Cosine Filter

We have:

$\large \displaystyle y(k) = \sum_{n=0}^7 \frac14 \cos\left [ \frac {2 \pi (n + 1/2)}{8} \right ] x(k-n)$

Using the referred time-shift property and considering that $x(k) = 0$ for $k \leq 0$ , i.e., assuming the time begins to count when the signal begins, and denoting $X(z)$ and $Y(z)$ as the Z-transforms of $x(k)$ and $y(k)$ respectively:

$\large \displaystyle Y(z) = X(z) \cdot \left [ \sum_{n=0}^7 \frac14 \cos\left [ \frac {2 \pi (n + 1/2)}{8} \right ] z^{-n} \right ]$

The transfer function $T(z)$ is the ratio between output and input:

$\large \displaystyle T(z) = \frac14 \left [ \sum_{n=0}^7 \cos\left [ \frac {2 \pi (n + 1/2)}{8} \right ] z^{-n} \right ]$

Since we're sampling at $480$ Hz, to convert from Z-transform to Laplace Transform we make $z = e^{s\cdot T_s}$ , where $T_s$ is the sampling period:

$\large \displaystyle T(s) = \frac14 \left [ \sum_{n=0}^7 \cos\left [ \frac {2 \pi (n + 1/2)}{8} \right ] e^{-\frac{s \cdot n}{480}} \right ]$

In the time domain, if we input a signal $x(t)$ through a transfer function $p(t)$ , the output signal $y(t)$ is a convolution, i.e.:

$\large \displaystyle y(t) = \int_0^{\infty} p(\tau) x(t - \tau) d\tau$

If $x$ happens to be a sinusoid, i.e., a signal in the form $e^{j \omega t}$ ( $j$ is the imaginary unit) or a linear combination of such signals (recall that cosine and sine are indeed linear combinations of signals like this), then we have:

$\large \displaystyle y(t) = \int_0^{\infty} p(\tau)e^{j \omega (t -\tau)} d\tau$

$\large \displaystyle y(t) = e^{j \omega t} \int_0^{\infty} p(\tau)e^{- j \omega \tau} d\tau$

This integral is the exact definition of Laplace transform of $p(t)$ (denoted $P(s)$ ) evaluated at $s = j \omega$ (also known as Fourier Transform ):

$\large \displaystyle y(t) = e^{j \omega t} P(j \omega)$

This means that if we input a sinusoid with frequency $\omega$ through a transfer function $p(t)$ , it will receive a gain equal to the absolute value of $P( j \omega)$ and a phase change equal to the angle of $P( j \omega)$ .

So, the gain at linear frequency $f$ (angular frequency $2\pi f$ ) is equal to the absolute value of $T(j2\pi f)$ . Thus:

$\large \displaystyle H(f) = |T(j2\pi f)| = \left | \frac14 \left [ \sum_{n=0}^7 \cos\left [ \frac {2 \pi (n + 1/2)}{8} \right ] e^{-\frac{n \cdot j2\pi f}{480}} \right ] \right |$

Integrating it numerically from $f=0$ to $300$ gives:

$\color{#3D99F6} \large \displaystyle I = \int_0^{300} H(f) df \approx 103.064 \rightarrow \boxed{ \large \displaystyle \lfloor I \rfloor = 103 }$

The convolution integral is

$y(k) = \displaystyle \sum_{n = 0}^7 h(n) x(k - n) \hspace{12pt} (1)$

where $h(n)$ is the impulse response of the filter, and is referred to in the problem as $\alpha_n$ .

If we plug in a complex sinusoidal input of frequency $f$ , as our input, then

$x(k) = e^{j 2 \pi f k} \hspace{12pt} (2)$

Plugging in (2) into (1) results in

$y(k) = \displaystyle \sum_{n = 0}^7 h(n) e^{ j 2 pi f (k - n) }$ $\hspace{12pt} = e^{ j 2 \pi f k} \left( \displaystyle \sum_{n = 0}^7 h(n) e^{- j 2 \pi f n } \right) \hspace{12pt} (3)$

The term between parantheses in (3) is exactly the complex transfer function (The fourier transform of $h(n)$ ). That is,

$H(f) = \displaystyle \sum_{n=0}^7 h(n) e^{ - j 2 pi f n } \hspace{12pt} (4)$

Hence, we can evaluate $| H(f) |$ , the magnitude of $H(f)$ at each frequency $f$ .

What remains is to numerically integrate that over the range of $f$ from $0$ to $300 \hspace{2pt} Hz$ .