GATE 2018 EC Q.14: solution and answer

GATE 2018 · ECQ.14 · 1 mark · MCQ

Let $x (t)$ be a periodic function with period $T = 10$ . The Fourier series coefficients for this series are denoted by $a_{k}$ , that is

x (t) = k = - \infty \sum \infty a_{k} e^{j k \frac{2 π}{T} t}

The same function $x (t)$ can also be considered as a periodic function with period $T^{'} = 40$ . Let $b_{k}$ be the Fourier series coefficients when period is taken as $T^{'}$ . If $\sum_{k = - \infty}^{\infty} ∣ a_{k} ∣ = 16$ , then $\sum_{k = - \infty}^{\infty} ∣ b_{k} ∣$ is equal to

(A)
256
(B)
64
(C)
16
(D)
4

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Answer and solution

ANSWER

(C) 16. When the period is taken as $T^{'} = 4 T$ , the coefficients are $b_{4 k} = a_{k}$ and every other $b_{m}$ is $0$ . So $\sum ∣ b_{k} ∣ = \sum ∣ a_{k} ∣ = 16$ .

Official answer: CSource: Official answer key, GATE 2018 EC (IIT Guwahati)

Fundamental frequencies. With $T = 10$ : $ω_{0} = \frac{2 π}{10}$ . With $T^{'} = 40$ : $ω_{0}^{'} = \frac{2 π}{40} = \frac{ω _{0}}{4}$ . Taking $T^{'} = 40$ is legitimate because $40 = 4 \times 10$ is a multiple of the period, so $x (t + 40) = x (t)$ .
Rewrite the given series on the new base frequency. Since $ω_{0} = 4 ω_{0}^{'}$ , $x (t) = k = - \infty \sum \infty a_{k} e^{j k ω_{0} t} = k = - \infty \sum \infty a_{k} e^{j (4 k) ω_{0}^{'} t}$
Compare with the $T^{'}$ series. By definition $x (t) = \sum_{m = - \infty}^{\infty} b_{m} e^{j m ω_{0}^{'} t}$ . Fourier series coefficients over a given period are unique (and the series converges absolutely here, because $\sum ∣ a_{k} ∣ = 16$ is finite), so matching the exponentials term by term gives $b_{m} = {a_{m /4}, 0, m = 0, \pm 4, \pm 8, \dots otherwise$
Check by direct integration (optional). $b_{m} = \frac{1}{40} \int_{0}^{40} x (t) e^{- j m \frac{2 π}{40} t} d t$ . Split $[0, 40]$ into the four intervals $[10 n, 10 n + 10]$ , $n = 0, 1, 2, 3$ , and put $t = τ + 10 n$ . Since $x (τ + 10 n) = x (τ)$ and $e^{- j m \frac{2 π}{40} (10 n)} = e^{- j mπ n /2}$ , $b_{m} = \frac{1}{40} (n = 0 \sum 3 e^{- j mπ n /2}) \int_{0}^{10} x (τ) e^{- j m \frac{2 π}{40} τ} d τ$ The bracket equals $4$ when $m$ is a multiple of $4$ . Otherwise $e^{- j mπ /2} \neq = 1$ , and the geometric sum is $\frac{1 - e ^{- j 2 π m}}{1 - e ^{- j mπ /2}} = 0$ . For $m = 4 k$ : $b_{4 k} = \frac{4}{40} \int_{0}^{10} x (τ) e^{- j k \frac{2 π}{10} τ} d τ = \frac{1}{10} \int_{0}^{10} x (τ) e^{- j k \frac{2 π}{10} τ} d τ = a_{k}$
Sum of magnitudes. The nonzero $b_{m}$ are exactly the $a_{k}$ , moved to the indices $m = 4 k$ . Every other $b_{m}$ is zero. Hence $m = - \infty \sum \infty ∣ b_{m} ∣ = k = - \infty \sum \infty ∣ a_{k} ∣ = 16$
Hence the answer is (C) 16.

KEY IDEA

The signal and its spectrum do not change when you choose a longer period. With $T^{'} = 4 T$ the new fundamental is $ω_{0} /4$ . The original harmonics sit at every 4th index of the new series, so $b_{4 k} = a_{k}$ and the three indices in between are zero. The nonzero coefficients keep their values, so any sum over them, including $\sum ∣ b_{k} ∣$ , stays the same.

WATCH OUT FOR

Scaling the coefficients by the period ratio. Seeing $\frac{1}{T ^{'}} = \frac{1}{40}$ in place of $\frac{1}{10}$ and dividing by 4 gives (D) 4. Thinking there are 4 times as many coefficients and multiplying gives (B) 64, and squaring gives (A) 256. Both errors forget two things. First, the integral over $T^{'} = 40$ covers four copies of the period, and that cancels the extra $\frac{1}{4}$ . Second, the new in-between coefficients are exactly zero. This is not a time-scaling of $x (t)$ : the function itself is unchanged.

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GATE 2018 EC Q.14: solution and answer

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