- DSP - Miscellaneous Signals
- DSP - Classification of DT Signals
- DSP - Classification of CT Signals
- DSP - Basic DT Signals
- DSP - Basic CT Signals
- DSP - Signals-Definition
- DSP - Home
Operations on Signals
- Operations Signals - Convolution
- Operations Signals - Integration
- Operations Signals - Differentiation
- Operations Signals - Reversal
- Operations Signals - Scaling
- Operations Signals - Shifting
Basic System Properties
- DSP - Solved Examples
- DSP - Unstable Systems
- DSP - Stable Systems
- DSP - Time-Variant Systems
- DSP - Time-Invariant Systems
- DSP - Non-Linear Systems
- DSP - Linear Systems
- DSP - Anti-Causal Systems
- DSP - Non-Causal Systems
- DSP - Causal Systems
- DSP - Dynamic Systems
- DSP - Static Systems
Z-Transform
- Z-Transform - Solved Examples
- Z-Transform - Inverse
- Z-Transform - Existence
- Z-Transform - Properties
- Z-Transform - Introduction
Discrete Fourier Transform
- DFT - Solved Examples
- DFT - Discrete Cosine Transform
- DFT - Sectional Convolution
- DFT - Linear Filtering
- DTF - Circular Convolution
- DFT - Time Frequency Transform
- DFT - Introduction
Fast Fourier Transform
Digital Signal Processing Resources
Selected Reading
- Who is Who
- Computer Glossary
- HR Interview Questions
- Effective Resume Writing
- Questions and Answers
- UPSC IAS Exams Notes
DSP - Computer Aided Design
FIR filters can be useful in making computer-aided design of the filters. Let us take an example and see how it works. Given below is a figure of desired filter.
![Computer Aided Design](/digital_signal_processing/images/computer_aided_design.jpg)
While doing computer designing, we break the whole continuous graph figures into discrete values. Within certain pmits, we break it into either 64, 256 or 512 (and so on) number of parts having discrete magnitudes.
In the above example, we have taken pmits between -π to +π. We have spanided it into 256 parts. The points can be represented as H(0), H(1),….up to H(256). Here, we apply IDFT algorithm and this will give us pnear phase characteristics.
Sometimes, we may be interested in some particular order of filter. Let us say we want to reapze the above given design through 9th order filter. So, we take filter values as h0, h1, h2….h9. Mathematically, it can be shown as below
$$H(e^{jomega}) = h_0+h_1e^{-jomega}+h_2e^{-2jomega}+.....+h_9e^{-9jomega}$$Where there are large number of dislocations, we take maximum points.
For example, in the above figure, there is a sudden drop of slopping between the points B and C. So, we try to take more discrete values at this point, but there is a constant slope between point C and D. There we take less number of discrete values.
For designing the above filter, we go through minimization process as follows;
$H(e^{jomega1}) = h_0+h_1e^{-jomega1}+h_2e^{-2jomega1}+.....+h_9e^{-9jomega1}$
$H(e^{jomega2}) = h_0+h_1e^{-jomega2}+h_2e^{-2jomega2}+.....+h_9e^{-9jomega2}$
Similarly,
$(e^{jomega1000}) = h_0+h_1eH^{-jomega1000}h_2e^{-2jomega1000}+.....+h_9+e^{-9jomega1000}$
Representing the above equation in matrix form, we have −
$$egin{bmatrix}H(e^{jomega_1})\.\.\H(e^{jomega_{1000}}) end{bmatrix} = egin{bmatrix}e^{-jomega_1} & ... & e^{-j9omega_1} \. & & . \. & & . \e^{-jomega_{1000}} &... & e^{j9omega_{1000}} end{bmatrix}egin{bmatrix}h_0\.\.\h_9end{bmatrix}$$Let us take the 1000×1 matrix as B, 1000×9 matrix as A and 9×1 matrix as $hat{h}$.
So, for solving the above matrix, we will write
$hat{h} = [A^TA]^{-1}A^{T}B$
$= [A^{*T}A]^{-1}A^{*T}B$
where A* represents the complex conjugate of the matrix A.
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