which of the analog sinusoidal frequency can not pass through the filter?

Question

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?

Answer 1

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.