1
$\begingroup$

Given a system, that behaves as a 1st order filter with network function $H(s)$. We input: $$v_1 (t)=1+3\cos(10^4 t)$$

And we obtain as output: $$v_2(t)=1+1.5\cos \left(10^4 t -\dfrac{\pi}{3}\right)$$

Say what kind of filter it is and find its network function $H(s)$.

I'm trying to solve this problem and I have solved the first part by saying that the filter is a low-pass filter because the continuous term is conserved in the input as well as in the output. And I have supposed that the network function will be of the form: $$H(s) = \dfrac{\kappa}{s+a}$$

My question is what do I have to do in order to compute the constants $\kappa, a$ according to the filter characteristics?

I have tried to compute the amplification for the following frequencies: $$\text{Amplification}(\omega=0) =\dfrac{1}{1} = 1$$ $$\text{Amplification}(\omega=10^4) = \dfrac{1.5}{3} = 0.5$$

But I don't know if this is correct since the phase of cosine in the output is different.

$\endgroup$
3
$\begingroup$

You need to consider the system's frequency response

$$H(j\omega)=\frac{\kappa}{j\omega+a}\tag{1}$$

Now you know that $$H(0)=1\tag{2}$$ and $$|H(j\omega_0)|=\frac12\tag{3}$$ (with $\omega_0=10^4$). Note that $H(0)$ is real-valued, whereas $H(j\omega_0)$ will generally be complex-valued. From (1) and (2) you immediately get $\kappa=a$. Furthermore, since the filter is stable we know that $a>0$ (i.e. the pole must lie in the left half-plane of the complex $s$-plane). Combining (3) with (1) gives

$$\frac{a}{\sqrt{\omega_0^2+a^2}}=\frac12\tag{4}$$

from which it should be easy for you to compute $a$.

Note that you don't need the phase value to specify the system. What should be checked is if the phase value is the actual phase value of the system that you just computed. In order to do this you need to verify that

$$\arg\{H(j\omega_0)\}=-\frac{\pi}{3}\tag{5}$$

Luckily that's the case, otherwise the assumption (1) would be wrong. I leave the proof of (5) up to you.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.