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, 2019 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, $\endgroup$ Jun 12, 2019 at 1:22

2 Answers 2


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 $\phi(f)=-2\pi ft_0$?

  • $\begingroup$ Plants face in hand - sometimes a little bit of insight beats a lot of mathematics :) $\endgroup$
    – David
    Jun 13, 2019 at 12:03

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

Which is the solution

  • $\begingroup$ it is the solution as long as $$ \big|2 \pi f t_0 \big| < \frac{\pi}{2} $$ $\endgroup$ Jun 12, 2019 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. $\endgroup$ Jun 12, 2019 at 1:13
  • $\begingroup$ @robert - his solution is in his question: "I know that the solution is: φ(f)=−2πft0". Unwrapping is not required? $\endgroup$ Jun 12, 2019 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. $\endgroup$ Jun 12, 2019 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? $\endgroup$ Jun 12, 2019 at 3:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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