Convolution                                                                                                                                

Example 1                                                                                                                                  

 From the file menu, open a new m file.

 Type the following MATLAB program

 Save this file with extension m (eg. dsp1.m)

 Return to command window and type "dsp1" (name of the saved file) and enter

 Enter the requested data as

 a= [-2 0 1 -1 3]

 b= [1 2  0 -1]

 Solution                                                                                                                                    

     a = input('Type in the first sequence = ');

     b = input('Type in the second sequence = ');

     xlabel('Time Index n')

     ylabel('Amplitude')  

 Result                                                                                                                                       

Out put sequence = 

c = -2   -4   1   3    1     5     1    -3               

Convolution                                                                                                                                

Exercise 1                                                                                                                                  

 Repeat example 1 with

a = [1 2 1 -1]

b = [1 2 3 1]

Result                                                                                                                                        

 Out put sequence = 

c = 1     4     8     8    3     -2    -1            

Signal y[n] is expanding and z[n] is not expanding (and found symmetry in +ve and -ve direction).

Impulse response                                                                                                                        

Example 2a                                                                                                                                 

 Find impulse response and step response of the following LTI system:

 y[n] + 0.7y[n-1] - 0.45y[n-2] - 0.6y[n-3] = 0.8x[n] - 0.44x[n-1] + 0.36x[n-2] + 0.02x[n-3]

 Solution                                                                                                                                    

p=[0.8 -0.44 0.36 0.02];

xlabel('Time Index n')

ylabel('Amplitude')

Result                                                                                                                                        

Step response                                                                                                                             

Example 2b                                                                                                                                

 To determine the step response replace in the above program the statement x=[1 zeros(1 ,N-1)] with x=[ones(1,N)].

 Solution                                                                                                                                    

     p=[0.8 -0.44 0.36 0.02];

xlabel('Time Index n')

ylabel('Amplitude')

Result                                                                                                                                        

Unit ramp response                                                                                                                     

Exercise 2a                                                                                                                                 

 Determine and sketch unit ramp response of the system given in example 2, i.e.

y[n] + 0.7y[n-1] - 0.45y[n-2] - 0.6y[n-3] = 0.8x[n] - 0.44x[n-1] + 0.36x[n-2] + 0.02x[n-3]

 Solution                                                                                                                                    

Result                                                                                                                                        

Impulse & Step response                                                                                                            

Exercise 2b                                                                                                                                

 Sketch the impulse and step response of the system described by the difference equation:

y[n] =  0.7y[n-1] - 0.1y[n-2] + 2x[n] - x[n-2]

 Solution                                                                                                                                    

Result                                                                                                                                        

Impulse response              

Step response 

Impulse  response                                                                                                                       

Exercise 2c                                                                                                                                 

 Sketch the response of the system characterized by the impulse response h[n] = (1/2)ⁿu[n] to the input signal.

            1  0<=n<=10

x[n] =

            0  otherwise

 Solution                                                                                                                                    

Result                                                                                                                                        

 HELP                                                                                                                                       

 CONV                                                                                                                                     

 CONV Convolution and polynomial multiplication.
C = CONV(A, B) convolves vectors A and B. The resulting vector is length LENGTH(A)+LENGTH(B)-1.
If A and B are vectors of polynomial coefficients, convolving them is equivalent to multiplying the two polynomials.

 LENGTH                                                                                                                                  

  LENGTH Length of vector.
LENGTH(X) returns the length of vector X. It is equivalent to MAX(SIZE(X)) for non-empty arrays and 0 for empty ones.

 FILTER                                                                                                                                    

 FILTER One-dimensional digital filter.
Y = FILTER(B,A,X) filters the data in vector X with the filter described by vectors A and B to create the filtered data Y. The filter is a "Direct Form II Transposed" implementation of the standard difference equation:
a(1)*y(n) = b(1)*x(n) + b(2)*x(n-1) + ... + b(nb+1)*x(n-nb)
- a(2)*y(n-1) - ... - a(na+1)*y(n-na)

If a(1) is not equal to 1, FILTER normalizes the filter coefficients by a(1).

When X is a matrix, FILTER operates on the columns of X. When X is an N-D array, FILTER operates along the first non-singleton dimension.

[Y,Zf] = FILTER(B,A,X,Zi) gives access to initial and final conditions, Zi and Zf, of the delays. Zi is a vector of length MAX(LENGTH(A),LENGTH(B))-1 or an array of such vectors, one for each column of X.

 ONES                                                                                                                                      

 ONES Ones array.
ONES(N) is an N-by-N matrix of ones.
ONES(M,N) or ONES([M,N]) is an M-by-N matrix of ones.
ONES(M,N,P,...) or ONES([M N P ...]) is an M-by-N-by-P-by-... array of ones.
ONES(SIZE(A)) is the same size as A and all ones.

 ZEROS                                                                                                                                     

 ZEROS Zeros array.
ZEROS(N) is an N-by-N matrix of zeros.
ZEROS(M,N) or ZEROS([M,N]) is an M-by-N matrix of zeros.
ZEROS(M,N,P,...) or ZEROS([M N P ...]) is an M-by-N-by-P-by-... array of zeros.
ZEROS(SIZE(A)) is the same size as A and all zeros.

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