This document discusses linear wave shaping, which is the process where a non-sinusoidal signal is altered by transmission through a linear network. It specifically focuses on low-pass RC circuits and how different input signals like step, pulse, square wave, and ramp are processed. For a step input through an RC circuit, the output rises exponentially towards the final value with a time constant of RC. The rise time is defined as the time taken to rise from 10% to 90% of the final value, which is approximately 2.2RC. For a square wave input, the output depends on whether RC is much less than, approximately equal to, or much greater than the period of the square wave. Mathematical expressions are also derived

1. Electronic Devices and Circuits
UNIT – I
SVEC19

2. Linear wave shaping
M.Balaji, Department of ECE, SVEC
2
A linear network is a circuit made up of linear elements only.
The process where a non- sinusoidal signal is altered by transmission
through a linear network is called linear wave shaping.
There are a number of waveforms, which appear very frequently. The
most important of these are sinusoidal, step, pulse, square wave and
ramp waveforms.
Low-pass RC circuit
High-pass RC circuit

3. The low-pass RC Circuit
𝑋 𝐶 =
1
2𝜋𝑓𝐶
When f= 0; XC = ∞ (Capacitor is open
circuited);
Vo(t) = Vi(t); then gain A=
𝑉𝑜(𝑡)
𝑉 𝑖(𝑡)
= 1.
When f increases, XC decreases, then Vo(t)
=
decreases; then gain 𝑎𝑙𝑠𝑜 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑒𝑠.
When f= ∞; XC = 0 (Capacitor is short
circuited);
M.Balaji, Department of ECE, SVEC 3
The circuit which transmits only low- frequency signals and attenuates or
stops high frequency signals.

4. Sinusoidal input
M.Balaji, Department of ECE, SVEC 4
Laplace transformed low- pass RC circuit and its frequency response

5. 𝐴 =
𝑉𝑂 𝑠
𝑉𝑖 𝑠
=
1
𝐶𝑠
𝑅 +
1
𝐶𝑠
=
1
1 + 𝑅𝐶𝑠
=
1
1 + 𝑗𝜔𝑅𝐶
=
1
1 + 𝑗2𝜋𝑓𝑅𝐶
𝐴 =
1
1 + 2𝜋𝑓𝑅𝐶 2
𝐴𝑡 𝑡ℎ𝑒 𝑢𝑝𝑝𝑒𝑟 𝑐𝑢𝑡𝑜𝑓𝑓 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑓2, 𝐴 =
1
2
1
2
=
1
1 + 2𝜋𝑓2 𝑅𝐶 2
Squaring both sides and equating the denominators,
2 = 1 + (2𝜋𝑓2 𝑅𝐶)2
M.Balaji, Department of ECE, SVEC 5

6. :. The upper cut-off frequency, 𝑓2 =
1
2𝜋𝑅𝐶
So 𝐴 =
1
1+𝑗
𝑓
𝑓2
𝐴 =
1
1 +
𝑓
𝑓2
2
The angle θ by which the output leads the input is given by
𝜃 = 𝑡𝑎𝑛−1
𝑓
𝑓2
M.Balaji, Department of ECE, SVEC 6

7. Step Voltage input
A step signal is one which maintains the value
zero for all times t < 0; and maintains the value
V for all times t > 0.
M.Balaji, Department of ECE, SVEC 7
t =
0
-
t =
0
+
t < 0 t > 0
t = 0
The transition between the two voltage
levels takes place at t = 0.
Vi = 0, immediately before t = 0 (to be
referred to as time t= 0-) and Vi = V,
immediately after t= 0 (to be referred to as
time t= 0+).

8. M.Balaji, Department of ECE, SVEC 8

9. Charge Q = C V
𝑑𝑄
𝑑𝑡
= 𝐶
𝑑𝑉
𝑑𝑡
𝑖 =
𝑑𝑄
𝑑𝑡
𝑖 = 𝐶
𝑑𝑉
𝑑𝑡
𝑖 = 𝐶
𝑑𝑉
𝑑𝑡
------------------------ (1)
Where i = instantaneous current through the capacitor
C = Capacitance (Farads)
dV/dt = instantaneous rate of change of voltage (Volts / Sec)
Integrate equation (1) on both sides we get,
𝑖 = 𝐶
𝑑𝑉
𝑑𝑡
𝑉 =
1
𝐶
𝑖 𝑡 𝑑𝑡
M.Balaji, Department of ECE, SVEC 9

10. Let V' be the initial voltage across the capacitor
Writing KVL around the loop, we get
𝑉𝑖 𝑡 = 𝑅𝑖 𝑡 +
1
𝐶
𝑖(𝑡)𝑑𝑡
Differentiating this equationwith respect to t, we get
𝑑𝑉𝑖(𝑡)
𝑑𝑡
= 𝑅
𝑑𝑖(𝑡)
𝑑𝑡
+
1
𝐶
𝑖(𝑡)
We know, for a step input, 𝑉𝑖 𝑡 = 𝑉, 𝑡ℎ𝑒𝑛
𝑑𝑉 𝑖(𝑡)
𝑑𝑡
= 0
M.Balaji, Department of ECE, SVEC 10

11. ∴ 0 = 𝑅
𝑑𝑖 𝑡
𝑑𝑡
+
1
𝐶
𝑖(𝑡)
Taking the Laplace transform on both sides,
0 = 𝑅 𝑠𝐼 𝑠 − 𝐼 0+ +
1
𝐶
𝐼 𝑠
∴ 𝐼 0+ = 𝐼 𝑠 𝑠 +
1
𝑅𝐶
----------- (1)
The initial current 𝐼 0+
is given by
𝐼 0+ =
𝑉 − 𝑉′
𝑅
From equation (1) we get, 𝐼 𝑠 =
𝐼 0+
𝑠+
1
𝑅𝐶
=
𝑉−𝑉′
𝑅 𝑠+
1
𝑅𝐶
M.Balaji, Department of ECE, SVEC 11
Step voltage, V
Initial voltage across
the capacitor is V'

12. 𝑉0 𝑠 = 𝑉𝑖 𝑠 − 𝐼(𝑠)𝑅
=
𝑉
𝑠
−
(𝑉 − 𝑉′)𝑅
𝑅 𝑠 +
1
𝑅𝐶
=
𝑉
𝑠
−
(𝑉 − 𝑉′)
𝑠 +
1
𝑅𝐶
Taking the inverse Laplace transform on both sides,
𝑉0 𝑡 = 𝑉 − (𝑉 − 𝑉′)𝑒− 𝑡 𝑅𝐶
Where V’ is the initial voltage across the capacitor (Vinitial) and V is the final
voltage (Vfinal) to which the capacitor can charge.
M.Balaji, Department of ECE, SVEC 12

13. The expression for the voltage across the capacitor of an RC circuit excited
by a step input is given by
𝑉0 𝑡 = 𝑉𝑓𝑖𝑛𝑎𝑙 − (𝑉𝑓𝑖𝑛𝑎𝑙 − 𝑉𝑖𝑛𝑖𝑡𝑖𝑎𝑙)𝑒− 𝑡 𝑅𝐶
If the capacitor is initially uncharged
𝑉0 𝑡 = 𝑉 1 − 𝑒− 𝑡 𝑅𝐶
M.Balaji, Department of ECE, SVEC 13

14. Expression for rise time
M.Balaji, Department of ECE, SVEC 14

15. Expression for rise time
At 𝑡 = 𝑡1, 𝑉0 𝑡 = 10% 𝑜𝑓 𝑉 = 0.1𝑉
∴ 0.1𝑉 = 𝑉 1 − 𝑒− 𝑡1 𝑅𝐶
∴ 𝑒− 𝑡1 𝑅𝐶
= 0.9
𝑒 𝑡1 𝑅𝐶
=
1
0.9
= 1.11
𝑡1 = 𝑅𝐶 ln 1.11 = 0.1𝑅𝐶
At 𝑡 = 𝑡2, 𝑉0 𝑡 = 90% 𝑜𝑓 𝑉 = 0.9𝑉
∴ 0.9𝑉 = 𝑉 1 − 𝑒− 𝑡2 𝑅𝐶
∴ 𝑒− 𝑡2 𝑅𝐶
= 0.1
𝑒 𝑡2 𝑅𝐶 =
1
0.1
= 10
𝑡2 = 𝑅𝐶 ln 10 = 2.3𝑅𝐶
M.Balaji, Department of ECE, SVEC 15
𝑅𝑖𝑠𝑒 𝑡𝑖𝑚𝑒 𝑡 𝑟 = 𝑡2 − 𝑡1 = 2.2𝑅𝐶

16. Relation between rise time and upper 3-dB frequency
M.Balaji, Department of ECE, SVEC 16
The upper 3-dB frequency of a low-pass circuit (same as bandwidth) is
𝑓2 =
1
2𝜋𝑅𝐶
𝑜𝑟 𝑅𝐶 =
1
2𝜋𝑓2
∴ 𝑅𝑖𝑠𝑒 𝑡𝑖𝑚𝑒 𝑡 𝑟 = 2.2𝑅𝐶
=
2.2
2𝜋𝑓2
=
0.35
𝑓2
=
0.35
𝐵𝑊
Thus the rise time is inversely proportional to the upper 3-dB frequency.
The time constant (τ=RC) of a circuit is defined as the time taken by the output to rise to
63.2% of the amplitude of the input step.

17. Pulse input
 The pulse is equivalent to a positive step followed by a delayed negative step.
M.Balaji, Department of ECE, SVEC 17
Pulse waveform Pulse waveform in terms of step

18. M.Balaji, Department of ECE, SVEC 18

19. A pulse shape will be preserved if the 3-dB frequency is approximately equal to the
reciprocal of the pulse width
To pass a 0.25µs pulse reasonably well requires a circuit with an upper cut-off
frequency of the order of 4MHz.
M.Balaji, Department of ECE, SVEC 19

20. Square wave input
 A square wave is a periodic waveform which maintains itself at one constant
level V’ with respect to ground for a time T1, and then changes abruptly to
another level V’’, and remains constant at that level for a time T2, and repeats
itself at regular intervals of time T1+T2.
 A square wave may be treated as a series of positive and negative steps.
M.Balaji, Department of ECE, SVEC 20

21. (a) Square wave input waveform (b) output waveform for
RC≪ 𝑇 M.Balaji, Department of ECE, SVEC 21

22. (c) Output waveform for RC ≈ T and (d) RC ≫ 𝑇
M.Balaji, Department of ECE, SVEC 22

23. M.Balaji, Department of ECE, SVEC 23

24. In the above figure when RC ≈ T, the equation for the rising portion is
𝑉0 𝑡 = 𝑉𝑓𝑖𝑛𝑎𝑙 − (𝑉𝑓𝑖𝑛𝑎𝑙 − 𝑉𝑖𝑛𝑖𝑡𝑖𝑎𝑙)𝑒− 𝑡 𝑅𝐶
𝑣 𝑜1 = 𝑉′ − 𝑉′ − 𝑉2 𝑒− 𝑡 𝑅𝐶
Where V2 is the voltage across the capacitor at t=0, and V’ is the maximum level to which
the capacitor can charge.
The equation for the falling portion is
𝑣 𝑜2 = 𝑉′′ − 𝑉′′ − 𝑉1 𝑒−( 𝑡−𝑇1) 𝑅𝐶
Where V1 is the voltage across the capacitor at t= T1 and V’’ is the minimum level to which
the capacitor can discharge.M.Balaji, Department of ECE, SVEC 24

25. Setting 𝑣01 = 𝑉1 𝑎𝑡 𝑡 = 𝑇1
𝑉1 = 𝑉′ − 𝑉′ − 𝑉2 𝑒− 𝑇1 𝑅𝐶
= 𝑉′
− 𝑉′
𝑒− 𝑇1 𝑅𝐶
+ 𝑉2 𝑒− 𝑇1 𝑅𝐶
= 𝑉′
1 − 𝑒− 𝑇1 𝑅𝐶
+ 𝑉2 𝑒− 𝑇1 𝑅𝐶
-------- (1)
M.Balaji, Department of ECE, SVEC 25
Setting 𝑣02 = 𝑉2 𝑎𝑡 𝑡 = 𝑇1 + 𝑇2
𝑉2 = 𝑉′′ − 𝑉′′ − 𝑉1 𝑒−( 𝑇1+𝑇2−𝑇1) 𝑅𝐶
= 𝑉′′
− 𝑉′′
𝑒−( 𝑇1+𝑇2−𝑇1) 𝑅𝐶
+ 𝑉1 𝑒−( 𝑇1+𝑇2−𝑇1) 𝑅𝐶
= 𝑉′′(1 − 𝑒− 𝑇2 𝑅𝐶) + 𝑉1 𝑒− 𝑇2 𝑅𝐶
----- (2)

26. Substituting this value of V2(eq 2) in the expression for V1(1), we get
𝑉1 = 𝑉′ 1 − 𝑒− 𝑇1 𝑅𝐶 + 𝑉′′(1 − 𝑒− 𝑇2 𝑅𝐶) + 𝑉1 𝑒− 𝑇2 𝑅𝐶 𝑒− 𝑇1 𝑅𝐶
𝑉1 − 𝑉1 𝑒− 𝑇2 𝑅𝐶 𝑒− 𝑇1 𝑅𝐶 = 𝑉′ 1 − 𝑒− 𝑇1 𝑅𝐶 + 𝑉′′(1 − 𝑒− 𝑇2 𝑅𝐶)𝑒− 𝑇1 𝑅𝐶
𝑉1 1 − 𝑒−(𝑇1+𝑇2) 𝑅𝐶 = 𝑉′ 1 − 𝑒− 𝑇1 𝑅𝐶 + 𝑉′′(1 − 𝑒− 𝑇2 𝑅𝐶)𝑒− 𝑇1 𝑅𝐶
𝑉1 =
𝑉′ 1 − 𝑒− 𝑇1 𝑅𝐶 + 𝑉′′(1 − 𝑒− 𝑇2 𝑅𝐶)𝑒− 𝑇1 𝑅𝐶
1 − 𝑒−(𝑇1+𝑇2) 𝑅𝐶
Similarly substituting the value of V1 in the expression for V2, we get
𝑉2 =
𝑉′′ 1 − 𝑒− 𝑇2 𝑅𝐶 + 𝑉′(1 − 𝑒− 𝑇1 𝑅𝐶)𝑒− 𝑇2 𝑅𝐶
1 − 𝑒−(𝑇1+𝑇2) 𝑅𝐶
M.Balaji, Department of ECE, SVEC 26

27. For a symmetrical square wave with zero average value,
𝑇1 = 𝑇2 =
𝑇
2
𝑎𝑛𝑑 𝑉′ = −𝑉′′ =
𝑉
2
So, V2 will be equal to –V1
𝑉1 =
𝑉′ 1 − 𝑒− 𝑇1 𝑅𝐶 + 𝑉′′(1 − 𝑒− 𝑇2 𝑅𝐶)𝑒− 𝑇1 𝑅𝐶
1 − 𝑒−(𝑇1+𝑇2) 𝑅𝐶
𝑉1=
𝑉
2
1 − 𝑒− 𝑇 2𝑅𝐶
−
𝑉
2
1 − 𝑒− 𝑇 2𝑅𝐶
𝑒− 𝑇 2𝑅𝐶
1 − 𝑒− 𝑇 𝑅𝐶
=
𝑉
2
1 − 𝑒− 𝑇 2𝑅𝐶
− 𝑒− 𝑇 2𝑅𝐶
+ 𝑒− 𝑇 𝑅𝐶
1 − 𝑒− 𝑇 𝑅𝐶
M.Balaji, Department of ECE, SVEC 27

28. =
𝑉
2
(1 − 𝑒− 𝑇 2𝑅𝐶
)2
1 + 𝑒− 𝑇 2𝑅𝐶 (1 − 𝑒− 𝑇 2𝑅𝐶)
=
𝑉
2
1 − 𝑒− 𝑇 2𝑅𝐶
1 + 𝑒− 𝑇 2𝑅𝐶
=
𝑉
2
𝑒 𝑇 2𝑅𝐶 − 1
𝑒 𝑇 2𝑅𝐶 + 1
=
𝑉
2
𝑒2𝑥
− 1
𝑒2𝑥 + 1
=
𝑉
2
𝑡𝑎𝑛ℎ𝑥
Where 𝑥 =
𝑇
4𝑅𝐶
𝑎𝑛𝑑 𝑇 is the period of the square wave. Now,
𝑉2 = −𝑉1 = −
𝑉
2
1 − 𝑒− 𝑇 2𝑅𝐶
1 + 𝑒− 𝑇 2𝑅𝐶
=
𝑉
2
1 − 𝑒− 𝑇 2𝑅𝐶
1 + 𝑒− 𝑇 2𝑅𝐶
M.Balaji, Department of ECE, SVEC 28

29. Ramp input
M.Balaji, Department of ECE, SVEC 29
When a low- pass RC circuit is excited with a ramp input, i.e
𝑣𝑖 𝑡 = 𝛼𝑡, 𝑤ℎ𝑒𝑟𝑒 𝛼 is the slope of the ramp
We have, 𝑉𝑖 𝑠 =
𝛼
𝑠2
For the frequency domain circuit of a low-pass RC circuit, the output is given by
𝑉0 𝑠 = 𝑉𝑖 𝑠
1
𝐶𝑠
𝑅 +
1
𝐶𝑠
=
𝛼
𝑠2
1
1 + 𝑅𝐶𝑠
=
𝛼
𝑅𝐶
1
𝑠2(𝑠 +
1
𝑅𝐶
)

30. =
𝛼
𝑅𝐶
𝑠2(𝑠+
1
𝑅𝐶
)
=
𝐴
𝑠
+
𝐵
𝑠2 +
𝐶
𝑠+
1
𝑅𝐶
=
𝐴𝑠 𝑠 +
1
𝑅𝐶
+ 𝐵 𝑠 +
1
𝑅𝐶
+ 𝐶𝑠2
𝑠2 𝑠 +
1
𝑅𝐶
𝛼
𝑅𝐶
= 𝐴𝑠 𝑠 +
1
𝑅𝐶
+ 𝐵 𝑠 +
1
𝑅𝐶
+ 𝐶𝑠2
Put s= 0; then B=α
Put 𝑠 = −
1
𝑅𝐶
; then C= αRC
To get the value of A,
𝛼
𝑅𝐶
= 𝐴𝑠2 +
𝐴𝑠
𝑅𝐶
+ 𝐵𝑠 +
𝐵
𝑅𝐶
+ 𝐶𝑠2
𝛼
𝑅𝐶
= 𝐴 + 𝐶 𝑠2
+
𝐴
𝑅𝐶
+ 𝐵 𝑠 +
𝐵
𝑅𝐶
M.Balaji, Department of ECE, SVEC 30

31. Equating the coefficients of s2 on both sides
A+C=0
Therefore A= - C
As C= αRC then A= - αRC
Substituting the values of A, B and C, we get
M.Balaji, Department of ECE, SVEC 31
𝑉0 𝑠 =
−𝛼𝑅𝐶
𝑠
+
𝛼
𝑠2
+
𝛼𝑅𝐶
𝑠 +
1
𝑅𝐶
𝑉0 𝑡 = −𝛼𝑅𝐶 + 𝛼𝑡 + 𝛼𝑅𝐶𝑒− 𝑡 𝑅𝐶

32. Taking the inverse Laplace transform on both sides, we get
𝑉0 𝑡 = −𝛼𝑅𝐶 + 𝛼𝑡 + 𝛼𝑅𝐶𝑒− 𝑡 𝑅𝐶
= 𝛼 𝑡 − 𝑅𝐶 + 𝛼𝑅𝐶𝑒− 𝑡 𝑅𝐶
If the time constant RC is very small, then
𝑒− 𝑡 𝑅𝐶
≈ 0
𝑉0 𝑡 = 𝛼 𝑡 − 𝑅𝐶
M.Balaji, Department of ECE, SVEC 32

33. When the time constant is very small relative to the total ramp time T,
the ramp will be transmitted with minimum distortion
(a) Response of a low-pass RC circuit for a
ramp circuit for (a) RC/T<< 1
(b) Response of a low-pass RC circuit for a
ramp circuit for (a) RC/T>> 1
M.Balaji, Department of ECE, SVEC 33

34. 𝑉0 𝑡 = −𝛼𝑅𝐶 + 𝛼𝑡 + 𝛼𝑅𝐶𝑒− 𝑡 𝑅𝐶
= 𝛼 𝑡 − 𝑅𝐶 + 𝛼𝑅𝐶𝑒− 𝑡 𝑅𝐶
Expanding 𝑒− 𝑡 𝑅𝐶 in to an infinite series in t/RC in the above equation for V0(t),
𝑉0 𝑡 = 𝛼 𝑡 − 𝑅𝐶 + 𝛼𝑅𝐶 1 −
𝑡
𝑅𝐶
+
𝑡
𝑅𝐶
2 1
2!
−
𝑡
𝑅𝐶
3 1
3!
+ ⋯
= 𝛼𝑡 − 𝛼𝑅𝐶 + 𝛼𝑅𝐶 − 𝛼𝑡 +
𝛼𝑡2
2𝑅𝐶
− ⋯
≈
𝛼𝑡2
2𝑅𝐶
≈
𝛼
𝑅𝐶
𝑡2
2
The above expression shows a quadratic response obtained for a linear input and
hence the circuit acts as an integrator for RC/T >> 1
M.Balaji, Department of ECE, SVEC 34

35. Low pass RC circuit as an integrator
For the low-pass RC circuit to behave as an integrator, RC >>T.
For T to be small when compared to RC , the frequency has to be
high.
At high frequencies, XC is very small when compared to R.
Therefore, the voltage drop across R is very large when compared to
the drop across C.
Hence, vi ∼= iR
M.Balaji, Department of ECE, SVEC 35
RC>>T
When f is high,
then T will be small
then RC will be much
greater than T
At high frequencies,
Xc =
1
2𝜋𝑓𝑐
= 𝑙𝑜𝑤
Drop across C is very
less. Then the entire
drop will be at R

36. 𝑉𝑖 𝑡 = 𝑖 𝑡 𝑅 +
1
𝐶
𝑖 𝑡 𝑑𝑡
𝑉𝑖 𝑡 = 𝑖 𝑡 𝑅
𝑖 𝑡 =
𝑉𝑖(𝑡)
𝑅
𝑉0 𝑡 =
1
𝐶
𝑖 𝑡 𝑑𝑡 =
1
𝐶
𝑣𝑖(𝑡)
𝑅
𝑑𝑡 =
1
𝑅𝐶
𝑣𝑖(𝑡)𝑑𝑡
𝑉0 𝑡 =
1
𝑅𝐶
𝑉𝑖 𝑡 𝑑𝑡
The output is proportional to the integral of the input signal
M.Balaji, Department of ECE, SVEC 36

37. If the time constant of an RC low-pass circuit is very large in
comparison with the time required for the input signal to make an
appreciable change, the circuit acts as an integrator.
The low-pass circuit acts as an integrator provided the time constant
of the circuit RC > 15T, where T is the period of the input signal.
When RC > 15T, the input sinusoid will be shifted at least by 89.4⁰
when it is transmitted through the network.
M.Balaji, Department of ECE, SVEC 37