# Question about impulse response of an ideal lowpass filter

Consider an ideal lowpass filter with a cutoff frequency of $$\omega_c$$. Following shows its impulse response:

$$h[n] = \frac{1}{\pi n}\sin(\omega_cn)$$ Also, some references shows $$h[n]$$ as: $$h[n] = \frac{\omega_c}{\pi}\textrm{sinc}\left(\frac{\omega_c}{\pi}n\right)$$ How can I show these two are equivalent? I searched the web but unfortunately found nothing. Also, I tried the following: $$h[n] = \frac{1}{\pi n}\sin(\omega_cn) = \frac{\omega_c}{\pi}\frac{\sin(\omega_cn)}{\omega_cn}=\frac{\omega_c}{\pi}\mbox{sinc}(\omega_cn)$$ I think the last equation is correct. But, it gives wrong results. Why is this happening?

Matlab uses the normalized sync function.

So we have

$$h[n] = \frac{\omega_c}{\pi}\mbox{sinc}( \frac{\omega_c}{\pi}n) = \frac{\omega_c}{\pi} \frac{\sin(\pi \frac{\omega_c}{\pi}n)}{\pi \frac{\omega_c}{\pi}n} = \frac{1}{\pi n} \sin(\omega_c n)$$

This checks out as long as you use the normalized sinc function.

If you want to go the other way, we have

$$\mbox{sinc}(x) = \frac{\sin(\pi x)}{\pi x}$$ which we can arrange into $$\sin(y) = y \cdot \mbox{sinc}(\frac{y}{\pi})$$

Then we get

$$\frac{1}{\pi n} \sin(\omega_c n) = \frac{1}{\pi n} (\omega_c n) \mbox{sinc}(\frac{\omega_c n}{\pi}) = \frac{\omega_c}{\pi} \mbox{sinc}(\frac{\omega_c}{\pi}n)$$

There are two common definitions of the Sinc function:

• the unnormalized Sinc function: $$\textrm{sinc}(x)=\displaystyle\frac{\sin x}{x}$$

• the normalized Sinc function: $$\textrm{sinc}(x)=\displaystyle\frac{\sin (\pi x)}{\pi x}$$

In signal processing the normalized form of the Sinc is more common, and I can assure you that the second definition uses the normalized form. Now it should only be a matter of basic algebra to show their equivalence. I trust that you can take it from here.

Your last equation is also correct, but it uses the unnormalized form of the Sinc function, which is different from Matlab's usage.

• Dear Matt L. I tried both equations in MATLAB and both give the same result. This means that the second form is also correct. But I cannot figure out how. The second equation is given in an online course on digital signal processing offered by COURSERA. Jun 25 at 12:54
• Would you just take a look on my question again? I just added some more details. Jun 25 at 13:01
• @Parham: The second equation is wrong, there shouldn't be an $n$ in de denominator because that's already taken care of by the Sinc function. Your last equation is correct IF you use the unnormalized definition of the Sinc function. If you use the normalized definition you must add a factor in the argument of the Sinc function. Jun 25 at 13:09
• Check the Matlab documentation to find out which definition of the Sinc function is used in Matlab. Jun 25 at 13:11
• I am so sorry. It was just a typo. I edited the second equation. But my question is still in place. Jun 25 at 13:30