GATE 2023 EE Q.49: solution and answer

Official answer: 48Source: Official answer key, GATE 2023 EE (IIT Kanpur)

1. The constant $1$ has the same value for every $n$, so it does not affect the period. The phase shifts ($\frac{3\pi }{4}$ and $-\frac{\pi }{4}$) and the amplitudes ($3$ and $-5$) do not affect it either. Only the two frequencies matter.

2. Periodicity test for $\sin ({\Omega }_{0}n)$: the sequence is periodic only when $\frac{{\Omega }_0}{2\pi }=\frac{k}{N}$ is a rational number. Write the fraction in lowest terms; its denominator $N$ is the fundamental period.

3. First term: ${\Omega }_1=\frac{15\pi }{8}$, so $\frac{{\Omega }_1}{2\pi }=\frac{15}{16}$. This is already in lowest terms, so $N_1=16$. (Note that $\frac{2\pi }{{\Omega }_1}=\frac{16}{15}$ is not an integer, so it cannot be the period of a discrete-time signal.)

4. Second term: ${\Omega }_2=\frac{\pi }{3}$, so $\frac{{\Omega }_2}{2\pi }=\frac{1}{6}$, which gives $N_2=6$.

5. For a sum of periodic sequences, a common period must be a multiple of every individual period, so the smallest one is the least common multiple: $$N=\text{LCM}(16,6)=\text{LCM}(2^4,2\cdot 3)=2^4\cdot 3=48$$

6. Check: $\frac{15\pi }{8}\cdot 48=90\pi =45(2\pi )$ and $\frac{\pi }{3}\cdot 48=16\pi =8(2\pi )$. Both are integer multiples of $2\pi$. Any smaller $N$ would have to be a multiple of both $16$ and $6$, and $48$ is the smallest such number, so $N=48$.