Step 1: Set up the periodicity argument. An $N$-point DFT is equivalent to sampling one period of the discrete-time Fourier transform, which forces the corresponding time-domain signal to be treated as periodic with period $N$. Multiplying two $N$-point DFTs and taking the inverse DFT therefore always yields the $N$-point circular convolution of the two sequences, never the linear convolution, unless the sequences are first padded with zeros. Step 2: Show the wraparound mechanically with a short example. Suppose $x[n]$ has 4 nonzero samples and $h[n]$ has 4 nonzero samples. Their true linear convolution has $4+4-1=7$ nonzero samples. If we take only a 4-point DFT of each (no padding) and multiply, the inverse DFT gives only 4 output samples; the last three samples of the true 7-sample linear result wrap around and add into the first three output samples. This corruption is precisely the wraparound error, and it is a form of aliasing because it is caused by the periodic extension of the sequences. Step 3: Test convolution is linear (option C) against this example. Because the 4-point circular result differs from the correct 7-point linear convolution (three samples are corrupted), the DFT product does not reproduce the linear convolution when no zero padding is used, so describing the operation as giving linear convolution is incorrect here. Step 4: Test aliasing can be avoided (option D). The standard fix is to zero-pad both $x[n]$ and $h[n]$ to length $N_1+N_2-1$ (7 in the example) before taking the DFT; only then does circular convolution equal linear convolution with no wraparound. Since the question fixes the condition as without zero padding, aliasing is unavoidable in this scenario, so option (D) does not hold. Step 5: Final answer. Only the statements that aliasing occurs and wraparound error occurs are correct.\[ \boxed{\text{Options (A) and (B)}} \]
