# Help understanding magnitude and phase of sinc functions?

• asdf12312
In summary, the student is having difficulty understanding the magnitude and phase of sinc functions. He is unclear on how to get the amplitude and phase graphs to look the way they are supposed to, and he is also confused about why there is a discontinuity at ω=0 in the phase graph. He is also wondering why the e-j(w+\frac{\pi}{2}) causes the sawtooth curve in the phase graph.
asdf12312
help understanding magnitude and phase of sinc functions??

## Homework Equations

pτ(t) ⇔ τsinc($\frac{τω}{2\pi}$)
Δτ(t) ⇔ $\frac{τ}{2}$sinc2($\frac{τω}{4\pi}$)
sinc(x) = $\frac{sin(\pi x)}{\pi x}$

## The Attempt at a Solution

My instructor had posted solutions to the problems:

I understand how to get the Fourier transforms, I was just having problems finding magnitude and phase graphs. For a) I don't know how they got amplitude near 1.5, but I know that in general for a single sinc function sinc(x), you will have zeroes at x-1. I also know that in general phase graphs of single sinc function will be +180 deg (doesnt really matter what the sign is, I guess) when you have zeroes. I just don't understand how this works when there are multiple sinc or sinc2 functions. And why the phase graphs are all sawtooth etc., when I was expecting just + 180deg.

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The way it's expressed, it's not easy to see how the amplitude and phase vary with ##\omega##, so you want to rewrite it:
\begin{align*}
X(\omega) &= -\mathrm{sinc }\left(\frac{\omega}{2\pi}\right) e^{-j\omega/2} + \mathrm{sinc }\left(\frac{\omega}{2\pi}\right) e^{-j3\omega/2} \\
&= \mathrm{sinc }\left(\frac{\omega}{2\pi}\right) (e^{-j3\omega/2} - e^{-j\omega/2}) \\
&= \mathrm{sinc }\left(\frac{\omega}{2\pi}\right) (e^{-j\omega/2} - e^{j\omega/2}) e^{-j\omega} \\
&= \mathrm{sinc }\left(\frac{\omega}{2\pi}\right) \left[-2j \sin\left(\frac{\omega}{2}\right)\right] e^{-j\omega} \\
&= 2\mathrm{sinc }\left(\frac{\omega}{2\pi}\right) \sin\left(\frac{\omega}{2}\right) e^{-j(\omega+\frac{\pi}{2})}
\end{align*} The factors out front are real, and the complex behavior arises from the exponential. Now it should be clearer that how you can get a maximum amplitude near 1.5. Also, you can see how the phase varies with ##\omega##. Do you understand why there's a discontinuity in the phase at ##\omega=0##?

This is example is relatively easy to analyze, but sometimes you're just going to have to brute force it and resort to figuring out the real and imaginary parts of X separately.

I am guessing the discontinuity at ω=0 is because of the sin function, sin(0)=0. In the phase at ω=0 is probably because of the exponential, there is imaginary value j. I still don't understand amplitude of 1.5 though (don't know how the exp changes amplitude/phase graph). If I set w=$\pi$ (since that's where it seems amplitude is highest) and do sinc-> sin I have:

= 2(2sin($\frac{ω}{2}$))$\times$sin($\frac{ω}{2}$)$\times$e-j$\frac{3\pi}{2}$
= j4sin2($\frac{ω}{2}$)

but maybe I'm doing this wrong? I'm also wondering how the e-j(w+$\frac{\pi}{2}$)causes the sawtooth curve in the phase graph.

asdf12312 said:
I am guessing the discontinuity at ω=0 is because of the sin function, sin(0)=0.
Why would the fact that sin 0 = 0 imply the existence of a discontinuity?

In the phase at ω=0 is probably because of the exponential, there is imaginary value j.
This sentence makes no sense.

I still don't understand amplitude of 1.5 though (don't know how the exp changes amplitude/phase graph).
You need to review polar notation of complex numbers. You seem to be missing some pretty basic facts about complex numbers, which is probably why you're so confused.

If I set w=$\pi$ (since that's where it seems amplitude is highest) and do sinc-> sin I have:

= 2(2sin($\frac{ω}{2}$))$\times$sin($\frac{ω}{2}$)$\times$e-j$\frac{3\pi}{2}$
= j4sin2($\frac{ω}{2}$)
I'm not sure why you're replacing sinc with sin, and if you're setting ##\omega=\pi##, why does ##\omega## still appear in the final line?

but maybe I'm doing this wrong? I'm also wondering how the e-j(w+$\frac{\pi}{2}$)causes the sawtooth curve in the phase graph.

I guess I was trying to replace sinc with sin because to understand it easier, my book shows how a basic sinc function looks but I guess i was confused because at ω=0 in this problem it is 0, however in the book the amplitude is highest at ω=0 for sinc. so I thought this could be because of multiplying by sin, at ω=0 sin is 0 so the entire eq. would be 0 (maybe I don't understand it at all). I think I understand the exponential, its magnitude would be 1, its phase is everything after the j. But I guess I am confused because I don't know how curve of e^-jω would look, my calc won't let me graph that because it has imaginary parts. so I am unsure how it affects both graphs.

I think the main trouble I am having: how does that phase in exp (ω+$\frac{\pi}{2}$) change the phase graph so its look like sawtooth instead of horizontal lines at 180. If I had to guess it is subtracting and adding 90 deg so that is why it looks like that. Also at ω=$\pi$ it seems an amplitude of 1.5 (I still don't understand this). If I plug in $\pi$ as ω for the exponential I get 'j' which I am unable to graph.

Last edited:
asdf12312 said:
I guess I was trying to replace sinc with sin because to understand it easier, my book shows how a basic sinc function looks but I guess i was confused because at ω=0 in this problem it is 0, however in the book the amplitude is highest at ω=0 for sinc. so I thought this could be because of multiplying by sin, at ω=0 sin is 0 so the entire eq. would be 0 (maybe I don't understand it at all). I think I understand the exponential, its magnitude would be 1, its phase is everything after the j. But I guess I am confused because I don't know how curve of e^-jω would look, my calc won't let me graph that because it has imaginary parts. so I am unsure how it affects both graphs.
Why don't you write down expressions for the amplitude and phase of ##X(\omega)##?

I think the main trouble I am having: how does that phase in exp (ω+$\frac{\pi}{2}$) change the phase graph so its look like sawtooth instead of horizontal lines at 180. If I had to guess it is subtracting and adding 90 deg so that is why it looks like that.
Why would the phase be a horizontal line at 180 degrees? That would imply that X(ω) is a negative real number.

Also at ω=$\pi$ it seems an amplitude of 1.5 (I still don't understand this). If I plug in $\pi$ as ω for the exponential I get 'j' which I am unable to graph.

## 1. What is the significance of the magnitude and phase of sinc functions?

The magnitude and phase of sinc functions are important in understanding the behavior and properties of these functions. The magnitude represents the amplitude or strength of the function, while the phase represents the shift or delay in the function. These two properties help determine the shape, frequency, and other characteristics of the sinc function.

## 2. How do I calculate the magnitude and phase of a sinc function?

The magnitude of a sinc function can be calculated by taking the absolute value of the function. The phase can be determined by finding the angle of the complex number in the function, which can be done using trigonometric functions such as arcsine or arctangent.

## 3. What is the difference between magnitude and phase in sinc functions?

The magnitude and phase are two distinct properties of sinc functions. While the magnitude represents the amplitude or strength of the function, the phase represents the shift or delay in the function. In other words, the magnitude determines the height of the function, while the phase determines the location of the peaks and valleys.

## 4. How does the magnitude and phase affect the behavior of sinc functions?

The magnitude and phase have a significant impact on the behavior of sinc functions. The magnitude affects the overall shape and amplitude of the function, while the phase determines the location and spacing of the peaks and valleys. Changes in these properties can result in shifts, stretches, or compressions of the function, as well as changes in its frequency and other characteristics.

## 5. Can the magnitude and phase of sinc functions be changed?

Yes, the magnitude and phase of sinc functions can be changed by multiplying the function with a constant or adding a phase shift to the function. These changes can alter the shape, frequency, and other characteristics of the sinc function. Additionally, certain properties of the function, such as the bandwidth, can also affect the magnitude and phase.

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