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WAVEFORMS AND SIGNALS
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EXAMPLE 6.15 If the switch in Fig. 6-8(a) is moved to position 2 at t 0 and then moved back to position 1 at t 5 s, express vAB using the step function. vAB V0 u t u t 5 EXAMPLE 6.16 Express v t , graphed in Fig. 6-9, using the step function.
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Fig. 6-9 v t u t u t 2 sin t
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THE UNIT IMPULSE FUNCTION
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Consider the function sT t of Fig. 6-10(a), which is zero for t < 0 and increases uniformly from 0 to 1 in T seconds. Its derivative dT t is a pulse of duration T and height 1=T, as seen in Fig. 6-10(b). 8 for t < 0 <0 30 for 0 < t < T dT t 1=T : 0 for t > T If the transition time T is reduced, the pulse in Fig. 6-10(b) becomes narrower and taller, but the area under the pulse remains equal to 1. If we let T approach zero, in the limit function sT t becomes a unit step u t and its derivative dT t becomes a unit pulse  t with zero width and in nite height. The unit impulse  t is shown in Fig. 6-10(c). The unit impulse or unit delta function is de ned by 1  t 0 for t 6 0 and
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An impulse which is the limit of a narrow pulse with an area A is expressed by A t . The magnitude A is sometimes called the strength of the impulse. A unit impulse which occurs at t t0 is expressed by  t t0 .
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EXAMPLE 6.17 The voltage across the terminals of a 100-nF capacitor grows linearly, from 0 to 10 V, taking the shape of the function sT t in Fig. 6-10(a). Find (a) the charge across the capacitor at t T and (b) the current iC t in the capacitor for T 1 s, T 1 ms, and T 1 ms. (a) At t T, vC 10 V. The charge across the capacitor is Q CvC 10 7 10 10 6 . b ic t C dvC dt
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WAVEFORMS AND SIGNALS
Fig. 6-10 From Fig. 6-10, 8 <0 iC t I0 10 6 =T A : 0 for t < 0 for 0 < t < T for t > T 32
For T 1 s, I0 10 6 A; for T 1 ms, I0 10 3 A; and for T 1 ms, I0 1 A. In all the preceding cases, the charge accumulated across the capacitor at the end of the transition period is T Q iC t dt I0 T 10 6 C
The amount of charge at t T is independent of T.
It generates a voltage vC 10 V across the capacitor.
EXAMPLE 6.18 Let dT t t0 denote a narrow pulse of width T and height 1=T, which starts at t t0 . Consider a function f t which is continuous between t0 and t0 T as shown in Fig. 6-11(a). Find the limit of integral I in (33) when T approaches zero. 1 I dT t t0 f t dt 33
dT t t0 Substituting dT in (33) we get I 1 T
1=T 0 t0 T
t0 < t < t0 T elsewhere
f t dt
34a
where S is the hatched area under f t between t0 and t0 T in Fig. 6.11(b). Assuming T to be small, the function f t may be approximated by a line connecting A and B. S is the area of the resulting trapezoid. S 1 f t0 f t0 T T 2 I 1 f t0 f t0 T 2 As T ! 0, dT t t0 !  t t0 and f t0 T ! f t0 and from (34c) we get 34b 34c
WAVEFORMS AND SIGNALS
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Fig. 6-11 lim I lim 1 f t0 f t0 T 2
34d
We assumed f t to be continuous between t0 and t0 T.
T!0 1
Therefore, 34e (34f) (34g) It is also used as another de nition for
lim I f t0  t t0 f t dt
But and so
lim I 1
 t t0 f t dt f t0
The identity (34g) is called the sifting property of the impulse function.  t .
THE EXPONENTIAL FUNCTION
The function f t est with s a complex constant is called exponential. It decays with time if the real part of s is negative and grows if the real part of s is positive. We will discuss exponentials eat in which the constant a is a real number. The inverse of the constant a has the dimension of time and is called the time constant  1=a. A decaying exponential e t= is plotted versus t as shown in Fig. 6-12. The function decays from one at t 0 to zero at t 1. After  seconds the function e t= is reduced to e 1 0:368. For  1, the function e t is called a normalized exponential which is the same as e t= when plotted versus t=.
EXAMPLE 6.19 Show that the tangent to the graph of e t= at t 0 intersects the t axis at t  as shown in Fig. 6-12. The tangent line begins at point A v 1; t 0 with a slope of de t= =dtjt 0 1=. The equation of the line is vtan t t= 1. The line intersects the t axis at point B where t . This observation provides a convenient approximate approach to plotting the exponential function as described in Example 6.20. EXAMPLE 6.20 Draw an approximate plot of v t e t= for t > 0. Identify the initial point A (t 0; v 1 of the curve and the intersection B of its tangent with the t axis at t . Draw the tangent line AB. Two additional points C and D located at t  and t 2, with heights of 0.368 and
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