Question 2 of 6: Periodic convolution from a product of discrete Fourier series

Question 2: Periodic convolution from a product of discrete Fourier series (12 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Both sequences are periodic with $N = 7$. Read from the printed stem plots, one period of $\tilde{x}_1[n]$ is the descending ramp $\{6,5,4,3,2,1,0\}$ for $n = 0,1,\ldots,6$ (the tallest stem sits at $n = 0$ and the sequence is zero at $n = 6$); $\tilde{x}_2[n]$ is a single unit stem per period at $n = 2$ (equivalently at $n = -5$ and $n = 9$); and $\tilde{x}_3[n]$ carries unit stems at $n = 0$ and $n = 4$ of every period.

Find. One period of each of $\tilde{y}_1[n]$ and $\tilde{y}_2[n]$, sketched on the same axes as the data.

Approach. A product of discrete Fourier series corresponds to periodic (circular) convolution over one period, so each answer is obtained by convolving $\tilde{x}_1$ with a sequence that consists only of unit impulses — and convolving with an impulse is simply a shift.

1. State the property being used. For two $N$-periodic sequences, the aid sheet gives
   $$\sum_{m=0}^{N-1} \tilde{x}_1[m]\,\tilde{x}_2[n-m] \;\longleftrightarrow\; \tilde{X}_1[k]\,\tilde{X}_2[k],$$

   i.e. multiplication of the DFS coefficients is periodic convolution of the sequences. Every answer below is therefore a periodic convolution, not a linear one, and the result is automatically $7$-periodic.

2. Recognise $\tilde{x}_2$ as a shifted impulse train. One period of $\tilde{x}_2$ has a single unit sample at $n = 2$, so $$\tilde{x}_2[n] = \sum_{r=-\infty}^{\infty}\delta[n - 2 - 7r] = \tilde{\delta}[n-2].$$

   Periodic convolution with a periodic unit impulse located at $n = m$ is a circular shift by $m$, which is the shift property $\tilde{x}[n-m] \leftrightarrow W_N^{km}\tilde{X}[k]$ read backwards.

3. Write down part (a). Therefore
   $$\boxed{\ \tilde{y}_1[n] = \tilde{x}_1[n-2] \ }$$

   and one period, tabulated for $n = 0,\ldots,6$, is $\{1,0,6,5,4,3,2\}$: the ramp bodily delayed by two samples, with the two samples that leave the right of the period wrapping around to the left.

y1[n] is the ramp delayed by two samples, wrapped circularly inside one period of length 7.
4. Decompose $\tilde{x}_3$ the same way. One period of $\tilde{x}_3$ has unit samples at $n = 0$ and $n = 4$, so $$\tilde{x}_3[n] = \tilde{\delta}[n] + \tilde{\delta}[n-4].$$

   Because periodic convolution is linear, convolving with a sum of two impulses produces the sum of two shifted copies.

x3[n]: unit stems at n = 0 and n = 4 of every period.
5. Write down part (b) and evaluate it. Hence $$\boxed{\ \tilde{y}_2[n] = \tilde{x}_1[n] + \tilde{x}_1[n-4] \ }$$

   Adding the ramp $\{6,5,4,3,2,1,0\}$ to its own four-sample circular shift $\{3,2,1,0,6,5,4\}$ gives one period

   $$\tilde{y}_2[n]\Big|_{n=0}^{6} = \{9,\,7,\,5,\,3,\,8,\,6,\,4\}.$$

   The result is no longer monotonic, because the wrapped copy re-injects the large samples of the ramp part-way through the period.

6. Check both answers at $k = 0$. The $k = 0$ DFS coefficient is the sum over one period, and the property forces $\tilde{Y}[0] = \tilde{X}_1[0]\tilde{X}[0]$. Here $\sum\tilde{x}_1 = 21$, $\sum\tilde{x}_2 = 1$ and $\sum\tilde{x}_3 = 2$, so the sums of the answers must be $21$ and $42$. Indeed $1+0+6+5+4+3+2 = 21$ and $9+7+5+3+8+6+4 = 42$, which confirms both results without evaluating a single complex exponential.