


&


,
II. Determine
using the above results.


III. Determine and sketch
,
and
for R=3Ro.
, R=3Ro=27875.9Ω, C = 4.7 nF, L = 0.1 H
The plots are as in experimental work part.
1e. I.
(critically damped response)

,
Hence the roots are equal to each other s1,2 = s0
,
Vch = c1es0t + c2tes0t
Vc (t) = c1es0t + c2tes0t +Vp
and Vx is a constant DC term put into eqn., to get;
d2
Rd
1
1
1
V+
Vx +
Vx =
Vin (t) =
Vpeak
2 x
dt
L dt
LC
LC
LC
As Vx is a constant term,
1
1
Vx =
Vpeak
LC
LC
Vc (t) = c1es0t + c2tes0t +Vpeak


;
II.
s0 = -
III.
R
= -46126.5
2L
, R=Ro=9225.3Ω, C = 4.7 nF, L = 0.1 H
Vc (t) = Vpeak e-46126.5t (-46126.5t -1)+Vpeak
1f. I.
(underdamped response)

,

Complex roots:
VC (t) = Vcp +Vch
and Vx is a constant DC term put into eqn., to get;
d2
Rd
1
1
1
V+
Vx +
Vx =
Vin (t) =
Vpeak
2 x
dt
L dt
LC
LC
LC
As Vx is a constant term,
1
1
Vx =
Vpeak
LC
LC

C
dVC
= IC
dt
IC (0+ ) = 0 
II.
III.
, R=Ro/4=2306.3Ω,

2a.
R1 = 10 KΩ
C = 0.1 μF
L = 0.1 H
R= 10 kΩ potentiometer
Vin(t) = Vpeak. u(t) volts
Firstly,

b. Characteristic eqn.
,
For critically damped;
R1=10 kΩ, C = 0.1 μF, L=0.1 H then R0=2207 Ω
c. For
For
,
,
, (it is not possible to change capacitor voltage suddenly without use of an
impusive source)
For
, so at steady-state circuit diagram will behave as
Nothing was said about the initial condition. Since there is no impulsive source on the circuit, the
capacitor voltage would be continuous.