# control systems

Discussion in 'Homework Help' started by vvkannan, Jan 15, 2009.

1. ### vvkannan Thread Starter Active Member

Aug 9, 2008
138
11
Hello ,
Iam stuck with this question

A second order system has a transfer function given by G(s) = 25 / (s^2+8s +25).If the system initially at rest is subjected to a unit step input at t=0,the second peak in response will occur at ?

using inverse laplace transforms we get the response as
y(t) = 1 - 0.6(exp(-4t) sin(3t + 37))

From this we have a formula to find the time taken to reach the first peak but i dont know how to find the time taken to reach second peak

Thank you

2. ### KL7AJ AAC Fanatic!

Nov 4, 2008
2,182
415
How's 2t work for ya?

3. ### mik3 Senior Member

Feb 4, 2008
4,846
69
The first positive peak will occur after 1.35 sec and the second positive peak after 2.01 sec of the first peak. It's the time the first peak needs to occur plus the time of the period of the sine wave.

4. ### vvkannan Thread Starter Active Member

Aug 9, 2008
138
11
Thank you mik3.But how did you find first peak?
Using the formua Tp = 3.14/damped frequency i get 1.047 seconds for peak time and plus time period gives the time of second peak as 3.14seconds

sorry KL7AJ i cant understand what you mean

5. ### mik3 Senior Member

Feb 4, 2008
4,846
69
The first peak will occur when the angle in the sine is 270 degrees (sin(270)=-1) but because it has a phase shift of +37 the angle becomes 270-37=233. In rad is 4.06, you know the angular velocity which is 3 rad/s and thus the time t is:

ωt=φ thus

t=φ/ω=4.06/3=1.35

where

6. ### vvkannan Thread Starter Active Member

Aug 9, 2008
138
11
Thanks mik3 but isnt pi/damped frequency the general procedure to find the peak time ?.Moreover the answer given is 3.14 seconds which we dont get when peak time = 1.35 seconds.On the other hand we get the same answer if we use the formula

Feb 4, 2008
4,846
69