# Filter design techniques for a Low-pass FIR (LTI) System

I came across certain interesting Digital Signal Processing techniques past week. One of them been a problem statement and its solution for a finite impulse response filter design. I thought it would be great to put up the same as an article on the blog. Problem Statement Calculate the filter coefficients for a low-pass FIR filter with cut-off frequency 1000Hz where the sampling frequency is 5Khz. Consider the filter passband gain as unity, and the impulse response sequence length is 7. Use a Hamming window to remove spectral leakage.

Pass band gain is given to be unity. $$ie,$$

$$H_d(e^{j\omega}) = 1 \tag1$$

Cut-off value of  is expressed as

$${\omega}c_1 = \frac{2\pi f_{c_1}}{F} \tag2$$

Where $$f_{c_1}$$ is the cut-off frequency and $$F$$ is the sampling frequency. In the problem statement the following data is given.

fc1 = 1000Hz

F = 5000Hz

Now, $${}c_1$$ can be calculated as follows.

$${\omega}c_1 = \frac{2\pi.1000}{5000} = \frac{2\pi}{5} \tag3$$

From the above calculation,

$$H_d(e^{j\omega}) = \left\{ \begin{array}{ll} 1 & \mbox{for } \frac{-2\pi}{5} \leq n \leq \frac{2\pi}{5} \\ 0 & \mbox otherwise \end{array} \right.$$

The inverse DTFT (Discrete time Fourier transform)

\begin{align} \mathcal h_d(n) &= \frac1{2\pi}\int_{-\pi}^{\pi}H(e^{j\omega}).e^{j{\omega}n}d\omega \\ & = \frac1{2\pi}\int_{-\frac{2\pi}5}^{\frac{2\pi}5}1* e^{j{\omega}n}d\omega \tag4 \\ & = \frac1{2\pi}\left[\frac{e^{j\omega{n}}}{jn}\right]_{\frac{-2\pi}5}^{\frac{2\pi}5} \\ & = \frac1{2\pi{jn}}\left[e^{jn\frac{2\pi}5}-e^{-jn\frac{2\pi}5}\right] \\ & = \frac1{\pi n}sin\left(\frac{2\pi n}5\right) \end{align}

Now substitute values for $$n$$ such that the length of the resulting sequence is 7,

− 3 ≤ n ≤ 3

$$h_d(-3) = -0.062$$

$$h_d(-2) = 0.0935$$

$$h_d(-1) = 0.302$$

$$h_d(0) =$$

$$h_d(1) = 0.302$$

$$h_d(2) = 0.0935$$

$$h_d(3) = -0.062$$

To calculate a finite value of $$h_d(0)$$, substitute the value $$n = 0$$ in equation $$(4)$$

\begin{align} \mathcal h_d(0) &= \frac1{2\pi}\int_{-\frac{2\pi}5}^{\frac{2\pi}5}1* e^0d\omega \\ &= \frac1{2\pi}\int_{-\frac{2\pi}5}^{\frac{2\pi}5}d\omega \\ &= \frac1{2\pi}\left[\omega\right]_{\frac{-2\pi}5}^{\frac{2\pi}5} \\ &= \frac1{2\pi}\left[\frac{2\pi}5 - \frac{-2\pi}5\right] \\ &= \frac1{2\pi}\left[\frac{2\pi}5 + \frac{2\pi}5\right] \\ &= \frac1{2\pi}\left[\frac{4\pi}5\right] \\ & = \frac25 \\ &= 0.4 \end{align}

The window to be used is a Hamming window, $$(= 0.5)$$

$$W_{Hm}(n) = \left\{ \begin{array}{ll} 0.5 + 0.5cos\left(\frac{2\pi n}{N-1}\right) & \mbox{for } -\frac{N-1}{2} \leq n \leq \frac{N-1}{2} \\ 0 & \mbox otherwise \end{array} \right.$$

The length of the sequence, $$N = 7$$. The Hamming window sequence is calculated as follows,

$$W_{Hm}(-3) = 0$$

$$W_{Hm}(-2) = 0.25$$

$$W_{Hm}(-1) = 0.75$$

$$W_{Hm}(0) = 1$$

$$W_{Hm}(1) = 0.75$$

$$W_{Hm}(2) = 0.25$$

$$W_{Hm}(3) = 0$$

Apply the window over the sequence $$h_d(n)$$ such that,

h(n) = hd(n) * WHm(n)

$$h(-3) = 0$$

$$h(-2) = 0.0234$$

$$h(-1) = 0.2265$$

$$h(0) = 1$$

$$h(1) = 0.2265$$

$$h(2) = 0.0234$$

$$h(3) = 0$$

These are the required filter coefficients. Further, this sequence can be converted to a realisable system by calculating transfer functions.