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WAVEFORMS AND SIGNALS
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THE AVERAGE AND EFFECTIVE (RMS) VALUES A periodic function f t , with a period T, has an average value Favg given by 1 T 1 t0 T f t dt f t dt Favg h f t i T 0 T t0
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The root-mean-square (rms) or e ective value of f t during the same period is de ned by  t0 T 1=2 1 Feff Frms f 2 t dt T t0
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2 It is seen that Feff h f 2 t i. Average and e ective values of periodic functions are normally computed over one period.
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EXAMPLE 6.9 Find the average and e ective values of the cosine wave v t Vm cos !t  . Using (11), 1 T V V cos !t  dt m sin !t  T 0 Vavg 0 T 0 m !T and using (12),
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2 Veff
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1 2T
2 2 Vm 1 cos 2 !t  dt Vm =2
from which
p Veff Vm = 2 0:707Vm
(14) In other words,
Equations (13) and (14) show that the results are independent of the frequency and phase angle . the average of a cosine wave and its rms value are always 0 and 0.707 Vm , respectively. EXAMPLE 6.10 Find Vavg and Veff of the half-recti ed sine wave  Vm sin !t when sin !t > 0 v t 0 when sin !t < 0 T=2 Vm sin !t dt
15
From (11), Vavg and from (12),
2 Veff
Vm cos !t T=2 Vm = 0 !T
16
T =2
2 Vm sin2 !t dt
1 2T
T=2
2 2 Vm 1 cos 2!t dt Vm =4
from which EXAMPLE 6.11
Veff Vm =2 Find Vavg and Veff of the periodic function v t where, for one period T,  V0 for 0 < t < T1 v t Period T 3T1 V0 for T1 < t < 3T1 Vavg
2 Veff
(17)
18
We have and from which
V0 V0 T 2T1 3T 1 3
(19)
2 V0 2 T 2T1 V0 3T 1
Veff V0 If jv t j V0 then Veff V0 .
(20)
The preceding result can be generalized as follows.
WAVEFORMS AND SIGNALS
[CHAP. 6
EXAMPLE 6.12 Compute the average power dissipated from 0 to T in a resistor connected to a voltage v t . Replace v t by a constant voltage Vdc . Find Vdc such that the average power during the period remains the same. p vi v2 =R Pavg 1 RT T
v2 t dt
1 2 V2 Veff dc R R
Vdc Veff
EXAMPLE 6.13 The current i t shown in Fig. 6-6 passes through a 1-mF capacitor. Find (a) vac the voltage across the capacitor at t 5k ms (k 0; 1; 2; 3; . . . and (b) the value of a constant current source Idc which can produce the same voltage across the above capacitor at t 5k ms when applied at t > 0. Compare Idc with hi t i, the average of i t in Fig. 6-6, for a period of 5 ms after t > 0.
Fig. 6-6 (a) At t 5 ms 1 vac C 5 10 3
i t dt 10 10
" 3 10 3 4 dt
5 10 3
3 10 3
# 2 dt 12 4 8 V
This is the net charging e ect of i t during each 5-ms interval. Every 5 ms the above amount is added to the capacitor voltage. Therefore, at t 5k ms, v 8k (V). (b) With a constant current Idc , the capacitor voltage vdc at t 5k ms is 3 1 5k 10 vdc Idc dt 106 Idc 5k 10 3 103 5k Idc C 0 Since vdc vac at 5k ms, we obtain 103 5k Idc 8k or Idc 8k= 5k 103 1:6 10 3 A 1:6 mA
Note that Idc hi t i of Fig. 6-6 for any period of 5 ms at t > 0.
NONPERIODIC FUNCTIONS
A nonperiodic function cannot be speci ed for all times by simply knowing a nite segment. Examples of nonperiodic functions are  0 for t < 0 (21) (a) v1 t 1 for t > 0 8 for t < 0 <0 (22) (b) v2 t 1=T for 0 < t < T : 0 for t > T  0 for t < 0 (c) v3 t (23) for t > 0 e t=
CHAP. 6]
WAVEFORMS AND SIGNALS
 (d) (e) (f) (g) (h) v4 t  v5 t
0 sin !t
for t < 0 for t > 0 for t < 0 for t > 0
(24) (25) (26) (27) (28)
0 e t= cos !t for all t for all t cos !t
v6 t e t= v7 t e ajtj v8 t e
ajtj
for all t
Several of these functions are used as mathematical models and building blocks for actual signals in analysis and design of circuits. Examples are discussed in the following sections.
THE UNIT STEP FUNCTION The dimensionless unit step function, is de ned by  0 for t < 0 u t 1 for t > 0 The function is graphed in Fig. 6-7. Note that the function is unde ned at t 0.
29
Fig. 6-7
To illustrate the use of u t , assume the switch S in the circuit of Fig. 6-8(a) has been in position 1 for t < 0 and is moved to position 2 at t 0. The voltage across A-B may be expressed by vAB V0 u t . The equivalent circuit for the voltage step is shown in Fig. 6-8(b).
Fig. 6-8
EXAMPLE 6.14 The switch in the circuit of Fig. 6-8(a) is moved to position 2 at t t0 . Express vAB using the step function. The appearance of V0 across A-B is delayed until t t0 . Replace the argument t in the step function by t t0 and so we have vAB V0 u t t0 :
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