Download Tutorial 2_Solutions_EMT212_2010-2011

TUTORIAL QUESTIONS – OSCILLATOR
Q1 With the aid of a suitable diagram, explain
briefly what are meant by the following terms;
a)
open-loop gain;
b)
Loop gain
c)
closed-loop gain
TUTORIAL QUESTIONS – OSCILLATOR
Solution
a)
Ve
Vs
The open-loop
gain is the gain of the
operational amplifier
without feedback.
Referring to the figure,
the open-loop gain is
A and is expressed as;
+
Vf
Vo
A
Ve
A
b
Vo
TUTORIAL QUESTIONS – OSCILLATOR
Solution (cont’d)
b)
Ve
Vs
The loop gain is the
+
amplification (or attenuation)
experienced by the signal as it
travels from the input to the
operational amplifier to the
output of the feedback network.
From the figure, the loop gain is
Ab and is expressed as;
Vf
Ab 
Vf
Ve
A
b
Vo
TUTORIAL QUESTIONS – OSCILLATOR
Solution (cont’d)
Ve
Vs
c)
+
Vf
A
Vo
b
The closed-loop gain is the amplification of the
input signal by the amplifier with feedback. From
the figure, the closed-loop gain is Af and is
expressed as;
Vo
Af 
Vs
TUTORIAL QUESTIONS – OSCILLATOR
Q2 With the aid of suitable figures, describe the
term Barkhausen criterion as applied to
sinusoidal oscillators
TUTORIAL QUESTIONS – OSCILLATOR
Solution
Ve
Vs
In the figure, it can be
shown that the closedloop gain, Af, is given
by the expression;
Vo
A
Af 

Vs 1  bA
+
Vf
A
b
Vo
TUTORIAL QUESTIONS – OSCILLATOR
Solution (cont’d)
Ve
Vs
When the magnitude
of the loop gain is
unity and the phase is
zero i.e. when;
bA  10
+
Vf
A
b

the system will produce an output for zero input.
Vo
TUTORIAL QUESTIONS – OSCILLATOR
Solution (cont’d)
Ve
Vs
+
Vf
A
b
Barkhausen criterion states that in order to
start and sustain an oscillation, the loop gain
must be unity and the phase shift through the
loop must be 0
Vo
TUTORIAL QUESTIONS – OSCILLATOR
Q3
The following figure
shows a Wien-Bridge
oscillator employing
an ideal operational
amplifier A. Derive
an expression for the
frequency of
oscillation o in
terms of R and C.

R1
R2
vO
A
.
+
C
C
R
R
TUTORIAL QUESTIONS – OSCILLATOR
Solution
1
1
  sC
Zp R
1  sCR

R
Zp C
R
Zp 
sCR  1

R1 v1
v2
R2
A
.
+
R
C
R
vO
Zs
TUTORIAL QUESTIONS – OSCILLATOR
Solution (cont’d)

R1 v1
v2
Zp C
R2
A
.
+
R
C
R
vO
Zs
1
Zs  R 
sC
sCR  1

sC
TUTORIAL QUESTIONS – OSCILLATOR
Solution (cont’d)
R1
v2
b
vo

Zp
Z p  Zs
Zp C

v1
v2
R2
A
.
+
R
C
R
vO
Zs
TUTORIAL QUESTIONS – OSCILLATOR
Solution (cont’d)
b
Zp
Z p  Zs
R / sCR  1

R / sCR  1  sCR  1 / sC
1

1
sCR  3 
sCR
TUTORIAL QUESTIONS – OSCILLATOR
Solution (cont’d)
R2
A  1
R1
R1
Zp C

v1
v2
R2
A
.
+
R
C
R
vO
Zs
TUTORIAL QUESTIONS – OSCILLATOR
Solution (cont’d)
1
b
and
sCR  3  1 / sCR 
R2
A 1
R1
The loop gain;

 R2 
1
1  
bA  

 sCR  3  1 / sCR  R1 
TUTORIAL QUESTIONS – OSCILLATOR
Solution (cont’d)
Substituting for s;

 R2 
1
1  
bA  

 jCR  3  1 / jCR R1 
Since bA must be real at the oscillation
frequency o, it follows that;
1
jo CR 
0
jo CR
TUTORIAL QUESTIONS – OSCILLATOR
Solution (cont’d)
or;
1
o 
RC
TUTORIAL QUESTIONS – OSCILLATOR
Q4
For the relaxation
oscillator shown in the
figure, sketch and
label the waveforms
of vC and vR2 and
indicate in your
sketch, the relevant
mathematical
equations describing
various sections of
the waveform.
TUTORIAL QUESTIONS – OSCILLATOR
Solution
TUTORIAL QUESTIONS – OSCILLATOR
Solution (cont’d)
TUTORIAL QUESTIONS – OSCILLATOR
Q5
Design a phase shift oscillator in the following
figure, to obtain a sinusoidal wave of 3 kHz.
Use C = 22 nF and R1 = 50 k.
TUTORIAL QUESTIONS – OSCILLATOR
Solution
From the expression;
we obtain;
6
o 
RC
6
R
oC
Substituting values;
6
R
 5.9 kΩ
3
-9
2π  3  10  22  10
Q1 A single-pole low-pass filter as shown in Fig.5-1 has a RC network
with the resistance R of 2 kΩ and the capacitance C of 0.04μF. If
the resistances of resistor R1 and R2 are 20 and 4 Ω respectively,
determine:
i) cutoff frequency, fC
ii) pass band voltage gain or the gain of non-inverting amplifier, Acl
iii) expression of output voltage, Vo
a
Figure 5-1.
Sol1
(a)
The cutoff frequency for a single-pole low-pass filter is
1
1
fc 

2RC 2(3.14)( 2 x103 )( 4 x10 8 F )
 1.99 x10 Hz  2kHz
3
(b)
The pass band voltage gain or the gain of non-inverting
amplifier is calculated as follows:
Acl ( NI )
R1
20

1 
1  6
R2
4
Sol1_Cont’d
Applying the voltage divider rule, the input voltage, which is the
voltage across the capacitor
C, is determined as follows:

 jX C
va  vin
 vin

R  jX C
R  XC
XC
Multiplying the numerator and the denominator by j and substituting for
XC results in the following:
XC
1
va  vin
 vin
R
X C  jR
1 j
XC
1
1
 vin
 vin
1  jRC
1  j 2fRC
Sol1_Cont’d
At the cutoff frequency fc, the magnitude of the capacitive reactance
XC equals the resistance of the resistor R.
1
 R or
2f c C
1
 2RC
fc
Substituting for 2πRC in Eq. results in the following equation for va:
va  vin
1
f
1 j
fc
Where f is the operating frequency and fc is the cutoff frequency.
Sol1_Cont’d
The output vo is the amplified version of vi.
vo  va Acl ( NI )
Acl ( NI )
 vin
f
1 j
fc
where Acl(NI) is the pass-band voltage gain or the gain of the noninverting amplifier.
Q2
Determine the critical frequency, pass-band voltage gain,
and damping factor for the second-order low-pass active filter
in Fig. 5-2 with the following circuit components:
RA = 5 kΩ, RB = 8 kΩ, CA = 0.02 μF, CB = 0.05 μF, R1 = 10 kΩ,
R2 = 20 kΩ
Figure 5-2.
Sol2
The critical frequency for the second-order low-pass active
filter is
1
fc 
2 RA RB C ACB
1

3
3
8
8
2(3.14) (5 x10 )(8 x10 )( 2 x10 )(5 x10 )
 796.18Hz
Sol2_Cont’d
The pass-band voltage gain set by the values of R1 and R2 is
Acl ( NI )
R1
10

1 
 1  1.5
R2
20
Q3
Design a multiple-feedback band-pass filter with the maximum gain,
Ao = 8, quality factor, Q = 25 and center frequency, fo =10 kHz.
Assume that C1 = C2 = 0.01µF.
Draw the circuit design of the active band-pass
Sol3
Q
25
R1 

 4.97k
2f oCAo 2 (10k )(0.01 )(8)
Q
25
R2 

 79.58k
f o C  (10k )(0.01 )
Q
25
R3 

 32
2
2
2f o C (2Q  Ao ) 2 (10k )(0.01 )( 2(25 )  8)
Sol3_Con’t
The drawing of the circuit diagram :
C1
R2
R1
C2
V in
V out
R3
Q4
Determine the center frequency, fc, quality factor, Q and bandwidth, BW
for the band-pass output of the state-variable filter in following figure.
Given that R1 = R2 = R3 = 25 kΩ, R4 = R7 =2.5 kΩ, R5 = 150 kΩ,
R6 = 2.0 kΩ, C1 = C2 = 0.003µF
Sol4_Con’t
i) Center frequency, fc
1
1
1
fc 


 21.22kHz
2R4 C1 2R7 C 2 2 (2.5k )(0.03 )
ii) Quality factor, Q
 1  150k 
1  R5
Q  
 1  
 1  25.67
3  R6

 3  2.0k
Sol4_Con’t
iii) Bandwidth, BW
The critical frequency, fo of the integrators usually made equal
to the critical frequency, fc
fo  fc
f o 21.22k
BW 

 826.65Hz
Q
25.67