Question

Consider two signals $x(n)=\{0,1,1,2\}$ and $h(n)=\{a,b,c\}$. The signal $y(n)=x(n)\otimes h(n)$. The DFT of $y(n)$ is given as $Y(e^{j\omega})$. If $Y(e^{j0})$ is 36, then which of the following relation is correct?

• A. \( a \times b \times c = 27 \)
• B. \( ab + c = 9 \)
• C. \( a + b + c = 9 \)
• D. \( a + b - c = 12 \)

Hint:

Remember these crucial properties in Discrete-Time Fourier Transform (DTFT) and Discrete Fourier Transform (DFT):

• DC component The value of the DTFT/DFT at $\omega=0$ (or $k=0$ for DFT) is the sum of all samples of the time-domain signal. That is, $X(e^{j0}) = \sum_n x(n)$.
• Convolution Property Convolution in the time domain corresponds to multiplication in the frequency domain. If $y(n) = x(n) \otimes h(n)$, then $Y(e^{j\omega}) = X(e^{j\omega}) H(e^{j\omega})$.

These two properties combined are often used to solve problems involving sums of sequences or DC components.

Answer 1

The Correct Option is C

To solve this problem, we need to calculate the Discrete Fourier Transform (DFT) of the convolution of the signals \( x(n) \) and \( h(n) \). This will help us relate the given information about \( Y(e^{j0}) = 36 \) to the correct expression involving \( a \), \( b \), and \( c \).

1. Understanding the Concepts:

- Discrete Fourier Transform (DFT): The DFT of a sequence \( y(n) \) is given by:

\[ Y(e^{j\omega}) = \sum_{n=0}^{N-1} y(n) e^{-j\omega n} \]

- Convolution: The convolution of \( x(n) \) and \( h(n) \) is defined as:

\[ y(n) = (x(n) \otimes h(n)) = \sum_{k=0}^{M-1} x(k) h(n-k) \]

For this case, \( x(n) = \{0, 1, 1, 2\} \) and \( h(n) = \{a, b, c\} \). The DFT of the convolution can be computed using the DFTs of \( x(n) \) and \( h(n) \), leveraging the fact that the DFT of a convolution is the product of the DFTs of the individual sequences.

2. Given Values:

The DFT at \( \omega = 0 \) (DC component) is:

\[ Y(e^{j0}) = \sum_{n=0}^{N-1} y(n) = 36 \]

Thus, the sum of the sequence \( y(n) \) equals 36. Since \( y(n) = x(n) \otimes h(n) \), we need to compute this sum explicitly. The sequence \( y(n) \) can be found by performing the convolution operation:

\[ y(n) = \{ x(0)h(0) + x(1)h(0) + x(2)h(0) + x(3)h(0), x(0)h(1) + x(1)h(1) + x(2)h(1) + x(3)h(1), x(0)h(2) + x(1)h(2) + x(2)h(2) + x(3)h(2) \} \]

3. Calculating the Sum of \( y(n) \):

Let’s compute the sum of \( y(n) \). We know that:

\[ Y(e^{j0}) = x(0)a + x(1)a + x(2)a + x(3)a + x(0)b + x(1)b + x(2)b + x(3)b + x(0)c + x(1)c + x(2)c + x(3)c \] \[ Y(e^{j0}) = a(0 + 1 + 1 + 2) + b(0 + 1 + 1 + 2) + c(0 + 1 + 1 + 2) \] \[ Y(e^{j0}) = a \times 4 + b \times 4 + c \times 4 = 4(a + b + c) \] Since \( Y(e^{j0}) = 36 \), we have: \[ 4(a + b + c) = 36 \] \[ a + b + c = 9 \]

Final Answer:

The correct relation is \( a + b + c = 9 \).