Ideal Filters

The questions below are due on Thursday April 10, 2025; 02:00:00 PM.
 
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"Ideal" Low-Pass Filter

A low-pass filter passes low frequencies and attenuates high frequencies. An ideal low-pass filter is parameterized by its "cutoff" frequency \Omega_c. It passes frequencies "below" the cutoff frequency and eliminates frequencies "above" the cutoff frequency. The plot below shows the frequency response of an ideal low-pass filter. Notice that H(\Omega) = 1 when |\Omega|<\Omega_c, and H(\Omega) = 0 for frequencies near \pi.

The ideal low-pass filter shown above is periodic in 2\pi. Explain why.

The low-pass filter described above is linear and time-invariant. Therefore, it's output should be given by the convolution of it input with the unit-sample response of the filter.

Enter an expression for the n^\text{th} sample of the unit sample response of this system, h[n], in terms of n, OMEGA_c (notice capitalization!), sin, cos, e, j, and/or other mathematical constants.

h[n] =~

Approximating an Ideal Low-Pass Filter

Notice that the unit-sample response of the ideal low-pass filter described in the previous section has infinite extent. Computing the output using convolution would require an infinite number multiplies and an infinite number of additions! Fortunately, the values of h[n] tend to get smaller as |n| gets large. This observation motivates approximating h[n] as \hat{h}[n]:

\hat{h}[n] = \begin{cases} h[n] & \text{if}~-M\leq n\leq M\\ 0 & \text{otherwise} \end{cases}

which can be thought of as resulting from multiplication of the original h[n] by a rectangular window w[n]

\hat{h}[n] = w[n]h[n]

where w[n] is given by the following:

w[n] = \begin{cases} 1 & \text{if}~-M\leq n\leq M\\ 0 & \text{otherwise} \end{cases}

Determine an expression for W(\Omega), which is the DTFT of a rectangular window.

Hint:The sum of a finite number of geometric terms is

\sum_{n=0}^{N-1} a^n = {1-a^N\over1-a}

provided a\ne1.

Enter your expression for W(\Omega) in terms of M, OMEGA, and necessary constants in the box below.

Notice that the above form is ill-defined at \Omega=0. Enter an expression for W[0] in terms of M:

Make plots of the frequency response of the windowed ideal low-pass filter with \Omega_c = \pi/4 and \Omega_c = \pi/10, each with M=50 and M=500. How do \Omega_c and M affect the frequency response of the system?

"Ideal" Band-Pass Filter

Next we'll consider a different type of filter called a band-pass filter. This type of filter passes frequencies in a particular range (not necessarily centered around 0) and eliminates frequencies outside that range.

A band-pass filter is parameterized by the center frequency \Omega_b and half width \Omega_c of the band of frequencies that is passed, as illustrated below.

Enter a closed form expression for the n^\text{th} sample of the unit sample response of this system, h[n], in terms of n, OMEGA_b, OMEGA_c (notice capitalization!), sin, cos, e, j, and/or other mathematical constants.

h[n] =~

To get a sense of the shape of this system's unit sample response, generate plots of the unit sample response for \Omega_b = \pi/ 2, for both \Omega_c = \pi / 4 and \Omega_c = \pi / 10, each plotted from from n = -100 to n = 100.

Approximating an Ideal Band-Pass Filter

The unit sample response of an ideal band-pass filter is also infinite. Once again, we will consider approximating this filter with a windowed version:

\hat{h}[n] = \begin{cases} h[n] & \text{if}~-M\leq n\leq M\\ 0 & \text{otherwise} \end{cases}

Make plots of the frequency response of the windowed ideal band-pass filter with \Omega_b = \pi/2 and \Omega_2 = \pi/4; \Omega_c = \pi/10 and \Omega_c = \pi/30; and M=50 and M=500.

Explain the effect of \Omega_b, \Omega_c, and M on the frequency response of the filter.