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I had the following question on edX:

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?

I had the following question on edX:

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?

I had the following question on edX:

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?

I had the following question on edX:

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?