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We have seen how jitter is generated and the way it accumulates in a string of regenerators, but how does jitter create errors in the transmission process Consider the operation of a simple decision circuit based on a D-type bistable as shown in Figure 23.10. The D-type makes a decision on whether a 1 or 0 is present on the data input on the rising edge of the recovered clock pulse. The possibility of making an error does not depend on the absolute amplitude of jitter on the incoming bit stream, but on the difference between the amplitude of the jitter on the recovered clock and the data signal. This is sometimes called misalignment jitter. If there were no other impairments affecting the shape of the waveform (such as noise or incorrect equalization), the maximum misalignment jitter that could be tolerated would be 0.5 UI. In practice, other impairments can reduce this to 0.2 UI or less.
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The Thevenin equivalent circuit is built by putting the Thevenin equivalent voltage in series with the Thevenin equivalent resistance. This is shown in Fig. 3-17. EXAMPLE 3-5 In the circuit shown in Fig. 3-18, nd the voltage across the load resistor if R1 = 2 , R2 = 3 , R3 = R4 = 6 , the voltage source is Vs = 15 V and , the current is I L = 3 A. SOLUTION The rst step is to remove the load resistor and calculate the Thevenin equivalent voltage across the resulting open circuit. This is shown in Fig. 3-19. We take the positive reference for the voltage across each resistor to be at the top of the resistor. This looks complicated, but it really isn t. If we ignore the two inside resistors R1 and R3 for the moment, it should be easy to deduce that resistors R2 and R4 are in series. Then we can apply KVL around the external
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#include <stdlib.h> #include <stdio.h> int main(void) { char *str[100]; int i; for(i=0; i<100; i++) { if((str[i]=(char *)malloc(128))==NULL) { printf("Allocation error."); exit(0); } gets(str[i]); } /* now free the memory */ for(i=0; i<100; i++) free(str[i]); return 0; }
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Both the relational and logical operators are lower in precedence than the arithmetic operators. This means that an expression like 10 > 1 + 12 is evaluated as if it were written 10 > (1 + 12). The result is, of course, false. Several operations can be combined in one expression, as shown here: 10>5 && !(10<9) || 3<=4 which will evaluate true.
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Apply your distortions using the Interactive Distort Tool, shown next, which is found in the Toolbox grouped with other effects tools and is used together with these Property Bar options.
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