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First time I am encountering this type of question so i just tried but not getting whether my logic correct or not.

First let the sinusoidal signal be $X(t)=\cos(2\pi ft)$.

After sampling this signal I will get $X(nT_s)=\cos(2\pi fnT_s)$; now at sampling rate $f_s=18\, \text{kHz}$ it becomes $X(n)=\cos(\frac{n\pi f}{9})$.

Now I am checking optionS $(C)$ (given correct answer is also C) to see if the output of filter will zero or not at $f=12\,\text{kHz}$ and at 12kHz X(n) becomes $X(nT_s)=\cos\left(\frac{4\pi n}{3}\right)$ .

$H(z)=1+z^{-1}+z^{-2} $ can be written as $H(e^{jw})=1+e^{-jw}+e^{-2jw}$, so from eigenvalue concept the value of $H(e^{jw})$ at $w=\frac{4\pi}{3}$ is $H(e^{jw})=1+e^{\frac{-j4\pi }{3}}+e^{\frac{-j8\pi }{3}}=0$, so it's proven that output at 12 kHz will be zero.

Therefore, 12 kHz frequency will not pass.

I am doing by checking options, option B also satisfying this condition.

Is there any alternate method to solve this?

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  • $\begingroup$ That was barely readable. Seriously, I know that English might not be your native language, but you definitely understand how English sentences work (your English is actually pretty good). So add "." at the end of your sentences. I had to add eleven punctuation marks to make this readable. Also, there's exactly one correct way to write "kHz", it's "kHz", not "Khz" nor "khz". Luckily, you even have added a picture that contains the right spelling! $\endgroup$ – Marcus Müller Sep 3 '17 at 19:11
  • $\begingroup$ H of Hertz ! V of Volt B of Bell... these are derived from the names of people... and capitalized. $\endgroup$ – Fat32 Sep 3 '17 at 19:37
  • $\begingroup$ @MarcusMüller sorry sir :( and Thanks... next time i will try to improve it. $\endgroup$ – Rohit Sep 4 '17 at 6:12
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The impulse response of the given transfer function $H(z)= 1 + z^{-1} + z^{-2}$ is $h[n] = \delta[n] + \delta[n-1] + \delta[n-2]$.

This is a rectangular window with three taps and its Frequency response magnitude is given by:

$$ H(w) = \frac{\sin( 3 w/2 )}{\sin(w/2)} $$

It'll have zeros (nulls) at the frequencies given by:

$$ \sin(3w/2) = 0 \longrightarrow 3 w_k / 2 = \pi k \longrightarrow w_k = 2\pi k / 3 ~~~~, \text{for} ~~~k = 0,1,2 $$

Note that the zero for $k=0$ is cancelled by the denominator and therefore there exists only two zeros $w_1 = 2\pi/3$ and $w_2 = 4\pi/3$

Now note that those two zeros are at mirror locations about $w=\pi$ which means that they will be the same, indistinguishable, discrete time frequencies.

Also you shall consider only the first zero , $w_1 = 2\pi/3$ to be the frequency of the analog sinusoidal that will be nulled by the filter.

The given sampling frequency of $f_s = 18$ kHz maps $w_1= 2\pi/3$ to the analog frequency of

$$ f_1 = \frac{f_s w_1}{2\pi} = \frac{18k 2\pi/3}{2\pi} = 6 kHz$$

The answer therefore is $f_1 = 6$ kHz which cannot pass through your filter when sampled properly at the sampling rate of $f_s = 18 $ kHz. This assumes that there is no aliasing during the sampling process which is prevented by an analogue anti-alasing filter before sampling. If this filter is omitted and sampling is performed then a sinusoid at the frequency of $f_2 = 12$ kHz will be aliased into a frequency of $f_1 = 6$ kHz and therefore it won't also pass through the filter. Note that there will be infinetely many new frequencies which will map into $f_1 = 6$ kHz and won't pass through the filter in such a case.

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  • $\begingroup$ @Rohit Did you deleted your comment here? You're asking for some details. I didn't have time then. You can ask it again if you need... $\endgroup$ – Fat32 Sep 6 '17 at 22:26

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