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

We are given two discrete-time signals $x(n)$ and $h(n)$: $x(n) = \{0, 1, 1, 2\}$ $h(n) = \{a, b, c\}$ 
The signal $y(n)$ is the convolution of $x(n)$ and $h(n)$: $y(n) = x(n) \otimes h(n)$ 
We are also given that $Y(e^{j\omega})$ is the Discrete Fourier Transform (DFT) of $y(n)$, and $Y(e^{j0}) = 36$. 
A key property of the DFT (or DTFT) is that its value at $\omega = 0$ corresponds to the sum of the samples of the time-domain signal. So, $Y(e^{j0}) = \sum_{n} y(n)$. 
Another important property is the convolution property of DFT: If $y(n) = x(n) \otimes h(n)$, then in the frequency domain, their DFTs multiply: $Y(e^{j\omega}) = X(e^{j\omega}) H(e^{j\omega})$ 
Now, let's evaluate this at $\omega = 0$: $Y(e^{j0}) = X(e^{j0}) H(e^{j0})$ 
We know that $X(e^{j0}) = \sum_{n} x(n)$ and $H(e^{j0}) = \sum_{n} h(n)$. 
Let's calculate the sum of samples for $x(n)$: $\sum_{n} x(n) = 0 + 1 + 1 + 2 = 4$. 
So, $X(e^{j0}) = 4$. Let's calculate the sum of samples for $h(n)$: $\sum_{n} h(n) = a + b + c$. So, $H(e^{j0}) = a + b + c$. 
Now, substitute these into the convolution property at $\omega = 0$: $Y(e^{j0}) = X(e^{j0}) H(e^{j0})$ $36 = 4 \times (a + b + c)$ 
Now, solve for $(a + b + c)$: $a + b + c = \frac{36}{4}$ $a + b + c = 9$ 
Therefore, the correct relation is $a + b + c = 9$.