The output y[n] of a particular LTI-system can be obtained by:
The previous equation is called Convolution between discrete-time signals x[n] and h[n].
By convention, the convolution between x[n] and h[n] is expressed as follows:
Let the following rectangular pulse x[n] be an input to an LTI system with impulse response h[n]:
Determine the output y[n] of the system.
In Elementary sequences we have designed a function stepseq for plotting the unit step function in discrete time, or any combination as example 1. Next Script allows plotting x[n].
Next Script allows plotting h[n].
Now, we use conv Matlab function to determine y[n]:
Another approach is by using the filter function (see Solving discrete-time differential equations):
There is a difference in the outputs of these two implementations that should be noted. As you can see in Figure 3, the output sequence from conv(x,h) function has a longer length than both x[n] and h[n] sequences. On the other hand, the output sequence from the filter(h,1,x) function in Figure 4 has exactly the same length as the input x[n] sequence. In practice, the use of the filter function is encouraged.
Watch out: The filter function can be used to compute the convolution indirectly. That was possible because of the impulse response in example 1 was a one-side exponential sequence for wich we could determine a difference equation representation. Not all infinite-lenght impulse responses can be converted into difference equations.
We can do the convolution of x[n] and h[n] Analytically:
Applying the given equation for convolution:
We now use an expression for the sum of any finite number of terms of a geometric series (The Geometric Series in DSP), and that is given by:
So, we can express equation (1) as follows:
Equation (3) is almost a geometric series sum as equation (2) except that the terms u(n-k) takes different values depending on n and k. There are three possible conditions under u(n-k) which can be evaluated:
Case 1. n<0. Then u(n-k)=0 for 0 ≤k ≤9. In consequence:
Case 2. In this case the nonzero values of and do not overlap. So, for 0≤n<9 then u(n-k)=1 for 0≤k≤n. In consequence:
Applying formula of equation (2):
Case 3. In this case the impulse response h[n] partially overlap the input x[n]. So, for 9 ≤n. Then u(n-k)=1 for 0 ≤k ≤9. In consequence:
In this last case h[n] completely overlaps the input x[n].
We can also use a graphical method as in the following example.
The input signal x[n] to an LTI system with impulse response h[n]:
Determine graphically y[n] through:
Sequences x[k] and h[n-k], and the convolution of both signals can be seen as follows:
We can also use convolution properties as follows.
Other interesting properties are:
With an input x[n], the response of a LTI system y[n] is:
The impulse response h[n] of the system is:
Determine x[n]. Choose the right answer from:
We express the impulse response in terms of displaced Dirac deltas:
If we select:
So right answer is letter a). Demonstration: Using the previous equation, Let´s plot y[n] in Matlab.
The impulse response h[n] and option a) for input x[n] can be plotting in Matlab using:
Now, y[n] can be plotting by using:
Which we would also have been able to get by using:
- Digital Signal Processing Using Matlab, 3erd ed
- Fundamentos_de_Señales_y_Sistemas_usando la Web y Matlab
- Oppenheim – Señales y Sistemas
- Análisis de Sistemas Lineales Asistido con Scilab – Un Enfoque desde la Ingeniería Eléctrica.
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Prof. Larry Francis Obando – Technical Specialist – Educational Content Writer
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