Given is the impulse response: $$h(t)=\delta(t-t_0)$$.

I calculated $$H(f)=e^{-j2 \pi ft_0}=|H(f)|\cdot e^{j\varphi(f)}$$.

Now, the magnitude response of $H(f)$ is:

$$\begin{align} |H(f)| &=\sqrt{\Re\{H(f)\}^2+\Im\{H(f)\}^2} \\ &= \sqrt{\cos^2(2{\pi}ft_0)+\sin^2(2{\pi}ft_0)} \\ &= 1 \\ \end{align}$$.

Using: $$e^{-j \theta}=\cos(\theta)- j \cdot \sin(\theta)$$ and $$\cos^2(\theta)+\sin^2(\theta) = 1$$.

But now I wonder about the solution of the phase response. I know that the solution is:

$$\varphi(f)=-2 \pi f t_0$$.

I would be thankful if someone could show me how the phase response was calculated, since following

$$\varphi(f) = \arctan \left(\frac{\Im\{H(f)\}}{\Re\{H(f)\}} \right) = \arctan \left(\frac{-\sin(2 \pi f t_0)}{\cos(2 \pi f t_0)} \right)$$

I instead obtain $$\varphi(f)=-\arctan\big(\tan(2{\pi}ft_0)\big)$$ which differs from the given sample solution.

  • $\begingroup$ try plotting both in matlab and see if there is a difference $\endgroup$ – user28715 Jun 12 '19 at 0:34
  • $\begingroup$ $$\varphi(f) = \arctan \left(\frac{\Im\{H(f)\}}{\Re\{H(f)\}} \right)$$ is not always correct. it is correct when $\Re\{H(f)\}>0$ but is off by an amount of $\pm \pi$ when $\Re\{H(f)\}<0$. you need to check out the complex argument, $\endgroup$ – robert bristow-johnson Jun 12 '19 at 1:22

Wouldn't it be much easier to just compare the left-hand side and the right-hand side of

$$e^{-j2\pi ft_0}=|H(f)|e^{j\phi(f)}\tag{1}$$

to see that $|H(f)|=1$ and $\phi(f)=-2\pi ft_0$?

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  • $\begingroup$ Plants face in hand - sometimes a little bit of insight beats a lot of mathematics :) $\endgroup$ – David Jun 13 '19 at 12:03

$\varphi(f)=-\arctan(\tan(2{\pi}ft_0)) = -2{\pi}ft_0$

Which is the solution

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  • $\begingroup$ it is the solution as long as $$ \big|2 \pi f t_0 \big| < \frac{\pi}{2} $$ $\endgroup$ – robert bristow-johnson Jun 12 '19 at 1:11
  • $\begingroup$ i might suggest that both Zelda and Eddy review the notion of phase wrapping and unwrapping phase, which is also the difference between the Principal Value of $\operatorname{Arg}(\cdot)$ and the unwrapped $\arg(\cdot)$. that might solve the problem. $\endgroup$ – robert bristow-johnson Jun 12 '19 at 1:13
  • $\begingroup$ @robert - his solution is in his question: "I know that the solution is: φ(f)=−2πft0". Unwrapping is not required? $\endgroup$ – DamienBradley Jun 12 '19 at 3:01
  • $\begingroup$ yeah, but the question he asks is essentially "How does... $$\begin{align} \varphi(f) &= \arctan \big(-\tan(2 \pi f t_0) \big) \\ &= -2 \pi f t_0 \\ \end{align} $$ "? and the answer will be about both what the complex $\arg\{ \cdot \}$ is and what phase wrapping and unwrapping are. $\endgroup$ – robert bristow-johnson Jun 12 '19 at 3:06
  • $\begingroup$ @Robert ah I see, I have not seen this question formatted this way (with real frequency), usually this would be set in H(e^jw) form - in which case that explanation would be extraneous to the solution. I would like to just ask your opinion additionally, do you agree with the question terminology as originally set? It says H(f) and not H(e^jw) or H(e^jf....), is this interchangable in this way? $\endgroup$ – DamienBradley Jun 12 '19 at 3:08

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