Problems Based on Convolution Sum

Problem 1:

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Solution:

Linear convolution is given as,

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Step 1: For n = 0, above equation will be,

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Figure 5.28 shows sketches of x(k) and time folded sequence h(-k) are multiplied and added as per above equation.

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Hence,

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Step 2: For n = 1 equation (1) becomes

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Here h(1-k) = h[-(k-1)]. This means h(-k) is delayed by '1' sample figure 5.29 shows sketches of x(k) and h(1 - k).

 

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Multiplying the sequence and adding them we get,

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Step 3: For n = 2, equation (1) becomes,

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Figure 5.30 shows the sketches of x(k) and h(2-k). Multiplying and adding the sequences,

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Step 4: for n = 3, equation (1) becomes,

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From figure 5.31 we can write,

y(3) = (1 × 0) + (1 × 0) + (1 × 2) + (1 × 2) = 0 + 0 + 2 + 2 = 4

Step 5: For n = 4, equation (1) becomes

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Figure 5.32 we can write,

y(4) = (1 × 0) + (1 × 0) + (1 × 0) + (1 × 2) = 0 + 0 + 0 + 2 = 2

for n ≥ 5 and n ≤ 0 there will be no overlap between x(k) and h(n - k). Hence convolution will be zero. Thus,

y(n) = {2, 4, 4, 4, 2}

↑

Important comment:

If there are M. samples in x(n) and N samples in h(n), then number of samples in y(n) will be M + N - 1

Since M = 4 and N = 2, y(n) contains M + N - 1 = 4 + 2 - 1 = 5 samples.

 

Problem 2:

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Solution:

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linear convolution is given as,

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Here x(k) has nonzero values from k = 0 to k = 3. Hence x(k). h(n - k) will be non zero only for k = 0 to k = 3. Then above equation will be,

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Here.

the lowest index of x(n) is n_xl = 0

the highest index of x(n) is n_xh = 3

and

the lowest index of h(n) is n_hl = 0

the highest index of h(n) is n_hh = 1

The range of 'n' is given as,

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Putting values,

0 + 0 ≤ n ≤ 3 + 1

or 0 ≤ n ≤ 4

This 'n' will vary from 0 to 4.

 

For n = 0 in equation (1)

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For n = 1 in equation (1)

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For n = 2 in equation (1)

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For n = 3 in equation (1)

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For n = 4 in equation (1)

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Thus, y(n) = {2,4,4,4,2}

↑

 

Problem 3:

Calculate the convolution of x(n) and h(n) using method of multiplication,

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Solution:

The method is relatively easy and is based on the technique similar to multiplication. The two sequences are multiplied as we multiply multiple digit numbers. The result of this as we multiplication is nothing but the convolution of two sequences. The compute procedure is illustrated in figure.

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Thus the result of convolution is obtained in figure 5.33 as,

y (n) = {1, −1, −5, 2, 3, -5, 1, 4}

↑

 

Problem 4:

Calculate the convolution of x(n) and h(n) using method of tabulation

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Solution:

The values of x(n) and h(n) can be written as follows:

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The above values of x(n) and h(n) are tabulated as shown in fig. 5.34 & 5.35.

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From given sequence x(n) and h(n) we have,

lowest index of x(n) ⇒ n_xl = -2

lowest index of h(n) ⇒ n_hl = -3

Hence lowest index of y(n) = n_yl = n_xl + n_hl = -2-3 = -5

The first element in y(n) will be will be y (-5). This element is equal to top left diagram array. It contains only one multiplication. i.e.

y(-5) = x(-2) h(-3)

The other diagonal arrays are successively y(-4), y(-3) y(-2),... as shown figure 5.35. Finally the last element in the array in y(2) and it is the bottom right element in table i.e.

y(2) = x(2) h(0).

Let is put the values in table of figure 5.31. Such table is shown in figure 5.32 from this table the sequence y(n) is,

y(n) - {1, -1, -5, 2, 3, -5, 1, 4}

↑

 

Problem 5:

The impulse response of the relaxed LTI system is given as,

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Determine the response of this system if it is excited by unit step sequence.

Solution:

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Putting for x(k) and h(k),

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Here let us use the standard relation

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Problem 6:

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Solution:

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Problem 7:

Determine the response of the system with impulse response h(n) = u(n+2) - u (n -9) for the input x(n) = u(n) - 2 u(n-3) + u(n-6) by performing convolution graphically.

Solution:

Figure 5.36 Shows the sketches of h(n) and x(n) as per the waveforms.

 

The signals are,

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The linear convolution of x(n) and h(n) gives the output y(n). it is,

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Here the index 'n' of y(n) will have following range

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Hence equation (1) can be calculated for n = -2 to n = 13

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Note that h(k) = 0 for k ≤ -2

Hence lower index will be always - 2. The product of h(k) x(n - k) will be always zero for k < -2 since x(k) = 0 for k < 0:

x(n-k) = 0 for k > n

Hence y(n) = eeeeeeeeeeeee

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Similarly y(n) can be calculated for 'n' upto 13. The result is as follows:

y(n) = {1, 2, 3, 1, 0, 0, 0, 0, 0, 0, −1, −2, -3, -2, -1}

↑

Problem 8:

Prove the convolution with any sequence with the unit sample sequence results in the same sequence.

OR

x(n) * δ(n) = x(n)

OR

x(n) * h(n) = x(n) if h(n) = {1}

Solution:

we know that convolution of x(n) and h(n) is given as,

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Since convolution is commutative, above equation can be written as,

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Hence the summation of equation (1) is evaluated at k = 0. i.e.,

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Thus x(n) * h(n) = x(n) * δ(n) = x(n) if h(n) = δ(n)

 

Problem 9:

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Solution:

The convolution of x(n) and h(n) is shown below:

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Comparing above sequence with given y(n) we get,

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Problem 10:

Determine the output sequence of LTI system whose input and unit sample response are given as,

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Solution:

Here both x(n) and h(n) are infinite duration sequences.

 

By definition of convolution

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Putting the given sequences in above equation,

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Hence lower limit summation in equation (1) becomes k = 0 and u(k) = 1, i.e.,

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Hence upper limit of summation in equation (2) becomes 'n' and u(n-k) = 1,

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