I had the following question on edX:

enter image description here

I'm failing to understand why the second signal has $M(\omega)=A(\omega)$. First I find the DTFT of the signal:

$$\mathcal{F}\{\delta [n]+\delta [n-1]\}\ =\ 1\ +\ e^{-j\omega}$$

I then deduce the $A(\omega)e^{\ j\phi_A(\omega)}$ representation:

$$1\ +\ e^{-j\omega}\ =\ \big[\cos (\omega)+1\big ]\ -\ j\cdot\sin (\omega)\ =\ 2\cos \bigg (\dfrac{\omega}{2}\bigg )e^{-j\frac{\omega}{2}}$$

The magnitude spectrum is:

$$M(\omega)\ =\ 2\cdot \bigg |\cos \bigg (\dfrac{\omega}{2}\bigg )\bigg |$$

Therefore $M(\omega)\neq A(\omega)$. As a result, $\phi_M(\omega)$ will equal $\phi_A(\omega)$ except contain discontinuities of size $\pm\pi$ at values of $\omega$ where $A(\omega)$ has a zero crossing.

The only scenario I can envisage where $M(\omega)=A(\omega)$ is if $A(\omega)\geq 0$ however as I got the question wrong there must be something wrong with my logic, could somebody clarify?


1 Answer 1


You are right that also for the second signal you have $M(\omega)\neq A(\omega)$, but the point here probably is that in the fundamental interval $\omega\in (-\pi,\pi)$ the two are identical. It is only at $\omega=\pm\pi$ that the phase $\phi_M(\omega)$ jumps and that $A(\omega)$ changes its sign. So what they probably mean is

$$M(\omega)=A(\omega)\quad\text{and}\quad\phi_M(\omega)=\phi_A(\omega)\quad\bbox[5px,border:2px solid red]{\text{for }\omega\in(-\pi,\pi)}$$

which is correct (and which is not the case for the first signal).

Remember that such quizzes are made by fallible humans.

  • $\begingroup$ Thanks so much, defining the interval makes all the difference! $\endgroup$
    – JamOrJelly
    Dec 3, 2016 at 16:22

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