# Find frequency response and circular frequency

1. Jan 8, 2016

### Hepic

1. The problem statement, all variables and given/known data
We are given that system of 1st grade
1/2 * dy(t)/dt + y(t) = x(t)

1) We need to find frequency response { H(w) }
2) We need to find circular frequency -3dB
3) We need to find the absolute value of H(w) in dB for frequency f = 3/π Hz

2. Relevant equations
(log is always with base 10)

Convertion to db => db = 20log(|H(w)|)
circular frequency = -3db => |H(w)|{w=-3db} = 1/sqrt(2) * H(w){max w}
Y(w) = H(w) * X(w)

3. The attempt at a solution
I solved the 1st one with that way.
(take Fourier) => 1/2* jw*Y(w) + Y(w) = X(w)
(divide with X(w)) => 1/2* j*w*H(w) + H(w) = 1 => H(w) = 1 / (1/2 * j*w + 1) => H(w) = 2 / (2 + jw)

For 2nd I dont know what is w = max, and for 3rd I did an attempt but I found wrong result.
My book gives as answer for 2nd the value = 2rad/sec, and for 3rd the value -20dB.
If anyone can help me with the procedure of solving I would be happy.

2. Jan 8, 2016

### BvU

For a first order system the "maximum response" is if dy/dt = 0 so for $\ \omega = 0$.

3. Jan 8, 2016

### Staff: Mentor

Hi Hepic. You should be able to recognize your transfer function H(ω) as that of a typical first order low pass filter.

As such you can find the -3dB frequency by inspection of H(ω) itself, or if you want to do the math, determine ω that makes |H(ω)|2 = 1/2, i.e., find the half-power point.

I find myself disagreeing with the book's answer of -20 dB for the response at 6 rad/sec. A first order low pass filter should decline at a rate of 20 dB per decade after the corner frequency (using the straight-line curve fit approximation). Now 6 rad/sec is not a decade past 2 rad/sec. I'd expect something closer to -10 dB for the response at 6 rad/sec.

4. Jan 8, 2016

### LvW

Yes - it is exactly a value of -10 dB.