Write a program that creates a plot of the power consumed by a 1000 resistor as the voltage across it is varied from 1 V to 200
V. Create two plots: one showing power in watts and one showing power in dBW (dB power levels with respect to a 1 W reference).
1 answer:
Answer:
Program to Plot the power in Watts
voltage=1:200;
resistance=1000;
current=voltage/resistance;
power=current.*voltage;
plot(voltage,power);
xlabel('Voltage in Volts');
ylabel('Power in Watts'); //plot of this program is attached
Program for power in dBW
voltage=1:200;
resistance=1000;
current=voltage/resistance;
power=current.*voltage;
powerdB=10*log10(power);
plot(voltage,powerdB);
xlabel('Voltage in Volts');
ylabel('Power in dBW'); // plot output is also attached
I hope it will help you!
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Answer:
a) Speedup gain is 1.428 times.
b) Speedup gain is 1.81 times.
Explanation:
in order to calculate the speedup again of an application that has a 60 percent parallel component using Anklahls Law is speedup which state that:
Where S is the portion of the application that must be performed serially, and N is the number of processing cores.
(a) For N = 2 processing cores, and a 60%, then S = 40% or 0.4
Thus, the speedup is:
Speedup gain is 1.428 times.
(b) For N = 4 processing cores and a 60%, then S = 40% or 0.4
Thus, the speedup is:
Speedup gain is 1.81 times.