pdc Chapter 3 problems

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






1 2 f 2

=


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 

1010 6 6

) = V (1  e 0.79510 ) =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 


2



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


3


T 2


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)   e0.1   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  10e 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



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


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.


5



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 .210

11

T 2



e



3

0.11103

=10+ V 1  10 0.1623 V 2 = 0.1623V 1 +8.3767 And V 1   10  V 2  10 e

(1)  0 . 310

3

0 . 11 10  3

=  10  V 2  100.0653 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.


(2)

6



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

1510 6

1510 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.


7



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


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 e20  3 V

(2)

V1  vo (T )  2  V2  2  e20  2 V


8




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.


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


(2)

9




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.


vo (t )  V f   vi  V f  e  Here



 t p  10  s

At t  t p , vo (t p )  10   0  10  e1  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


10



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  1012 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


11



(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 11

 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 11

 106  (75  50)  1012  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 11

 0.5 V .

Rise time constant  2 = RC  2 =

R1 R2 R1  R2

(C1  C 2 ) 

1 1 11

 106  (25  50)  1012  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


12



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 /

 V1et

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


13


V 1  V 2 e

T 2


 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 


14


vo (min) 

V

 0.14 


V

 (1  0.869)  0.076V  0.869 2 2  0.07V  0.065V  0.066V

vo (min)  0.071V


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  103 1

2  3  103  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  106

 3.5 kHz. © Dorling Kindersley India Pvt. Ltd 2010

15


Also f 2 

 R 


1 2 RC

1 2 f 2 C



1 2    3.5  103  0.28  106

 162.4 Ω.


16

pdc Chapter 3 problems

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