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Page 526 void enter(void), qstore(char *q), review(void), delete_ap(void); int main(void) { char s[80]; register int t; for(t=0; t < MAX; ++t) p[t] = NULL; /* init array to nulls */ for(;;) { printf(''Enter, List, Remove, Quit: "); gets(s); *s = toupper(*s); switch(*s) { case 'E': enter(); break; case 'L': review(); break; case 'R': delete_ap(); break; case 'Q': exit(0); } } return 0; } /* Enter appointments in queue */ void enter(void) { char s[256], *p; do { printf("Enter appointment %d: ", spos+1); gets(s); if(*s==0) break; /* no entry */ p = (char *) malloc(strlen(s)+1); if(!p) { printf("Out of memory\n");



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Looking at h(t) it immediately becomes clear that if a > 0 the function is stable We saw an example of this type of function in the beginning of the chapter where we met h(t) = e 2t cos 20t in Fig 14-3

Page 527 return; } strcpy(p, s); if(*s) qstore(p); } while(*s); } /* See what's in the queue */ void review(void) { register int t; for(t=rpos; t < spos; ++t) printf(''%d %s\n", t+1, p [t]); } /* Delete an appointment from the queue */ void delete_ap(void) { char *p; if((p=qretrieve()) ==NULL) return; printf("%s\n", p); } /* Store an appointment */ void qstore(char *q) { if(spos==MAX) { printf ("List Full\n"); return; } p[spos] = q; spos++; } /* Retrieve an appointment */ char *qretrieve(void) { if(rpos==spos) { printf("No more appointments\n"); return NULL; }





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installation of optical fiber by a communications carrier to a specific building, such as a house or an apartment or office building In this section we first briefly examine the history of fiber to the home and the reason why its deployment to date has been underwhelming Once this is accomplished, we will focus on the one area where fiber to the home represents a practical technology This area concerns the provision of large buildings and their occupants with the ability to bypass the local telephone company and, as you might expect, has economics as its dri, ving force

Page 528 rpos++; return p[rpos-1]; }

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Now we consider the stability of a circuit in the zero-input response case, that is, for a circuit with initial values for currents and voltages but no sources We can determine zero-input response stability by again looking at the poles of the transfer function H (s) The rst case we consider is when the poles of H (s)are real and negative In that case the circuit is stable, which indicates that the currents and voltages in the circuit will decay to zero as t EXAMPLE 14-5 Consider a series RLC circuit with R = 2, C = 1/2, and L = 4 If the initial voltage across the capacitor is v c (0) = 2 and the initial current owing through the capacitor is i c (0) = 0, determine whether the circuit is zero input stable SOLUTION As usual, KVL around the loop of a series RLC circuit gives LC dv c d 2vc + RC + vc = 0 2 dt dt

The Circular Queue In studying the preceding appointment-scheduler program, an improvement may have occurred to you Instead of having the program stop when the limit of the array used to store the queue is reached, you could have both the store index (spos) and the retrieve index (rpos) loop back to the start of the array In this way, any number of items could be placed on the queue, so long as items were also being taken off This implementation of a queue is called a circular queue because it uses its storage array as if it were a circle instead of a linear list To create a circular queue for use in the scheduler program, you need to change the functions qstore ( ) and qretrieve( ) as shown here:

void qstore(char *q) { /* The queue is full if either spos is one less than rpos or if spos is at the end of the queue array and rpos is at the beginning */ if(spos+1= =rpos || (spos+1==MAX && !rpos)) { printf(''List Full\n"); return; } p[spos] = q; spos++; if(spos==MAX) spos = 0; /* loop back */ } char *qretrieve(void) { if(rpos==MAX) rpos = 0; /* loop back */ if(rpos==spos) { printf("No events to retrieve\n"); return NULL; } rpos++; return p[rpos-1]; }

Taking the Laplace transform with the initial conditions speci ed gives the s-domain equation LC(s 2 Vc (s) sv c (0)) + RC(sVc (s) v c (0)) + Vc (s) = 0 Solving we have Vc (s) = RCv c (0) LCs 2 + (RC LCv c (0))s + 1

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