# Phase response of an LTI-system

I've got an LTI-system such as follows: $$y'(t)+{2} y(t)={5} x(t-{3}), t>0, \\ \\ y(0)=0 \ \text{ ja } x(t)=0, \ t<0.$$ From this i've already figured out the transfer function: $$H(f)=\frac{5\cdot e^{\left(-\mathrm{i}\right)\cdot 2\cdot \pi\cdot f\cdot 3}}{2+\mathrm{i}\cdot 2\cdot \pi\cdot f}$$ And the amplitude response: $$\frac{5}{\sqrt{4+\left(2\cdot \pi\cdot f\right)^2}}$$ However, I seem to have trouble with the phase response. I know the formula for phase response is as follows: $$θ(f)=arg[H(f)]$$ But I still have some trouble using it. The denominator would go as follows I think: $$arg(2+i*2*pi*f)=arctan(\frac{Im}{Re})=arctan(\frac{2*pi*f}{2})$$ But i'm a bit confused about the numerator, and the exponential within it. Is the imaginary part of the numerators argument pi times 6, or does the exponential stay? Or perhaps there is no real part at all?

• $\arg e^{i \theta} = \theta$. Does that help? – TimWescott Nov 15 '19 at 22:57
• further help: Do you see how your numerator has magnitude 5 and phase versus frequency $2\pi f 3$? If $H(f)$ was just your numerator the answer should be very clear. So now if you know the phase and magnitude versus frequency of your denominator and you know how to divide the two (What is the phase of $e^{j\phi_1}/e^{j\phi_2}$???), you will have your answer. – Dan Boschen Nov 16 '19 at 15:08
• When you have three factors of functions of frequency, let's say $H(f)G(f)E(f)$, their phase function is $\angle H(f) + \angle G(f) + \angle E(f)$. Does that help any further? Keep in mind that positive real numbers have zero phase. – GKH Nov 17 '19 at 7:53
• @TootsieRoll OK, let's see: $\angle 5 = 0$ and $\angle e^{-j2\pi 3f} = -6\pi f$. Since the other factor is $1/(2+j2\pi f)$, its phase would be $\angle 1 - \angle (2+j2\pi f)$. You can verify this by checking the argument of a complex number $1/z = 1/(re^{j\theta}) = (1/r)e^{-j\theta}$, that is, $\angle z = -\theta$. So, given that, what is the total phase in your case? – GKH Nov 17 '19 at 18:39
• @TootsieRoll This one looks much better ;) – GKH Nov 18 '19 at 21:40