Frequency response FIR

Introduction

Besides the impulse response, denoted by $h[n]$, and the Difference Equation (DE) we will see that the frequency response is an alternative description of an FIR filter. Obviously all these different descriptions are related to each other and each of the descriptions has its own advantage. The advantage of the frequency response description is that we can fairly easily see the response of the FIR to frequencies.

Screencast video [⯈]



Module overview

This module covers the following topics:

  1. Frequency response - This section will introduce the frequency response and will derive the frequency response of an arbitrary FIR filter.
  2. Properties [⯈] - As will all other transforms, the frequency response has some properties which ease the calculations.



Exercises

In this section several exercises are available, including their answers. The exercises marked in blue are explained by means of more extensive pencast videos.

Video quiz

The input signal is $x[n] = \cos(0.5\pi n)$. This signal is applied to a filter with impulse response $h[n]$ and frequency response $H(e^{j\theta}) = 2\sin(\theta)$. Give a description of the output signal $y[n]$.






Have a look at the following system:
FIR filter, question 2.
Signal $x[n]$ consists out of two frequencies $\theta_1 = \frac{\pi}{3}$ and $\theta_2=\pi$. Which frequencies does the output signal $y[n]$ contain?







What is the response of the following filter to a DC signal?
FIR filter, question 3.






What is the response of the following cascaded filters to a DC signal?
cascaded FIR filter, question 4.







What is does the system with the following phase response represent?
Phase response, question 5.








Exercise bundle

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Answers

Download the answers here.

Pencast videos [⯈]

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MATLAB lab

Accompanied to this modules are some exercises in MATLAB, which will test your knowledge of the module and will help improve your MATLAB skills.

Lab assignment

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MATLAB demo [⯈]



Summary

Relationship impulse response and frequency response: $$ \boxed{ \begin{eqnarray*} \text{Time domain} & \circ \hspace{-2.3mm} - \hspace{-1.1mm} \circ & \text{Frequency domain} \newline h[n]= \sum_{k=0}^{M-1} h[k] \delta[n-k] & \circ \hspace{-2.3mm} - \hspace{-1.1mm} \circ & H(e^{j{\theta}})= \sum_{k=0}^{M-1} h[k] e^{-j{\theta k}} \end{eqnarray*} } $$


Basic FIR property:
When applying a sinusoidal signal with a single frequency $\theta=\theta_1$ to an FIR filter, the output is a sinusoidal signal with the same frequency $\theta_1$ as the input signal, only the amplitude and phase have changed. The change in amplitude is described by the magnitude response evaluated at frequency $\theta_1$, thus $|H(e^{j{\theta}})\mid_{\theta=\theta_1}$, while the change in phase is described by the phase response evaluated at frequency $\theta_1$, thus $\angle{H(e^{j{\theta}})}\mid_{\theta=\theta_1}$.

When applying a sum of frequencies $x[n] = A_0 + \sum_{k=1}^{N} A_k \cos ( \theta_k n + \phi_k)$ to an FIR filter with frequency response $H(e^{j{\theta}}) = |H(e^{j\theta})| e^{j{\angle{H(e^{j{\theta}})}}}$ then the output can be written as follows: $$ y[n] = |H(e^{j0})| + \sum_{k=1}^{N} |H(e^{j\theta_k})| A_k \cos (\theta_k n + \phi_k + \angle{H(e^{j\theta_k})}) $$
Properties frequency response $$ \boxed{ \begin{eqnarray*} \text{Periodic } & : & H(e^{j\theta}) = H(e^{j(\theta + l \cdot 2 \pi)}) \newline \text{Complex conjugated } & : & H(e^{-j\theta}) = (H(e^{j\theta}))^\ast \end{eqnarray*} } $$
Graphical representation:
  • Both Magnitude- and phase- response are depicted in the Fundamental Interval (FI): $|\theta| \leq \pi$.
  • The phase response is depicted in the range $|\angle{H(e^{j\theta})}| \leq \pi$.
  • A phase jump of $\pm \pi$ is applied in the phase response plot for each zero crossing of the magnitude response.

Cascading FIR filters:
When cascading two different FIR filters, one with frequency response $H_1(e^{j\theta})$ and the other with frequency response $H_2(e^{j\theta})$, we may combine these two FIR filters to one FIR filter of which the frequency response is given by the product of the individual frequency responses: $H(e^{j\theta})= H_1(e^{j\theta}) \cdot H_2(e^{j\theta})$.

Convolution in time- is equivalent to multiplication in frequency-domain: $$ \boxed{ h_1[n] * h_2[n] \hspace{3mm} \circ \hspace{-2.3mm} - \hspace{-1.1mm} \circ \hspace{3mm} H_1(e^{j\theta}) \cdot H_2(e^{j\theta}) } $$
Steady state and transient behaviour:
  • Transient region: The complex multiplier $\left ( \sum_{k=0}^{{\color{red}{n}}} h[k] e^{-j\theta_1 k} \right )$ depends on the index $n$.
  • Steady state region: The complex multiplier $\sum_{k=0}^{{\color{blue}{M-1}}} h[k] e^{-j\theta_1 k}$ is constant. Output contains only input frequency ($\theta_1$ in this example).
  • If for $n > M-1$ the input changes to zero or to another frequency $\theta_2 \neq \theta_1$, then there will be a new transition- and steady state- region.
Last updated on May 30, 2020