/H TI IIIIIIIIIIIII . as tfrey are conrfl@d. Thefi rtrton the,!xt lcltton. L Determineth e rransfer...

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/H TI IIIIIIIIIIIII . as tfrey are conrfl@d. Thefi rtrton the,!xt lcltton. L Determineth e rransfer that i (l) is the excitation ( l ) G(s) : , ., , tL . ZR + sL (2) G(s' ) )rt+sZ (3) G(s) = 'L R+sl function of the circuir and v(l) is the response (4) G(s) = s l of Fie. tr-1. Assume resultingir omi (l). (5) G(s) = (Hint: Use see L 2R transforms er"r, 7-2) to find 81, rtg. E-| 2. Consider ingth ati r ( l ) is theo urpura nd u, ( l ) oi Fig. E-2,d etermineth ei mpulser esponse. (l) g(t) : e-rrsin t is the input of the circuit (Hint: Topic '7-2 ex.7-Z) l t . . I ie ( t ) 8n Fig.56hhFig. E-12. A current generator that produces a current that increases at a rate of 3 A per sec is switched on at t = 2 sec. Determine the equation for the resulting current. (Hint: Use Eq XXXXXXXXXXi(t) = 3 t u(t XXXXXXXXXXi(t)= 2(t - 3) u(t - 3) l(2) i(t) = 3(t - 2) u(t XXXXXXXXXXKO= t u(3t XXXXXXXXXXi(t) = 3 t u(t + 2)3. Determine the equation for the waveform of Fig. E-2. (Hint: Eq XXXXXXXXXXf(t)= u(t + 4) Ve(3) f(t)= -2(t + 2)u(t XXXXXXXXXXf(t) = -2tu(t XXXXXXXXXXf(t) = -4tu(t + 2)4. A voltage of 10 V is switched on at t = 4 sec. This voltage is applied to the input of an integrator. What is the output voltage? (Hint: Eq. 2-68 and Topic XXXXXXXXXXe(t)= 10t u(t - 4) yi(3) e(t) = 10(t - 4) u(t XXXXXXXXXXe(t) = 4t u XXXXXXXXXXe(t) =1012 u(t + 4)5. A current source that produces a current equal to i(t) = 5 + t2 u(t) is applied to the input of a differentiator. Determine the output current of the differentiator. (1) i'(t) = (5 + t) u(t) (3) t(t)= t u(t + 5) fr. (2) F(t) = 2t u(t) (4) /1(t)= 3t2u(t) 6. A certain voltage can be expressed as v(t) = (5 + t2) u(t - 1). What is the value of this voltage at t = 3 sec? XXXXXXXXXXV XXXXXXXXXXV XXXXXXXXXXV7. A certain function is expressed as f(t) = 2e-005. What is the value of f(t) at t = 1.5 sec? (Hint: To find e-x, refer to lesson XXXXXXXXXX XXXXXXXXXX0.10
	
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-1uF*l)
Fig. E-1
generator produces 
2. A cument that a current that increases at a iat" of 3 A
= 
per sec is the equation 
switched on at f 2 sec. Determine for the
resulting (Hint: Eq.2-52)
current. Use 
- - - 
= = 
(r) i(t) u(t XXXXXXXXXXi(t) 2(t 3) u(t 3)
3t 
- - - 
= = 
(, 2) u(t XXXXXXXXXXi(t) t u(3t 2)
3(r 
X2) 
' 
= 
(3) i(t) + 2)
3t u(t 
waveform Fig. E-2. (Hinc Eq.2-52)
3. Determine the equation for the of 
-2(t 
= = 
(l) f(t) + 4) f(0 + Z)u(t + 2)
u(t 
*

David · Accepted Answer

2.  A current generator that produces a current that increases at a rate of 3A/s is switched 
on at 2s.t   We wish to determine the equation for the resulting current.
We have
    3 2  and 0 for 2.
di
u t i t t
dt
   
Thus we have
 
   
   
3 2
3 2 2
i t u t dt
t u t C
 
   


for some constant C. Note that  0 0 ,i C   whence 0C   and hence 
 
      3 2 2 .i t t u t    

3. We wish to determine the equation for the waveform  f t  from the following figure: 
 
 

 For 2t    we have   0f t  , while for 2t    we have    2 2 .f t t    Thus we have  
 
       2 2 2 .f t t u t     

4. A voltage of 10 V is switched on at 4s.t  The voltage is applied to the input of an 
integrator. What is the output voltage?
We have
   
 
   
in
10 4
10 4 4 .
t
t
e t e s ds
u s ds
t u t



 
  


5. A current source that produces a current equal to    25i t t u t   is applied to the input 
of a differentiator. We wish to determine the output current of the differentiator.
The output current is given by
   
   
 
 
2
2
' 5
5
0 2
2 .
d
i t t u t
dt
d d
t u t
dt dt
t u t
t u t
   
    
 


6. A certain voltage can be expressed as      25 1 .v t t u t    What is the voltage at 
3s?t 
We have
     
  
23 5 3 2
5 9 1
14V.
v u 
 


7. A certain function is expressed as   /0.52 .tf t e  We wish to compute the value of 
 f t  at 1.5s.t   
 
We have
  1.5/0.5
3
1.5 2
2
0.1.
f e
e






8. A voltage source,  ,e t  which has a value of 2 V at 0,t   rises exponentially to a final 
value of 6 V with a time constant of 3 s. We wish to determine the expression for  .e t  
 
We have
  
/3
0 if 0;
6 4  if 0,t
t
e t
e t

 
 

whence      /36 4 .te t e u t   

9. A current is defined by    /32 .ti t e u t  We wish to compute the total area A under the 
curve for 0 .t 
We have
 
 
 
0
/3
0
/3
0
/3
0
2
2
6 / 3
6.

/H TI IIIIIIIIIIIII . as tfrey are conrfl@d. Thefi rtrton the,!xt lcltton. L Determineth e rransfer that i (l) is the excitation ( l ) G(s) : , ., , tL . ZR + sL (2) G(s' ) )rt+sZ (3) G(s) = 'L R+sl...

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