POJ 1045:Bode Plot

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Bode Plot
Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 13392 Accepted: 8462

Description

Consider the AC circuit below. We will assume that the circuit is in steady-state. Thus, the voltage at nodes 1 and 2 are given by v1 = VS coswt and v2 = VRcos (wt + q ) where VS is the voltage of the source,w is the frequency (in radians per second), and t is time. VR is the magnitude of the voltage drop across the resistor, and q is its phase. 

You are to write a program to determine VR for different values of w. You will need two laws of electricity to solve this problem. The first is Ohm's Law, which states v2 = iR where i is the current in the circuit, oriented clockwise. The second is i = C d/dt (v1-v2) which relates the current to the voltage on either side of the capacitor. "d/dt"indicates the derivative with respect to t. 

Input

The input will consist of one or more lines. The first line contains three real numbers and a non-negative integer. The real numbers are VS, R, and C, in that order. The integer, n, is the number of test cases. The following n lines of the input will have one real number per line. Each of these numbers is the angular frequency, w

Output

For each angular frequency in the input you are to output its corresponding VR on a single line. Each VR value output should be rounded to three digits after the decimal point.

Sample Input

1.0 1.0 1.0 90.010.0316230.10.316231.03.162310.031.623100.0

Sample Output

0.0100.0320.1000.3020.7070.9530.9951.0001.000

水题,公式推导。话说真不能再总纠结这水题了,越做越水了现在。

 v1 = VS

v2 = VRcos (wt + q ) 

 v2 = iR

C=d/dt (v1-v2) 即(v1-v2) 对t求导

已知Vs,C,R,w,求 VR。因为任意的t,等式都成立,所以对t取特殊值,解方程即可。

代码:

#include <iostream>#include <cmath>#include <algorithm>using namespace std;int main(){double Vs,R,C,w;int n,i;cin>>Vs>>R>>C>>n;for(i=1;i<=n;i++){cin>>w;printf("%.3f",sqrt(1.0/(1+C*C*w*w*R*R))*C*w*R*Vs);cout<<endl;}    return 0;}


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