1) How is this a low pass filter?

$0.99 \leq |H(e^{j\omega})|\leq1.01 {\rm\ for\ } 0\leq|\omega|\leq0.19\pi$

$|H(e^{j\omega}\;)|\leq0.01{\rm\ for\ } 0.21\pi\leq|\omega|\leq\pi$

2) What kind of filter is this? Low pass, high pass, band pass, band reject or what? $0.99 \leq |H(e^{j\omega})|\leq1.01 {\rm\ for\ } 0\geq|\omega|\geq0.19\pi$

$|H(e^{j\omega}\;)|\leq0.01{\rm\ for\ } 0.21\pi\leq|\omega|\leq\pi$

I am really confused on how to identify if a filter is high pass, band pass, low pass, band reject from given specifications. Can you make me understand this concepts?

  • 2
    $\begingroup$ If you have trouble reading the math, why don't you just draw it ? It becomes blatantly obvious then. $\endgroup$
    – Hilmar
    Commented Sep 5, 2021 at 6:47
  • $\begingroup$ Yeah I got it now(the first one). Can you tell me about the second question? How can sth be lesser than 0 yet greater than $0.19\pi$, is that second question wrong? $\endgroup$
    – achhainsan
    Commented Sep 5, 2021 at 8:20
  • 1
    $\begingroup$ The second one is clearly a typo, that's pretty obvious. $\endgroup$
    – Matt L.
    Commented Sep 5, 2021 at 10:51
  • $\begingroup$ oh thanks a lot. i was geting confused due to that. $\endgroup$
    – achhainsan
    Commented Sep 5, 2021 at 10:52

1 Answer 1


Here is the trick-: I just found it out somewhere enter image description here

  • $\begingroup$ Please provide additional details in your answer. As it's currently written, it's hard to understand your solution. $\endgroup$
    – Community Bot
    Commented Sep 5, 2021 at 4:59
  • 1
    $\begingroup$ I think it is easy and self explanatory to understand. So I didn't explain it. For eg-: LPF means high for -wc to wc, otherwise 0. And so on. It is very intiutive to understand. $\endgroup$
    – achhainsan
    Commented Sep 5, 2021 at 6:42

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