Pulse and Digital Circuits
Venkata Rao K., Rama Sudha K. and Manmadha Rao G.
Chapter 3 Linear Waveshaping: Low-pass Circuits, Attenuators and RLC Circuits
with zero rise rise time, an amplitud amplitudee of 10 V and duration duration 10 s is applied applied to an amplifier amplifier 1. A pulse with through a low-pass coupling network shown in Fig.3p.1. Plot the output waveform to scale under the following conditions: (i) f (i) f 2 = 20 MHz and (ii) f (ii) f 2 = 0.2 MHz.
Fig.3p.1 The given coupling network with input Solution: When f When f 2 is specified, the given circuit is a low-pass circuit. Case i:
f 2 = 20 MHz f 2 =
1 2 RC
, RC = =
1 2 f 2
=
1 2 20 10
6
= 0. 0.00795 s.
As, T the capacitor charges rapidly with a rise time t r r = 2.2 RC. t r r = 2.2R 2.2RC C = 2.2 2.2 0.007 0.00795 95 = 0.017 0.01749 49 s The output is shown in Fig.1.
Fig.1.1 Output response for f for f 2 = 20 MHz Case ii: f ii: f 2 = 0.2 MHz.
© Dorling Kindersley India Pvt. Ltd 2010
1
Pulse and Digital Circuits
1 2 f 2
=
Venkata Rao K., Rama Sudha K. and Manmadha Rao G.
1 2 0 . 2 10
6
= 0.795 s . As T the capacitor charges with a rise time
t r = 2.2 = 1.75 s v o (T ) = V (1 e
T
1010 6 6
) = V (1 e 0.79510 ) =V
The output is shown in Fig.1.2.
Fig.1.2 Output response for f 2 = 0.2 MHz
2. A ramp shown in Fig.3p.2 is applied to a low-pass RC circuit. Draw to scale the output waveform for the cases (i) T = 0.2 RC,and (ii) T = 10 RC .
Fig. 3p.2 The given input to the low-pass RC circuit Solution: t
For low-pass circuit, vo (t ) (t ) e , where
V T T
At t = T , vo (T ) (T ) e
RC
i. T = 0.2
© Dorling Kindersley India Pvt. Ltd 2010
2
Pulse and Digital Circuits
Venkata Rao K., Rama Sudha K. and Manmadha Rao G.
0.2
vo (T ) (0.2 ) e
V 0.3 0.8
(0.8 )
V
0.2 0.818 V V 0.2 0.2 vo (T ) 0.09V
e
0.2
ii) T = 10 10
v o (T ) (10 ) e
V
(9 )
V
10 10 v o (T ) 0.9V
e
10
3. The input to low-pass RC circuit is periodic and trapezoidal as shown in Fig. 3p.3. Find and sketch the steady-state output if RC = 10T 1 = 10T 2
Fig. 3p.3 Input to the low-pass RC circuit
Solution: Given =10T 1 = 10T 2 As the time constant τ is very large when compared to T 2 /2, upto T 2 /2, the output cannot follow the input and at T 2 /2 let this value be V 1 . t
For ramp input vo (t ) (t ) e where
V T
and V = 10 V
© Dorling Kindersley India Pvt. Ltd 2010
3
Pulse and Digital Circuits
T 2
Venkata Rao K., Rama Sudha K. and Manmadha Rao G.
T
2 V T 2 2 RC for t , V 1 = vo ( ) ( ) e T 2 2 2 2 2
T 2
T2
20V
20V
1
1
) ( 1) e 2 20 V 1 = 20V (–0.95)+19.02V = 0.02V 0.02 10 0.2 V vo (
20
t
For pulse input to low-pass circuit, vo (t ) V f vi V f e Here, V f 10 V and v i 0.2 V T 1
for t T 1 , V 2 = vo (T1 ) 10 0.2 10 e V 2 10 9.8e 0.1 =1.13 V
T V V RC for t T 2 , V 3 = vo (T2 ) (T2 ) e T 2 T2 1 200 200 V3 vo (T2 ) ( 1) e0.1 0.9 V 10 Here, V f 10 V and vi 0.9 V 2
T 1
for t T 1 , V 4 = vo (T1 ) 10 0.9 10 e V 4 10 0.9 10e 0.1 1.76 V
T
T2
2 V T 2 2 RC for t , V 5 = vo ( ) ( ) e T 2 2 2 2 2
T 2
vo (
T 2
)
200
(
1
1)
200
2 20 V 5 = 200(–0.95)+190.2 = 0.2 V
1
e 20
Fig. 3 The output waveform
4. A 1 kHz symmetrical square wave of peak-to-peak voltage 20 V, as shown in Fig. 3p.4, is applied to a low-pass RC circuit having R = 100 , C = 1 F . Sketch the output waveform to scale by determining the corner voltages. © Dorling Kindersley India Pvt. Ltd 2010
4
Pulse and Digital Circuits
Venkata Rao K., Rama Sudha K. and Manmadha Rao G.
Fig. 3p.4 The given input to the low-pass RC circuit Solution: As the input is a symmetric square wave, we have: T V T1 T2 , V1 V2 and V ' V '' 2 2 V T V2 tanh 2 4 Here 0.1 ms and T = 1 ms 20 1 V2 tanh 10tanh (2.5) 9.866 V 2 4 0.1 Hence V 1 V 2 9.866 V
Fig. 4 The output waveform
5. The input waveform shown in Fig.3p.5 is applied to a low-pass RC network. Sketch the waveform of the voltage across the capacitor to scale for two cycles. The time constant of the RC circuit is 0.11 ms.
© Dorling Kindersley India Pvt. Ltd 2010
5
Pulse and Digital Circuits
Venkata Rao K., Rama Sudha K. and Manmadha Rao G.
Fig.3p.5 The given input to a low-pass RC circuit Solution: RC =
vo1 V 1 V1 V 1 e
t
At t T1 , V2 V 1 V1 V 1 e vo2 V
11
V2 V
11
e
T 2
At t T2 , V1 V V2 V 11
V2 10 V1 10 e
T 1
0 .210
11
T 2
e
3
0.11103
=10+ V 1 10 0.1623 V 2 = 0.1623V 1 +8.3767 And V 1 10 V 2 10 e
(1) 0 . 310
3
0 . 11 10 3
= 10 V 2 100.0653 V 1 = 0.0653V 2 9.346 Solving (1) and (2), we get V 2 = 0.1623(0.0653V2 –9.346) + 8.3767 V 2 = 6.933V V 1 =–8.89V The out waveform is shown in Fig.5.
© Dorling Kindersley India Pvt. Ltd 2010
(2)
6
Pulse and Digital Circuits
Venkata Rao K., Rama Sudha K. and Manmadha Rao G.
Fig.5. Output of the low-pass circuit for the input in Fig.3p.5
6. The periodic waveform shown in Fig.3p.6 is applied to an RC integrating network whose time constant is 15 µs. Sketch the output and calculate the maximum and minimum values of output voltage with respect to the ground.
Fig.3p.6 The given input to a low-pass RC circuit Solution:
V1 V2 e
T 1
V2 e
1
15
0.935V2
V2 200 (200 V1 )e
T 2
200 (200 V1 )e
1510 6
1510 6
126.42 0.344V 2 V 2
126.42
1 0.344 V 1 180.32 V
192.75
The output waveform is shown in Fig.6.
© Dorling Kindersley India Pvt. Ltd 2010
7
Pulse and Digital Circuits
Venkata Rao K., Rama Sudha K. and Manmadha Rao G.
Fig.6 Output of the low-pass RC circuit for the input in Fig.3p.6
7. The waveform shown in Fig. 3p.7 is applied to a low-pass RC circuit. Sketch the output waveform to scale by determining the corner voltages for the following cases: (1) RC = 20T , (2) RC = T /20
Fig. 3p.7 The given input to the low-pass RC circuit Solution: t
vo (t ) V f vi V f e Case 1: RC = 20T 1
V2 vo (T ) 3 V1 3 e 20 V 2 0.95V 1 0.14
(1) 1
V1 vo (T ) 2 V2 2 e 20 V 1 0.95V 2 0.1 Solving Eqs (1) and (2) V 1 2.39 V and V 2 2.41 V Case 2: RC = T /20 V2 vo (T ) 3 V1 3 e20 3 V
(2)
V1 vo (T ) 2 V2 2 e20 2 V
© Dorling Kindersley India Pvt. Ltd 2010
8
Pulse and Digital Circuits
Venkata Rao K., Rama Sudha K. and Manmadha Rao G.
Fig. 7 The output waveform
8. A square wave extends 0.5 V with respect to ground. The duration of the positive section is 0.1 s and the negative duration is 0.2 s, as shown in Fig. 3p.8. This waveform is applied as an input to low-pass RC circuit whose time constant is 0.2 s. Sketch the steady-state output waveform to scale and find the maximum and minimum values of the output.
Fig. 3p.8 The given input to the low-pass RC circuit Solution: t
vo (t ) V f vi V f e and 0.2 s 0.1
V2 vo (0.1) 0.5 V1 0.5 e 0.2 V 2 0.6V 1 0.19
(1) 0.2
V1 vo (0.2) 0.5 V2 0.5 e 0.2 V 1 0.36V 2 0.317 Solving Eqns. (1) and (2) V 1 0.317 V and V 2 0 V
© Dorling Kindersley India Pvt. Ltd 2010
(2)
9
Pulse and Digital Circuits
Venkata Rao K., Rama Sudha K. and Manmadha Rao G.
Fig. 8 The output waveform
9. A pulse of amplitude 10 V and pulse width of 10 µs, as shown in Fig. 3p.9, is applied to an RC circuit with R = 100 k and C = 0.1 F. Sketch the capacitor voltage waveform to scale.
Fig. 3p.9 The given input to the low-pass RC circuit Solution: t
vo (t ) V f vi V f e Here
t p 10 s
At t t p , vo (t p ) 10 0 10 e1 10 9.9999 9.9 mV ( t t p ) 3
At t t p , vo (t ) 9.9 10 e
Fig. 9 The capacitor voltage form
© Dorling Kindersley India Pvt. Ltd 2010
10
Pulse and Digital Circuits
Venkata Rao K., Rama Sudha K. and Manmadha Rao G.
10. An oscilloscope has an input impendence of 10 MΩ in parallel with 25 pF. Design a 10:1 voltage divider for the oscilloscope. Determine the values of the input resistance and the capacitance for the compensated divider. Solution:
For perfect compensation R1C 1 R2 C 2 For step input to the perfect compensation attenuator vo (0) vo () vo (0) vo () vo vi vo vi 1 10
R2 R1 R2
R2 R1 R2 10 106 R1 10 106
0.1( R1 10 106 ) 10 106 R1 90 MΩ C 1
R2C 2 R1
10 106 25 1012 90 106
2.7 pF
11. The attenuator shown in Fig.3p.11 has R 1 = R 2 = 1 MΩ, C 2 = 50 pF. Find the magnitudes of initial and final response and draw the output waveform to scale for C 1 = 50, 75 and 25 pF for given unit step voltage.
Fig. 3p.11 The given attenuator circuit Solution: For perfect compensation R 1 C 1 = R 2 C 2 Here R 1 = R 2. (i) When C 1 = 50 pF, then the attenuator is perfectly compensated. The rise time of the output waveform is zero. R2 1 Attenuation constant, 0.5 R1 R2 1 1
vo (0 ) = v o ( ) =
v i =
0.5 1 = 0.5 V
© Dorling Kindersley India Pvt. Ltd 2010
11
Pulse and Digital Circuits
Venkata Rao K., Rama Sudha K. and Manmadha Rao G.
(ii) When C 1 = 75 pF, then attenuator is overcompensated, hence vo (0 ) > v o ( ) The output at t = 0 , vo (0 ) = vi
C 1
+
C1 C 2
The output at t = , v o ( ) = vi
R2 R1 R2
1
75 75 50
1
1 11
0.6 V
0.5 V
Fig.11.1 Equivalent circuit to get the time constant where R
R1 R2 R1 R2
and C C 1 C 2 .
Fall time constant, 1 = RC =
R1 R2 R1 R2
(C1 C 2 )
1 1 11
106 (75 50) 1012 62.5 s
(iii) When C 1 = 25 pF, then the attenuator is undercompensated. C 1 25 + The output at t = 0 , vo (0 ) = vi 1 0.33 V C1 C 2 25 50 The output at t = , v o ( ) = vi
R2 R1 R2
1
1 11
0.5 V .
Rise time constant 2 = RC 2 =
R1 R2 R1 R2
(C1 C 2 )
1 1 11
106 (25 50) 1012 37.5 s
Rise time, t r = 2.2 RC R R tr 2.2 1 2 (C1 C 2 ) 2.2 37.5 10 6 82.5 s R1 R2
Fig. 11.2 The output waveform
© Dorling Kindersley India Pvt. Ltd 2010
12
Pulse and Digital Circuits
Venkata Rao K., Rama Sudha K. and Manmadha Rao G.
12. The periodic ramp with T 1 = T 2 = 2τ , shown in Fig.3p.12, is applied to a low-pass RC circuit. Find the equations to determine the steady-state output waveform. The initial voltage on the condenser is V 1 . Find the maximum and minimum value of the voltage and plot the waveform.
Fig. 3p.12 The given periodic ram p input and the low-pass circuit Solution: We have the output for RC low-pass circuit for ramp input as
vo t t e
t /
If there is an initial voltage of V 1 on C , then
vo t t e
t /
V1et
For the ramp input, the slope =
vo (t ) =
V T 1
t –
t
V T 1
(1 e
V T 1
. t
) V 1e
(1)
The capacitor charges from V 1 to V 2 in time T 1 . During T 2 when the input is zero, the capacitor discharges from V 2 to V 1 . Given T 1 = T 2 = 2 At t = T 1 vo (t ) = V 2 Using (1) V 2 =
V T 1
T 1 –
V T 1 T 1
2
(1 e 2 ) V 1e 2
0.5V 0.5Ve 2 V 1e 2 V 2 0.56V 0.135V 1
(2)
When the ramp input is reduced to zero, V 2 decays to V 1
© Dorling Kindersley India Pvt. Ltd 2010
13
Pulse and Digital Circuits
V 1 V 2 e
T 2
Venkata Rao K., Rama Sudha K. and Manmadha Rao G.
V 2 e 2 0.135V 2
(3)
Substituting (3) in (2), V 2 0.56V 0.135V 1 0.56V 0.135 0.135V 2 V 2 (1 0.018) 0.56V V 2 0.57V V 1 0.135 0.57V 0.076V Minimum value of output occurs when t
vo (t ) = t + V 1e dvo (t )
= –
dt V T 1 V T 1
–
T 1 t
e (
t
e –
2V 1 T 1
+
0
(1 e )
t
e
2V 1
dt
t
V 1
dvo (t )
t
e 0
V T 1 V T 1
t
e 0
)=0
t
( 2V 1 V )e V t
e t
V ( 2V 1 V )
V 0.15V V
0.869
= 0.14.
At t = 0.14 , v o is v o (min) Substituting this value in v o (t ),
vo (t ) =
V T 1
t –
V T 1
t
(1 e
t
) V 1e
© Dorling Kindersley India Pvt. Ltd 2010
14
Pulse and Digital Circuits
vo (min)
V
0.14
Venkata Rao K., Rama Sudha K. and Manmadha Rao G.
V
(1 0.869) 0.076V 0.869 2 2 0.07V 0.065V 0.066V
vo (min) 0.071V
Fig. 12 The output waveform
13. For a low-pass RC circuit, it is desired to pass a 3 ms sweep for a ramp input with less than 0.4 per cent transmission error. Calculate the upper 3-dB frequency. Solution: T = 3 ms
per cent e t (max) 0.4 per cent et (max) 0.004 e(t )
1 2 f 2 T
0.004
f 2
1 2 f 2 3 103 1
2 3 103 0.004
13.26 kHz.
14. A step input of 20 V is applied to an RC integrating circuit. Calculate the upper 3-dB frequency and the value of resistance, if the rise time and capacitor values are 100 sand 0.28 F, respectively. Solution: Rise time is the time taken for the waveform to rise from 10 to 90 per cent of its maximum value. 2.2 Rise time t r . 2 f 2
Upper 3-dBfrequency f 2
2.2 2 t r
2.2 2 100 106
3.5 kHz. © Dorling Kindersley India Pvt. Ltd 2010
15
Pulse and Digital Circuits
Also f 2
R
Venkata Rao K., Rama Sudha K. and Manmadha Rao G.
1 2 RC
1 2 f 2 C
1 2 3.5 103 0.28 106
162.4 Ω.
© Dorling Kindersley India Pvt. Ltd 2010
16