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Quiz: DSP - Discrete Fourier transform

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Number of Questions: 26

Question: 21 -

Find the number of smallest DFTs required to compute the linear convolution of length 40 sequences with a length of 900 another sequences using 64 DFT.

Options:

  1. 28

  2. 64

  3. 36

  4. 54

    5. Answer:

    36
    Solution:

    Let the two sequences be M and N.

    M = 40

    N = 900

    Number of DFT = 64

    The number of smaller DTS required = L + M - 1 = Number of given DFT points

    L + M - 1 = 64

    L + 40 - 1 = 64

    L = 25

    Total blocks = N / L = 900/25 = 36

    Hence, the number of smallest DFTs required to compute the linear convolution is 36.

Question: 22 -

Which of the following statement is/are correct about linear convolution?

1. The Input and output sequence is Aperiodic.

2. It requires zero padding.

3. The length of the input and output sequence is the same.

4. The length of output sequence is greater than the input sequence.

Options:

  1. 1 and 2

  2. Only 1

  3. 1, 2, 3, and 4

  4. 1 and 4

    5. Answer:

    1 and 4
    Solution:

    The linear convolution does not require the use of zero padding. The length of output sequence is greater than the input sequence length.

Question: 23 -

Which of the following statement is incorrect about DIT- FFT?

Options:

  1. The output sequence is represented in bit-reversal order.

  2. The input sequence is represented in bit-reversal order.

  3. It requires complex additions of 'N log2N.'

  4. The number of input samples is given by 2^i.

    5. Answer:

    The output sequence is represented in bit-reversal order.
    Solution:

    The output sequence of the DIT-FFT is represented in regular order instead of bit-reversal order.

Question: 24 -

IDFT of the sequence {1, 0, 1, 0} is:

Options:

  1. {0.5, 1, 0.5, 0}

  2. {0.5, 0, 0.5, 0}

  3. {1, 0, 0, 1}

  4. None of the above

    5. Answer:

    {0.5, 0, 0.5, 0}
    Solution:

    Step 1: For, n = 0

    x(0) = ¼ [ x(0) + x(1) + x(2) + x(3)]

    = ¼[1 + 0 + 1 + 0]

    = 2/4

    = 1/2

    = 0.5

    Step 2: For, n = 1

    x(1) = ¼ [ x(0) + x(1) + x(2) + x(3)]

    = ¼[1 + 0(j)+ 1(-1) + 0(-j)]

    = ¼ [1 +0 -1 + 0]

    = 0

    Step 3: For, n = 2

    x(2) = ¼ [ x(0) + x(1) + x(2) + x(3)]

    = ¼[1 + 0(-1)+ 1(1) + 0(-1)]

    = ¼[1 + 0 + 1 + 0]

    = 2/4

    = 1/2

    = 0.5

    Step 4: For, n = 3

    x(3) = ¼ [ x(0) + x(1) + x(2) + x(3)]

    = ¼[1 + 0(-j)+ 1(-1) + 0(j)]

    = ¼ [1 +0 - 1 + 0]

    = 0

    Thus, x(n) = {0.5, 0, 0.5, 0}

Question: 25 -

An analog signal has a bandwidth of 5KHz. If we are using an N-point DFT to compute the signal spectrum with a resolution less than or equal to 25Hz. Find the minimum length of the signal.

Options:

  1. 0.04s

  2. 0.2s

  3. 0s

  4. 0.02s

    5. Answer:

    0.04s
    Solution:

    Given, Bandwidth = 5 KHz.

    N = 2^m, where m is the integer

    Minimum length of the signal (T) is given by:

    T = L/Fs

    Where,

    L is the minimum number of requires samples

    Fs is the minimum sampling rate

    Fs = 2fm

    It means that the sampling rate is twice the bandwidth.

    Fs = 2 x 5 = 10 KHz.

    L = Fs/Resolution

    So, T = (Fs/ Resolution)/ Fs

    T = 1/Fs

    T = 1/25Hz

    T = 0.04s